AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 4 సమీకరణ వాదం Exercise 4(d) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Exercise 4(d)

అభ్యాసం – 4(డి)

I.

ప్రశ్న 1.
x3 + 2x2 – 4x + 1 = 0 సమీకరణపు మూలాలకు 3 రెట్లున్న మూలాలు గల బీజీయ సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x3 + 2x2 – 4x + 1 = 0 అనుకొనుము.
∴ కావలసిన సమీకరణం f(\(\frac{x}{3}\)) = 0
\(\left(\frac{x}{3}\right)^3+2\left(\frac{x}{3}\right)^2-\frac{4 x}{3}+1=0\)
\(\frac{x^3}{27}+\frac{2}{9} x^2-\frac{4}{3} x+1=0\)
27 గుణించగా
కావలసిన సమీకరణం x3 + 6x2 – 36x + 27 = 0

ప్రశ్న 2.
x5 – 2x4 + 3x3 – 2x2 + 4x + 3 = 0 సమీకరణపు మూలాలకు 2 రెట్లున్న మూలాలు గల బీజీయ సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x5 – 2x4 + 3x3 – 2x2 + 4x + 3 = 0
f(\(\frac{x}{2}\)) = 0 సమీకరణం కావలసిన లక్షణాలలో ఉంటుంది.
కావలసిన సమీకరణం f(\(\frac{x}{3}\)) = 0
⇒ \(\left(\frac{x}{2}\right)^5-2\left(\frac{x}{2}\right)^4+3\left(\frac{x}{2}\right)^3-2\left(\frac{x}{2}\right)^2+4\left(\frac{x}{2}\right)\) + 3 = 0
⇒ \(\frac{x^5}{32}-2 \cdot \frac{x^4}{16}+3 \cdot \frac{x^3}{8}-2 \cdot \frac{x^2}{4}+4 \cdot \frac{x}{2}+3=0\)
32 చే గుణించగా
కావలసిన సమీకరణం x5 – 4x4 + 12x3 – 16x2 + 64x + 96 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 3.
x4 + 5x3 + 11x + 3 = 0 సమీకరణ మూలాలకు వ్యతిరేక గుర్తులు కలిగిన సంఖ్యలు మూలాలుగా గల రూపాంతర సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x4 + 5x3 + 11x + 3 = 0
1, -α2, -α3, -α4 లు మూలాలుగా గల సమీకరణం f(-x) = 0
⇒ (-x)4 + 5(-x)3 + 11(-x) + 3 = 0
⇒ x4 – 5x3 – 11x + 3 = 0

ప్రశ్న 4.
x7 + 3x5 + x3 – x2 + 7x + 2 = 0 సమీకరణం మూలాలకు వ్యతిరేక గుర్తులు కలిగిన సంఖ్యలు మూలాలుగా గల రూపాంతర సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x7 + 3x5 + x3 – x2 + 7x + 2 = 0
1, -α2, …….., -α7 లు మూలాలుగల
సమీకరణం f(-x) = 0
⇒ (-x)7 + 3(-x)5 + (-x)3 – (-x)2 + 7(-x) + 2 = 0
⇒ -x7 – 3x5 – x3 – x2 – 7x + 2 = 0
⇒ x7 + 3x5 + x3 + x2 + 7x – 2 = 0

ప్రశ్న 5.
x4 – 3x3 + 7x2 + 5x – 2 = 0 సమీకరణ మూలాల వ్యుత్కమాలు మూలాలుగా గల బహుపది సమీకరణాన్ని కనుక్కోండి. [Mar. ’11]
సాధన:
దత్త సమీకరణం f(x) = x4 – 3x3 + 7x2 + 5x – 2 = 0
కావలసిన సమీకరణం f(\(\frac{1}{x}\)) = 0
⇒ \(\frac{1}{x^4}-\frac{3}{x^3}+\frac{7}{x^2}+\frac{5}{x}-2=0\)
x4 చే గుణించగా
⇒ 1 – 3x + 7x2 + 5x3 – 2x4 = 0
⇒ 2x4 – 5x3 – 7x2 + 3x – 1 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 6.
x5 + 11x4 + x3 + 4x2 – 13x + 6 = 0 సమీకరణం మూలాల వ్యుత్కమాలు మూలాలుగా గల బహుపది సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x5 + 11x4 + x3 + 4x2 – 13x + 6 = 0
కావలసిన సమీకరణం f(\(\frac{1}{x}\)) = 0
\(\frac{1}{x^5}+\frac{11}{x^4}+\frac{1}{x^3}+\frac{4}{x^2}-\frac{13}{x}+6=0\)
x5 చే గుణించగా
⇒ 1 + 11x + x2 + 4x3 – 13x4 + 6x5 = 0
⇒ 6x5 – 13x4 + 4x3 + x2 + 11x + 1 = 0

II.

ప్రశ్న 1.
x4 + x3 + 2x2 + x + 1 = 0 సమీకరణ మూలాల వర్గాలు మూలాలుగా గల బహుపది సమీకరణాన్ని రూపొందించండి.
సాధన:
దత్త సమీకరణం f(x) = x4 + x3 + 2x2 + x + 1 = 0
కావలసిన సమీకరణం f(√x) = 0
⇒ x2 + x√x + 2x + √x + 1 = 0
⇒ √x(x + 1) = -(x2 + 2x + 1)
వర్గం చేయగా
⇒ x(x + 1)2 = (x2 + 2x + 1)2
⇒ x(x2 + 2x + 1) = x4 + 4x2 + 1 + 4x3 + 4x + 2x2
⇒ x3 + 2x2 + x = x4 + 4x3 + 6x2 + 4x + 1
⇒ x4 + 3x3 + 4x2 + 3x + 1 = 0

ప్రశ్న 2.
x3 + 3x2 – 7x + 6 = 0 సమీకరణ మూలాల వర్గాలు మూలాలుగా గల బహుపది సమీకరణాన్ని రూపొందించండి.
సాధన:
దత్త సమీకరణం f(x) = x3 + 3x2 – 7x + 6 = 0
కావలసిన సమీకరణం f(√x) = 0
⇒ x√x + 3x – 7√x + 6 = 0
⇒ √x(x – 7) = -(3x + 6)
వర్గం చేయగా
⇒ x(x – 7)2 = (3x + 6)2
⇒ x(x2 – 14x + 49) = 9x2 + 36 + 36x
⇒ x3 – 14x2 + 49x – 9x2 – 36x – 36 = 0
⇒ x3 – 23x2 + 13x – 36 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 3.
x3 + 3x2 + 2 = 0 సమీకరణ మూలాల ఘనాలు మూలాలుగా గల బహుపది సమీకరణాన్ని రూపొందించండి.
సాధన:
దత్త సమీకరణం x3 + 3x2 + 2 = 0
y = x3 అయిన x = \(y^{1 / 3}\) అవుతుంది
∴ y + 3\(y^{2 / 3}\) + 2 = 0
3y\(y^{2 / 3}\) = -(y + 2)
ఘనం చేయగా
27y2 = -(y + 2)3 = -(y3 + 6y2 + 12y + 8)
∴ y3 + 6y2 + 27y2 + 12y + 8 = 0
⇒ y3 + 33y2 + 12y + 8 = 0
కావలసిన సమీకరణం x3 + 33x2 + 12x + 8 = 0

III.

ప్రశ్న 1.
-2 తో మార్పు చెందిన x4 – 5x3 + 7x2 – 17x + 11 = 0 సమీకరణ మూలాల విలువలు మూలాలుగా గల బీజీయ సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x4 – 5x3 + 7x2 – 17x + 11 = 0
కావలసిన సమీకరణం f(x + 2) = 0
⇒ (x + 2)4 – 5(x + 2)3 + 7(x + 2)2 – 17(x + 2) + 11 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q1
కావలసిన సమీకరణం x4 + 3x3 + x2 – 17x – 19 = 0

ప్రశ్న 2.
-3 తో మార్పు చెందిన x5 – 4x4 + 3x2 – 4x + 6 = 0 సమీకరణ మూలాల విలువలు మూలాలుగా గల బహుపది సమీకరణాన్ని కనుక్కోండి. [T.S. Mar. ’16]
సాధన:
దత్త సమీకరణం f(x) = x5 – 4x4 + 3x2 – 4x + 6 = 0
కావలసిన సమీకరణం f(x + 3) = 0
(x + 3)5 – 4(x + 3)3 + 3(x + 3)2 – 4(x + 3) + 6 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q2
∴ కావలసిన సమీకరణం x5 + 11x4 + 42x3 + 57x2 – 13x – 60 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 3.
2 తో మార్పు చెందిన x4 – x3 – 10x2 + 4x + 24 = 0 సమీకరణ మూలాల విలువలు మూలాలుగా గల బహుపది సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = x4 – x3 – 10x2 + 4x + 24 = 0
కావలసిన సమీకరణం f(x – 2) = 0
(x – 2)4 – (x – 2)3 – 10(x – 2)2 + 4(x – 2) + 24 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q3
∴ కావలసిన సమీకరణం x4 – 9x3 + 20x2 = 0

ప్రశ్న 4.
4తో మార్పు చెందిన 3x5 – 5x3 + 7 = 0 సమీకరణ మూలాల విలువలు మూలాలుగా గల బహువది సమీకరణాన్ని కనుక్కోండి.
సాధన:
దత్త సమీకరణం f(x) = 3x5 – 5x3 + 7 = 0
కావలసిన సమీకరణం f(x – 4) = 0
3(x – 4)5 – 5(x – 4)3 + 7 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q4
∴ కావలసిన సమీకరణం 3x5 – 60x4 + 475x3 – 1860x2 + 3600x – 2745 = 0

ప్రశ్న 5.
x యొక్క రెండో అత్యధిక ఘాత గుణకం సున్నా అయ్యే విధంగా కింది సమీకరణాలను పరివర్తన చేసి రూపాంతర సమీకరణాలను కనుక్కోండి.
(i) x3 – 6x2 + 10x – 3 = 0
సాధన:
x యొక్క రెండో అత్యధిక ఘాత గుణకం లుప్తం అయ్యే విధంగా సమీకరణ మూలాలను \(\frac{-a_1}{n \cdot a_0}=\frac{-(-6)}{(3)(1)}\) = 2 తో మూలాల విలువలను పరివర్తనము చేయాలి.
అంటే f(x + 2) = 0 ను కనుక్కోవాలి.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q5(i)
∴ కావలసిన సమీకరణము x3 – 2x + 1 = 0

(ii) x4 + 4x3 + 2x2 – 4x – 2 = 0
సాధన:
దత్త సమీకరణం x4 + 4x3 + 2x2 – 4x – 2 = 0
మూలాలను h = \(-\frac{a_1}{n a_0}=\frac{-4}{4}\) = -1 తో మార్పు చెందించాలి.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q5(ii)
∴ కావలసిన సమీకరణం x4 – 4x2 + 1 = 0

(iii) x3 – 6x2 + 4x – 7 = 0
సాధన:
దత్త సమీకరణం x3 – 6x2 + 4x – 7 = 0
మూలాలను h = \(-\frac{a_1}{n a_0}=\frac{6}{3}\) = 2 తో మార్పు చెందించాలి.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q5(iii)
∴ కావలసిన సమీకరణం x3 – 8x – 15 = 0

(iv) x3 + 6x2 + 4x + 4 = 0
సాధన:
దత్త సమీకరణం x3 + 6x2 + 4x + 4 = 0
రెండవ పదాన్ని లోపింపచేయటానికి మూలాలను h = \(-\frac{a_1}{n a_0}=-\frac{6}{3}\) = -2 కు మార్పు చెందించాలి.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q5(iv)
∴ కావలసిన సమీకరణం x3 – 8x + 12 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 6.
x యొక్క మూడో అత్యధిక ఘాత గుణకం సున్నా అయ్యే విధంగా కింది సమీకరణాలను పరివర్తన చేయండి.
(i) x4 + 2x3 – 12x2 + 2x – 1 = 0
సాధన:
దత్త సమీకరణం
f(x) = x4 + 2x3 – 12x2 + 2x – 1 = 0
x యొక్క మూడో అత్యధిక ఘాత గుణకం సున్నా కావాలి.
అంటే, దత్త సమీకరణ, మూలాలను h కు మార్పు చెందించాలి.
ఇచ్చట h అనేది \(f^{(4-3+1)}(h)\) = 0 ⇒ \(f^{(2)} \text { (h) }\) = 0 నుండి వస్తుంది.
f'(x) = 4x3 + 6x2 – 24x + 2
f”(x) = 12x2 + 12x – 24
f”(h) = 0
⇒ 12h2 + 12h – 24 = 0
⇒ h2 + h – 2 = 0
⇒(h + 2) (h – 1) = 0
⇒ h = -2 (లేదా) 1
సందర్భము (i):
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q6(i)
∴ కావలసిన సమీకరణం x4 – 6x3 + 42x – 53 = 0
సందర్భము (ii):
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q6(i).1
కావలసిన సమీకరణము x4 + 6x3 – 12x – 8 = 0
∴ కావలసిన సమీకరణాలు x4 – 6x3 + 42x – 53 = 0 (లేదా) x4 + 6x3 – 12x – 8 = 0

(ii) x3 + 2x2 + x + 1 = 0
సాధన:
f(x) = x3 + 2x2 + x + 1 అనుకోండి.
x యొక్క మూడో అత్యధిక గుణకం సున్నా కావాలి అంటే, దత్త సమీకరణ మూలాలను ‘h’ తో మార్పు చెందించాలి.
ఇచ్చట h అనేది f'(h) = 0 నుండి వస్తుంది.
f'(x) = 3x2 + 4x + 1
f'(h) = 0
⇒ 3h2 + 4h + 1 = 0
⇒ (3h + 1) (h + 1) = 0
⇒ h = -1, \(-\frac{1}{3}\)
సందర్భము (i):
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q6(ii)
∴ కావలసిన సమీకరణం x3 – x2 + 1 = 0
సందర్భము (ii):
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q6(ii).1
∴ కావలసిన సమీకరణం x3 + x2 + \(\frac{23}{27}\) = 0
⇒ 27x3 + 27x2 + 23 = 0
∴ కావలసిన సమీకరణాలు x3 – x2 + 1 = 0 (లేదా) 27x3 + 27x2 + 23 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(d)

ప్రశ్న 7.
కింది సమీకరణాలను సాధించండి.
(i) x4 – 10x3 + 26x2 – 10x + 1 = 0
సాధన:
దత్త సమీకరణం ఒకటో కోవకు చెందిన సరిఘాత వ్యుత్కమ సమీకరణం
x2 చే భాగించగా x2 – 10x + 26 – \(\frac{10}{x}+\frac{1}{x^2}\) = 0
\([latex]\frac{10}{x}+\frac{1}{x^2}\)[/latex] …….(1)
a = x + \(\frac{1}{x}\) అనుకుంటే
\(x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2-2\) = a2 – 2
(1) లో వ్రాయగా a2 – 2 – 10a + 26 = 0
⇒ a2 – 10a + 24 = 0
⇒ (a – 4) (a – 6) = 0
⇒ a = 4 (లేదా) 6
సందర్భము (i): a = 4
x + \(\frac{1}{x}\) = 4
⇒ x2 + 1 = 4x
⇒ x2 – 4x + 1 = 0
⇒ x = \(\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm 2 \sqrt{3}}{2}\)
⇒ x = 2 ± √3
సందర్భము (ii): a = 6 అయిన
x + \(\frac{1}{x}\) = 6
⇒ x2 + 1 = 6x
⇒ x2 – 6x + 1 = 0
⇒ x = \(\frac{6 \pm \sqrt{36-4}}{2}=\frac{6 \pm 4 \sqrt{2}}{2}\)
⇒ x = 3 ± 2√2
∴ దత్త సమీకరణానికి మూలాలు 3 ± 2√2, 2 ± √3

(ii) 2x5 + x4 – 12x3 – 12x2 + x + 2 = 0 [A.P. Mar ’16, Mar. ’08, ’07]
సాధన:
f(x) = 2x5 + x4 – 12x3 – 12x2 + x + 2 = 0
ఒకటవ కోవకు చెందిన బేసి తరగతి వ్యుత్కమ సమీకరణం
∴ -1 మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) III Q7(ii)
f(x) ను (x + 1) చే భాగించగా
2x4 – x3 – 11x2 – x + 2 = 0
x2 చే భాగించగా
2x2 – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0
\(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)-11=0\) ………(1)
a = x + \(\frac{1}{x}\) అయిన x2 + \(\frac{1}{x^2}\) = a2 – 2
(1) లో వ్రాయగా
2(a2 – 2) – a – 11 = 0
⇒ 2a2 – 4 – a – 11 = 0
⇒ 2a2 – a – 15 = 0
⇒ (a – 3) (2a + 5) = 0
⇒ a = 3 లేదా \(\frac{-5}{2}\)
సందర్భము (i): a = 3 అయిన
x + \(\frac{1}{x}\) = 3
⇒ x2 + 1 = 3x
⇒ x2 – 3x + 1 = 0
⇒ x = \(\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\)
సందర్భము (ii): a = \(\frac{-5}{2}\) అయిన
\(x+\frac{1}{x}=-\frac{5}{2}\)
⇒ \(\frac{x^2+1}{x}=-\frac{5}{2}\)
⇒ 2x2 + 2 = -5x
⇒ 2x2 + 5x + 2 = 0
⇒ (2x + 1) (x + 2) = 0
⇒ x = \(\frac{-1}{2}\), -2
∴ దత్త సమీకరణానికి మూలాలు -1, \(\frac{-1}{2}\), -2, \(\frac{3 \pm \sqrt{5}}{2}\)

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Ex 1.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.1

Question 1.
Multiply the following.
(i) 5 × 7
Answer:
5 × 7 = 35

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

(ii) (-9) × (6) .
Answer:
(-9) × (6)
= -(9 × 6) = – 54

(iii) (9) × (-4)
Answer:
9 × – 4
= – (9 × 4) = – 36

(iv) (8) × (-7)
Answer:
8 × (-7) = – (8 × 7) = – 56

(v) (-124) × (-1)
Answer:
(-124) × (-1) = +(124 × 1) = + 124

(vi) (-12) × (-7)
Answer:
(-12) × (-7) = + (12 × 7) = + 84

(vii) (-63) × (7)
Answer:
(-63) × (7) = -(63 × 7) = – 441

(viii) (7) × (-15)
Answer:
7 × (-15) = – (7 × 15) = – 105

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

Question 2.
Which is greater?
(i) 2 × (-5) or 3 × (-4)
Answer:
2 × (-5) or 3 × (-4)
– (2 × 5) or – (3 × 4)
– 10 or – 12
– 10 greater than – 12
∴(-10) > (-12)

(ii) (-6) × (-7) or (-8) × 5
Answer:
(-6) × (-7) or (-8) × 5
6 × 7 or -(8 × 5)
42 or – 40
42 greater than – 40
∴ 42 > (-40)

(iii) (- 6) × 10 or (- 3) × (- 21)
Answer:
(-6) × 10 or (- 3) × (-21)
– (6 × 10) or 3 × 21
– 60 or 63
(- 60) less than 63
(or)
63 greater than (-60)
∴ 63 > (- 60)

(iv) 9 × (-11) or 6 × (-16)
Answer:
9 × (-11) or 6 × (-16)
-(9 × 11) or – (6 × 16)
(-99) or (- 96)
(- 99) less than (- 96)
(- 99) < (- 96)
(or)
(- 96) greater than (- 99)
∴ (-96) > (-99)

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

(v) (-8) × (-5) or (-9) × (-4)
Answer:
(-8) × (-5) or (-9) × (-4) + (8 × 5) or + (9 × 4)
40 or 36
40 greater than 36
∴ 40 > 36

Question 3.
Write the pair of integers whose product will give
(i) A negative integer
Answer:
AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1 1

(ii) A positive integer
Answer:
AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1 2

(iii) Zero
Answer:
AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1 3

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

Question 4.
A frog is slipping into a well from upper surface at a rate of 3 meters per minute, after 5 minutes what is the position of the frog in the well?
AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1 4
Answer:
No. of meters slipped by frog per minute = 3m = – 3 m
No. of meters slipped by frog per 5 minutes = (-3) × 5
= – (3 × 5)
= – 15 m
That is the frog is 15 m down from the upper surface.

Question 5.
During the summer, the level of water in a pond decreases by 5 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks ?
AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1 5
Answer:
Decrease in water level per every week = 5 inches = (- 5)
Decrease in water level per 6 weeks = (-5) × 6 .
= – (5 × 6)
= – 30 inches
Change in the level of the water in the pond per 6 weeks is 30 inches decreased.

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

Question 6.
A shop keeper earns a profit of ₹ 5 on one note book and loss of ₹ 3 on one pen by selling in the month of July. He sells 1500 books and 1500 pens. Find out what is his profit or loss.
Answer:
Profit on each notebook = ₹5
Profit on 1500 notebooks = 1500 × 5
Total profit on 1500 note books = ₹ 75 007/-
Loss on each pen = ₹3
Which we denoted by – 3.
Loss on 1500 pens = 1500 × -3
= – ₹ 4500
profit > loss
So, he will get profit.
Total profit = 7500 – 4500
= ₹ 3000/-

Question 7.
A cement company earns a profit of ₹ 8 per bag of white cement and a loss of per bag of grey cement by selling.
The company sells 2,000 bags of white cement and 3,000 bags of grey cement in a month. Find out what is its profit or loss.
Answer:
Profit on each white cement bag = ₹ 8
Profit on each white cement bag
= 2000 × 8
= ₹ 16000
Loss on each grey cement bag = ₹ 6
Which we denoted by – 6.
Loss on 3000 grey cement bags = 3000 × – 6
= – ₹ 18000
Loss is more than profit.
So, he will get loss.
∴ Total loss = 16000 + (-18000)
= + 16000 – 18000
= – 2000/-

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

Question 8.
Fill in the blanks with suitable integer to make the statement true.
(i) (-4) × _________ = – 20
Answer:
(- 4) × 5 = – 20
(- 4) × 5 = – 20
– (4 × 5) = – 20
– 20 = – 20

(ii) __________ × 5 = – 35
Answer:
-7 × 5 = – 35
(-7) × 5 = – 35
– (7 × 5) = – 35
– 35 = – 35

(iii) (-6) × __________ = 48
Answer:
(-6) × (-8) = 48
(-6) × (-8) = 48
6 × 8 = 48
48 = 48

(iv) __________ × (- 9) = 45
Answer:
(-5) × (-9) = 45
(-5) × (-9) = 45
5 × 9 = 45
45 = 45

(v) ___________ × 7 = – 42
Answer:
– (6) × 7) = – 42
– (6) × 7) = – 42
– (6 × 7) = – 42
– 42 = – 42

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.1

(vi) 8 × = – 8
Answer:
8 × (-1) = – 8
(8) × (-1) = – 8 – (8 × 1) = – 8
– 8 = – 8

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.2

10th Class Maths 5th Lesson Quadratic Equations Ex 5.2 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations by factorisation,
i) x2 – 3x – 10 = 0
Answer:
Given: x2 – 3x – 10 = 0
x2 – 5x + 2x- 10 = 0
⇒ x(x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
⇒ x = 5 or -2
are the roots of the given Q.E.

ii) 2x2 + x – 6 = 0
Answer:
Given: 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ (x + 2) or 2x – 3 = 0
⇒ x = -2 or 2x = 3
⇒ x = -2 or \(\frac{3}{2}\)
are the roots of the given Q.E.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

iii) √2x2 + 7x + 5√2 =0
Answer:
Given: √2x2 + 7x + 5√2 =0
⇒ √2x2 + 5x + 2x + 5√2 = 0
⇒ x(√2x + 5)+ √2(√2x + 5) = 0
⇒ (√2x + 5) (x + √2) = 0
⇒ √2x + 5 = 0 or x + √2 = 0
⇒ √2x = -5 or x = -√2
⇒ x = \(\frac{-5}{\sqrt{2}}\) = or -√2
are the roots of √2 the given Q.E.

iv) 2x2 – x + \(\frac{1}{8}\) = 0
Answer:
Given: 2x2 – x + \(\frac{1}{8}\) = 0
⇒ \(\frac{16 x^{2}-8 x+1}{8}\) = 0
⇒ 16x2 – 8x + 1 =0
⇒ 16x2 – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) – l(4x – 1) = 0
⇒ (4x – 1) (4x – 1) – 0
⇒ 4x – 1 = 0
⇒ 4x = l
⇒ x = \(\frac{1}{4}\), \(\frac{1}{4}\)
are the roots of given Q.E.

v) 100x2 – 20x + 1 = 0
Answer:
Given : 100x2 – 20x + 1 =0
⇒ 100x2 – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – l(10x – 1) = 0
⇒ (10x – 1) (10x – l) = 0
⇒ 10x – 1 = 0
⇒ 10x = 1
⇒ x = \(\frac{1}{10}\), \(\frac{1}{10}\)
are the roots of the given Q.E.

vi) x(x + 4) = 12
Answer:
Given: x(x + 4) = 12
⇒ x2 + 4x = 12
⇒ x2 + 4x – 12 = 0
⇒ x2 + 6x – 2x – 12 = 0
⇒ x(x + 6) – 2(x + 6) = 0
⇒ (x + 6) (x – 2) = 0
⇒ x + 6 = 0 or x – 2 = 0
⇒ x = -6 or x = 2
⇒ x = -6 or 2
are the roots of the given Q.E.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

vii) 3x2 – 5x + 2 = 0
Answer:
Given: 3x2 – 5x + 2 = 0
⇒ 3x2 – 3x – 2x + 2 = 0
⇒ 3x(x – 1) – 2(x – 1) = 0
⇒ (x – 1) (3x – 2) = 0
⇒ x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or \(\frac{2}{3}\),
⇒ x = 1 or \(\frac{2}{3}\) are the roots of the given Q.E.

viii) x – \(\frac{3}{x}\) = 2
Answer:
Given: x – \(\frac{3}{x}\) = 2
⇒ \(\frac{x^{2}-3}{x}\) = 2
⇒ x2 – 3 = 2x
⇒ x2 – 2x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + l(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ (x – 3) = 0 or (x + 1) = 0
⇒ x = 3 or x = -1
⇒ x = 3 or -1 are the roots of the given Q.E.

ix) 3(x – 4)2 – 5(x – 4) = 12
Answer:
Take (x – 4) = a, then the given Q.E. reduces to 3a2 – 5a = 12
⇒ 3a2 – 5a – 12 = 0
⇒ 3a2 – 9a + 4a – 12 = 0
⇒ 3a(a – 3) + 4(a – 3) = 0
⇒ (a – 3) (3a + 4) = 0
⇒ a – 3 = 0 or 3a + 4 = 0
⇒ a = 3 or a = \(\frac{-4}{3}\)
but a = x – 4
x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)
⇒ x = 7 or x = 4 – \(\frac{-4}{3}\) = \(\frac{8}{3}\)
∴ x = 7 or \(\frac{8}{3}\)
are the roots of the given Q.E.

Question 2.
Find two numbers whose sum is 27 and product is 182.
Answer:
Let a number be x.
Then the other number = 27 – x
Product of the numbers = x(27 – x) = 27x – x2
By problem 27x – x2 = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 14x – 13x + 182 = 0
⇒ x(x- 14) – 13(x – 14) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 or x – 14 = 0
⇒ x = 13 or 14.
∴ The numbers are 13; 27 – 13 = 14 or 14 and 27 – 14 = 13.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 3.
Find two consecutive positive integers, sum of whose squares is 613.
Answer:
Let a positive integer be x.
Then the second integer = x + 1
Sum of the squares of the above integers = x2 + (x + 1)2
= x2 + x2 + 2x + 1
= 2x2 + 2x + 1
By problem 2x2 + 2x + 1 = 613
⇒ 2x2 + 2x – 612 = 0
⇒ x2 + x – 306 = 0
⇒ x2 + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17) (x + 18) = 0
⇒ x – 17 = 0 (or) x + 18 = 0
⇒ x = 17 (or) -18,
we do not consider -18
Then the numbers are (17, 17 + 1)
i.e., 17, 18 are the required two consecutive positive integers.

Question 4.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer:
Let the base of the right triangle = x cm
Then its altitude = x – 7 cm
By Pythagoras Theorem
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2 1
(base)2 + (height)2 = (hypotenuse)2
⇒ x2 + (x – 7)2 = 132
⇒ x2 + x2 – 14x + 49 = 169 .
⇒ 2x2 – 14x + 49 – 169 = 0
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
⇒ x – 12 = 0 (or) x + 5 = 0
⇒ x = 12 (or) x = -5 But x can’t be negative.
∴ x = 12
x – 7 = 12 – 7 = 5
The two sides are 12 cm and 5 cm.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 5.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Answer:
Let the number of articles produced be x.
Then the cost of each article = 2x + 3
Total cost of the articles produced = x [2x + 3] = 2x2 + 3x
By problem 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x – 12x – 90 = 0
⇒ x (2x + 15) – 6 (2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ 2x + 15 = 0 (or) x – 6 = 0
⇒ x = \(\frac{-15}{2}\) or x = 6
But x can’t be negative.
∴ x = 6
2x + 3 = 2 × 6 + 3 = 15
∴ Number of articles produced = 6 Cost of each article = Rs. 15.

Question 6.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Answer:
Let the length of the rectangle = x
Given perimeter = 2(1 + b) = 28
⇒ (1 + b) = \(\frac{28}{2}\) = 14
Breadth of the rectangle = 14 – x
Area = length . breadth = x (14 – x)
= 14x – x2
By problem, 14x – x2 = 40.
⇒ x2 – 14x + 40 = 0
⇒ x2 – 10x – 4x + 40 = 0
⇒ x(x – 10) – 4(x – 10) = 0
⇒ (x – 10) (x – 4) = 0
⇒ x – 10 = 0 (or) x – 4 = 0
⇒ x = 10 (or) 4
∴ Length = 10 m or 4 m
Then breadth = 14 – 10 = 4 m (or) 14 – 4 = 10 m

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 7.
The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Answer:
Let the altitude of the triangle h = x cm
Then its base ‘b’ = x + 4.
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)(x + 4)(x)
= \(\frac{x^{2}+4 x}{2}\)
By problem \(\frac{x^{2}+4 x}{2}\) = 48
⇒ x2 + 4x = 2 × 48
⇒ x2 + 4x – 96 = 0
⇒ x2 + 12x – 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12)(x – 8) = 0
⇒ x + 12 = 0 (or) x – 8 = 0
⇒ x = -12 (or) x = 8
But x can’t be negative.
∴ x = 8 and x + 4 = 8 + 4 = 12
Hence altitude = 8 cm and base = 12 cm.

Question 8.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Answer:
Let the speed of the slower train = x kmph
Then speed of the faster train = x + 5 kmph.
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2 2
Distance = Speed × Time
Distance travelled by the first train = 2(x + 5) = 2x + 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)2 = (side)2 + (side)2
⇒ (2x)2 + (2x + 10)22 = 502
⇒ 4x2 + (4x2 + 40x + 100) = 2500
⇒ 4x2 + 4x2 + 40x + 100 = 2500
⇒ 8x2 + 40x – 2400 = 0
⇒ x2 + 5x – 300 = 0
⇒ x2 + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) – 20
But x can’t be negative.
∴ Speed of the slower train x = 15 kmph.
Speed of the faster train x + 5 = 15 + 5 = 20 kmph.

AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2

Question 9.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs. 1600, how many boys are there in the class?
Answer:
Let the number of boys in the class = x
Then number of girls in the class = 60 – x [∵ total students = 60]
Money contributed by the boys = x(60 – x) = 60x – x2 [∵ given]
Money contributed by the girls = (60 – x)x = 60x – x2
∴ Money contributed by the class = 120x – 2x2
By problem 120x -2x2 = 1600
⇒ 2x2– 120x + 1600 = 0
⇒ x2 – 60x + 800 = 0
⇒ x2 – 40x – 20x + 800 = 0
⇒ x(x – 40) – 20 (x – 40) = 0
⇒ (x – 40) (x – 20) = 0
⇒ x = 40 (or) 20
∴ Boys = 40 or 20 Girls = 20 or 40.

Question 10.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed ?
Answer:
Let the speed of the boat in still water be x kmph.
Speed of the current = 3 kmph
Then speed of the boat in upstream = (x – 3) kmph
Speed of the boat in downstream = (x + 3) kmph
By problem total time taken = 6h.
AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.2 3
⇒ 24(2x) = 6(x2 – 9)
⇒ 8x = x2 – 9
⇒ x2 – 8x – 9 = 0
⇒ x2 – 9x + x-9 = 0
⇒ x (x – 9) + 1 (x – 9) = 6
⇒ (x – 9) (x + 1) = 0
⇒ x – 9 = 0 or x + 1 = 0
x can’t be negative,
∴ x = 9
i.e., speed of the boat in still water = 9 kmph.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 4 సమీకరణ వాదం Exercise 4(c) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Exercise 4(c)

అభ్యాసం – 4(సి)

I.

ప్రశ్న 1.
క్రింది మూలాలు గల బహుపది సమీకరణాలను రూపొందించండి.
(i) 2 + 3i, 2 – 3i, 1 + i, 1 – i
సాధన:
కావలసిన సమీకరణం [x – (2 + 3i)] [x – (2 – 3i)] [x – (1 + i)] [x – (1 – i)] = 0
⇒ [(x – 2) – 3i)] [(x – 2) + 3i] [(x – 1) – i] [(x – 1) + i] = 0
⇒ [(x – 2)2 – 9i2] [(x – 1)2 – i2] = 0
⇒ (x2 – 4x + 4 + 9) (x2 – 2x + 1 + 1) = 0
⇒ (x2 – 4x + 13) (x2 – 2x + 2) = 0
⇒ x4 – 4x3 + 13x2 – 2x3 + 8x2 – 26x + 2x2 – 8x + 26 = 0
⇒ x4 – 6x3 + 23x2 – 34x + 26 = 0

(ii) 3, 2, 1 + i, 1 – i
సాధన:
కావలసిన సమీకరణం (x – 3) (x – 2) [x – (1 + i)] [x – (1 – i)] = 0
⇒ (x2 – 5x + 6) [(x – 1) – i] [(x – 1) + i) = 0
⇒ (x2 – 5x + 6) [(x – 1)2 – i2] = 0
⇒ (x2 – 5x + 6) (x2 – 2x + 1 + 1) = 0
⇒ (x2 – 5x + 6) (x2 – 2x + 2) = 0
⇒ x4 – 5x3 + 6x2 – 2x3 + 10x2 – 12x + 2x2 – 10x + 12 = 0
⇒ x4 – 7x3 + 18x2 – 22x + 12 = 0

(iii) 1 + i, 1 – i, -1 + i, -1 – i
సాధన:
కావలసిన సమీకరణం [x – (1 + i)] [x – (1 – i)] [x – (-1 + i)] [x – (-1 – i)] = 0
⇒ [(x – 1) – i] [(x – 1) + i] [(x + 1) – i] [(x + 1) + i) = 0
⇒ [(x – 1)2 – i2] [(x + 1)2 – i2] = 0
⇒ (x2 – 2x + 1 + 1) (x2 + 2x + 1 + 1) = 0
⇒ (x2 – 2x + 2) (x2 + 2x + 2) = 0
⇒ x4 – 2x3 + 2x2 + 2x3 – 4x2 + 4x + 2x2 – 4x + 4 = 0
⇒ x4 + 4 = 0

(iv) 1 + i, 1 – i, 1 + i, 1 – i
సాధన:
కావలసిన సమీకరణం [x – (1 + i)] [x – (1 – i)]
⇒ [x – (1 + i)] [x – (1 – i)] = 0
⇒ [(x – 1) – i]2 [(x – 1) + i]2 = 0
⇒ [(x – 1)2 – i2] = 0
⇒ (x2 – 2x + 1 + 1)2 = 0
⇒ x4 + 4x2 + 4 – 4x3 + 4x2 – 8x = 0
⇒ x4 – 4x3 + 8x2 – 8x + 4 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

ప్రశ్న 2.
కింది మూలాలు గల అకరణీయ గుణకాలు గల బహుపది సమీకరణాన్ని రూపొందించండి.
(i) 4√3, 5 + 2i
సాధన:
బహుపది సమీకరణ గుణకాలు అకరణీయ సంఖ్యలైన, దాని మూలాలు సంయుగ్మ కరణులు మరియు సంయుగ్మసంకీర్ణ సంఖ్యలు.
α = 4√3 అయిన β = -4√3 మరియు γ = 5 + 2i అయిన δ = 5 – 2i
α, β, γ, δ లు మూలాలు
α + β = 0, αβ = -48
γ + δ = 10, γδ = 25 + 4 = 29
కావలసిన సమీకరణం [x2 – (α + β)x + αβ] [x2 – (γ + δ)x + γδ] = 0
⇒ (x2 – 48) (x2 – 10x + 29) = 0
⇒ x4 – 10x3 + 29x2 – 48x2 + 480x – 1932 = 0
⇒ x4 – 10x3 – 19x2 + 480x – 1932 = 0

(ii) 1 + 5i, 5 – i
సాధన:
బహుపది సమీకరణ గుణకాలు అకరణీయ సంఖ్యలైన, దాని మూలాలు సంయుగ్మ కరణులు మరియు సంయుగ్మసంకీర్ణ సంఖ్యలు.
α = 1 + 5i అయిన β = 1 – 5i
మరియు γ = 5 + i అయిన δ = 5 – i లు మూలాలు.
α + β = 2, αβ = 26
γ + δ = 10, γδ = 26
కావలసిన సమీకరణం [x2 – (α + β)x + αβ] [x2 – (γ + δ)x + γδ] = 0
⇒ (x2 – 2x + 26) (x2 – 10x + 26) = 0
⇒ x4 – 12x3 + 72x2 – 312x + 676 = 0

(iii) i – √5
సాధన:
బహుపది సమీకరణ గుణకాలు అకరణీయ సంఖ్యలైన, దాని మూలాలు సంయుగ్మ కరణులు మరియు సంయుగ్మసంకీర్ణ సంఖ్యలు.
α = i – √5, β = i + √5, γ = -i – √5, δ = -i + √5 లు మూలాలు
α + β = 2i, αβ = -6
γ + δ = -2i, γδ = -6
కావలసిన సమీకరణం [x2 – (α + β)x + αβ] [x2 – (γ + δ)x + γδ] = 0
⇒ (x2 – 2ix – 6) (x2 + 2ix – 6) = 0
⇒ [(x2 – 6) – 2ix] [(x2 – 6) + 2ix] = 0
⇒ (x2 – 6)2 + 4x2 = 0
⇒ x4 + 36 – 12x2 + 4×2 = 0
⇒ x4 – 8x2 + 36 = 0

(iv) -√3 + i√2
సాధన:
α = -√3 + i√2, β = -√3 – i√2, γ = √3 – i√2, δ = √3 + i√2 లు మూలాలు
α + β = -2√3
αβ = (-√3)2 – (i√2)2
= 3 – i2 (2)
= 5
γ + δ = 2√3, γδ = 5
కావలసిన సమీకరణము [x2 – (α + β)x + αβ] [x2 – (γ + δ)x + γδ] = 0
⇒ (x2 + 2√3x + 5) (x2 – 2√3x + 5) = 0
⇒ (x2 + 5)2 – (2√3x)2 = 0
⇒ x4 + 25 + 10x2 – 12x2 = 0
⇒ x4 – 2x2 + 25 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

II.

ప్రశ్న 1.
x4 + 2x3 – 5x2 + 6x + 2 = సమీకరణపు ఒక మూలం 1 + i అయిన, సమీకరణాన్ని సాధించండి.
సాధన:
1 + i ఒక మూలం ⇒ 1 – i ఇంకొక మూలం అవుతుంది.
1 ± i మూలాలుగా గల సమీకరణం
x2 – 2x + 2 = 0
∴ x2 – 2x + 2 ఒక కారణాంకము
x4 + 2x3 – 5x2 + 6x + 2 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q1
x = -2 ± √3
∴ మూలాలు 1 ± i, -2 ± √3

ప్రశ్న 2.
3x3 – 4x2 + x + 88 = 0 సమీకరణపు ఒక మూలం 2 – √-7 అయిన, సమీకరణాన్ని సాధించండి.
సాధన:
2 – √-7 ⇒ 2 – √7i ఒక మూలం
⇒2 + √7i ఇంకొక మూలం
2 ± √7i మూలాలుగా గల సమీకరణం x2 – 4x + 11 = 0
∴ x2 – 4x + 11 దత్త సమీకరణానికి ఒక కారణాంకము
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q2
3x + 8 = 0
⇒ x = \(\frac{-8}{3}\)
∴ దత్త సమీకరణానికి మూలాలు 2 ± √7i, \(\frac{-8}{3}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

ప్రశ్న 3.
x4 – 4x2 + 8x + 35 = 0 సమీకరణపు ఒక మూలం 2 + i√3 అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
2 + i√3 ఒక మూలం ⇒ 2 – i√3 ఇంకొక మూలం
2 ± i√3 మూలాలుగాగల సమీకరణం x2 – 4x + 7 = 0
∴ x2 – 4x + 7 దత్త సమీకరణానికి ఒక మూలం
x4 – 4x2 + 8x + 35
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q3
∴ దత్త సమీకరణానికి మూలాలు 2 ± i√3, -2 ± i

ప్రశ్న 4.
x4 – 6x3 + 11x2 – 10x + 2 = 0 సమీకరణపు ఒక మూలం 2 + √3 అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
2 + √3 ఒక మూలం ⇒ 2 – √3 ఇంకొక మూలం.
2 ± √3 మూలాలుగాగల సమీకరణం x2 – 4x + 1 = 0
∴ x2 – 4x + 1 ఒక కారణాంకము
x4 – 6x3 + 11x2 – 10x + 2 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q4
∴ దత్త సమీకరణానికి 2 ± √3, 1 ± i

ప్రశ్న 5.
x4 + 2x2 – 16x + 77 = 0 సమీకరణపు ఒక మూలం -2 + √-7 అయితే, సమీకరణాన్ని పూర్తిగా సాధించండి.
సాధన:
-2 – √-7 (i.e.) -2 + i√7 ఒక మూలం.
⇒ -2 – i√7 ఇంకొక మూలం -2 + i√7
-2 ± i√7 మూలాలుగా గల సమీకరణం x2 + 4x + 11 = 0
∴ x2 + 4x + 11 ఒక కారణాంకము
x4 + 2x2 – 16x + 77 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q5
∴ దత్త సమీకరణానికి మూలాలు -2 ± i√7, 2 ± √3i

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

ప్రశ్న 6.
x4 + 2x3 – 16x2 – 22x + 7 = 0 సమీకరణపు ఒక మూలం 2 – √3 అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
2 – √3 ఒక మూలం ⇒ 2 + √3 ఇంకొక మూలం
2 ± √3 లు మూలాలుగా గల వర్గ సమీకరణం
x2 – (2 + √3 + 2 – √3)x + (2 + √3) (2 – √3) = 0
⇒ x2 – 4x + 1 = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q6
∴ దత్త సమీకరణానికి మూలాలు 2 ± √3, -3 ± √2

ప్రశ్న 7.
3x5 – 4x4 – 42x3 + 56x2 + 27x – 36 = 0 సమీకరణానికి ఒక మూలం √2 + √5 అయితే, సమీక రణాన్ని సాధించండి.
సాధన:
√2 + √5 ఒక మూలం
⇒ √2 – √5, -√2 + √5, -√2 – √5 లు కూడా దత్తసమీకరణానికి మూలాలు.
√2 ± √5 మూలాలుగా గల వర్గ సమీకరణం
x2 – (√2 + √5 + √2 – √5)x + (√2 + √5) (√2 – √5) = 0
⇒ x2 – 2√2x – 3 = 0
-√2 ± √5 లు మూలాలుగా గల వర్గ సమీకరణం
x2 – (-√2 + √5 – √2 – √5)x + (-√2 + √5)(-√2 – √5) = 0
⇒ x2 + 2√2x – 3 = 0
±√2±√5 లు మూలాలుగా గల సమీకరణం
(x2 + 2√2x – 3) (x2 – 2√2x – 3) = 0
⇒ (x2 – 3)2 – (2√2x)2 = 0
⇒ x4 – 6x2 + 9 – 8×2 = 0
⇒ x4 – 14x2 + 9 = 0
3x5 – 4x4 – 42x3 + 56x2 + 27x – 36 = 0
⇒ 3x(x4 – 14x2 + 9) – 4(x4 – 14x2 + 9) = 0
⇒ (x4 – 14x2 + 9) (3x – 4) = 0
⇒ x = ±√2 ± √5 లేదా \(\frac{4}{3}\)
∴ దత్త సమీకరణానికి మూలాలు ±√2 ± √5, \(\frac{4}{3}\)

ప్రశ్న 8.
x4 – 9x3 + 27x2 – 29x + 6 = 0 సమీకరణపు ఒక మూలం 2 – √3 అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
2 – √3 ఒక మూలం ⇒ 2 + √3 ఇంకొక మూలం.
2 ± √3 లు మూలాలుగా గల సమీకరణం
x2 – 4x + 1 = 0
∴ x2 – 4x + 1 ఒక కారణాంకము
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q8
x2 – 5x + 6 = 0
⇒ (x – 2) (x – 3) = 0
⇒ x = 2, 3
∴ దత్త సమీకరణానికి మూలాలు 2 ± √3, 2, 3

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c)

ప్రశ్న 9.
a, b, c…. k, m, a’, b’, c’….k’ లు అన్నీ వాస్తవ సంఖ్యలైనపుడు \(\frac{a^2}{x-a^{\prime}}+\frac{b^2}{x-b^{\prime}}+\frac{c^2}{x-c^{\prime}}\) +…..+ \(\frac{k^2}{x-k^{\prime}}\) = m సమీకరణం వాస్తవేతర మూలాన్ని కలిగి ఉండదని చూపండి.
సాధన:
దత్త సమీకరణానికి α + iβ ఒక మూలం అనుకోండి.
β ≠ 0 అనుకుందాం.
అపుడు α – iβ కూడా దత్త సమీకరణానికి మూలం అవుతుంది.
దత్త సమీకరణంలో α + iβ వ్రాయగా
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(c) II Q9
= 0
⇒ β = 0
ఇది అనుకొన్నదానికి విరుద్ధం.
∴ దత్త సమీకరణానికి వాస్తవేతర మూలాలు ఉండవు.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.3

10th Class Maths 6th Lesson Progressions Ex 6.3 Textbook Questions and Answers

Question 1.
Find the sum of the following APs:
i) 2, 7, 12,…, to 10 terms.
Answer:
Given A.P: 2, 7, 12, …… to 10 terms
a = 2; d = a2 – a1 = 7 – 2 = 5; n = 10
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S10 = \(\frac{10}{2}\)[2 × 2 + (10 – 1)5]
= 5 [4 + 9 × 5]
= 5 [4 + 45]
= 5 × 49
= 245

ii) -37, -33, -29,…, to 12 terms.
Answer:
Given A.P: -37, -33, -29,…, to 12 terms.
a = -37; d = a2 – a1 = (-33) – (-37) = -33 + 37 = 4; n = 12
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S12 = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6 [-74 + 11 × 4]
= 6 [-74 + 44]
= 6 × (-30)
= -180

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

iii) 0.6, 1.7, 2.8,…, to 100 terms.
Answer:
Given A.P : 0.6, 1.7, 2.8,…. S100
a = 0.6; d = a2 – a1 = 1.7 – 0.6 = 1.1; n = 100
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S100 = \(\frac{100}{2}\)[2 × 0.6 + (100 – 1)1.1]
= 50 [1.2 + 99 × 1.1]
= 50 [1.2 + 108.9]
= 50 × 110.1
= 5505

iv) \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\),…, to 11 terms.
Answer:
Given A.P: \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\),…, S11
a = \(\frac{1}{15}\); d = a2 – a1 = \(\frac{1}{12}\) – \(\frac{1}{15}\) = \(\frac{5-4}{60}\) = \(\frac{1}{60}\); n = 11
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 1

2. Find the sums given below =:
i) 7 + 10\(\frac{1}{2}\) + 14 + …. + 84
Answer:
Given A.P : 7 + 10\(\frac{1}{2}\) + 14 + …. + 84
a = 7; d = a2 – a1 = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\) and the last term l = an = 84
But, an = a + (n – 1) d
∴ 84 = 7 + (n – 1) 3\(\frac{1}{2}\)
⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\)
⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22
⇒ n = 22 + 1 = 23
Now, Sn = \(\frac{n}{2}\)(a + l) where a = 7; l = 84
S23 = \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91
= \(\frac{2093}{2}\)
= 1046\(\frac{1}{2}\)

ii) 34 + 32 + 30 + … + 10
Answer:
Given A.P: 34 + 32 + 30 + … + 10
a = 34; d = a2 – a1 = 32 – 34 = -2 and the last term l = an = 10
But, an = a + (n – 1) d
∴ 10 = 34 + (n – 1) (-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) = 13
∴ n = 13
Also, Sn = \(\frac{n}{2}\)(a + l)
where a = 34; l = 10
S13 = \(\frac{13}{2}\)(34 + 10)
= \(\frac{13}{2}\) × 44
= 13 × 22
= 286

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

iii) -5 + (-8) + (-11) + … + (-230)
Answer:
Given A.P: -5 + (-8) + (-11) + … + (-230)
Here first term, a = -5;
d = a2 – a1 = (-8) – (-5) = -8 + 5 = -3 and the last term l = an = 10
But, an = a + (n – 1) d
∴ (-230) = -5 + (n – 1) (-3)
⇒ -230 + 5 = -3n + 3
⇒ -3n + 3 = -225
⇒ -3n = -225 – 3
⇒ 3n = 228
⇒ n = \(\frac{228}{3}\) = 76
∴ n = 76
Now, Sn = \(\frac{n}{2}\)(a + l)
where a = -5; l = -230
S76 = \(\frac{76}{2}\)((-5) + (-230))
= 38 × (-235)
= -8930

Question 3.
In an AP:
i) Given a = 5, d = 3, an = 50. find n and Sn.
Answer:
Given :
a = 5; d = 3;
an = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1) 3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
Now, Sn = \(\frac{n}{2}\)(a + l)
S16 = \(\frac{16}{2}\)(5 + 50)
= 38 × 55
= 440

ii) Given a = 7, a13 = 35, find d and S13.
Answer:
Given: a = 7;
a13 = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ n = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, Sn = \(\frac{n}{2}\)(a + l)
S13 = \(\frac{13}{2}\)(7 + 35)
= \(\frac{13}{2}\) × 42
= 13 × 21
= 273

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

iii) Given a12 = 37, d = 3 find a and S12.
Answer:
Given:
a12 = a + 11d = 37
d = 3
So, a12 = a + 11 × 3 = 37
⇒ a + 33 = 37
⇒ a = 37 – 33 = 4
Now, Sn = \(\frac{n}{2}\)(a + l)
S12 = \(\frac{12}{2}\)(4 + 37)
= 6 × 41
= 246

iv) Given a3 = 15, S10 = 125, find d and a10.
Answer:
Given:
a3 = a + 2d = 15
⇒ a = 15 – 2d ……… (1)
S10 = 125 but take S10 as 175
i.e., S10 = 175
We know that,
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 2
⇒ 35 = 2 (15 – 2d) + 9d [∵ a = 15 – 2d]
⇒ 35 = 30 – 4d + 9d
⇒ 35 – 30 = 5d
⇒ d = \(\frac{5}{5}\) = 1
Substituting d = 1 in equation (1) we get
a = 15 – 2 × 1 = 15 – 2 = 13
Now, an = a + (n – 1) d
a10 = a + 9d = 13 + 9 × 1 = 13 + 9 = 22
∴ a10 = 22; d = 1

v) Given a = 2, d = 8, Sn = 90, find n and an.
Answer:
Given a = 2, d = 8, Sn = 90
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 3
⇒ 90 = 2n [2n – 1]
⇒ 4n2 – 2n = 90
⇒ 4n2 – 2n – 90 = 0
⇒ 2(2n2 – n – 45) = 0
⇒ 2n2 – n – 45 = 0
⇒ 2n2 -10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (n – 5)(2n + 9) = 0
⇒ n – 5 = 0 (or) 2n + 9 = 0
⇒ n = 5 (or) n = \(\frac{-9}{2}\) (discarded)
∴ n = 5
Now an = a5 = a + 4d = 2 + 4 x 8
= 2 + 32 = 34

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

vi) Given an = 4, d = 2, Sn = -14, find n and a.
Answer:
Given an = a + (n – 1) d = 4 ……. (1)
d = 2; Sn = – 14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2
a = 6 – 2n
Given a = 2, d = 8, Sn = 90
Sn = \(\frac{n}{2}\)[a + an]
-14 = \(\frac{n}{2}\)[(6-2n) + 4] [∵ a = 6 – 2n]
-14 × 2 = n (10 – 2n)
⇒ 10n – 2n2 = – 28
⇒ 2n2 – 10n – 28 = 0
⇒ n2 – 5n – 14 = 0
⇒ n2 – 7n + 2n – 14 = 0
⇒ n (n – 7) + 2 (n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 (or) n = – 2
∴ n = 7
Now a = 6 – 2n = 6 – 2 × 7
= 6 – 14 = -8
∴ a = – 8; n = 7

vii) Given l = 28, S = 144, and there are total 9 terms. Find a.
Answer:
Given:
l = a9 = a + 8d = 28 and S9 = 144 But,
Now, Sn = \(\frac{n}{2}\)(a + l)
144 = \(\frac{9}{2}\)(a + 28)
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 4

Question 4.
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
Given A.P in which a = 17
Last term = l = 350
Common difference, d = 9
We know that, an = a + (n – 1) d
350 = 17 + (n- 1) 9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, Sn = \(\frac{n}{2}\)(a + l)
S38 = \(\frac{38}{2}\)(17 + 350)
= 19 × 367 = 6973
∴ n = 38; Sn = 6973

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 5.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Given A.P in which
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 4
Substituting d = 4 in equation (1),
we get a + 4 = 14
⇒ a = 14 – 4 = 10
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 5

Question 6.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer:
Given :
A.P such that S7 = 49; S17 = 289
We know that,
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 6
Substituting d = 2 in equation (1), we get,
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 2
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 7
∴ Sum of first n terms Sn = n2.
Shortcut: S7 = 49 = 72
S17 = 289 = 172
∴ Sn = n2

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 7.
Show that a1, a2 …,an, …. form an AP where an is defined as below:
i) a = 3 + 4n
ii) an = 9 – 5n. Also find the sum of the first 15 terms in each case.
Answer:
Given an = 3 + 4n
Then a1 = 3 + 4 × l = 3 + 4 = 7
a2 = 3 + 4 × 2 = 3 + 8 = 11
a3 = 3 + 4 × 3 = 3 + 12 = 15
a4 = 3 + 4 × 4 = 3 + 16 = 19
Now the pattern is 7, 11, 15, ……
where a = a1 = 7; a2 = 11; a3 = 15, ….. and
a2 – a1 = 11 – 7 = 4;
a3 – a2 = 15 – 11 = 4;
Here d = 4
Hence a1, a2, ….., an ….. forms an A.P.
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 8
ii) an = 9 – 5n
Given: an = 9 – 5n.
a1 = 9 – 5 × l = 9 – 5 = 4
a2 = 9 – 5 × 2 = 9 – 10 = -1
a3 = 9 – 5 × 3 = 9 – 15 = -6
a4 = 9 – 5 × 4 = 9 – 20 = -11
Also
a2 – a1 = -1 – 4 = -5;
a3 – a2 = -6 – (-1) = – 6 + 1 = -5
a4 – a3 = -11 – (-6) = -11 + 6 = -5
∴ d = a2 – a1 = a3 – a2 = a4 – a3 = …. = -5
Thus the difference between any two successive terms is constant (or) starting from the second term, each term is obtained by adding a fixed number ‘-5’ to its preceding term.
Hence {an} forms an A.P.
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 9

Question 8.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (remember the first term is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Answer:
Given an A.P in which Sn = 4n – n2
Taking n = 1 we get
S1 = 4 × 1- 12 = 4 – 1 = 3
n = 2; S2 = a1 + a2 = 4 × 2 – 22 = 8 – 4 = 4
n = 3; S3 = a1 + a2 + a3 = 4 × 3 -32 = 12 – 9 = 3
n = 4; S4 = a1 + a2 + a3 + a4 = 4 × 4 – 42 = 16 – 16 = 0
Hence, S1 = a1 = 3
a2 = S2 – S1 = 4 – 3 = 1
a3 = S3 – S2 = 3 – 4 = -1
a4 = S4 – S3 = 0 – 3 = -3
So, d = a2 – a1 = l – 3 = -2
Now, a10 = a + 9d  [∵ an = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
an = 3 + (n – 1) × (-2)
= 3 – 2n + 2
= 5 – 2n

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 9.
Find the sum of the first 40 positive integers divisible by 6.
Answer:
The given numbers are the first 40 positive multiples of 6
⇒ 6 × 1, 6 × 2, 6 × 3, ….., 6 × 40
⇒ 6, 12, 18, ….. 240
Sn = \(\frac{n}{2}\)(a + l)
S40 = \(\frac{40}{2}\)(6 + 240)
= 20 × 246
= 4920
∴ S40 = 4920

Question 10.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer:
Given:
Total/Sum of all cash prizes = Rs. 700
Each prize differs by Rs. 20
Let the prizes (in ascending order) be x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120
∴ Sum of the prizes = S7 = \(\frac{n}{2}\)(a + l)
⇒ 700 = \(\frac{7}{2}\)[x + x + 120]
⇒ 700 × \(\frac{2}{7}\) = 2x + 120
⇒ 100 = x + 60
⇒ x = 100 – 60 = 40
∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Question 11.
In a school, students thought of plant¬ing trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer:
Given: Classes: From I to XII
Section: 3 in each class.
∴ Trees planted by each class = 3 × class number
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 10
∴ Total trees planted = 3 + 6 + 9 + 12 + …… + 36 is an A.P.
Here, a = 3 and l = 36; n = 12
∴ Sn = \(\frac{n}{2}\)(a + l)
S12 = \(\frac{12}{2}\)[3 + 36]
= 6 × 39
= 234
∴ Total plants = 234

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 12.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in figure. What is the total length of such a spiral made up of thirteen
consecutive semicircles? (Take π = \(\frac{22}{7}\))
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 11
[Hint: Length of successive semicircles is l1, l2, l3, l4,….. with centres at A, B, A, B,…, respectively.]
Answer:
Given: l1, l2, l3, l4,….., l13 are the semicircles with centres alternately at A and B; with radii
r1 = 0.5 cm [1 × 0.5]
r2 = 1.0 cm [2 × 0.5]
r3 = 1.5 cm [3 × 0.5]
r4 = 2.0 cm [4 × 0.5] [∵ Radii are in A.P. as aj = 0.5 and d = 0.5]
……………………………
r13 = 13 × 0.5 = 6.5
Now, the total length of the spiral = l1 + l2 + l3 + l4 + ….. + l13 [∵ 13 given]
But circumference of a semi-cirle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ………. + π × 6.5
= π × \(\frac{1}{2}\)[l + 2 + 3 + ….. + 13]
[∵ Sum of the first n – natural numbers is \(\frac{n(n+1)}{2}\)
= \(\frac{22}{7} \times \frac{1}{2} \times \frac{13 \times 14}{2}\)
= 11 × 13
= 143 cm.

Question 13.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 12
Answer:
Given: Total logs = 200
Number of logs stacked in the first row = 20
Number of logs stacked in the second row = 19
Number of logs stacked in the third row = 18
The number series is 20, 19, 18,….. is an A.P where a = 20 and
d = a2 – a1 = 19 – 20 = -1
Also, Sn = 200
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 13
400 = 41n – n2
⇒ n2 – 41n + 400 = 0
⇒ n2 – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n – 25 (or) 16
There can’t be 25 rows as we are starting with 20 logs in the first row.
∴ Number of rows must be 16.
∴ n = 16

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

Question 14.
In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 14
A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Answer:
Given: Balls are placed at an equal distance of 3 m from one another.
Distance of first ball from the bucket = 5 m
Distance of second ball from the bucket = 5 + 3 = 8 m (5 + 1 × 3)
Distance of third ball from the bucket = 8 + 3 = 11 m (5 + 2 × 3)
Distance of fourth ball from the bucket = 11 + 3 = 14 m (5 + 3 × 3)
………………………………
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32 m.
1st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2nd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3rd ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
………………………………
10th ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 32 = 64 m.
Total distance = 10 m + 16 m + 22 m + …… + 64 m.
Clearly, this is an A.P in which a = 10; d = a2 – a1 = 16 – 10 = 6 and n = 10.
∴ Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S10 = \(\frac{10}{2}\)[2 × 10 + (10 – 1)6]
= 5 [20 + 54]
= 5 × 74
= 370 m
∴ Total distance = 370 m.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 4 సమీకరణ వాదం Exercise 4(b) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Exercise 4(b)

అభ్యాసం – 4(బి)

I.

ప్రశ్న 1.
x3 – 3x2 – 16x + 48 = 0 సమీకరణం రెండు మూలాల మొత్తం సున్నా అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
x3 – 3x2 – 16x + 48 = 0 కు మూలాలు α, β, γ లు
α + β + γ = 3
α + β = 0 (∵ రెండు మూలాల మొత్తం సున్న)
∴ γ = 3
i.e., x – 3 అనేది
x3 – 3x2 – 16x + 48 కు కారణాంకం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) I Q1
x2 – 16 = 0
⇒ x2 = 16
⇒ x = ±4
∴ మూలాలు -4, 4, 3

ప్రశ్న 2.
x3 – px2 + qx – r = 0 సమీకరణం యొక్క రెండు మూలాల మొత్తం సున్న కావటానికి నియమాన్ని కనుక్కోండి.
సాధన:
మూలాలు α, β, γ లు అనుకుందాం
అప్పుడు α + β + γ = p …….(1)
αβ + βγ + γα = q ……(2)
αβγ = r …….(3)
రెండు మూలాల మొత్తం సున్న కనుక α + β = 0 అనుకోండి.
(1) నుండి γ = p
‘γ’ దత్త సమీకరణానికి మూలం కనుక
γ3 – pγ2 + qγ – r = 0
⇒ p3 – p(p2) + q(p) – r = 0
⇒ r = pq
∴ కావలసిన నియమం r = pq

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 3.
x3 + 3px2 + 3qx + r = 0 సమీకరణపు మూలాలు
(i) అంకశ్రేఢిలో వుంటే 2p3 – 3pq + r = 0
(ii) గుణశ్రేఢిలో వుంటే p3r = q3
(iii) హరాత్మక శ్రేఢిలో వుంటే 2q3 = r(3pq – r) అని చూపండి.
సాధన:
దత్త సమీకరణం x3 + 3px2 + 3qx + r = 0
(i) మూలాలు అంకశ్రేఢిలో వున్నవి కనుక అవి
a – d, a, a + d అనుకుందాం. అప్పుడు
(a – d) + a + (a + d) = -3p
⇒ 3a = -3p
⇒ a = -p
‘a’ దత్త సమీకరణానికి మూలం, కనుక
a3 + 3pa2 + 3qa + r = 0
⇒ (-p)3 + 3p(-p)2 + 3q(-p) + r = 0
⇒ -p3 + 3p3 – 3pq + r = 0
⇒ 2p3 – 3pq + r = 0
∴ కావలసిన నియమం 2p3 – 3pq + r = 0

(ii) మూలాలు గుణశ్రేఢిలో వున్నవి కనుక అవి \(\frac{a}{R}\), a, aR అనుకుందాం.
అపుడు మూలాల లబ్ధం = (\(\frac{a}{R}\)) (a) (aR) = -r
⇒ a3 = -r
⇒ a = \((-r)^{1 / 3}\)
‘a’ దత్త సమీకరణానికి మూలం కనుక
a3 + 3pa2 + 3qa + r = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) I Q3(ii)
∴ కావలసిన నియమం p3r = q3

(iii) దత్త సమీకరణం x3 + 3px2 + 3qx + r = 0 …….(1)
x = \(\frac{1}{y}\) వ్రాయగా
\(\left(\frac{1}{y}\right)^3+3 p\left(\frac{1}{y}\right)^2+3 q\left(\frac{1}{y}\right)+r=0\)
⇒ 1 + 3py + 3qy2 + ry3 = 0
⇒ ry3 + 3qy2 + 3py + 1 = 0 ………(2)
(1) యొక్క మూలాలు హరాత్మక శ్రేఢిలో వుంటే, (2) యొక్క మూలాలు అంకశ్రేఢిలో వుంటాయి.
కనుక అవి a – d, a, a + d అనుకుందాం.
(a – d) + a + (a + d) = \(\frac{-3q}{r}\)
⇒ 3a = \(\frac{-3q}{r}\)
⇒ a = \(\frac{-q}{r}\)
‘a’ అనేది (2) కు ఒక మూలం కనుక
ra3 + 3qa2 + 3pa + 1 = 0
⇒ \(r\left(\frac{-q}{r}\right)^3+3 q\left(\frac{-q}{r}\right)^2+3 p\left(\frac{-q}{r}\right)+1=0\)
⇒ \(\frac{-q^3}{r^2}+\frac{3 q^3}{r^2}+\frac{3 p q}{r}+1=0\)
⇒ -q3 + 3q3 – 3pqr + r2 = 0
⇒ r2 – 3pqr + 2q3 = 0
⇒ 2q3 = r(3pq – r)
∴ కావలసిన నియమం 2q3 = r(3pq – r)

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 4.
x3 – px2 + qx – r = 0 సమీకరణ మూలాలు గుణ శ్రేఢిలో ఉండటానికి నియమాన్ని రాబట్టుము.
సాధన:
మూలాలు గుణశ్రేఢిలో వున్నవి. కనుక అవి \(\frac{a}{R}\), a, aR అనుకోండి.
అపుడు మూలాల లబ్దం = (\(\frac{a}{R}\)) (a) (aR) = r
⇒ a3 = r
⇒ a = \(r^{1 / 3}\)
‘a’ దత్త సమీకరణానికి ఒక మూలం కనుక
a3 – pa2 + qa – r = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) I Q4
∴ కావలసిన నియమం p3r = q3

II.

ప్రశ్న 1.
9x3 – 15x2 + 7x – 1 = 0 సమీకరణపు రెండు మూలాలు సమానమైతే, సమీకరణాన్ని సాధించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q1
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q1.1

ప్రశ్న 2.
2x3 + 3x2 – 8x + 3 = 0 సమీకరణపు ఒక మూలం, మరోదానికి రెట్టింపు అయిన, ఆ సమీకరణాన్ని సాధించండి.
సాధన:
α, β, γ లు 2x3 + 3x2 – 8x + 3 = 0 మూలాలు అనుకుందాం.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q2
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q2.1
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q2.2

ప్రశ్న 3.
x3 – 9x2 + 14x + 24 = 0 సమీకరణపు రెండు మూలాలు 3 : 2 నిష్పత్తిలో ఉంటే, ఆ సమీకరణాన్ని సాధించండి.
సాధన:
మూలాలు α, β, γ లు అనుకుందాం.
α + β + γ = 9 …….(1)
αβ + βγ + γα = 14 ……..(2)
αβγ = -24 ………(3)
α : β = 3 : 2 అనుకుందాం.
2α = 3β
⇒ β = \(\frac{2}{3}\)α
(1) నుండి α + \(\frac{2}{3}\)α + γ = 9
γ = 9 – \(\frac{5 \alpha}{3}\) …….(4)
(2) నుండి (α) (\(\frac{2}{3}\)α) + γ(α + β) = 14
⇒ \(\frac{2}{3} \alpha^2+\left(9-\frac{5 \alpha}{3}\right)\left(\alpha+\frac{2}{3} \alpha\right)=14\)
⇒ \(\frac{2}{3} \alpha^2+\left(9-\frac{5}{3} \alpha\right)\left(\frac{5}{3} \alpha\right)=14\)
⇒ 6α2 + (27 – 5α) (5α) = 126
⇒ 6α2 + 135α – 25α2 – 126 = 0
⇒ 19α2 – 135α + 126 = 0
⇒ 19α2 – 114α – 21α + 126 = 0
⇒ 19α(α – 6) – 21(α – 6) = 0
⇒ (α – 6) (19α – 21) = 0
⇒ α = 6 (లేదా) α = \(\frac{21}{19}\)
సందర్భము (i): α = 6
β = \(\frac{2}{3}\)α
= \(\frac{2}{3}\) × 6 = 4
γ = 9 – \(\frac{5 \alpha}{3}\)
γ = 9 – \(\frac{2}{3}\) × 6
= 9 – 10
= -1
α = 6, β = 4, γ = -1
αβγ = -24
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q3
కానీ αβγ ≠ -24
∴ మూలాలు = 6, 4, -1

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 4.
మూలాలు అంకశ్రేఢిలో ఉన్న క్రింది సమీకరణాలను సాధించండి.
(i) 8x3 – 36x2 – 18x + 81 = 0
సాధన:
మూలాలు అంకశ్రేఢిలో వున్నవి కనుక అవి a – d, a, a + d అనుకొనుము.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q4(i)

(ii) x3 – 3x2 – 6x + 8 = 0
సాధన:
మూలాలు అంకశ్రేఢిలో ఉన్నవి కనుక మూలాలు a – d, a, a + d అనుకుందాం.
(a – d) + a + (a + d) = -(-3)= 3
⇒ 3a = 3
⇒ a = 1
(a – d) (a) (a + d) = -8
⇒ a(a2 – d2) = -8
⇒ 1(1 – d2) = -8
⇒ d2 = 1 + 8 = 9
⇒ d = 3
మూలాలు a – d, a, a + d
= 1 – 3, 1, 1 + 3
= -2, 1, 4

ప్రశ్న 5.
మూలాలు గుణశ్రేఢిలో ఉన్న క్రింది సమీకరణాలను సాధించండి.
(i) 3x3 – 26x2 + 52x – 24 = 0
సాధన:
మూలాలు గుణశ్రేఢిలో ఉన్నవి కనుక అవి \(\frac{a}{r}\), a, ar అనుకుందాం.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q5(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q5(i).1

(ii) 54x3 – 39x2 – 26x + 16 = 0
సాధన:
మూలాలు గుణశ్రేఢిలో వున్నవి కనుక అవి \(\frac{a}{r}\), a, ar అనుకుందాం.
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q5(ii)
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q5(ii).1

ప్రశ్న 6.
మూలాలు హరాత్మకశ్రేఢిలో ఉన్న క్రింది సమీకరణాలను సాధించండి.
(i) 6x3 – 11x2 + 6x – 1 = 0
సాధన:
దత్త సమీకరణం 6x3 – 11x2 + 6x – 1 = 0 ………(1)
y = \(\frac{1}{x}\) వ్రాయగా
\(6\left(\frac{1}{y}\right)^3-11\left(\frac{1}{y}\right)^2+6\left(\frac{1}{y}\right)-1=0\)
⇒ 6 – 11y + 6y2 – y3 = 0
⇒ y3 – 6y2 + 11y – 6 = 0 ……(2)
(1) యొక్క మూలాలు హరాత్మక శ్రేఢిలో వుంటే, (2) యొక్క మూలాలు అంకశ్రేఢిలో వుంటాయి.
(2) యొక్క మూలాలు a – d, a, a + d అనుకుంటే,
(a – d) + a + (a + d) = \(-\frac{(-6)}{1}\) = 6
⇒ 3a = 6
⇒ a = 2
మరియు (a – d) (a) (a + d) = \(-\frac{(-6)}{1}\) = 6
⇒ a(a2 – d2) = 6
⇒ 2(22 – d2) = 6
⇒ 4 – d2 = 3
⇒ d2 = 1
⇒ d = 1
(2) మూలాలు a – d, a, a + d
= 2 – 1, 2, 2 + 1
= 1, 2, 3
∴ దత్త సమీకరణానికి మూలాలు = 1, \(\frac{1}{2}\), \(\frac{1}{3}\)

(ii) 15x3 – 23x2 + 9x – 1 = 0
సాధన:
దత్త సమీకరణం 15x3 – 23x2 + 9x – 1 = 0 …….(1)
x = \(\frac{1}{y}\) వ్రాయగా
\(15\left(\frac{1}{y}\right)^3-23\left(\frac{1}{y}\right)^2+9\left(\frac{1}{y}\right)-1=0\)
⇒ 15 – 23y + 9y2 – y3 = 0
⇒ y3 – 9y2 + 23y – 15 = 0 ………(2)
(1) యొక్క మూలాలు హరాత్మక శ్రేఢిలో (H.P.) వుంటే, (2) యొక్క మూలాలు అంకశ్రేఢీలో ఉంటాయి.
కనుక ఆ మూలాలు a – d, a, a + d అనుకుందాం.
అప్పుడు (a – d) + (a) + (a + d) = -(-9) = 9
⇒ 3a = 9
⇒ a = 3
మరియు (a – d) (a) (a + d) = -(-15)
⇒ a(a2 – d2) = 15
⇒ 3(32 – d2) = 15
⇒ 9 – d2 = 5
⇒ d2 = 4
⇒ d = 2
∴ (2) యొక్క మూలాలు a – d, a, a + d
= 3 – 2, 3, 3 + 2
= 1, 3, 5
∴ దత్త సమీకరణానికి = 1, \(\frac{1}{3}\), \(\frac{1}{5}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 7.
పునరావృత మూలాలున్న క్రింది సమీకరణాలను సాధించండి.
(i) x4 – 6x3 + 13x2 – 24x + 36 = 0
సాధన:
f(x) = x4 – 6x3 + 13x2 – 24x + 36
f'(x) = 4x3 – 18x2 + 26x – 24
ఇప్పుడు f'(3) = 4(3)3 – 18(3)2 + 26(3) – 24
= 108 – 162 + 78 – 24
= 0
ఇట్లే f(3) = (3)4 – 6(3)3 + 13(3)2 – 24(3) + 36
= 81 – 162 + 117 – 72 + 36
= 0
కనుక x – 3; f(x), f'(x) లకు కారణాంకం
∴ f(x) = 0 కు 3 ఆవృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q7(i)
కనుక x2 + 4 = 0
⇒ x2 = -4
⇒ x = ±2i
∴ దత్త సమీకరణానికి 3, 3, 2i, -2i

(ii) 3x4 + 16x3 + 24x2 – 16 = 0
సాధన:
f(x) = 3x4 + 16x3 + 24x2 – 16
f'(x) = 12x3 + 48x2 + 48x
= 12x(x2 + 4x + 4)
= 12x(x + 2)2
f'(-2) = 0
f(-2) = 3(-2)4 + 16(-2)3 + 24(-2)2 – 16
= 48 – 128 + 96 – 16
= 0
కనుక f(x), f'(x) లకు (x + 2) కారణాంకం
∴ f(x) = 0 కు -2 ఆవృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) II Q7(ii)
3x2 + 4x – 4 = 0
⇒ 3x2 + 6x – 2x – 4 = 0
⇒ 3x(x + 2) – 2(x + 2) = 0
⇒ (x + 2) (3x – 2) = 0
⇒ x = -2, x = \(\frac{2}{3}\)
∴ దత్త సమీకరణానికి -2, -2, -2, \(\frac{2}{3}\)

III.

ప్రశ్న 1.
x4+ x3 – 16x2 – 4x + 48 = 0 సమీకరణపు రెండు మూలాల లబ్ధం 6 అయితే ఆ సమీకరణాన్ని సాధించండి.
సాధన:
మూలాలు α, β, γ, δ లు అనుకుందాం.
α + β + γ + δ = -1 …….(1)
αβ + αδ + αγ + βγ + βδ + γδ = -16 …….(2)
αβγ + αβδ + βγδ + αγδ = -(-4) = 4 ……….(3)
αβγδ = 48 …….(4)
∵ రెండు మూలాల లబ్దం = 6 కనుక
αβ = 6 అనుకుందాం.
(4) నుండి 6γδ = 48 ⇒ γδ = 8
(3) నుండి 6γ + 6δ + 8β + 8α = 4
⇒ 6(γ + δ) + 8(α + β) = 4
6(γ + δ) + 6(α + β) = -6 – (1) × 6
2(α + β) = 10
α + β = 5
(1) నుండి γ + δ = -1 – 5 = -6
α + β = 5, αβ = 6
(α – β)2 = (α + β)2 – 4αβ
= (5)2 – 4(6)
= 1
α – β = 1
α + β = 5
2α = 6
⇒ α = 3, β = 2
ఇదే విధంగా γ + δ = -6, γδ = 8
(γ – δ)2 = (γ + δ)2 – 4γδ
= (-6)2 – 4(8)
= 36 – 32
= 4
γ – δ = 2
γ + δ = -6
2γ = -4
⇒ γ = -2, δ = -4
∴ దత్త సమీకరణానికి = 3, 2, -2, -4

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 2.
8x4 – 2x3 – 27x2 + 6x + 9 = 0 సమీకరణ రెండు మూలాలు ఒకే పరమమూల్యాన్నీ, వ్యతిరేక గుర్తులను కలిగి వుంటే, సమీకరణాన్ని సాధించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q2
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q2.1

ప్రశ్న 3.
18x3 + 81x2 + 121x + 60 = 0 సమీకరణపు ఒక మూలం తక్కిన రెండు మూలాల మొత్తంలో సగమైతే, సమీకరణాన్ని సాధించండి. [May ’11, Mar. ’05]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q3
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q3.1

ప్రశ్న 4.
ax4 + 4bx3 + 6cx2 + 4dx + c = 0 సమీకరణపు మూలాల్లో రెండు జతలు సమానంగా ఉండటానికి నియమాలను రాబట్టండి.
సాధన:
దత్త సమీకరణం ax4 + 4bx3 + 6cx2 + 4dx + e = 0
⇒ \(x^4+4 \frac{b}{a} x^3+6 \frac{c}{a} x^2+4 \frac{d}{a} x+\frac{e}{a}=0\)
మూలాలు α, α, β, β లు అనుకుందాం
అపుడు మూలాల మొత్తం 2(α + β) = -4\(\frac{b}{a}\)
⇒ α + β = -2\(\frac{b}{a}\)
αβ = k అనుకుంటే α, β లు మూలాలుగా గల వర్గ సమీకరణం
x2 – (α + β)x + αβ = 0
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q4
⇒ ad2 = eb2 ఇది మరొక నియమము
∴ కావలసిన నియమాలు 3abc = 2b3 + a2d, ad2 = eb2

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 5.
(i) x5 – 5x3 + 5x2 – 1 = 0 సమీకరణానికి 3 సమాన మూలాలు ఉంటాయని చూపండి. ఆ మూలాన్ని కనుక్కోండి.
సాధన:
f(x) = x5 – 5x3 + 5x2 – 1
f'(x) = 5x4 – 15x2 + 10x = 5x(x3 – 3x + 2)
f'(1) = 5(1) (1 – 3 + 2) = 0
f(1) = 1 – 5 + 5 – 1 = 0
కనుక f(x), f'(x) లకు (x – 1) కారణాంకం
⇒ f(x) = 0 కు 1 ఆవృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q5(i)
∴ x3 + 2x2 – 2x – 1 = 0
x = 1 పై సమీకరణాన్ని తృప్తిపరుస్తుంది.
కనుక f(x) = 0 కు 1 అనేది 3 సార్లు ఆవృత మూలము అవుతుంది.

(ii) x5 – 3x4 – 5x3 + 27x2 – 32x + 12 = 0 సమీకరణం యొక్క పునరావృత మూలాలను కనుక్కోండి
సాధన:
f(x) = x5 – 3x4 – 5x3 + 27x2 – 32x + 12 అనుకోండి.
f'(x) = 5x4 – 12x3 – 15x2 + 54x – 32
f'(2) = 5(2)4 – 12(2)3 – 15(2)2 + 54(2) – 32
= 80 – 96 – 60 + 108 – 32
= 188 – 188
= 0
f(2) = 25 – 3(24) – 5(23) + 27(22) – 32(2) + 12
= 32 – 48 – 40 + 108 – 64 + 12
= 152 – 152
= 0
కనుక f(x), f'(x) లకు x – 2 కారణాంకం
∴ f(x) = 0 కు 2 ఆవృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q5(ii)
g(x) = x3 + x2 – 5x + 3 అనుకుందాం
g'(x) = 3x2 + 2x – 5
g'(1) = 3(1)2 + 2 – 5 = 0
g(1) = 1 + 1 – 5 + 3 = 0
కనుక g(x), g'(x) లకు x – 1 కారణాంకం
∴ g(x) = 0 కు 1 ఆవృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q5(ii).1
x + 3 = 0 ⇒ x = -3
∴ మూలాలు = 1, 1, 2, 2, -3

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b)

ప్రశ్న 6.
8x3 – 20x2 + 6x + 9 = 0 సమీకరణం యొక్క పునరావృత మూలాలు కనుక్కోండి.
సాధన:
f(x) = 8x3 – 20x2 + 6x + 9 అనుకోండి
f'(x) = 24x2 – 40x + 6
= 2(12x2 – 20x + 3)
= 2[12x2 – 18x – 2x + 3]
= 2[6x(2x – 3) – 1(2x – 3)]
= 2(2x – 3) (6x – 1)
f'(x) = 0
⇒ x = \(\frac{3}{2}\), x = \(\frac{1}{6}\)
\(f\left(\frac{3}{2}\right)=8\left(\frac{3}{2}\right)^3-20\left(\frac{3}{2}\right)^2+6\left(\frac{3}{2}\right)+9\)
= 27 – 45 + 9 + 9
= 0
కనుక x – \(\frac{3}{2}\), f(x), f'(x) లకు కారణాంకం
∴ f(x) = 0 కు \(\frac{3}{2}\) పునరావృత మూలం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(b) III Q6
8x + 4 = 0 ⇒ x = \(-\frac{4}{8}=-\frac{1}{2}\)
∴ f(x) = 0 యొక్క మూలాలు \(\frac{3}{2}, \frac{3}{2},-\frac{1}{2}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 4 సమీకరణ వాదం Exercise 4(a) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Exercise 4(a)

అభ్యాసం – 4(ఎ)

I.

ప్రశ్న 1.
క్రింద ఇచ్చిన మూలాలు గల కనిష్ఠ తరగతి బహుపది సమీకరణాలను రూపొందించండి.
(i) 1, -1, 3
సాధన:
కావలసిన బహుపది సమీకరణం
(x – 1) (x + 1) (x – 3) = 0
⇒ (x2 – 1) (x – 3) = 0
⇒ x3 – 3x2 – x + 3 = 0

(ii) 1 ± 2i, 4, 2
సాధన:
కావలసిన బహుపది సమీకరణం
[x – (1 + 2i)] [x – (1 – 2i)] (x – 4) (x – 2) = 0
⇒ (x – 1 – 2i) (x – 1 + 2i) (x – 4) (x – 2) = 0
⇒ [(x – 1)2 – 4i2] (x2 – 6x + 8) = 0
⇒ (x2 – 2x + 1 + 4) (x2 – 6x + 8) = 0
⇒ (x2 – 2x + 5) (x2 – 6x + 8) = 0
⇒ x4 – 8x3 + 25x2 – 46x + 40 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

(iii) 2 ± √3, 1 ± 2i
సాధన:
కావలసిన బహుపది సమీకరణం
[x – (2 + √3)] [x – (2 – √3)] [x – (1 + 2i)] [x – (1 – 2i)] = 0
⇒ [(x – 2) – √3] [(x – 2) + √3] [(x – 1) – 2i] [(x – 1) + 2i] = 0
⇒ [(x – 2)2 – 3] [(x – 1)2 – (2i)2] = 0
⇒ (x2 – 4x + 4 – 3) (x2 – 2x + 1 – 4i2) = 0
⇒ (x2 – 4x + 1)(x2 – 2x + 5) = 0
⇒ x4 – 4x3 + x2 – 2x3+ 8x2 – 2x + 5x2 – 20x + 5 = 0
⇒ x4 – 6x3 + 14x2 – 22x + 5 = 0

(iv) 0, 0, 2, 2, -2, -2
సాధన:
కావలసిన బహుపది సమీకరణం
(x – 0) (x – 0) (x – 2) (x – 2) (x + 2) (x + 2) = 0
⇒ x2 (x – 2)2 (x + 2)2 = 0
⇒ x2 (x2 – 4)2 = 0
⇒ x2 (x4 + 16 – 8x2) = 0
⇒ x6 – 8x4 + 16x2 = 0

(v) 1 ± √3, 2, 5
సాధన:
కావలసిన బహుపది సమీకరణం
[x – (1 + √3)] [x – (1 – √3)] [(x – 2) (x – 5)] = 0
⇒ [(x – 1)- √3] [(x – 1) + √3] (x2 – 7x + 10) = 0
⇒ [(x – 1)2 – (√3)2] (x2 – 7x + 10) = 0
⇒ (x2 – 2x + 1 – 3) (x2 – 7x + 10) = 0
⇒ (x2 – 2x – 2) (x2 – 7x + 10) = 0
⇒ x4 – 2x3 – 2x2 – 7x3 + 14x2 + 14x + 10x2 – 20x – 20 = 0
⇒ x4 – 9x3 + 22x2 – 6x – 20 = 0

(vi) 0, 1, \(\frac{-3}{2}\), \(\frac{-5}{2}\)
సాధన:
కావలసిన బహుపది సమీకరణం
(x – 0) (x – 1) (x + \(\frac{3}{2}\)) (x + \(\frac{5}{2}\)) = 0
⇒ x(x – 1) (2x + 3) (2x + 5) = 0
⇒ (x2 – x) (4x2 + 16x + 15) = 0
⇒ 4x4 – 4x3 + 16x3 – 16x2 + 15x2 – 15x = 0
⇒ 4x4 + 12x3 – x2 – 15x = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

ప్రశ్న 2.
4x3 – 6x2 + 7x + 3 = 0 మూలాలు α, β, γ అయితే, αβ + βγ + γα విలువను కనుక్కోండి.
సాధన:
α, β, γ లు 4x3 – 6x2 + 7x + 3 = 0 మూలాలు
α + β + γ = \(-\frac{a_1}{a_0}=\frac{6}{4}\)
αβ + βγ + γα = \(\frac{a_2}{a_0}=\frac{7}{4}\)

ప్రశ్న 3.
x3 – 6x2 + 9x – 4 = 0 కు 1, 1, α లు మూలాలైన α విలువను కనుగొనుము. [May ’11]
సాధన:
1, 1, α లు x3 – 6x2 + 9x – 4 = 0 కు మూలాలు కనుక
మూలాల మొత్తం = 1 + 1 + α = \(-\left(-\frac{6}{1}\right)\) = 6
⇒ 2 + α = 6
⇒ α = 6 – 2 = 4

ప్రశ్న 4.
2x3 + x2 – 7x – 6 = 0 మూలాలు -1, 2, α అయితే, α ను కనుక్కోండి. [Mar. ’14]
సాధన:
-1, 2, α లు 2x3 + x2 – 7x – 6 = 0 కు మూలాలు
కనుక -1 + 2 + α = \(\frac{-1}{2}\)
α + 1 = \(\frac{-1}{2}\)
⇒ α = -1 – \(\frac{1}{2}\) = \(\frac{-3}{2}\)

ప్రశ్న 5.
x3 – 2x2 + ax + 6 = 0 కు మూలాలు 1, -2, 3 అయితే a ను కనుక్కోండి. [Mar. ’04]
సాధన:
x3 – 2x2 + ax + 6 = 0 కు 1 మూలం కనుక
(1)3 – 2(1)2 + a(1) + 6 = 0
⇒ a + 5 = 0
⇒ a = -5

ప్రశ్న 6.
4x3 + 16x2 – 9x – a = 0 సమీకరణ మూలాల లబ్ధం 9 అయిన a విలువను కనుగొనుము. [T.S. Mar. ’16]
సాధన:
α, β, γ మూలాల లబ్దం
4x3 + 16x2 – 9x – a = 0
αβγ = \(\frac{a}{4}\) = 9
⇒ a = 36

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

ప్రశ్న 7.
క్రింది సమీకరణాలకు s1, s2, s3, s4 లను కనుగొనుము.
(i) x4 – 16x3 + 86x2 – 176x + 105 = 0
(ii) 8x4 – 2x3 – 27x2 + 6x + 9 = 0
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) I Q7
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) I Q7.1

II.

ప్రశ్న 1.
x3 – 2x2 – 5x + 6 = 0 కు మూలాలు α, β, 1 అయితే α, β లను కనుక్కోండి. [Mar. ’08]
సాధన:
x3 – 2x2 – 5x + 6 = 0 కు α, β, 1 లు మూలాలు
α + β + 1 = 2
⇒ α + β = 1
లబ్ధం = αβ = -6
(α – β)2 = (α + β)2 – 4αβ
= 1 + 24
= 25
α – β = 5
α + β = 1
కలుపగా 2α = 6
⇒ α = 3
α + β = 1
⇒ β = 1 – α
= 1 – 3
= -2
∴ α = 3, β = -2

ప్రశ్న 2.
x3 – 2x2 + 3x – 4 = 0 మూలాలు α, β, γ అయితే, (i) Σα2β2 (ii) Σαβ(α + β) లను కనుక్కోండి. [May ’07]
సాధన:
x3 – 2x2 + 3x – 4 = 0 మూలాలు α, β, γ కనుక
α + β + γ = 2
αβ + βγ + γα = 3
αβγ = 4
(i) Σα2β2 = α2β2 + β2γ2 + γ2α2
= (αβ + βγ + γα)2 – 2αβγ(α + β + γ)
= 9 – 2 (2) (4)
= 9 – 16
= -7

(ii) Σαβ(α + β) = α2β + β2γ + γ2α + αβ2 + βγ2 + γα2
= (αβ + βγ + γα) (α + β + γ) – 3αβγ
= 2(3) – 3(4)
= 6 – 12
= -6

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

ప్రశ్న 3.
x3 + px2 + qx + r = 0 మూలాలు α, β, γ అయితే, క్రింది వాటిని కనుక్కోండి.
సాధన:
x3 + px2 + qx + r = 0 మూలాలు α, β, γ లు
కనుక α + β + γ = -p
αβ + βγ + γα = q
αβγ = -r

(i) \(\sum \frac{1}{\alpha^2 \beta^2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) II Q3(i)

(ii) \(\frac{\beta^2+\gamma^2}{\beta \gamma}+\frac{\gamma^2+\alpha^2}{\gamma \alpha}+\frac{\alpha^2+\beta^2}{\alpha \beta}\) లేదా \(\Sigma \frac{\beta^2+\gamma^2}{\beta \gamma}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) II Q3(ii)

(iii) (β + γ – 3α) (γ + α – 3β) (α + β – 3γ)
సాధన:
(β + γ – 3α) (γ + α – 3β) (α + β – 3γ)
= (α + β + γ – 4α) (α + β + γ – 4β) (α + β + γ – 4γ)
= (-p – 4α) (-p – 4β) (-p – 4γ)
= -(p + 4α) (p + 4β) (p + 4γ)
= -[(p3 + 4p2 (α + β + γ) + 16p (αβ + βγ + γα) + (64αβy)]
= -(p3 – 4p3 + 16pq – 64r)
= 3p3 – 16pq + 64r

(iv) Σα3β3
సాధన:
Σα3β3 = α3β3 + β3γ3 + γ3α3
(αβ + βγ + γα)2 = α2β2 + β2γ2 + γ2α2 + 2αβγ (α + β + γ)
q2 = α2β2 + β2γ2 + γ2α2 + 2pr
α2β2 + β2γ2 + γ2α2 = q2 – 2pr
∴ α3β3 + β3γ3 + γ3α3 = (α2β2 + β2γ2 + γ2α2) (αβ + βγ + γα) – αβγ Σα2β
= (q2 – 2pr) . q + r[(αβ + βγ + γα) (α + β + γ) – 3αβγ]
= q3 – 2pqr + r(-pq + 3r)
= q3 – 2pqr – pqr + 3r2
= q3 – 3pqr + 3r2

III.

ప్రశ్న 1.
x3 – 6x2 + 11x – 6 = 0 సమీకరణ మూలాలు α, β, γ అయితే, α2 + β2, β2 + γ2, γ2 + α2 మూలాలుగా గల సమీకరణాన్ని కనుక్కోండి.
సాధన:
1వ పద్ధతి:
α, β, γ లు x3 – 6x2 + 11x – 6 = 0 కు మూలాలు,
∴ α + β + γ = 6, αβ + βγ + γα = 11
y = α2 + β2 = α2 + β2 + γ2 – γ2 అనుకొనుము.
= (α + β + γ)2 – 2(αβ + βγ + γα) – x2 (∵ α, β, γ లు మూలాలు)
= 36 – 22 – x2
⇒ x2 = 14 – y
⇒ x = \(\sqrt{14-y}\) ను x3 – 6x2 + 11x – 6 = 0 లో వ్రాయగా
⇒ (\(\sqrt{14-y}\))3 – 6(\(\sqrt{14-y}\))2 + 11(\(\sqrt{14-y}\)) – 6 = 0
⇒ (14-y) \(\sqrt{14-y}\) – 6(14 – y) + 11\(\sqrt{14-y}\) – 6 = 0
⇒ -6(14 – y + 1) = \(\sqrt{14-y}\) [-11 – 14 + y]
⇒ -6(15 – y) = (\(\sqrt{14-y}\)) (y – 25)
ఇరువైపులా వర్గం చేయగా
i.e., [-6(15 – y)]2 = [\(\sqrt{14-y}\) (y – 25)]2
⇒ 36(225 – 30y + y2) = (14 – y) (y2 – 50y + 625)
⇒ 8100 – 1080y + 36y2 = 14y2 – 700y + 8750 – y3 + 50y2 – 625y
⇒ 8100 – 1080y + 36y2 = -y3 + 64y2 – 1325y + 8750
⇒ y3 – 28y2 + 245y – 650 = 0
∴ కావలసిన సమీకరణం x3 – 28x2 + 245x – 650 = 0
2వ పద్ధతి :
α, β, γ లు x3 – 6x2 + 11x – 6 = 0 కు మూలాలు,
ఇది రెండవ కోవకు చెందిన వ్యుతమ సమీకరణం
∴ x – 1 అనేది x3 – 6x2 + 11x – 6 కు ఒక కారణాంకం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) III Q1
∴ x3 – 6x2 + 11x – 6 = (x – 1) (x2 – 5x + 6) = (x – 1) (x – 2) (x – 3)
∴ x3 – 6x2 + 11x – 6 = 0 కు మూలాలు
α = 1, β = 2, γ = 3
ఇప్పుడు α2 + β2 = 12 + 22 = 5
β2 + γ2 = 22 + 32 = 13
γ2 + α2 = 32 + 12 = 10
α2 + β2, β2 + γ2, γ2 + α2 లు మూలాలుగా గల ఘన సమీకరణం (x – 5) (x – 13) (x – 10) = 0
⇒ x3 – (5 + 13 + 10)x2 +(65 + 130 + 50)x – 650 = 0
⇒ x3 – 28x2 + 245x – 650 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

ప్రశ్న 2.
x3 – 7x + 6 = 0, సమీకరణ మూలాలు α, β, γ అయితే (α – β)2, (β – γ)2, (γ – α)2 మూలాలుగా గల సమీకరణం కనుక్కోండి.
సాధన:
1వ పద్ధతి :
x3 – 7x + 6 = 0 …..(1) మూలాలు α, β, γ
కనుక α + β + γ = 0, αβγ = -6
y = (α – β)2 = (α + β)2 – 4αβ అనుకొనుము.
= (-γ)2 – 4(\(\frac{-6}{\gamma}\))
= γ2 + \(\frac{24}{\gamma}\)
= x2 + \(\frac{24}{x}\)
⇒ xy = x3 + 24
⇒ xy = 7x – 6 + 24 [(1) నుండి]
⇒ x(y – 7) = 18
⇒ x = \(\frac{18}{y-7}\)
x3 – 7x + 6 = 0 లో x = \(\frac{18}{y-7}\) ను వ్రాయగా
\(\left(\frac{18}{y-7}\right)^3-7\left(\frac{18}{y-7}\right)+6=0\)
⇒ (18)3 – 7(18) (y – 7)2 + 6(y – 7)3 = 0
⇒ 5832 – 126(y2 – 14y + 49) + 6(y3 – 21y2 + 147y – 343) = 0
⇒ 972 – 21(y2 – 14y + 49) + (y3 – 21y2 + 147y – 343) = 0
⇒ y3 – 42y2 + 441y – 400 = 0
(α – β)2, (β – γ)2, (γ – α)2 మూలాలుగా గల సమీకరణం x3 – 42x2 + 441x – 400 = 0
2వ పద్ధతి :
x3 – 7x + 6 = 0 మూలాలు α, β, γ లు యత్నదోష పద్ధతిన x = 1 సమీకరణాన్ని ధృవీకరిస్తుంది.
x3 – 7x + 6 కు x – 1 ఒక కారణాంకం
AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a) III Q2
∴ x3 – 7x + 6 = (x – 1) (x2 + x – 6) = (x – 1) (x + 3) (x – 2)
∵ x3 – 7x + 6 = 0 మూలాలు
α = 1, β = 3, γ = 2
ఇప్పుడు (α – β)2 = [1 – (-3)]2 = (4)2 = 16
(β – γ)2 = [-3 – 2)2 = 25
(γ – α)2 = [2 – 1]2 = 1
∴ (α – β)2, (β – γ)2, (γ – α)2 మూలాలుగా గల సమీకరణం (x – 16) (x – 25) (x – 1) = 0
⇒ x3 – (16 + 25 + 1)x2 + (400 + 25 + 16)x – 400 = 0
⇒ x3 – 42x2 + 441x – 400 = 0

AP Inter 2nd Year Maths 2A Solutions Chapter 4 సమీకరణ వాదం Ex 4(a)

ప్రశ్న 3.
x3 – 3ax + b = 0 యొక్క సమీకరణం యొక్క మూలాలు α, β, γ అయితే, Σ(α – β) (α – γ) = 9a అని నిరూపించండి.
సాధన:
α, β, γ లు x3 – 3ax + b = 0 మూలాలు
α + β + γ = 0, αβ + βγ + γα = -3a, αβγ = -b
Σ(α – β) (α – γ) = Σ[α2 – α(β + γ) + βγ
= Σ(α2 + α2 + βγ)
= 2(α2 + β2 + γ2) + (βγ + γα + αβ)
= 2(α + β + γ)2 – 4(αβ + βγ + γα) + (αβ + βγ + γα)
= 0 – 4(-3a) + (-3a)
= 9a

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.2

Question 1.
Which of the following angles form a triangle?
(a) 60°, 70°, 80°
Answer:
Given angles are 60°, 70°, 80°.
Sum of the angles = 60° + 70° + 80°
= 210° >180°
So, 60°, 70°, 80° cannot form a triangle.

(b) 65°, 45°, 70°
Answer:
Given angles are 65°, 45°, 70°.
Sum of the angles = 65° + 45° + 70° = 180°
So, 65°, 45°, 70° can form.a triangle.

(c) 40°, 50°, 60°
Answer:
Given angles are 40°, 50°, 60°
Sum of the angles = 40° + 50° + 60°
= 150° <180°
So, 40°, 50°, 60° cannot form a triangle.

(d) 60°, 30°, 90°
Answer:
Given angles are 60°, 30°, 90°.
Sum of the angles = 60° + 30° + 90° = 180°
So, 60°, 30°, 90° can form a triangle.

(e) 38°, 102°, 40°
Answer:
Given angles are 38°, 102°, 40°
Sum of the angles = 38° + 102° + 40° = 180°
So, 38°, 102°, 40° can form a triangle.

(f) 100°, 30°, 45°
Sol. Given angles are 100°, 30°, 45°
Sum of the angles = 100° + 30° + 45° = 175° < 180°
So, 100°, 30°, 45° cannot form a triangle.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.2

Question 2.
Sum of two interior angles of a triangle is 105°. Find the third angle.
Answer:
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 1
In ∆ABC given sum of two angles is 105°.
Let ∠A + ∠B = 105°
We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ 105° + ∠C = 180°
⇒ 105° + ∠C – 105° – 180° – 105°
⇒ ∠C = 75°
∴ Third angle is 75°.

Question 3.
In ∆PQR, if ∠P=65° and ∠Q = 50°, then find ∠R.
Answer:
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 2
Given in ∆PQR, ∠P = 65°, ∠Q = 50°
We know that in ∆PQR,.
∠P + ∠Q + ∠R = 180°
⇒ 65° + 50° + ∠R =180°
⇒ 115° + ∠R = 180°
⇒ 115° + ∠R – 115°
⇒ 180° – 115°
⇒∠R = 65°
∴ ∠R = 65°

Question 4.
Find the missing angles in each of the following triangles.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 3
Answer:
Given in ∆SKV, ∠K = 60°, ∠V = 70°
We know that in ∆SKV,
∠S + ∠K + ∠V = 180°
⇒ ∠S + 60° + 70° = 180°
⇒ ∠S + 130° = 180°
⇒∠S + 130°- 130° = 180°- 130°
⇒ ∠S = 50°
∴ ∠S = 50°

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 4
Answer:
Given in ∆BUN, ∠B = 105°, ∠U – 55°
We know that in ABUN,
∠B + ∠U + ∠N = 180°
⇒ 105° + 55° + ∠N = 180°
⇒ 160° + ∠N = 180°
⇒ 160° + ∠N – 160° = 180° – 160°
∴ ∠N = 20°

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 5
Answer:
Given in ∆PAT, ∠A = 90°, ∠T = 38°
We know that in ∆PAT,
∠P + ∠A + ∠T = 180°
⇒ ∠P + 90° + 38° = 180°
⇒ ∠P + 128° = 180°
⇒ ∠P + 128°- 128° = 180° – 128°
∴ ∠P = 52°

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.2

Question 5.
Find the value of ‘x’ in each of the given triangles.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 6
Answer:
In ∆CUT, ∠C = 64°, ∠U = 46° and exterior angle ∠CTE = x =?
∠C + ∠U + ∠T= 180°
⇒ 64° + 46° + ∠T = 180°
⇒ 110° + ∠T= 180°
⇒ 110° + ∠T – 110° = 180°- 110°
∠UTC = ∠T = 70°
∠UTC + ∠CTE = 180° (linear pair of angles)
⇒ 70° + x° = 180°
⇒ 70° + x° – 70° = 180°-70°
∴ x = 110°

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 7
Answer:
In ∆NTE, ∠N = 78°, ∠T = x, ∠E = x,
We know that in ANTE, ‘
∠N + ∠T + ∠E = 180°
⇒ 78° + x + x = 180°
⇒ 78° + 2x – 78° = 180° – 78°
⇒ 2x = 102°
⇒ \(\frac{2 x}{2}=\frac{102^{\circ}}{2}\)
∴ x = 51°

Question 6.
Find the value of ‘x’ and ‘y’ in each of the following triangles.
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 8
Answer:
In ∆TOP given ∠T = 6O, ∠O = y°
∠OPT = x° and ∠RPQ = 68°
∠OFT = ∠RPQ (Vertically, opposite angles are equal)
x = 68°

We know that in ∆TOP
∠T + ∠O + ∠P= 1800
⇒ 60° + y° + x° = 180°
⇒ 60° + y° + 68° = 180°
⇒ 128° + y – 128° = 180°- 128°
∴ y = 52° ‘
∴ x = 68°and y = 52°

AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 9
Answer:
In ∆EFG, ∠E = 70°, ∠F = 74° and ∠EGF – x° and∠EGH = y°
We know that in ∆EFG,
∠E + ∠F + ∠G = 180°
⇒ 70° + 74° + ∠EGF – 180°
⇒ 144° + x – 144° = 180° – 144°
⇒ x = 36°
∴ ∠EGF .= 36°

∠EGF + ∠EGH = 180° (linear pair of angles)
⇒ 36° + ∠EGH = 180°
⇒ 36° + y° – 36° = 180° – 36°
∴ y = 144°
∴ x = 36° and y = 144°

Question 7.
In a right angled triangle one acute angle is 37°. Find the other acute angle.
Answer:
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 10
Given in ∆ADI, ∠A = 37°, ∠D = 90°, ∠I =?
We know that in ∆ADI.
∠A + ∠D + ∠I = 180°
⇒ 37° + 90° + ∠I = 180°
⇒ 127° + ∠I – 127° = 180°- 127°
∴ ∠I – 53°
∴ Other acute angle is 53°.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.2

Question 8.
If the three angles of a triangular sign-board are 2x°, (x – 10)° and (x + 30)° respectively. Then find it’s angles.
Answer:
Given the three angles of a triangular signboard are 2x°, (x – 10)° and (x +30)°.

We know that in a triangle,
2x + (x – 10) + (x + 30) = 180°
⇒ 2x + x- 10° + x + 30 = 180°
⇒ 4x + 20 = 180°
⇒ 4x + 20 – 20 = 180° – 20°
⇒ 4x = 160°
⇒ \(\frac{4 x}{4}=\frac{160^{\circ}}{4}\)
∴ x = 40°
Angles are 2x°, (x- 10)°, (x + 30)° 2(40°), 40° – 10, 40 + 30
Angles of signboard are : 80°, 30°, 70°.

Question 9.
If one angle of a triangle is 80°, find the other two angles which are equal.
Answer:
AP Board 7th Class Maths Solutions Chapter 5 Triangles Ex 5.2 11
Given one angle of a triangle is 80°,
In ∆SRI, ZS = 80° and ∠R = ∠I = x

We know that in ∆SRI,
∠S + ∠R + ∠I = 180°
⇒ 80° + x + x = 180°
⇒ 80° + 2x-80° = 180°-80°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°
Therefore angles of triangle are 80°, 50° and 50°.

Question 10.
State TRUE or FALSE for each of the following statements and write the reasons for the FALSE statement.
(i) A triangle can have two right angles.
Answer:
FALSE.

In triangle sum of three angles is 180°. In triangle, if two angles are two right angles (90° + 90° = 180°).
Then, sum of three angles is greater than 180°.

(ii) A triangle can have two acute angles.
Answer:
TRUE.

(iii) A triangle can have two obtuse angles.
Answer:
FALSE.

In triangle sum of three angles is 180°. In traingle, if two angles are two obtuse angles, then sum of three angles is greater than 180°.

AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.2

Question 11.
The angles of a triangle are in the ratio 2 : 4 : 3, then find the angles.
Answer:
Given the ratio of the angles of a ’ triangle are 2 : 4 : 3.
2x : 4x : 3x

Sum of the angles of a triangle is 180°.
⇒ 2x + 4x + 3x = 180°
⇒ 9x = 180°
⇒ \(\frac{9 x}{9}=\frac{180^{\circ}}{9}\)
∴ x = 20°
Angles are ⇒ 2x : 4x :,3x
2(20°) : 4(20°): 3(20°)
40°: 80° : 60°
∴ Angles of a triangle are 40°, 80°, 60°.

AP Board 7th Class Social Solutions 2nd Lesson Forests

SCERT AP 7th Class Social Study Material Pdf 2nd Lesson Forests Textbook Questions and Answers.

AP State Syllabus 7th Class Social 2nd Lesson Questions and Answers Forests

7th Class Social 2nd Lesson Forests Textbook Questions and Answers

Observe the given picture.
AP Board 7th Class Social Solutions 2nd Lesson Forests 1

Answer the following questions.

Question 1.
Look at the above figure and say what do you observe in the figure?
Answer:
Forest, Birds and Wild animals habitation.

Question 2.
What are the components you can see in the forest ? Complete the given diagram.
Answer:
AP Board 7th Class Social Solutions 2nd Lesson Forests 2

Question 3.
Express your views about the forest in your own words.
Answer:
In my words, a forest is a piece of land with many trees. Many animals need forests to live and survive.

Forests are very important and grow in many places around the world. They are an ecosystem which includes many plants and animals.

We depend on forests for our survival.

Improve Your Learning

I. Answer the following questions.

1. A) Mention the types of forests in India.
Answer:
For -administrative convenience, the Government of India divided forests into three types. They are :

  1. Reserved forests
  2. Protected forests
  3. Unclassified forests.

B) Into how many types forests are classified?
Answer:
Forests are divided into five types based on climate, rainfall and types of soils. They are :

  1. Evergreen forests
  2. Deciduous forests
  3. Thorny forests
  4. Mangrove forests
  5. Montane forests

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 2.
Describe briefly about the Evergreen forests.
Answer:
Evergreen Forests :

  1. Evergreen forests are grown in the areas with high rainfall more than 200 cm.
  2. Trees in this area are very tall and they contain broad leaves.
  3. The trees in these forests remain green throughout the year. _
  4. These forests are located in Himalayan region, North-eastern states and Western ghats.
  5. Mahogany, Ebony, Rosewood trees etc. are found.
  6. Different types of snakes, Lion-tailed macaque and a variety of insects found here.

Question 3.
Describe the features of deciduous forests. Explain about flora and fauna of these forests in India.
Answer:

  1. The deciduous forest are grown in areas with rainfall between 70 cm and 200 cm.
  2. These forests are located in Peninsular plateau.
  3. The trees shed their leaves during very dry months to minimise transpiration.

Flora of this region :
Teak, Sal, Bamboo, Rosewood, Sandalwood, Neem, Shisham, Khair, Kusum, Arjun and Mulberry trees are found.

Fauna of this region :
The most common animals are found in this region like, Deers, Hares, Elephants, Tigers, Leopards, Peacocks, several species of Birds, etc.

Question 4.
Explain in detail about the thorn forests in India?
Answer:

  1. The thorn forests are found in regions with less than 70 cm of rainfall.
  2. Due to the arid climate most of the trees in these forests are sharp, thorny and bushy.
  3. This type of vegetation is found in the north-western parts of the country.
  4. Trees are scattered and have long roots penetrating deep into the soil in order to get moisture.
  5. The stems are succulent to conserve water.
  6. These forests are found in Madhya Pradesh, Uttar Pradesh, Rajasthan and Haryana.

Question 5.
Prepare some slogans on “conservation of forests”.
Answer:

  1. Save forests – Save the climate and wild life.
  2. Plant a tree – Keeps the flood away.
  3. Don’t destroy the greenery and don’t spoil the scenery. Save mother earth.
  4. Trees are Green Gold.
  5. Save trees now – They will save you in future.
  6. Plant a tree – Predict the mother earth.

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 6.
“Forests are essential for us, but yve destroy them.” Respond on this.
Answer:
The uses of forests.

  1. Forests play a major role in our life.
  2. Forests prevent soil erosion and floods.
  3. Trees help to regulate the climate of a place.
  4. We get more products which are needed to us from forests.

Even though the forests are useful for us, but we destroy them due to following relfeons.

  1. The ever-growing human consumption and population is the biggest cause of forest destruction.
  2. Along with population increase we need resources, products and services, so we destroy the forest.
  3. For the development of our nation for mining, for infrastructure projects, for agriculture purpose there is no other alternative so we destroy the forests.

Conclusion :
Governments will invent alternative resources for the development of our nation without destroying the forests.

Question 7.
List out the resources of the forests used by you in your daily life.
Answer:
The following resources are used by ourselves in our daily life from the forests.

  1. Notebooks → Wood pulp is used
  2. Furniture → Wood is used
  3. Medicines → Herbs are used
  4. Food & Fodder → Roots, tubers, bean sprouts, etc.
  5. Beedi leaves → Making beedis
  6. Bamboo is used for making fences
  7. Essential oils → Eucalyptus tree, camphor
  8. Fruits → Coconut, pear
  9. Cane → Walking sticks, etc.

Question 8.
Read about policies of forests and Fill in the following table.

YearName of the policyObjectives
1894
1950
1952
1980
1988

Answer:

YearName of the policyObjectives
1894The Forest LawMeeting needs of local people and after meet­ing local needs maximum revenue collection.
1950Forest FestivalVana Mahotsav – Create awareness on about saving forests, and bad effects of deforestation.
1952National Forest PolicyShould maintain 33% of forest cover.
1980Forest Conservation Act1) To protect the forest, its flora, fauna and other diverse ecological component.

2) To protect the integrity, territory, and individuality of the forests.,

1988National Forest PolicyTaking steps to create massive people’s move­ment with involvement of women to achieve the objectives and minimise pressure on exist­ing forest.

II. Choose the correct answer.

1. Which forests are green throughout the year?
a) Deciduous forests
b) Evergreen forests
c) Tidal forests
d) Mangrove forests
Answer:
b) Evergreen forests

2. Which of the following is not the slogan of conservation of forests?
a) Save the trees save the earth.
b) Save nature save future.
c) Greenery for better environment.
d) Good food good life.
Answer:
d) Good food good life.

3. Which of the following one is not a forest product?
a) Timber
b) Honey
c) Plums
d) Bread
Answer:
d) Bread

4. In which year was the National Conservation Policy enacted by the Central government?
a) 1984
b) 1950
c) 1952
d) 1980
Answer:
d) 1980

5. Which of the following forests have a variety of snakes and insects?
a) Evergreen forests
b) Deciduous forests
c) Mangrove forests
d) Thorny forests
Answer:
a) Evergreen forests

III. Match the following.

1.

Group-AGroup-B
1. Evergreen forestsA) Snow Leopard.
2. Deciduous forestsB) Variety of fishes.
3. Montane forestsC) Lion tailed macaque.
4. Mangrove forestsD) Different kinds of deers.

Answer:

Group-AGroup-B
1. Evergreen forestsC) Lion tailed macaque.
2. Deciduous forestsD) Different kinds of deers.
3. Montane forestsA) Snow Leopard.
4. Mangrove forestsB) Variety of fishes.

2.

Group-AGroup-B
1. High rainfallA) Mangrove forest.
2. Little rainfallB) Montane forests.
3. Coastal lineC) Evergreen forests.
4. Mountain regionD) Thorny forests

Answer:

Group-AGroup-B
1. High rainfallC) Evergreen forests.
2. Little rainfallD) Thorny forests
3. Coastal lineA) Mangrove forest.
4. Mountain regionB) Montane forests.

Puzzle

Solve the puzzle with the words related to given hints.
AP Board 7th Class Social Solutions 2nd Lesson Forests 3
Cross:
1. Largest hills in Andhra Pradesh (9).
2. Flora in evergreen forests (8).
3. These are known as Selvas (9).
4. Product of forest (4).
5. Raw material for Paper (6).

Down :
1. Hills in Tamilnadu (7).
2. Product of forest (6).
3. Forest in coastal region (8).
4. Product of forest (4).
5. Flora in deciduous forest (4).
Answer:
AP Board 7th Class Social Solutions 2nd Lesson Forests 4

7th Class Social 2nd Lesson Forests InText Questions and Answers

7th Class Social Textbook Page No. 20

Question 1.
Locate a few important countries in various climatic regions in the world map.
Answer:
AP Board 7th Class Social Solutions 2nd Lesson Forests 5

Question 2.
Prepare a table with various climatic regions and important countries in those regions.
Answer:

Climatic regionsCountries
Equatorial / Tropical
Climatic Region
South America – Brazil
Asia – Indonesia
Africa – Kenya, etc.
The SavannaIndia-China
Australia                                                         ‘
Desert regionSahara – Egypt, Libya
Kalahari – Namibia
Australian Desert – Australia
Thar desert – India
Mediterranean ClimateEurope – Spain, Italy, Turkey, Israel, Greece, etc.
Steppe ClimateUkraine, China, Uzbekistan, etc.
Taiga RegionAlaska, Canada, Scandinavia, etc.
Tundra RegionRussia, Greenland, Iceland – Sub – Antarctic islands

7th Class Social Textbook Page No. 21

Question 3.
Locate the Ever-green forests in an outline map of India.
Answer:

7th Class Social Textbook Page No. 22

Question 4.
Locate the areas of Deciduous forests in an outline map of India.
Answer:

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 5.
Locate the areas of Thorny and shrub forests in an outline map of India.
Answer:

7th Class Social Textbook Page No. 23

Question 6.
Locate the areas of Montane forests in an outline map of India.
Answer:

Question 7.
Fill up the following table.
AP Board 7th Class Social Solutions 2nd Lesson Forests 6
Answer:
AP Board 7th Class Social Solutions 2nd Lesson Forests 7

7th Class Social Textbook Page No. 25

AP Board 7th Class Social Solutions 2nd Lesson Forests 9
Question 8.
Observe the above Andhra Pradesh map. Which districts have more forest cover and which districts have less forest cover.
Answer:
YSR Kadapada, East Godavari, Visakhapatnam districts have more forest cover and Krishna and Anantapur districts have less forest cover.

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 9.
Witch type of forests do you find in your district?
Answer:
(Example : I am living in Krishna district. Mangrove forests are in Krishna district.)

7th Class Social Textbook Page No. 27

Question 10.
Fill up the following table.

Types of forestExtentFlora
Moist deciduous forests
Dry deciduous forests
Tidal forests

Answer:

Types of forestExtentFlora
Moist deciduous forestsSrikakulam, Visakhapatnam and East GodavariVegi, Egisa, Bamboo, Maddi, Bandaru, Jittegi and Sal.
Dry deciduous forestsYSR Kadapa, Kurnooi, Ananthapur, and Chittoor.Maddi, Teak, Biliu, Velaga, Egisa, Neem, Buruga, Moduga and Red sandal.
Tidal forestsKorangi region of East GodavariUppu ponna, Boddu ponna, Urada, Mada, Tellamad-a, Patri Teega and Balabandi Teega.

7th Class Social Textbook Page No. 29

Question 11.
Prepare some slogans on social forestry.
Answer:
Slogans :

  1. A tree that stay, keep flood away
  2. Don’t cut a tree don’t cut a life.
  3. Don’t cut trees if you want cool air. ‘
  4. No Trees, No Mankind.
  5. Plant a tree today; make the life of the earth much longer.

Question 12.
Plant a tree on your birthday and take care of it.
Answer:

Question 13.
Gift a plant on important occasions to your friends and relatives.
Answer:

7th Class Social Textbook Page No. 30

Question 14.
Make a poster on tribal culture and tribal products.
Answer:
AP Board 7th Class Social Solutions 2nd Lesson Forests 8

Question 15.
Celebrate Vana Mahostav in your school/locality and plant a few plants and notice their growth.
Answer:

Think & Respond

7th Class Social Textbook Page No. 20

Question 1.
Compare the climate of the various climatic regions.
Answer:

Climatic regionsComparision of climatic condition
Tropical climatic regionHigh temperature and heavy precipitation
The SavannaThe climate is usually warm and temperature ranges from 20° to 30°C. Annual rainfall – 25 to 75 cm per year.
Desert RegionHot summers and cold winters, high diurnal range of temperatures.
Mediterranean ClimatePleasant climate with dry summer, and moderate to high rainfall in winter.
Steppe climateExtremes of heat & cold low amount of rajnfall.
Taiga climateWinters are extremely cold and long whereas summers are moderately hot and short.
Tundra climateThe climate is cold and windy and rainfall is scanty.

If we compare the different regions through above table, equatorial region has high temperature & high rainfall compare with Taiga and Tundra regions. Desert regions have high temperature. Compare with mediterranean climate. Steppes have low amount of rainfall.

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 2.
What is the impact of climate on natural vegetation in climatic regions ?
Answer:

  1. The amount of temperature, rainfall, moist in air and soil determine the type of vegetation in forest.
  2. Ex : Evergreen forests in high temperature and rainfall regions – Bushy plants in arid regions.

Question 3.
Name the plants in forest which are having medicinal values.
Answer:
The commonly used plants in India are :

1) Sarpagandha :
Used to treat blood pressure, it is found only in India.

2) Jamun :
The powder of the seed is used for controlling diabetes.

3) Arjun
: The fresh juice of leaves is cure for ear ache.

4) Babool :
Leaves are used as a cure for eye sores.

5) Neem :
Has high antibiotic and antibacterial properties.

6) Tulasi plant :
Is used to cure cough and cold.

7) Kachnar :
Is used to cure asthma and ulcers.

7th Class Social Textbook Page No. 21

Question 4.
Why is a variety of flora and fauna found in evergreen forests ?
Answer:

  1. Evergreen forests are located in tropical regions.
  2. They receive a lot of sunlight and rainfall.
  3. There is a lot of energy in these forests.
  4. This energy is stored in plant vegetation, which is eaten by animals.
  5. The thick, dense forests are abode of Herbivorous and Carnivorous animals. So, a variety of flora and fauna is found in evergreen forests.

7th Class Social Textbook Page No. 22

Question 5.
Trees in deciduous forests shed their leaves. When and why ?
Answer:
The trees in deciduous forests shed their leaves during very dry months to minimise transpiration.

Question 6.
Have you ever seen the beauty of deciduous forests ? Describe the beauty of the forest.
Answer:
Yes. I have seen the beauty of deciduous forests.

  1. Forests are the nature gift to human beings. The deciduous forests shed leaves during the months of February and March.
  2. This is to reduce the consumption of water that it releases through leaves in the form of transpiration
  3. At that time the trees in the forest look completely dry.
  4. But when it starts budding of new leaves in spring, the forest looks very shiny and fresh. Later all the trees turned into very beautiful scenic view with shiny leaves and blossoms.
  5. One should watch and admire the picturesque marvel of deciduous forests.

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 7.
Mangrove forests are natural protectors of sea coast. Discuss.
Answer:
Mangrove swamps protect coastal areas from erosion, storm surge and tsunamis.

Mangroves act as shock absorbers. They reduce high tides and waves and help prevent soil erosion.

The mangroves massive root systems are efficient at dissipating wave energy.

7th Class Social Textbook Page No. 24

AP Board 7th Class Social Solutions 2nd Lesson Forests 10
Question 8.
Observe the above map. Which state has more forest cover. Give reasons.
Answer:
Madhya Pradesh state in India has more forest area.

The following reasons are responsible for more forest area in Madhya Pradesh.

Madhya pradesh has two horizontal mountain ranges from East to West. Vindya and Satpur mountains. These two ranges make a buffer of heavy clouds coming from South.

Hence, it results in longer period of rains and creates many seasonal rivers and ponds, which help dense forests.

Question 9.
Which state has less forest cover. Give reasons.
Answer:

  1. Punjab has the lowest forest cover with respect to total geographical area in India at 6.79 percent. It is a small state. Most of the people depended on agriculture only. As most of the land is under the cultivation there is less forest cover.
  2. Gujarath and Rajasthan states have less forest cover due to adverse conditions in most of their area.

Question 10.
The Western side of the Western Ghats is covered with thick forest than the Eastern side. Find the reason.
Answer:
Western Ghats are layered with thick evergreen forests because they receive more rainfall as compared to the Eastern Ghats that are covered with deciduous forests.

The Western Ghats get their rainfall from the monsoon winds that blow from Arabian sea.

7th Class Social Textbook Page No. 26

Question 11.
What are the uses of Red Sandalwood and Sandalwood?
Answer:
Uses of Red Sandalwood.

  1. Red Sandalwood is a tree.
  2. The wood at the centre of the trunk is used as medicine.
  3. Red Sandalwood is used for treating digestive tract problems, fluid retention, coughs, and for blood purification.
  4. In manufacturing, red Sandalwood is used as a flavoring in alcoholic beverages.

Uses of Sandalwood :

  1. Sandalwood is used as a fragrance in incense, cosmetics, perfumes and soaps.
  2. It is also used as a flavour for foods and beverages.
  3. The wood has been valued in carving because of its dense character.
  4. It is also used for medicinal purpose for headache, stomachache and urinary and genital disorders.

7th Class Social Textbook Page No. 27

Question 12.
Why do tribal people reject to leave the forests?
Answer:

  1. The tribal people lived in forest area for generations.
  2. They cleared the forests and follow agriculture.
  3. They worshipped nature gods.
  4. They domesticated animals.
  5. Their entire life depends on forests. ,

So they reject to leave the forests.

7th Class Social Textbook Page No. 28

Question 13.
What is the role of forests in conservation of environment?
Answer:
Role of forests in conservation of environment : „

  1. Prevent soilerosion and help in maintaining fertility of soil.
  2. They provide shelter to wild animals and are areas that sustain biodiversity.
  3. They reduce atmospheric pollution.
  4. They increase humidity and frequency of rainfall.
  5. They check the increase of atmospheric temperature.

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 14.
Write a list of items made from forest products in your surroundings.
Answer:
The following items we are using in our daily life.
Foods : Honey, Fruits, Palm oil, Mushroom, etc.
Wood : Furniture, paper, decorative items, etc.’
Medicines : Natural Aspirin and Acne medication.
Other items : Chewing Gum, Sponges, Carnauba wax, Henna Dye, Rubber, etc.

7th Class Social Textbook PageNo.29

Question 15.
What are the reasons for deforestation?
Answer:

  1. Agriculture
  2. Urbanization
  3. Industrialization
  4. Forest fires
  5. Mining
  6. Construction of Dams and Reservoirs etc. are the reasons for deforestation.

Question 16.
What are the consequences of deforestation?
Answer:

  1. Flooding
  2. Loss of soil
  3. Loss of biodiversity
  4. Climatic change
  5. Health problems and
  6. Displacement of the indigenous people are the consequences of deforestation.
  7. Imbalance of ecosystem causing natural calamities.
  8. Loss of habitat of wild life.

Question 17.
Suggest a few measures for afforestation!
Answer:
Methods that can be used to enhance afforestation are :

  1. Increase the number of planting trees and protect them.
  2. Choose the barron lands, waste lands to plant trees under social forestry.
  3. Road side areas, industrial areas can be enriched with thick growth of trees.
  4. Encourage the plants that provide forest products to the needy so that they pay interest to conserve them.

Question 18.
Do you observe any plantation of trees in your surroundings? What are the uses of plantation?
Answer:
Yes. I observed.

Uses of plantation :

  1. The more trees are there, the cleaner the air will be.
  2. Trees and forests can provide natural filtration, resulting in cleaner wafer.
  3. Planting trees can help to slow down the process of heat trapping of carbondioxide in our atmosphere.
  4. Trees are essentially nature’s wind-breakers, providing protection and shade during hot weather and dazzling sunshine.

Explore

7th Class Social Textbook Page No. 20

Question 1.
Go through library books or internet to know more about climatic regions.
Answer:

AP Board 7th Class Social Solutions 2nd Lesson Forests

Question 2.
Know about Podu cultivation with the help of your teacher.
Answer:
Podu is a traditional system of cultivation used by tribes in India, whereby different areas of jungle forest are cleared by burning each year to provide land for crops.

Podu is a form of shifting agriculture using slash-and-burn methods. Traditionally used on the hill-slopes of Andhra Pradesh. It is also known as Jhum cultivation.

7th Class Social Textbook Page No. 29

Question 2.
Go throiigh library books or browse internet to know more information about the given forest acts.
Answer:
The objectives of Forests Acts :

  1. To consolidate all the previous laws regarding forests.
  2. To give the Government the power td create different classes of forests for their effective usage for the colonial purpose.
  3. To regulate movement and transit of forest produce, and duty leviable on timber and other forest produce.
  4. To define the procedure to be followed for declaring an area as Reserved Forest, Protected Forest or Village Forest.
  5. To define forest offences acts prohibited inside the Reserved Forest, and penalties leviable on the violation.
  6. To make conservation of forests and wildlife more accountable.

Project Work

Prepare a model of forest use with natural material. Ans. Student’s own activity.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

AP State Board Syllabus AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development.

AP State Syllabus SSC 10th Class Social Studies Important Questions 2nd Lesson Ideas of Development

10th Class Social 2nd Lesson Ideas of Development 1 Mark Important Questions and Answers

Question 1.
State reason for protesting against the establishment of Kudankulam Nuclear Power Project in Tamil Nadu.
Answer:

  1. Kudankulam people have protested on the grounds of safety, security and livelihood.
  2. They also want their coast and country protected from the radio active peril.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 2.
Why did the people oppose the establishment of Kudankulam Nuclear Power project in Tamil Nadu?
Answer:
On the grounds of safety, security and livelihood, the people protested against the project.
(OR)
To protect their cost from radio active peril people opposed it.

Question 3.
Observe the following table and answer the questions a, b, c and d.
Answer:

CountryH.D.I. Ranking 2012Average Life Span in 2012Average years of Schooling
Norway181.312.6
America378.713.3
Sri Lanka9275.19.3
China10173.77.5
India13665.84.4
Bangladesh14669.24.8
Pakistan14665.74.9
World Average70.17.5

a) Which two countries have more average schooling years?
Answer:
Norway, America.

b) Which two Asian countries have better HDI rank than that of India?
Answer:
Srilanka, China.

c) Which countries are lacking behind to the average life expectancy of World?
Answer:
India, Bangladesh, Pakistan.

d) What are the reasons for having less average schooling in India, Bangladesh and Pakistan?
Answer:
The reasons for having less average schooling in India, Bangladesh and Pakistan are

  1. Poverty,
  2. More rural population,
  3. No awareness with regard to literacy.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 4.
In the context of development, two persons or groups of persons may seek things which are conflicting. Give one example for this.
Answer:
Example – 1: To get more electricity industrialists may want more dams. But this may submerge the land and disrupt the lives of people such as the tribals who are dis-placed.
Example – 2: A girl expects as much freedom and opportunity as her brother and that he also shares in the household work. But brother may not like this.

Question 5.
What idea is the poster promoting?
AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development 1
Answer:
Importance of Education.

Question 6.
State any two goals of development other than income.
Answer:
Equal treatment, freedom, security and respect from others.

Question 7.
What is HDI?
Answer:
The index developed for comparing coun¬tries for measuring human development is called HDI (Human Development Index). It generally includes income, educational levels and health status of the people.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 8.
What is per capita income?
Answer:
Per capita income is “the total income of the country divided by its total population”.

Question 9.
Expand IMR.
Answer:
Infant Mortality Rate.

Question 10.
Which organization publishes HDR?
Answer:
UNDP publishes Human Development Report.

Question 11.
What is the main criteria for comparing the development of different countries?
Answer:
The main criteria for comparing the development of different countries are per capita income, life expectancy, average years of schooling, expected years of schooling, etc.

Question 12.
What is health?
Answer:
Health means a state of couple soundness – physical and mental.

Question 13.
Why do people look at a mix of goals?
Answer:
People look at a mix of goals for development.

Question 14.
Why are dams opposed?
Answer:
Dams are opposed because they will disrupt the lives of the people and submerge their own lands.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 15.
How are the women who work outside the homes?
Answer:
Women who work outside the homes are economically independent and self¬confident.

Question 16.
What is adult literacy rate?
Answer:
The rate of percentage of people aged 15 and above, who can understand, read and write a short and simple statement in their regional languages is known as adult literacy rate.

Question 17.
What is the main criterion for comparing the development of different countries?
Answer:
Average income is the main criterion for comparing the development of different countries.

Question 18.
Explain the calculation of BMI.
Answer:
BMI can be calculated by dividing the total weight of a person by the square of his height.

Question 19.
What is educational development?
Answer:
Education attained by the people of a country on an average basis is referred as educational development.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 20.
What isthe percentage of population living in the rural India?
Answer:
70% of the Indian population is living in rural India.

Question 21.
What is Infant Mortality Rate?
Answer:
The rate at which children in a country die within an age of 0-1 year is known as “Infant Mortality Rate.”

Question 22.
What are the factors of production?
Answer:
Land, Labour, Capital and Enterprise are the four factors of production.
Technology is also added to the factors of production.

Question 23.
How is the standard of living measured?
Answer:
Standard of living is measured by real GDP per capita.

Question 24.
What is development?
Answer:
Development refers to progress or improvement in lifestyle.

Question 25.
What is PDS?
Answer:
PDS is a system to distribute ration to the poor at a reasonable rate through govern¬ment ration shops. PDS – Public Distribution System.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 26.
A girl from a rich urban family has some development goals for her. State any one such goal.
Answer:
She can get as much freedom as a boy has.

Question 27.
Write any one advantage of public facilities.
Answer:
It develops national feelings and a sense of collective responsibility. Reduces expenditure.

Question 28.
What is NAR?
Answer:
Net Attendance Rate: Out of the total num¬ber of children in age group 6-17, the per¬centage of children attending schools.

Question 29.
Which is considered to be one of the most important attributes for comparing coun¬tries?
Answer:
Income is considered to be one of the most important attributes for comparing countries.

Question 30.
Which is not a useful measure for comparison between countries?
Answer:
Total income is not such a useful measure for comparison between countries.

Question 31.
What are called developed countries?
Answer:
The rich countries, excluding countries of West Asia and certain other small countries, are generally called developed countries.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 32.
Why, in some areas, children, particularly girls are not able to achieve secondary level schooling?
Answer:
Due to social restrictions imposed on girl child i.e., gender bias.

Question 33.
Why have some states lesser Infant Mortality Rate?
Answer:
Some states have a lesser Infant Mortality Rate because they have adequate provision of basic health and educational facilities.

Question 34.
What is the major consideration of Himachali women?
Answer:
One major consideration is that many Himachali women are themselves employed outside the home and hence show lesser gender bias.

Question 35.
How are the women who work outside the homes?
Answer:
Women who work outside the home are economically independent and self-confident.

10th Class Social 1st Lesson India: Relief Features 2 Marks Important Questions and Answers

Question 1.
What do you learn from the schooling revolution in Himachal Pradesh?
Answer:

  1. Both the government and the people of Himachal Pradesh were keen on education.
  2. They started many schools.
  3. They made sure that education was largely free.
  4. They allocated a good share to education in the government budget.
  5. They tried to ensure that the schools had all the facilities.
  6. Most of the students enjoy their schooling experience.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 2.
People generally do not like to work in unorganised sector. Why?
Answer:

  1. Unorganised sector is largely outside the control of the government.
  2. Rules and regulations are often not followed.
  3. Jobs are not regular.
  4. Jobs are low paid.
  5. It is difficult to avail the leaves.
  6. No provision for paid leave.
  7. Job is not secure.
  8. Working conditions are generally poor.
  9. No safety measures followed in work places.
  10. Health hazards would be there.
  11. No insurance
  12. No welfare schemes.

Question 3.
Give examples of the criterion for the measurement of human development.
Answer:
The examples of the criterion for the measurement of human development:

  1. Per capita income
  2. The education levels of people and health status.
  3. Standard of life of the people
  4. Availability of electricity
  5. Transportation
  6. Sanitation facilities
  7. Expected years of schooling
  8. Average years of schooling, etc.
    Ex: Sri Lanka, one of our neighbours is much ahead of India in every respect.

Question 4.
What are the different indicators in which development is measured? Which one do you agree with ?
Answer:

  1. Per capita income
  2. Literacy rate
  3. Average years of schooling 4) Expected years of schooling
  4. Life expectancy at birth 6) Health status
  5. Employment status 8) Equal distribution
  6. I agree with all the above things because those are useful for measuring the complete development.

Question 5.
Give examples for different persons can have different developmental goals.
Answer:

Category of personsDevelopmental goals
Landless rural labourers:More days of work and better wages, quality education for their children, no social discrimination.
Prosperous farmers:Higher support prices for crops, should be able to settle their children abroad.
Farmers who depend only on rains for growing crops:Adequate rainfall.
Urban unemployed youth:High salaried jobs
An Adivasi from mining fields :To protect their livelihoods.
Persons from fishing community in the coastal area.Good weather and a good catch of fisher.

Question 6.
What do the people desire other than income?
Answer:
People desire the following other than income.

  1. Equal treatment
  2. Freedom
  3. Security
  4. Respect from others.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 7.
Prepare a pamphlet on making mahila mandals active in villages.
Answer:
Mahila Mandals are voluntary service organizations that work for the betterment of women in the villages of India. To better their life the village women should have their say in the active involvement of Mahila Mandals’ activities. Active Mahila Mandals can be found In many villages. Suggestions to make Mahila Mandals active in the villages:

  • First priority should be given to girl education and to eradicate illiteracy.
  • Vocational training and credit facilities to women for self-employment should be provided.
  • Mahila Mandals should collectively work for the betterment of women who need nutrition, education and family welfare.
  • They should help the women in immunization of children, small savings, provision of bathrooms, women crafts centres and balwadis.
  • They should work towards the elimination of discrimination, inequality, intolerance, and violence-both, within and outside the home.
  • They should have comparatively high involvement in social life and village politics.

The government should provide basic equipment and stationery, etc. to the Mahila Mandals. It should create awareness among women regarding the Mahila Mandals.

Women empowerment leads to a strong nation.

Copies: 2000

Surya Printers.

Question 8.
What is Development ? Why do different people have different developemntal goals? Explain with two Examples.
Answer:
Meaning of Development:

  1. Growth plus change is called development. .
  2. All the persons may not have the same notion of development or progress.
  3. Each one of them seeks different things.
    Examples:
    i) A girl expects as much freedom and opportunity as her brother and that he also shares in the household work.
    ii) To get more electricity, industrialists, may want more dams. But this may submerge the land and disrupt the lives of people who are displaced as the tribals.

Question 9.
Create a few slogans on promoting girl education.
Answer:

  1. Girl with education – helpful to the family.
  2. Educate a girl – she educates a family.
  3. Encourage girl education – save the nation.
  4. An educated girl – serves the nation well.
    (Students can sit together and discuss to prepare a few slogans of their own.)

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 10.
Observe the following table and prepare a paragraph on it.
Answer:
Progress in Himachal Pradesh
The comparison across two different years is indicative of development that has happened. Clearly, there has been greater development in schooling and spread of education in Himachal Pradesh than India as a whole. Though there is still a lot of difference in the average levels of education among boys and girls, i.e., across genders, there has been some progress towards greater equal¬ity in the recent years.

Question 11.
Why was it necessary for government to run schools in Himachal Pradesh?
Answer:
The rich children can get the education in private sector schools. But the majority of Indian chil¬dren are enrolled in the government schools. Education has also been made free for children from 6 to 14 years of age or up to VIII class under the R.T.E. Act 2009. So, it was necessary for government to run schools in Himachal Pradesh.

Question 12.
‘Human development is the essence of social development’ – Explain.
Answer:

  1. Human development focuses on the people.
  2. It is concerned with the well-being of the people, their needs, choices and aspirations. All these help in building a right kind of society.
  3. It is all about the enlarging or widening the choices for the people. It is the building of human capabilities, such as education, information and knowledge, to have opportunities of livelihood.
  4. Human development focuses on the expansion of basic choices.

Question 13.
‘Money cannot buy all the goods and services that one needs to live well.’ Explain.
Answer:

  1. Even though per capita income is high in many states, education and health facilities are still lacking.
  2. Money or high per capita income cannot buy a pollution-free environment or good health.
  3. Money cannot buy peace and democracy.

10th Class Social 2nd Lesson Ideas of Development 4 Marks Important Questions and Answers

Question 1.
Study the following table and answer the questions that given below.
(a) What do you mean by literacy rate?
Answer:
The number of literates per every 100 persons in the population is known as literacy rate.

(b) In which state the net attendance is highest?
Answer:
Himachal Pradesh.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

(c) What could be the reason for the highest, Literacy rate in Himachal Pradesh?
Answer:

  1. Both the government of Himachal Pradesh and the people of the state were keen on education.
  2. The government started schools and made sure that education was largely free, or costs very little for parents.
  3. Further, it tried to ensure that these schools had at least the minimum facilities of teach¬ers, classrooms, toilets, drinking water, etc.

(d) In which state the IMR is least?
Answer:
Himachal Pradesh.

Question 2.
Read the following paragraph :
“In many parts of the country, girls’ are still given less priority by parents compared to boys.”
Comment on the gender bias in India.
Answer:

  1. Ours is a male-dominated society.
  2. Female literacy rate is low.
  3. Women who work outside their homes are less in number.
  4. Traditionally, in our society, women have less involvement in social life.
  5. Because of all these reasons, gender bias is still continuing.
  6. This is a hurdle for the development of society.
  7. Boys and girls should be treated equally.

Question 3.
“We should be able to integrate environmental concerns with the idea of progress”. Explain.
Answer:

  1. We must show concern on environmental issues while achieving development.
  2. The environmental source function will deplete while using the sources in a speedy way.
  3. When waste output exceeds the limit, it will cause long-term damage to the environment.
  4. The big projects may harm bio-diversity.
  5. Use of chemical fertilizers and pesticides in modern agriculture leads to a big loss to the environment.
  6. The fuel used as a part of industrial development causes a lot of air pollution.
  7. Ground water levels are being depleted.
  8. Deforestation is being occurred.
  9. The rights of low-income countries, future generations also should be viewed.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 4.
Plot the below information on a bar graph. Write your observation.

SI. No.StateLiteracy rate (%)
1.Punjab77
2.Himachal Pradesh84
3.Bihar64

Answer:
AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development 2Observation:
As Himachal Pradesh has high literacy rate, it can be considered as developed state.

Question 5.
Read the given paragraph and comment.
In many parts of the country, girls’ education is still given less priority by the parents as compared to boys’ education. While girls may study for a few years, they may not complete their schooling.
Answer:
According to this paragraph two things are clearly mentioned that there is a gender bias with regard to giving education to boys and girls among the people of the country and the second one is only the Himachali Pradesh Government is concentrating on girl education.
My opinion on these two issues is that the gender bias was there once in the society. As there is a vast awareness in the parents they send their daughters to the schools in many places. Even a rickshaw puller also wants to make his daughter study in a school.
He hopes his daughter becomes a professional. The parents are interested to send their children to English medium schools irrespective of their income and status. It shows their interest. A little bit fear about girls is there among them because of other reasons. They are afraid of the safety of their daughters. Just like in Himachal Pradesh other states are also spending much amount on education and schooling. It is accepted that the Himachal Pradesh state has taken the step earlier. I don’t say that other states are not taking steps to improve the conditions of schools for providing good education for the girls.
It is a sensitive issue and it is to be taken seriously to bring awareness among people to make their daughters admit in schools and the governments should consider the problems of girls in schools. Sufficient toilets and other facilities are to be provided so as to enroll all the girls in schools.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 6.
Observe the table and answer the following questions.
Some data regarding India and its neighbours for 2013

CountryPer capita  Income in $Life Expectancy at birth (Years)Average years of schoolingExpected years of schoolingHuman Development Index (HDI) rank in the world
Sri Lanka517075.19.312.792
India328565.84.410.7136
Pakistan256665.74.97.3146
Myanmar181765.73.99.4149
Bangladesh178569.24.88.1146
Nepal113769.13.28.9157
  1. Which country stands ahead of India in all aspects in HDI ?
    Answer:
    The country Sri Lanka stands ahead of India in all aspects in HDI.
  2. Mention the aspects that are considered in making Human Development Index.
    Answer:

    1. Percapita income
    2. Literacy Rate
    3. Average years of schooling
    4. Expected years of schooling
    5. Life expectancy at birth
    6. Health status, etc. are the aspects considered in making Human Development Index.
  3. Name the country that has the lowest per capita income.
    Answer:
    The country Nepal has the lowest per capita income.
  4. Give two suggestions for the improvement of the rank of India in HDI.
    Answer:

    1. Education should be improved and more skill development centres should be established.
    2. The poor and needy people should be provided cheap and better health facilities.

Question 7.
Observe the given table and analyse the HDI data of India and its neighbours.
Some data regarding India and its neighbours for 2016

CountryPer capita income in $Life expectancy at birthLiteracy rateHuman Development Index (HDI)
Sri Lanka10,78974.992.670
India5,66368.374.04131
Pakistan5,03166.260.0148
Myanmar4,94365.993.1146
Bangladesh3,34171.661.5140
Nepal2,33769.664.7144

Answer:
The given table is about Human Development data of India and some of the neighbouring countries pertaining to 2016. In this table per capita income in dollars, life expectancy at birth and literacy rate are considered and HDI Ranking is given. In per capita income Sri Lanka stands high and in the same of life expectancy but in literacy Myanmar is better than Srilanka. Pakistan is very poor in literacy rate and so Bangladesh. These countries do not show interest on literacy. In over all ranking Srilanka stands well, Pakistan’s ranking is least. Countries should concentrate on what (the people need proper medication, medical facilities to the poorer people, wide availabiltiy of
104,108 services are essential in rural areas. In Telangana to some extent these services are provided. Each one teach one programme is to be maintained so that all people will be literated. Schooling should be strengthened. Per capita income should be spent on Health facilities and education. Priorities are to be set first and proper planning for its implementation is essential. So that countries can be developed.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 8.
Read the text and interpret it.
The ongoing protest over the Kudankulam Nuclear Power Project in theTirunelveli district of Tamil Nadu is one such conflict. The government of India set up the nuclear power plant in this quiet coastal town of fisher folks. The aim is to generate nuclear power to meet the growing energy needs of the country. The people in the region have protested on the grounds of safety, security and livelihood.
Answer:

  1. The Government of India is going on with the Nuclear Power Project of the Kundankulum.
  2. The aim of the project is to generate nuclear power to meet the growing energy needs of the people.
  3. The fisher folks of this area are protesting against the project on the grounds of safety, security and livelihood.
  4. Hence the ideas on development are different for different people.
  5. Development for one may not be the development for other.
  6. Here the ideas on development of government is conflicting with the interests of the local people.

Question 9.
Read the following text and state your opinion on it.

Human Development Report

When we realise that even though the level of income is important, it is an inadequate measure of the level of development, we begin to think of other criterion. There could be a long list of such criteria but then it would not be so useful. What we need is a small number of the most important things. Health and education indicators, such as the ones we used in comparison of Kerala and Punjab, are among them. Over the past decade or so, health and education indicators have come to be widely used along with income as a measure of development. For instance, Human Development Report published by United Nations Development Programme (UNDP) compares countries based on the educational levels of the people, their health status and per capita Income.

Answer:

  1. The income is not the correct criteria to measure the level of development.
  2. Most of the time it hides disparities.
  3. So we begin to think of another criterion.
  4. There could be a long list of criteria.
  5. We selected some such as “health and education”.
  6. Health and Education indicators were used to compare Punjab and Kerala.
  7. Human Development Report was published by UNDP.
  8. The countries were compared on the basis of the education levels of the people, health status and per capita income.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 10.
Read the following paragraph and write your opinion.
If you get a job in afar off place, you would try to consider many factors, apart from income.
This could be facilities for your family, working atmosphere or opportunity to learn. In another case, a job may give you less pay but may offer regular employment that enhances your sense of security. Another job, however, may offer high pay but no job security and also leave no time for your family. This will reduce your sense of security and freedom.
Answer:

  1. According to this paragraph, people give preference to income as well as facilities.
  2. They also want to spend time with their families.
  3. They need job security and freedom. It is not possible in all the cases.
  4. Many workers from India are migrating even to foreign countries for getting work so as to earn something for their livelihoods.
  5. Some people are working in cities like Hyderabad by keeping their families at their hometowns.
  6. They are losing the opportunity of spending their time with their families. Their living conditions are also not good.
  7. Many of them are leading sedentary life. In some cases pay is less but job is secure. In some other cases payment may be high but job security is not there.
  8. Whatever it is, I am coming to the conclusion that there are many factors that affect livelihoods.

Question 11.
Read the following paragraph and write your comments.
When we looked at Individual aspirations and goals, we found that people not only think of better income but also have goals such as security, respect of others, equal treatment, freedom, etc. In mind. Similarly, when we think of a nation or a region, we may, besides average income, think of other equally important attributes.
Answer:

  1. This paragraph is about the aspirations and goals of the individuals.
  2. The people want their income and they want to be treated well.
  3. Though the wages are well and good, they don’t want to be ill-treated.
  4. Everyone in the society wants to live with dignity which our Constitution promises.
  5. Many of the labourers are now looking towards prestige.
  6. Some states in our country are getting more per capita income but they are lacking in providing other facilities to the people.
  7. Nowadays schooling is very importing to its children.
  8. For nations and states the literacy rate, net attendance rate, infant mortality rate are also considered in its development.
  9. They have to provide schools, pollution-free atmosphere, unadulterated medicines, to its people for better living.

Question 12.
Read the following paragraph and write your opinion on it.
In many parts of the country, girls’ education is still given less priority by parents compared to
boys’ education. While girls may study for a few classes, they may not complete their schooling.
A welcome trend in Himachal Pradesh is the lower gender bias. Himachali parents have ambitious educational goals for their girls, just as for their boys.
Answer:

  1. According to this paragraph two things are clearly mentioned that there is a gender bias with regard to giving education to boys and girls among the people of the country.
  2. Only the Himachali Pradesh Government is concentrating on girl education.
  3. My opinion on these two,issues is that the gender bias was there once in the society.
  4. As there is a vast awareness in the parents they send their daughters to the schools in many
    places.
  5. Even a rickshaw puller also wants to make his daughter study In a school.
  6. He hopes his daughter becomes a professional.
  7. The parents are interested to send their children to English medium schools Irrespective of their income and status.
  8. It is a sensitive issue and it is to be taken seriously to bring awareness among people to make their daughters admitted in schools and the governments should consider the problems of girls in schools.
  9. Sufficient toilets and other facilities are to be provided so as to enroll all the girls In schools.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 13.
Table: Read the following table and answer the following questions.
Some Comparative data of Selected States

stateIMR per 1000 (2006)Literacy rate (%) {2011)Net Attendance Rate (2006)
Punjab427776
Himachal Pradesh368490
Bihar626456
  1.  What does the table tell us?
    Answer:
    The table tells us about some comparative data of selected states.
  2. What are the three states compared here?
    Answer:
    The three states compared are Punjab, Himachal Pradesh and Bihar.
  3. What does IMR mean?
    Answer:
    IMR means Infant Mortality Rate.
  4. Which state ranks best in literacy rate?
    Answer:
    Himachal Pradesh ranks best in literacy rate with 84%.
  5. What is the Net Attendance Rate of Bihar in 2006?
    Answer:
    The Net Attendance Rate of Bihar in 2006 is 56.
  6. What is the position of Punjab in 2011 with regard to literacy rate?
    Answer:
    The position of Punjab in 2011 with regard to literacy rate is next to Himachal Pradesh.
  7. What is literacy rate?
    Answer:
    Literacy rate is that the percentage of literate people in the 7 and above years age group.
  8. Why is the number of children below the age of 6 not considered for counting of Net Atten¬dance Rate?
    Answer:
    The children below the age of 6 do not go to school.
  9. Which state ranks first when we consider the above data ?
    Answer:
    Himachal Pradesh ranks first.
  10. How do you say Bihar is an underdeveloped state ?
    Answer:
    Bihar has high infant mortality rate i.e., 62. It has less literacy rate and Net attendance rate i.e., 64 and 56 respectively. Hence we can say Bihar is a backward state.

Question 14.
Study the table given and answer the following questions.
Table: Some data regarding India and its neighbours for 2013

CountryPer Capita Income in $Life expectancy at birth (Years)Average years of schoolingExpected years of schoolingHuman Develop­ment Index (HDI) rank in the world
Sri Lanka517075.19.312.792
India328565.84.410.7136
Pakistan256665.74.97.3146
Myanmar181765.73.99.4149
Bangladesh178569.24.88.1146
Nepal113769.13.28.9157
  1. What is the table about?
    Answer:
    The table is about some data regarding India and its neighboring countries for 2013.
  2. What is the life expectancy at birth in India?
    Answer:
    The life expectancy at birth in India is 65.8 years.
  3. What components are considered in this table?
    Answer:
    The components considered in the table are per capita income, life expectancy at birth, aver¬age years of schooling, expected years of schooling and HDI.
  4. Which country ranks best in HDI?
    Answer:
    Sri Lanka ranks best in HDI with 92nd rank.
  5. Which country is lacking in expected years of schooling?
    Answer:
    Pakistan is lacking in expected years of schooling.
  6. Per capita income is shown in some symbol. What does it mean?
    Answer:
    The symbol given means dollar.
  7. Which country has the lowest average years of schooling?
    Answer:
    Nepal has the lowest average years of schooling.
  8. What is the per capita income of India ?
    Answer:
    The per capita income of India is $ 3,285.
  9. What is the lowest life expectancy at birth in the table?
    Answer:
    65.7 years is the lowest life expectancy at birth in the table.
  10. What is the difference between expected years of schooling and average years of schooling
    for India?
    Answer:
    The difference between expected years of schooling and average years of schooling for India is 6.3 years.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 15.
By studying the given map answer the following questions.
AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development 3

  1. Name two countries which have low income.
    Answer:
    Zimbabwe and Kenya are the two countries with low income.
  2. Which two countries have very high income?
    Answer:
    USA and Canada are the two countries which have very high income.
  3. Name two countries with high income.
    Answer:
    Brazil and Russia are the two countries with high income.
  4. Which two countries have medium income?
    Answer:
    India and Egypt are the two countries which have medium income.

Question 16.
Prepare a pamphlet on Promoting Girl Education.
Answer:

PROMOTING GIRL EDUCATION

Girls and boys in the society are equal but many of the parents give less importance to girl education compared to boys. This treatment of girls and boys in different ways is called gender bias. Some parents feel that boy is income and girl is expenditure. After marriage also the in-law’s family of the bride normally gives very less importance to her education. This is the wrong notion that the people have. It is to be removed.
The notion is to be changed. Many women have come forward to discharge their duties in political, educational and administrative areas. They are proving that they can do everything. In education also many girls are getting good results and ranks. Their number in civil services and other competitive examinations is rapidly increasing. If a girl is educated, she can manage her family herself well.
She can educate her children.
Many women now are district collectors or police officers and administrative officers and a few banks are being run under the leadership of female authorities. Many departments are under their control. For many years it has been a custom that the women have to work at kitchen but it is disproved, if they were given choice to do something they can do it as we expected.
All the parents should send their daughters to schools to study. They should be given an opportunity to show their intellect. All the parents should understand that their notion is to be
changed and think positively about their daughters’ future.

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

Question 17.
Write a paragraph after reading the table.
Answer:
Comparison of two countries

Name of the countryMonthly incomes of citizens in 2014 (in Rupees)
Country A1IIIIIIVVAverage
9,500105009,800100001020010,000
Country B5005005005004800010,000

According to this table in the country ‘A’, all the five persons are earning mostly equal monthly income. In the country ‘B’ the first four persons are earning only five hundred each but the fifth person is earning forty eight thousand rupees per month. If you are the fifth one it is OK but if you are one among the first four it will be worst to live in. If we consider the average, it will be the same as ten thousand per person per month. In these two countries the country A has more equitable distribution of income.
In many countries more income is there with a few persons. Many people in the countries are poor. They don’t have minimum amount of income for their livelihoods. In the above table the average income of the two countries is the same but in B it is not equally distributed. The Gross Domestic Product is to be distributed among the people of the country that means the poor also have to get their share in the country. The gap between the poor and the rich is to be removed and so the society of equality emerged.

Question 18.
Locate the following in the given map of India.

  1. Nuclear power plant in Tamil Nadu.
    Answer:
    Kudankulam
  2. Schooling Revolution took place in the state.
    Answer:
    Himachal Pradesh
  3. Draw the Indian standard time.
    Answer:
    82 1/2° E longitude.
  4. Sahyadri Range.
  5. Islands in Bay of Bengal.
    Answer:
    Andoman & Nicobar
  6. Locate any one of Hill station.
    Answer:
    Nainital
  7. River which is flowing through a rift valley.
    Answer:
    Narmada
  8. The Hill Station located near Nilgiris.
    Answer:
    Ooty
  9. The largest river in South India.
    Answer:
    Godavari
  10. The largest river in India.
    Answer:
    Ganga

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development 4

AP SSC 10th Class Social Studies Important Questions Chapter 2 Ideas of Development

AP SSC 10th Class Biology Solutions Chapter 2 Respiration – The Energy Releasing System

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 2 Respiration – The Energy Releasing System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 2nd Lesson Respiration – The Energy Releasing System

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System Textbook Questions and Answers

Improve your learning

Question 1.
Distinguish between
(a) Inspiration and Expiration
(b) Aerobic and Anaerobic respiration
(c) Respiration and Combustion
(d) Photosynthesis and Respiration
Answer:
(a) Inspiration and Expiration:

InspirationExpiration
1. It is also called inhalation.1. It is also called exhalation.
2. The air or water is taken into the respiratory organ.2. The air or water is sent out of the respiratory organ.
3. It is an active process.3. It is a passive process.
4. Rib cage moves forward and outward.4. Rib cage moves downward and inward.
5. Diaphragm contracts and becomes flattened.5. Diaphragm relaxes and becomes original dome shaped.
6. Increase in volume of thoracic cavity.6. Decrease in volume of thoracic cavity.
7. Air pressure in lungs is less than the atmospheric pressure.7. Air pressure in lungs is greater than the atmospheric pressure.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

(b) Aerobic respiration and Anaerobic respiration:
(OR)
Respiration is energy-producing process in the organisms. It takes place both in the presence and absence of oxygen. Laxmi said there are some differences between the two processes. How do you support her?
Answer:

Aerobic respirationAnaerobic respiration
1. It takes place in the presence of oxygen.1. It takes place in the absence of oxygen.
2. In aerobic respiration, complete oxidation of glucose takes place.2. In anaerobic respiration, the glucose molecule is incompletely oxidised.
3. End products are CO2 and water.3. End products are either ethyl alcohol or lactic acid and CO2.
4. Lot of energy is liberated (38 ATP).4. Relatively small energy is liberated (2 ATP).
5. It occurs in plant’s and animal’s cells.5. Occurs in many anaerobic bacteria and human muscle cells.
6.
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 1
6.
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 2
7. It has two stages – Glycolysis and Krebs cycle.7. It has two stages – Glycolysis and Fermentation.

(c) Respiration and Combustion:
(OR)
Even though both are oxidation processes, combustion and respiration are different in many aspects. Explain those differences.
(OR)
Combustion and respiration are oxidative processes but still there are differences between them. What are they?
(OR)
Write the differences between respiration and combustion.
Answer:

RespirationCombustion
1. It occurs in living cells.1. It is non – cellular.
2. Oxidation of food materials especially glucose to carbon dioxide and water is called respiration.2. When sugar burns CO2 and water are produced and energy is released as heat. This process is called combustion.
3. Oxidation of sugar molecules occurs at the body temperature of an organism.3. Heat is to be supplied for the sugar molecule to burn.
4. The energy is released in several stages.4. The energy is released at once as heat.
5. Several intermediate substances are formed.5. No intermediate substances are formed.
6. Enzymes are required for oxidation.6. Enzymes are not required for combustion.
7. Respiration occurs in the presence of water.7. Combustion occurs in the absence of water.
8. It is a controlled process.8. It is an uncontrolled process.
9. Energy is stored in ATP in the body.9. Energy is not stored and is released into the atmosphere.

(d) Photosynthesis and Respiration:

PhotosynthesisRespiration
1. Occurs only in all plants and photo­synthetic bacteria.1. Occurs in all living organisms.
2. Takes place in the presence of sunlight.2. Takes place throughout day and night.
3. A plant can survive without performing photosynthesis for few days.3. No organism can survive without respiration for few minutes even.
4. In plants, only few cells perform photosynthesis.4. All living cells in an organism perform this process.
5. Raw materials are C02 and water.5. Uses carbohydrates and oxygen.
6. Oxygen is liberated.6. Carbon dioxide is released.
7. It occurs in chloroplast.7. It takes place in cytoplasm and mitochondria.
8. Adds weight to the organism.8. Decrease weight of the organism.
9. It is an anabolic process.9. It is a catabolic process.
10.
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 3
10.
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 4

Question 2.
State two similarities between aerobic and anaerobic respiration.
Answer:
Similarities between aerobic and anaerobic respiration:

  1. Both are catabolic processes.
  2. Both aerobic and anaerobic respiration takes place in all cells.
  3. Energy is released in both the processes.
  4. CO2 is the end product of both processes.
  5. First stage of both respiration is glycolysis.
  6. Respiratory substances in both processes are glucose, fatty acids and amino acids.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 3.
Food sometimes enters the wind pipe and causes choking. How does it happen?
Answer:

  1. Pharynx is the common passage of both air and food.
  2. From here air enters into trachea and food enters into oesophagus.
  3. Pharynx is connected to larynx through glottis a slit like opening.
  4. A cartilagenous flap called EPIGLOTTIS act as a lid over glottis and prevents food from entering into trachea during swallowing.
  5. Any food particles enters the trachea it causes chocking.
  6. Sometimes the food particles are forced back by cough.

Question 4.
Why does the rate of breathing increase while walking uphill at a normal pace in the mountains? Give two reasons.
Answer:
The rate of breathing increases while walking uphill at a normal pace in the mountains.

  1. It is because as we go up the hill above sea level the concentration of oxygen is greatly reduced. So we have to breathe more to get required amount of oxygen.
  2. While walking uphill a lot of oxygen is used by our body to release energy from glucose.
  3. This leads to lack of oxygen in the cells.
  4. We take in oxygen when we breathe.
  5. Hence to increase the amount of oxygen intake there is an increase in breathing rate during walking uphill.

Question 5.
Air leaves the tiny sacs in the lungs to pass into capillaries. What modification is needed in the statement?
Answer:

  1. Gaseous exchange takes place within the lungs by diffusion from the alveoli to blood capillaries and vice versa.
  2. The carbondioxide in the blood is exchanged for oxygen in alveoli.
  3. This sentence may be modified as “Air that contains oxygen reaches the tiny sacs in the lungs to pass into capillaries.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 6.
Plants photosynthesize during daytime and respire during the night. Do you agree to this statement? Why? Why not?
Answer:

  1. No, I do not agree with this statement. Plants photosynthesize during daytime only and respire during the daytime as well as night time also.
  2. During daytime when photosynthesis occurs in the presence of sunlight. Oxygen is produced. The leaves use some of this oxygen for respiration and the rest diffuses into air.
  3. During daytime CO2 produced by respiration is all used up in photosynthesis by leaves.
  4. At night time no photosynthesis occurs and oxygen diffuses into leaves to carryout respiration.

Question 7.
Why does a deep sea diver carry oxygen cylinder on his/her back?
Answer:

  1. When we go deep into the sea, the oxygen level decreases. Oxygen is in dissolved state in water.
  2. Humans are adapted for utilizing oxygen in gaseous state. They cannot use dissolved oxygen for breathing.
  3. Only aquatic animals like fish can utilize the dissolved oxygen for breathing using gills.
  4. Human beings have lungs for respiration. Therefore, sea divers have to carry oxygen cylinders in their back so as to receive oxygen.
  5. If they do not carry them, they do not get oxygen and there is a chance even to die.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 8.
How are alveoli designed to maximise the exchange of gases?
Answer:
The human lungs have been designed to maximise the exchange of gases as follows.

  1. The interior of lung is divided into millions of small chambers called alveoli.
  2. The presence of millions of alveoli in the lungs provide a very large surface area.
  3. If all alveoli of our lungs are spread out they will cover an area of nearly 160 m2.
  4. Availability of large surface area maximises the exchange of gases.

Question 9.
Where will the release of energy from the glucose in respiration take place? Mala writes lungs, while Jiya writes muscles. Who is correct and why?
Answer:

  1. Respiration is the process of releasing energy from the breakdown of glucose.
  2. Respiration takes place in every living cell, all the time.
  3. All cells need to respire in order to produce the energy that they require.
  4. During respiration the release of energy from the glucose takes place in muscles but not in lungs.
  5. So Jiya is correct. The energy is released from the muscle cells during respiration. Only gaseous exchange takes place in lungs.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 10.
What is the role of epiglottis and diaphragm in respiration?
Answer:
Epiglottis:

  1. Epiglottis, a flap like muscular valve controls movement of air and food towards their respective passages.
  2. Epiglottis is partly closed when we swallow food and it opens more widely when we take a breath and air enters the lungs.
  3. Epiglottis allows air pass through the larynx and the respiratory system.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 6

Diaphragm :

  1. The diaphragm in the respiratory system is the dome shaped sheet of muscle that separates the chest from the abdomen.
  2. When the diaphragm contracts during inhalation it flattens out a bit. This results in the enlargement of the volume of the chest cavity.
  3. This reduces the pressure in the lungs and air enters into lungs from outside the body.
  4. During exhalation, the diaphragm relaxes and assumes its dome shape. This change increases the pressure on the lungs and squeezes the air through the nose to the atmosphere.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 7

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 11.
How does gaseous exchange take place at blood level?
Answer:

  1. Lungs are made up of several thousands of small chambers called alveoli.
  2. Within the alveoli, exchange of gases take place between the gases inside the alveoli and blood.
  3. Blood arriving in the alveoli has higher CO2 concentration which is produced during respiration by the body cells.
  4. At the same time air in the alveoli has a much lower concentration of CO2 and this allows the diffusion of CO2 out of the blood and to alveolar air.
  5. Similarly blood arriving in the alveoli has a lower oxygen concentration while air in the alveoli has a higher oxygen concentration.
  6. Therefore oxygen moves into the blood by diffusion.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 8

Question 12.
Explain the mechanism of gaseous exchange at bronchiole level.
Answer:

  1. the trachea (wind pipe) is divided into two tubes called BRONCHI. As there are two lungs each bronchus (singular) enters the lungs on the same side.
  2. In the lung, the bronchus divides into smaller and smaller branches called BRONCHIOLES which enters into each alveoli.
  3. When oxygen from outside reaches the alveoli through bronchioles and the carbondioxide from alveoli moves out.
  4. The inhaled air from outside enters into bronchioles through nostrils → nasal cavities → pharynx → larynx → trachea → bronchus.
  5. The exhaled air from alveoli enters bronchioles → pharynx → nasal cavities → nostrils → outside.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 9

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 13.
After a vigorous exercise or work we feel pain in muscles. What is the relationship between pain and respiration?
Answer:

  1. We obtain energy by oxidation of glucose molecule.
  2. In the absence of oxygen (anaerobic respiration) glucose is converted to latic acid.
  3. During vigorous exercise oxygen gets used up faster in the muscle cells that can be supplied by blood.
  4. When oxygen supply is inadequate the muscles use energy released during the anaerobic breakdown of glucose.
  5. The anaerobic respiration by muscles bring about partial breakdown of glucose to form lactic acid.
  6. The accumulation of lactic acid in the muscles causes muscular pains or cramps.

Question 14.
Raju said, “Stems also respire along with leaves in plants”. Can you support this statement? Give your reasons.
Answer:
Yes. I support the statement of Raju that stems also respire along with leaves in plants.
The reasons are

  1. The stems of herbaceous plants have stomata.
  2. So the exchange of respiratory gases in the stems of herbaceous plants takes place through stomata.
  3. The oxygen from air diffuses into the stem through stomata and reaches all the cells for respiration.
  4. The carbon dioxide released during respiration diffuses out into the air through the stomata.
  5. In woody stems the bark has lenticels for gaseous exchange. Through lenticels, oxygen diffuses in and carbon dioxide diffuses out into the air.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 10

Question 15.
What happens if diaphragm is not there in the body?
Answer:

  1. The lungs cannot draw in air or push it out by themselves. The chest wall muscles and the diaphragm helps the lungs in moving air into and out of them.
  2. If diaphragm is not there in the body, we would not be able to breathe.
  3. The diaphragm is the major muscle in the process of respiration.
  4. It separates the thoracic and abdominal cavities.
  5. In the absence of diaphragm, the relaxation and contraction of the chest wall muscles do not take place and thereby inspiration and expiration become difficult that leads to death of the person.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 16.
If you have a chance to meet pulmonologist, what questions are you going to ask about pulmonary respiration?
Answer:
If I have a chance to meet pulmonologist, I would like to ask the following questions:

  1. What is pulmonary respiration?
  2. What is the organ involved in pulmonary respiration?
  3. What is the name of blood vessel that brings deoxygenated blood to lungs?
  4. What is the name of the blood vessel that carries oxygenated blood from the lungs to the heart?
  5. Out of the two lungs which one is larger than the other?
  6. What type of diagnostic test will be performed to assess the function of lungs?
  7. What is pulmonary edema? How does it occur?
  8. Can all the diseases of the lungs be cured permanently?

Question 17.
What procedure do you follow to understand anaerobic respiration in your school laboratory?
(OR) (Lab Activity)
Write the procedure and observations of the experiment which you have conducted in your laboratory to prove that CO2 and heat are evolved during anaerobic respiration by using yeast.
(OR)
How do you prove that carbon dioxide is released during anaerobic respiration?
(OR)
How do yeast cells convert glucose solution to CO2 and ethyl alcohol?
Answer:
Aim : To prove that CO2 is released during anaerobic respiration.
Apparatus: Thermos flask, splitted corks, thermometer, wash bottle, glass tubes, liquid
paraffin, glucose solution, yeast cells, bicarbonate solution.
Procedure:

  1. Remove dissolved oxygen from glucose solution by boiling it for a minute and then cooling it without shaking.
  2. Now add some yeast to the glucose solution and fix a two-holed rubber stopper to the flask.
  3. The supply of oxygen from the air can be cut off by pouring a 1cm layer of liquid paraffin into the mixture through the holes.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 11
  4. Insert one end of the thermometer into the thermos flask. See the mercury bulb of thermometer keep inside the solution.
  5. Arrange for any gas produced by the yeast to escape through a wash bottle containing bicarbonate solution or lime water as shown in the figure.
  6. Add a few drops of diazine green (Janus Green B) solution to the yeast suspension before you pour liquid paraffin over it.
  7. The blue diazine green solution turns pink when oxygen is in short supply around it.
  8. Warm the apparatus to about 37° F in order to speed up the test.
  9. Keep the apparatus undisturbed for one or two days.

Observations :

  1. After two days it can be observed that lime-water of the wash bottle turns into milky white precipitate.
  2. Increase in temperature in thermometer.
  3. Alcohol smell given off from the flask.

Result: These observations indicate that yeast cells respire anaerobically converting glucose solution into CO2, ethyl alcohol and release heat energy.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 18.
What are your observations in the combustion of sugar activity?
Answer:
Observations in the combustion of sugar:

  1. When sugar is heated, first it melts, chars and later burns producing flames.
  2. When sugar is combusted, carbon dioxide and water are produced.
  3. Energy is also released in the form of heat and it released at once.
  4. We cannot control the combustion of sugar.
  5. Intermediate products are not formed.
  6. We can combust sugar in the absence of water and enzymes.
  7. When combustion of sugar, heat energy is released into the atmosphere and we cannot store it for further use.

Question 19.
Collect information about cutaneous respiration in frog. Prepare a note and display them in your classroom.
(OR)
How does frog respire with the help of skin?
Answer:

  1. Respiration through skin is called cutaneous respiration.
  2. In frog, skin is an additional or secondary or accessory respiratory organ.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 12
  3. Skin is a very important respiratory organ in both on land and water.
  4. One-third of the total oxygen taken up by frog is through the skin.
  5. Frog also keeps its skin moist. Frog skin has a large number of mucous glands which secrete mucous onto the surface of the skin.
  6. The mucous layer retains water and reduces evaporation of water from body.
  7. To keep the skin wet and moist frogs jump into water very frequently.
  8. Frog skin is supplied with a large number of blood vessels which help in absorbing oxygen from the water.
  9. The carbon dioxide produced during to respiration, diffuses out into the water through the blood vessels present in the skin of the frog.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 20.
Collect information about respiratory diseases (because of pollution, tobacco) .and discuss with your classmates.
Answer:
Respiratory diseases because of pollution:

  1. Irritation of eyes, nose, mouth and throat.
  2. Headaches and dizziness.
  3. Respiratory symptoms such as coughing and running nose.
  4. Respiratory and lung diseases including
    a) Asthma attacks
    b) Chronic obstructive pulmonary disease (COPD)
    c) Reduced lung function
    d) Pulmonary cancer caused by a series of carcinogen chemicals that through inhalation
    e) Mesothelioma: A particular type of lung cancer, usually associated with expo¬sure to asbestos (it usually occurs 20 – 30 years after the initial exposure)
    f) Pneumonia: Infection of lungs caused by bacteria.
    g) Bronchitis: It is inflammation or swelling of bronchial tubes.
    h) Emphysema: It is a lung condition in which tiny air sacs in lungs alveoli fill up with water.

Respiratory diseases due to tobacco :

  1. Chronic bronchitis: A long term inflammation of the bronchi is characterized by coughing.
  2. Lung cancer: An abnormal continuous multiplication of cells that can result in tumors in the lining of the bronchi.
  3. Emphysema: A chronic lung condition that affects the air sacs in the lungs characterized by shortness of breath, coughing, fatigue, sleep and heart problems.

Question 21.
What is the pathway taken by air in the respiratory system? Illustrate with a labelled diagram.
Answer:
The path way taken by air in the respiratory system:
Nostrils → Nasal cavity → Pharynx → Larynx → Trachea → Bronchus → Bronchioles → Alveolus → Blood.

  1. Nostrils : Air enters the body through the nostrils.
  2. Nasal cavity: Air is filtered and its temperature is also brought close to that of the body.
  3. Pharynx: It is the junction of respiratory and digestive system. Epiglottis – a flap like muscular valve controls movement of air and food towards their respective passages.
  4. Larynx: Also called voice box. This stiff box contains vocal cords. When air passes out of the lungs and over vocal cords, it causes them to vibrate. This produces sounds on the basis of our speech, song etc.
  5. Trachea: This is also called wind pipe. It channels air to lungs.
  6. Bronchi: Trachea at it’s lower end divides into two bronchi one leading to each lung.
  7. Bronchioles: The bronchi further divided into smaller and smaller branches called bronchioles.
  8. Alveoli: Clusters of air sacs called alveoli in the lungs which are very small and numerous. The gaseous exchange takes place here as blood capillaries take up oxygen and expel CO2.
  9. Blood: It carries oxygen, to each and every cell of the body and collects CO2 from them.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 13

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 22.
Draw a block diagram showing events in respiration. Write what you understood about cellular respiration.
Answer:
Events in respiration :
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 14AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 15

Cellular respiration :

  1. All living cells must carry out cellular respiration.
  2. Oxidation of glucose of fatty acids releasing energy takes place in cells, hence it is called cellular respiration.
  3. It can be in the presence of oxygen that is aerobic respiration or in its absence that is anaerobic respiration (fermentation).
  4. Cellular respiration in prokaryotic cells like that of bacteria occurs within the cytoplasm.
  5. In Eukaryotic cells cytoplasm and mitochondria are the site of cellular respiration.
  6. The energy released in cellular respiration is stored in a special compound called ATP.
  7. ATP is utilised for carrying out other functions in the cell.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 23.
How do you appreciate the mechanism of respiration in our body?
Answer:

  1. Respiration is essential for life because it provides energy for carrying out all the life processes which are necessary to keep the organism alive.
  2. The energy that is obtained from respiration is used to build the organism by way of cell growth, reproduction and cell repair, etc.
  3. All systems in living beings need energy to survive.
  4. Respiration helps to expel out the toxic carbon dioxide out of the cells. This CO2 will be utilised by the plants to produce food materials through the process of photosynthesis.
  5. The respiratory system goes into operation from the movement of our birth and works without ever stopping as long as we live our breath continues.
  6. During exhalation, the vocal cords in the larynx vibrate to produce sounds and help in speaking as we like.

Question 24.
Prepare an article on anaerobic respiration to present school symposium.
Answer:
Anaerobic respiration :

  1. Respiration that occurs without oxygen is known as anaerobic respiration.
  2. It is present in primitive organisms and muscular cells in higher animals.
  3. Alcohol, CO2 and H2O are end products in this process.
  4. AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 5

Merits:

  1. In the absence of oxygen it is good process.
  2. Mechanism is simple.
  3. Suitable to microorganisms

Demerits:

  1. Provides less energy
  2. Not suitable to higher animals

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 25.
Prepare a cartoon on discussion between haemoglobin and chlorophyll about respiration.
Answer:
Discussion between haemoglobin and chlorophyll about respiration:
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 16

Haemoglobin: Hello good morning chlorophyll. How are you?

Chlorophyll: Very good morning haemoglobin. I am fine.

Haemoglobin: I am also doing well. Let me know something about you.

Chlorophyll: I am a green coloured pigment present in leaves of plants.

Haemoglobin: How many types of chlorophylls are there ?

Chlorophyll: We are four types i.e., chlorophyll – a, chlorophyll – b, chlorophyll – c and chlorophyll – d.

Haemoglobin: May I know your job in leaves?

Chlorophyll: Yes. Why not? I am an essential factor required to prepare food through the process of photosynthesis in plants.

Haemoglobin: Oh! You are participating in the process of preparing food materials by green plants.

Chlorophyll: Now tell me about your presence.

Haemoglobin: I am present only in animal cells. That too in the red blood cells of the blood. The red colour of the blood is due to my presence.

Chlorophyll: Then tell me about your function in respiration?

Haemoglobin: During respiration, I carry oxygen to the cells in the body tissues.

Chlorophyll: How are you able to do this?

Haemoglobin: I have an oxygen binding element iron. It binds oxygen on four corners of it. I form oxy-haemoglobin with oxygen in the lungs.

Chlorophyll: What happens to the digested food in the cells?

Haemoglobin: In cells, oxygen breakdown the glucose molecule into C02 and H20 releasing large amount of energy. Around 38 ATP molecules are produced.

Chlorophyll: What is the use of this energy?

Haemoglobin: This energy is utilised by the cell to carry other functions.

Chlorophyll: Thank you haemoglobin. You have taught me everything about respiration. In our next meeting we will discuss another topic.

Haemoglobin: Thank you chlorophyll for your interest and patience.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Fill in the blanks.

  1. Exhaled air contains ———– and ———–.
  2. A flap like muscular valve controls movement of air and food is ———–.
  3. Energy currency of the cell is called ———–.
  4. Lenticels are the respiratory organs that exist in ———– part of the plant.
  5. Mangrove trees respire with their ———–.

Answer:

  1. carbon dioxide, water vapour
  2. Epiglottis
  3. ATP (Adenosine Tri Phosphate)
  4. stem
  5. aerial roots

Choose the correct answer.

6. We will find vocal cords in [ ]
A) Larynx
B) Pharynx
C) Nasal cavity
D) Trachea
Answer: A

7. Cluster of air sacs in lungs are called [ ]
A) Alveoli
B) Bronchi
Answer: A

8. Which of the following is correct ? [ ]
i) The diaphragm contracts – volume of chest cavity increased
ii) The diaphragm contracts – volume of chest cavity decreased
iii) The diaphragm expands – volume of chest cavity increased
iv) The diaphragm expands – volume of chest cavity decreased
A) i B) i and ii C) ii and iii D) iv
Answer: A

9. Respiration is a catabolic process because of [ ]
A) Breakdown of complex food molecules
B) Conversion of light energy
C) Synthesis of chemical energy
D) Energy storage
Answer: A

10. Energy is stored in [ ]
A) Nucleus
B) Mitochondria
C) Ribosomes
D) Cell wall
Answer: B

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System

Activity – 1

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

How do you test the presence of water vapour and heat in the exhaled air?
Answer:

  1. Keep your palm around an inch away from your nose.
  2. Feel you breathing out.
  3. Do not remove your palm until you have finished the activity.
  4. Breathe steadily for 1 – 2 minutes.
  5. Now take a piece of any fruit.
  6. Chew and before swallowing it keep the fingers of the other palm on your neck, now swallow it.

Questions:

  1. What did you notice? What happens to your breath as you try to swallow?
    Answer:
    We cannot swallow while breathing. We usually stop breathing when we swallow food.
  2. What is helping you to swallow without deflecting it to the windpipe?
    Answer:
    Epiglottis is helping me to swallow without deflecting food to the windpipe.

Activity – 2
Write an experiment to observe changes during combustion of sugar.
(OR)
What are your observations in combustion of sugar activity.
Answer:
Aim: To observe changes during combustion of sugar.
Apparatus: Wooden stand, test tubes, rubber stopper, delivery tube, glucose or sucrose powder, lime water, spirit lamp.
Procedure:

  1. Take a small amount of glucose in a small test tube.
  2. Arrange the apparatus as shown in the figure.
  3. Heat the test tube until the glucose catches fire.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 17

Observations :

  1. In the initial stage the glucose becomes liquid.
  2. Later it turns to black colour after catching fire.
  3. In this process carbon dioxide and water are produced.
  4. Energy is released as heat.
  5. The carbon dioxide released changes lime water to milky white.

Conclusion:

  1. From this experiment, we can conclude that carbon dioxide, water and heat are produced during combustion of glucose in the laboratory.
  2. The carbon dioxide changes lime water to milky white in nature.

Activity – 3

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

How can you prove that carbon dioxide is evolved during respiration?
(OR)
Write the experimental procedure and draw the arrangement of apparatus to show that CO2 is evoloved in respiration.
To understand that CO2 is evolved during respiration, what experiment you have performed in your laboratory? Explain the procedure.
(OR)
Write an experiment to prove that CO2 is released during respiration.
Answer:
Aim: To prove that CO2 is released during aerobic respiration.
Apparatus: Two wide mouthed plastic or glass bottles, germinating seeds, dry seeds, two small injection bottles or beakers with lime water.

  1. Take two wide mouthed glass bottles.
  2. Keep germinating bengal gram seeds in one bottle and dry seeds in another bottle.
  3. Keep two small beakers with lime water in each glass bottles.
  4. Close the glass bottles tightly.
  5. Keep both the sets undisturbed for one or two days.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 18

Observation:

  1. After two days it can be observed that lime water of the beaker placed in the bottle containing germinating seeds turns into milky white.
  2. And the lime water kept in the glass bottle containing dry seeds do not change its colour.
    Result: It indicates that germinating seeds liberated carbon dioxide which turns lime
    water into milky white.

Activity – 4

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Explain the procedure you have adopted in your school to prove that heat is liberated during respiration. What result we will get, if you perform this experiment with dry seeds?
(OR)
Write the procedure you have followed to observe “heat is evolved during respiration” in your laboratory. What precautions did you take during the activity?
Answer:
Aim: To prove that heat is liberated during respiration.
Apparatus: Two Thermos flasks, two thermometers, two rubber corks, dry seeds, germinating seeds.
Procedure:

  1. Take a handful of moong or bazra seeds.
  2. Soak the seeds in water a day before experiment.
  3. Keep these soaked seeds in a cloth pouch and tie with a string tightly.
  4. Next day collect the sprouts / germinated seeds from the pouch in a thermos flask and take dry seeds in another thermos flask.
  5. Remove the lid and prepare a cork through which you can bore a hole to insert thermometers into two flasks in such a way that the bulb of the thermometer should dip into the germinating and dry seeds.
  6. Close the thermos flasks with tight fitting rubber corks.
  7. Record the initial temperature in both the flasks and record it for every two hours for at least 24 hours.
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 19

Observation: Constant increase in the temperature is observed in thermometer placed in the germinating seeds.
Result: Hence it is proved that germinated seeds respire and liberate heat which is responsible for increase in the temperature.
Questions:

  1. Make a graph by using your observations.
    Answer:
    AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 20
  2. Is there any increase in temperature?
    Answer:
    Yes, there is increase in temperature.
  3. Does the temperature increase steadily or does it abruptly increase at a time of the day?
    Answer:
    The temperature in the thermometer increases steadily.
  4. Where does the heat come from?
    Answer:
    The sprouting or germinating seeds respire and liberate heat.

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System InText Questions and Answers

10th Class Biology Textbook Page No. 25

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 1.
Can it be said that Priestly’s experiment helped us to find out more about composition of air? How?
Answer:
Yes, Priestly’s experiments helped to find out the composition of air when burning of charcoal, carbon dioxide is produced which is the one of the composition of air done by Lavoisier.
Another experiment with phosphorus done by Lavoisier was cleared that a gas which is the respirable air that is helped in burning was oxygen also a component of air.
Lavoisier proved experimentally that carbon dioxide and oxygen were the components of air. Lavoisier confirmed the experiments of Priestly about the gases present in the air.

Question 2.
What gas was produced by combustion according to Lavoisier?
Answer:
The gas produced by combustion is carbon dioxide.

Question 3.
What did Lavoisier find out about air from his experiments?
Answer:
A fixed air carbon dioxide and respirable air oxygen which helped in burning were liberated during his experiments.

Question 4.
What conclusion can be drawn from Lavoisier’s experiments?
A. Oxygen and carbon dioxide are the compositions of air.

10th Class Biology Textbook Page No. 26

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 5.
Which gas do you think is Lavoisier talking about when he says chalky acid gas?
Answer:
Carbon dioxide.

Question 6.
Which gas according to Lavoisier is respirable air?
Answer:
Oxygen.

Question 7.
What steps in the process of respiration does Lavoisier mention as an inference of his experiments?
Answer:
Lavoisier mentioned that there were two steps in the respiration.
1. Inspiration: Breathing oxygen.
2. Expiration: Eliminating carbon dioxide from lungs.

Question 8.
It is a common observation that our breath is warmer than the air around us ; does respiration have anything to do with this?
Answer:
Our exhaled air is warmer than the air around us because heat is liberated during respiration.

10th Class Biology Textbook Page No. 27

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 9.
What does this experiment indicate?
Answer:
This experiment indicates that carbon dioxide is liberated during respiration.

Question 10.
Which gas turns lime water milky?
Answer:
Carbon dioxide.

Question 11.
Which gas do you think might be present in greater quantities in the air we breathe out as compared to air around us?
Answer:
Nitrogen (78%) is present in greater quantities in the air.

Question 12.
We are also aware of the fact that water vapour deposits on a mirror if we breathe out on it; where does this water vapour come from in Exhaled air?
Answer:
Water vapour is liberated during respiration.

10th Class Biology Textbook Page No. 29

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 13.
Why are we advised not to talk while eating food?
Answer:

  1. When we are eating food, epiglottis helps to avoid food entering into trachea.
  2. If we talk while we eat food, there is a chance of food entering into trachea and causes choking.
  3. As a result irritation and inflammation takes place in the respiratory tract.
  4. So we are advised not to talk while eating.

10th Class Biology Textbook Page No. 30

Question 14.
What is the role of diaphragm and ribs in respiration? Are both active in man and woman?
Answer:

  1. Diaphragm and ribs are helpful the chest cavity to increase or decrease the volume for inspiration and expiration in respiration.
  2. Diaphragm plays a major role in men and ribs play a major role in mechanism of respiration.

10th Class Biology Textbook Page No. 31

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 15.
What can be concluded from this?
Answer:
All movements of breathing is controlled by nerves leading from the brain.

Question 16.
What happens during the process of breathing?
Answer:
During the process of breathing, the patterns of breathing show a great range for they are coordinated with moment by moment needs of the body for supply of oxygen and removal of carbon dioxide.

Question 17.
Which gas needs to be removed from our body during exhalation? Where does the extra amount of gas come from?
Answer:
Carbon dioxide needs to be removed during exhalation. The extra amount of gas comes from the breakdown of glucose to release energy in the mitochondria. Carbon dioxide gas is released here.

Question 18.
What is the composition of inhaled air?
Answer:
Inhaled air contains oxygen -21%, C09 – 0.03%, Nitrogen – 78%.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 19.
When exhaled air is compared with inhaled air, is there any difference in composition?
Answer:
Yes, there is a difference between inhaled air and exhaled air.
The difference is
AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System 21

10th Class Biology Textbook Page No. 32

Question 20.
Why does the amount of oxygen vary between exhaled and inhaled air?
Answer:
Because some amount of oxygen will be utilised during cellular respiration in the body. Hence the difference in amount of oxygen occurs.

Question 21.
What has raised the percentage of carbon dioxide in exhaled air?
Answer:
CO2 is released from all the cells in the body in respiration and is added to the exhaled air.

10th Class Biology Textbook Page No. 34

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 22.
Do cells of alveoli or lungs also require oxygen to carry out cellular respiration? Why / Why not?
Answer:

  1. Alveoli are made of squamous epithelium tissue which is very thin and elastic
  2. Alveoli are so thin that oxygen can pass from air-filled alveoli to R.B.C inside the vessels.
  3. Simple squamous epithelial cells function as mediators of filtration and diffusion. As these cells are living tissue they also need oxygen.
  4. This is done through the exchange of gases in the alveoli.

Question 23.
After undergoing strenuous exercise we feel pain in muscles, does adequate oxygen reach the muscles?
Answer:
No. Adequate oxygen does not reach the muscles.

Question 24.
What is being formed in the muscles?
Answer:
Lactic acid.

10th Class Biology Textbook Page No. 42

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 25.
In which set does the colour change faster? Why?
Answer:
In the set which has germinating seeds the colour changes faster. Because CO2 is formed faster in aerobic respiration.

Think and Discuss

10th Class Biology Textbook Page No. 29

Question 1.
What will happen if the respiratory tract is not moist?
Answer:

  1. If the respiratory tract is not moist the dirt particles in the inhaled air will not be removed from air in the nasal cavities and reaches lungs and create problems to lungs.
  2. The temperature of the inhaled air is brought close to that of the body for the smooth passage in the respiratory tract. If it is dry, it is not possible.
  3. If the surface dries out, gas exchange will happen at a very reduced rate since fast moving gaseous oxygen molecules do not efficiently cross the alveoli membrane.
  4. The reduced gas exchange is most likely not enough to support blood oxygenation for vital functions of the body.
  5. Hence respiratory tract should be moist for smooth exchange of gases.

Question 2.
Are both lungs similar in size?
Answer:
No. Right lung is slightly bigger than left lung.

AP SSC 10th Class Biology Solutions Chapter 2 Respiration - The Energy Releasing System

Question 3.
Why are alveoli so small and uncountable in number?
Answer:

  1. The pouch-like air sacs at the ends of the smallest bronchioles are called alveoli.
  2. The walls of the alveolus are very thin and they are surrounded by very thin blood capillaries.
  3. It is in the alveoli that gaseous exchange takes place.
  4. There are millions of alveoli in the lungs. The presence of millions of alveoli in the lungs provides a very large area for the exchange of gases.
  5. And the availability of large surface area maximises the exchanges of gases.

 

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

AP State Board Syllabus AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security.

AP State Syllabus SSC 10th Class Social Studies Important Questions 11th Lesson Food Security

10th Class Social 11th Lesson Food Security 1 Mark Important Questions and Answers

Question 1.
What are the two ways which you would suggest for the eradication of malnutrition among the children?
Answer:
My suggestions:

  1. Sufficient food should be given to children.
  2. Children should be given nutritious food with low cost.

Question 2.
What is the active role of the judiciary in implementing food security in India? How is this role useful to children?
Answer:
In recent time, the Indian Judiciary also has become pro-active in ensuring food security. Through judicial verdict on court cases filed by non-governmental organizations, the Supreme Court directed all the state governments and central government to provide Mid-day-meals to all the young children studying in schools. The court also set up monitoring mechanisms and provided suggestions for better Implementation.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 3.
What is meant by Minimum Support Price (MSP)?
Answer:

  1. MSP: Minimum Support Price is a form of market intervention by the Govern¬ment of India to insure agricultural producers against any sharp fail in farm prices.
  2. With this price the government procures food grains through the FCI.

Question 4.
Give any two examples for ‘Nutricereals.
Answer:
Examples for the Nutri-cereals are

  1. Jowar,
  2. Ragi,
  3. Bajra.

Question 5.
Give any two examples, which give vitamins.
Answer:
Fruits, Leafy vegetables, sprouts, unpolished rice, etc.

Question 6.
Expand the term F.C.I.
Answer:
Food Corporation of India.

Question 7.
What is the reason for providing mid-day meals in government schools?
Answer:

  1. To increase the literacy rate in India and enroll more children in schools.
  2. To provide one meal for all the poor young children studying in government schools.

Question 8.
Write any two reasons for better implementation of the Public Distribution System in southern states.
Answer:
A universal Public Distribution System is there in southern states. Specific cards are issued to the needy people. Low cost foodgrains are available for them in the ration shops.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 9.
What is the need of food security?
Answer:
Food security is needed for various purposes. The main purpose of food security is ‘no person should go to bed with an empty stomach’.

Question 10.
Which caused large scale starvation deaths in pre-independent India?
Answer:
In pre-independent India, famines – situations of extreme scarcity of food, were a common cause of large-scale starvation deaths.

Question 11.
What was the loss of the Bengal famine?
Answer:
The Bengal famine in 1943-45, took away about 3 to 5 million people lives in and around Bengal, Assam and Orissa.

Question 12.
Why did the famines occur in Indian history?
Answer:
The famines occurred in Indian history because food grains supply was not organized by the rulers.

Question 13.
Name some organizations through which government today ensures food security.
Answer:
Ration shops and Anganwadis.

Question 14.
What is an important requirement of food security?
Answer:
Producing a sufficient amount of foodgrains is an important requirement of food security.

Question 15.
What are the results of this method?
Answer:
Soil degradation and depletion of groundwater resources.

Question 16.
How is the availability of foodgrains per person per day estimated?
Answer:
It is estimated as follows:
Availability of foodgrains per person per day = (Availability of foodgrains for the year T population)/ 365

Question 17.
What do consumers need?
Answer:
Consumers need a diverse food basket and a balanced diet.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 18.
What can farmers do to increase their incomes?
Answer:
Farmers producing foodgrains can go in for crop diversification in order to increase their incomes.

Question 19.
Give any one reason for farmers’ distress and even suicides.
Answer:
The conversation of food grain fields into cash crops such as cotton in Andhra Pradesh during the last two decades.

Question 20.
What do the nutritionists suggest?
Answer:
Nutritionists suggest that every person in India should eat 300 gms of vegetables and 100 grams of fruits in a day whereas per person availability of these food materials 180 and 58 gms respectively.

Question 21.
What do farmers require?
Answer:
Farmers require support in terms of inputs and market opportunities for diversification to other food items. They may have to. be supported and guarded against market risks that they face in the new situation.

Question 22.
Why may the foodgrain production come down?
Answer:
Since resources are diverted to non-food grain uses, foodgrain production may come down.

Question 23.
What is something to be worried about for India’s food security?
Answer:
The decline in the level of per capita availa¬bility of food grains is something to be worried about for India’s food security.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 24.
How are the national average calorie levels?
Answer:
The national average calorie levels in both rural and urban areas are below the needed calorie requirements.

Question 25.
What are used to examine the nutritional status of children?
Answer:
To examine the nutritional status of children, simple but accurate measurements of height and weight are used.

Question 26.
Why is buffer stock created by the government?
Answer:
Buffer stock has been created to meet any exigencies. The buffer stock can be utilized in case of drought or flood or any natural calamity.

Question 27.
What are Fair price shops?
Answer:
The ration shops which come under the PDS are called fair price shops.

Question 28.
Write short notes on MSP.
Answer:
MSP means Minimum Supportive Price. The FCI procures foodgrains and other farm produce by giving MSP for their produce. This ensures farmers of a minimum guarantee for their produce.

Question 29.
How many grams of vegetables and fruits every person eat a day?
Answer:
Every person in India should eat 300 grams of vegetables and 100 grams of fruits in a day.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 30.
Today what are called as “nutri-cereals”?
Answer:
The coarse cereals like jowar, ragi, bajra, etc., are today called as nutri-cereals.

Question 31.
What is meant by PDS system?
Answer:
PDS system means the government supplying low-cost foodgrains to people with ration cards.

Question 32.
What is meant by Nutrition Food?
Answer:
Nutritious food is that food that provides for energy, growth and capacity to remain healthy and fight illness.

Question 33.
How is BMI derived?
Answer:
BMI = Weight in legs/height in meters squared.

10th Class Social 11th Lesson Food Security 2 Marks Important Questions and Answers

Question 1.
The Indian government came out with a new law in 2013 called the National Food Security Act to legalize people’s right to food. What do you suggest to implement this Act?
Answer:
My suggestions:

  1. Subsidy rice should be given to the people of the low-income group.
  2. More rice should be given to very poor people.
  3. Dal, oil, tamarind and jaggery, etc. also should be supplied to these people through PDS.
  4. MDM scheme should be implemented with more efforts.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 2.
Our country could not achieve self-sufficiency in the production of foodgrains even today. Illustrate your reasons.
Answer:

  1. Foodgrain production is not proportional when compared to the rate of population growth.
  2. There is no proper water management.
  3. Farmers are following traditional methods.
  4. Fragmentation of landholdings.
  5. Improper manuring and low application of fertilizers.

Question 3.
State any two defects, which you find in the organization of the Public Distribution System.
Answer:

  1. Selling with higher prices.
  2. Cheating in the process of weighing.
  3. Selling ration goods in the black markets.
  4. Ineligible candidates holding white cards.
  5. Only few days distribution is following.

Question 4.
If the food grain production is affected because of natural calamity, in what ways can the Government ensure higher availability of food grains?
Answer:

  1. The FCI releases the buffer stock.
  2. By bringing supplies from other areas.
  3. Through a universal PDS system, low-cost food grains would be available.
  4. The Anthyodaya Cardholders are entitled to get 35 Kgs. of food grains per month per family.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 5.
“The Supreme Court of India directed to all the state governments and central government to provide mid-day-meal to the children studying in all schools.”
Prepare a pamphlet on better implementation of the Mid-day-meal program.
Answer:
Pamphlet on Mid-day-meal programme.

  1. Foodgrains available in local areas are to be used.
  2. Follow the menu regularly that meets the need of nutritional requirements of the children.
  3. Cooking in clean and hygienic environment.
  4. Involvement of staff and children for effective implementation.

Question 6.
Appreciate the benefits of the Mid-day meal programme implemented in Government schools.
Answer:

  1. It ensures food security.
  2. It provides nutritious food.
  3. It helps to increase net attendance rate.

Question 7.
Write suggestions to avoid food waste in Mid-day meal in your school.
Answer:

  1. Meals cooked should be tasty and healthy.
  2. Children should be educated on the consequences of food wastage.
  3. Teachers should supervise the programme.

Question 8.
Observe the bar diagram and answer the questions.
AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security 1a) What percentage of people in rural India are consuming more calories than are required?
Answer:
20%
b) What is the reason for consuming less calories than are required in rural India?
Answer:

  1. The major reason for low-calorie intake is the lack of purchasing power of the people.
  2. People don’t have adequate incomes to buy food.
  3. The reasons are poverty, unemployment, etc.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 9.
What are the measures you suggest for improving the present public distribution system?
Answer:
Suggestions:

  1. Provide ration cards only to the poor and vulnerable groups.
  2. Check every month whether they are rationed in an effective manner or not.
  3. Quantity supplied by the government should be increased.
  4. Remove the fake cards.

Question 10.
Analyze a week’s food habits of your family. And write how it impacts on Agriculture sector and the environment.
Answer:
Family food habits:

  1. Rice and redgram, vegetables, milk, curd, eggs, etc.
  2. For breakfast – idly, dosa, chapati, etc.
  3. Sometimes some members take non-vegetarian food.

Impact on agriculture:

  1. For increase of food grains and vegetables production farmers are using pesticides and fertilizers. Due to this soil loses its fertility.
  2. For increasing of milk production farmers follow artificial methods. Those methods cause harm to animals.

Impact on the environment:

  1. Excessive use of chemicals causes water, air pollutions.
  2. Bio-diversity is effected very highly.
  3. Deforestation.

Question 11.
Write a letter to the concerned officer for the proper implementation of the Public Distribution System in your area.
Answer:

Siricilla.
31st March 2017.

To
The Tasildar,
Mandal Revenue Office,
Siricilla.

Sub: PDS – irregular functioning of ration shop in our locality – Request for take action to regularize the functioning of the shop -reg.

Respected sir,

I am from Subhashnagar, Siricilla. I would like to bring the following to your notice and favourable action. The ration shop under the Public Distribution System running in our locality is not functioning well.

In our locality the ration shop is always kept closed. Many cardholders come eager to get their commodities. As the shop is closed they return scolding the government and the dealer. They are going to open market and purchase their commodities there. They lose the wages for that day also. Very often the shop is open but not all the commodities are given. Sometimes they are collecting more prices. Kerosene is being sold in open market.

Hence I request you to take necessary action to regulate the functioning of the ration shop and make it useful to the rural poor people. By doing so, the real purpose of PDS is served. I hope you take immediate and favourable action.

Thanking you sir,

Yours truly,
………………….
………………….
………………….

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security 2

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 12.
Prepare a pamphlet to bring awareness among the people on food security in India
Answer:
Pamphlet

Food Security in India

National Food Security Act 2013 legalizes the people’s right to food. According to this Act White, Pink, Anthyodaya cards are distributed to the needy on identification. Free cooked meal for pregnant women, lactating mothers, children 1-5 in Anganwadi is supplied.

Many ration shops are not open on proper timings. There will be no display of prices list and the stock of commodities. The card holders come to ration shops, and as they are closed, they get back to their home without commodities. They lose their wages for that day also. The shop dealers are diverting commodities to other shops in the market. Many times it is seen in newspapers and channels but no change is found with the dealers.

Mid day meals in schools is also like that. The minimum quantity of pulses and oil are not used. The watery curries are served. No one cares about it.

People should get aware about this. Many non Government organizations are working for the benefit of society. The government should organize a campaign about this.

No. of copies
5000

Published by
Society for people

Question 13.
What is the position of India’s per capita availability of foodgrains compared to other countries?
Answer:

  1. India’s per capita availability of foodgrains in 2010-11 was 463 grams.
  2. It was very low when compared to the same of countries in Europe (700grams) and USA (850 grams).
  3. Since resources are diverted to non-foodgrain uses, food grain production has come down.
  4. The decline in the level of per capita availability of foodgrains is something to be worried about for India’s food security.
  5. To avoid this, the policy of the government should aim to increase the production of foodgrains and other types of food simultaneously.

Question 14.
Read the passage and answer the question.

The State and Central Governments procure nearly one-third of foodgrains from farmers. These food grains are distributed to people through various mechanisms. In recent times, government agencies are procuring more foodgrains than what is required to meet the public distribution system. If government stocks keep increasing year after year, less is available (see year 2011 in Table on foodgrain availability). The government has been criticized that it is not distributing these foodgrains to the needy people. Sometimes, governments also exported these foodgrains to other countries.

Do you support this, when a large section of people within the country are not able to access to food grains?
Answer:

  1. The state government should procure foodgrains to distribute through PDS at lower prices.
  2. Needs of our people are to be considered.
  3. So I do not support the concept of exporting foodgrains to foreign countries.
  4. Instead, the foodgrains should be kept available to the needy of our country.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 15.
Read the passage and answer the question

“The prevalence of chronic energy deficiency (BMI<18.5) among men was about 35%, while overweight/obesity (BMI >25) was 10% …….
“About 35% of adult women had chronic energy deficiency andl4% were overweight/obese.
The prevalence of chronic energy deficiency was highest in the States of Odisha, Gujarat and Uttar Pradesh, followed by 33-38% in Karnataka, Telangana, Andhra Pradesh, Maharashtra, Madhya Pradesh and West Bengal…”

How is this related to food security?
Answer:

  1. The above information says that 35% of adult women had chronic energy deficiency.
  2. The prevalence of CED in our state is between 33% and 38%.
  3. The above information proves that people do not have access to good food.
  4. Hence “food security” should be implemented to avoid chronic energy deficiency.

Question 16.
How is food security ensured in India?
Answer:
Food security is being ensured in India by using the following methods:

  1. By improving production
  2. By maintaining buffer stock
  3. By the Public Distribution System.

Question 17.
What has to be done in order to increase the per hectare yield of a crop?
Answer:

  1. In order to increase the per hectare yield of a crop, necessary inputs have to provided in a judicious manner.
  2. One way is to expand irrigation but use water in a manner so that this vital resource is shared and made available to all.
  3. Modern methods of farming are to adopted.

Question 18.
Which have led to a continuous but unsustainable increase in the yield levels?
Answer:
Some scientists and people working in the field of agriculture report that the way rice and wheat are cultivated in India by intensive and unscientific application of chemical fertilizers and insecticides have led to a continuous but unsustainable increase in the yield levels.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 19.
How is the availability of foodgrains for the year estimated?
Answer:
It is estimated as follows.
Availability of foodgrains for the year = Production of foodgrains during the year (production – seed, feed and wastage) + net imports (imports-exports)-change in stocks with the government (closing stock at the year end – opening stock at the beginning)

Question 20.
Write your comments on the Food Security Bill.
Answer:
The Food Security Bill promises to alleviate hunger and guarantee very cheap food to India’s poor people. But there are concerns it has not been properly thought through and become unsustainable.
Under the programme everyone who qualifies for the subsidy will be entitled to 5 kg in total of rice, wheat, etc. The government intends to use the PDS for delivering subsidies to the poor.

Question 21.
Write your comments on P.D.S.
Answer:
The Public Distribution System (PDS) has helped in stabilizing food prices and making food available to consumers at affordable prices. It has helped in avoiding hunger and famine by supplying food from surplus regions of the country to deficient regions. The share of PDS in the consumption of rice and wheat has risen steeply. I think this PDS is working effectively throughout the nation.

10th Class Social 11th Lesson Food Security 4 Marks Important Questions and Answers

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 1.
Read the given paragraph and answer the question.

Studies indicate that Southern states of India have a good record in the public distribution system. Notably, these are the states that have followed a universal PDS system which means that low cost food grains would be available to all. This Is in contrast to other states where poor families have been Identified and foodgrains are sold to at different prices to poor and non-poor. Even among the poor, the very poor have different entitlements, or guarantees for access.

Interpret the relation between the PDS and food security.
Answer:

  1. People get food grains at reasonable prices through public distribution system.
  2. The government ensures it through Food Corporation of India.
  3. The Government should take suitable measures to reduce lapses in distribution and ensures benefit to the target groups.
  4. The various concepts of PDS like Anganwadis. Mid-day meals and fair price shops etc., provide food security to the different low income groups.

Thus there is a close relation between the PDS and food security.

Question 2.
Read the following paragraph and write your opinion on it.

“While Indian Parliament enacts various laws such as National Food Security Act and implements schemes such as Integrated Child Development Scheme (ICDS) in recent times, the Indian Judiciary also has become pro-active in ensuring food security. Through Judicial verdict as court cases filed by non-governmental organizations, the Supreme Court directed all the state governments and central government to provide mid-day meals to all the young children studying in schools.

Answer:

  1. The Indian government came out with a new law in 2013 called ‘the National Food Security Act1 to legalise people’s Right to Food.
  2. It applies to approximately 2/3rd of the population of India.
  3. Every person of low income families is entitled to 5kgs of foodgrains per month at subsidised rates.
  4. The poorest families are entitled to 35 kgs of foodgrains.
  5. For a few years, the central government supply rice, wheat and millets for Rs. 3/-, Rs. 2/- and Rs. 1/- respectively.
  6. If the government is not able to arrange foodgrains, It will give cash for the people to buy food grains.
  7. Providing free cooked meals for pregnant women, lactating mothers, children aged 1-6 coming to anganwadis and mid-day meals for children aged 6-14 years in schools.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 3.
Observe the given table and answer the question that follows.
Crop Production (Kilograms per Hectare)

Crop1950 – 19512000 – 2001
Paddy6681901
Wheat6552708
Pulses441544
Oilseeds481810
Cotton88190
Jute10432026

Write an analysis on the yielding trends of different crops.
Answer:

  1. The above table explains the crop production in 1950-51 and 2000-2001 kgs per hectare.
  2. Paddy production increases highly, wheat production also increased. Pulses production increased 100 kgs only. Oil seeds production increased double. Cotton production also increased double. Jute production also increased.

Due to Green Revolution, crop production increased gradually within the 50 years. Among the all wheat & paddy production increased because of both are benefited with Green Revolution. Food grain production also increased.

Question 4.
How far is ‘Food Security’ required today in India?
Answer:
Due to the increase of more population food security Is need nowadays. The following factors are responsible.

  1. Per person availability of food grain has actually not rise but declined in the recent years.
  2. Most people are in fact consuming fewer calories than required. This gap is severe for the poorest.
  3. Lack of employment or with low salaries majority of people are not able to purchase require food grains.
  4. In some areas PDS system is not working properly.
  5. A large section of people are malnourished, even when we have adequate food in the country.
  6. Majority of children and adults as being underweight in a chronic way.

On above issues food security is need today.

Question 5.
Observe the information given in the table and analyse.
Table: Per person availability of food grains in India

YearPopulation (in millions)Food grains production (in million tonnes)Per person availability per day (grams)
195136150.8395
196143982469
1971548108.4469
1991846176.4510
20111210232.07454

Answer:
Table Analysis:

  1. In 1951 the population of India was 361 million and the food grains production was 50.8 million and 395 grams of food grains were available to a person per a day.
  2. The population increased in 1961. It was 78 million. At the same time food grains production increased 32 million tons. The availability of food grains per person per a day was 469 grams.
  3. In 1971,1991 and 2011 the population increased. At the same time the food grains production also increased. But in 1961,1971 the availability of food grains was the same; not increased.
  4. From 1991 to 2011 the availability of food grains decreased due to floods, famines, natural calamities. More population was also another cause.

Till 1991 the population, food grains production and availability of food grains per person per day increased. But in 2011 only the population and food grains production increased. Availability of food grains drastically decreased.

The first requirement for a country is to be able to produce food grains for the whole population. Per person availability of food grains should be sufficient and also increasing over the years. From the information given above, it is evident that per person availability of food grains is decreasing since 2011.

So Government should take necessary measures in this regard. It should encourage farmers to produce more food grains. There should be control on exports also. Sometimes local people suffer for food grains but the government exports the food grains to other countries to import what they need. It is the minimum duty of the welfare government to keep food grains available to its people that to especially the poor.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 6.
Analyse how the Public Distribution System provides access to food to the poor?
Answer:
It is a joint responsibility of central government, state government and union territory administration to ensure the smooth functioning of the PDS.

While the responsibility of central government is to procure, store and transport it from purchase points to central godowns, the responsibility of state government and union territory administrations is to lift these commodities from the central godowns and distribute them to consumers through the network of fair price shops. FCI procures and distributes foodgrains, to provide subsidized food to the poor to mitigate regional inequalities through moving the surplus food to deficit areas and to stabilize agricultural prices.

The universal coverage of PDS was replaced by Targeted Real Public Distribution System in 1997 in order to achieve the objective of food security through sale of foodgrains to APL householders at economic cost and confirming the food subsidy bill toward the identified BPL families whose number crossed 97 million families in 2007.

Access to food through Public Distribution System :

  1. National Food Security Act legalizes people’s right to food.
  2. It is applicable to 2/3rd of the population of India.
  3. PDS ensures the availability of foodgrains to the poor at subsidized rates.
  4. Antyodaya cardholders are entitled to get 35 kgs of foodgrains (rice or wheat) per month per family.
  5. Every person of low-income families is entitled to 5 kilograms of foodgrains per month at subsidized rates.
  6. 75% of people living in rural areas purchase foodgrains through PDS.
  7. 50% of the urban population purchases foodgrains through PDS.
  8. PDS helps in maintaining the nutrition status of the people.
  9. Studies indicate that the southern states of India have a good record in PDS.

Question 7.
What would be the impact of the availability of other food items and food security?
Answer:

  1. It is said that there is a change in consumption pattern with people demanding more fruits, vegetables, milk, meat, poultry and fisheries.
  2. This is a good sign for the consumers as well as producers.
  3. Farmers producing foodgrains can go in for crop diversification in order to increase their incomes.
  4. Farmers require support in terms of inputs and market opportunities for diversification to other food items.
  5. Over the years, although there has been increasing in the production of other food items, it is not sufficient to meet the minimum dietary requirements.
  6. The policies should aim at increasing food crop production and other food items simultaneously.
  7. It will fill the gap and make food security more viable.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 8.
What are the salient features of Food Security Bill?
Answer:

  1. The Indian government came out with a new law in 2013 called ‘the National Food Security Act’ to legalise people’s Right to Food.
  2. It applies to approximately 2/3rd of the population of India.
  3. Every person of low-income families is entitled to 5kgs of foodgrains per month at subsidised rates.
  4. The poorest families are entitled to 35 kgs of foodgrains.
  5. For a few years, the central government supply rice, wheat and millets for Rs. 3/-, Rs. 2/- and Rs. 1/- respectively.
  6. If government is not able to arrange foodgrains, it will give cash for the people to buy foodgrains.
  7. Providing free cooked meals for pregnant women, lactating mothers, children aged 1-6 coming to anganwadis and mid-day meals for children aged 6-14 years in schools.

Question 9.
Why do we require food? How do we classify the food that we consume?
Answer:

  1. Food is required by the body for all its functions for energy, growth and the capacity to remain healthy and fight illness.
  2. The food that we consume is normally classified as –
    1. Carbohydrates: that provide energy, through wheat, rice, ragi, jo war, oils, sugar, fats, etc.
    2. Proteins: that help growth and regeneration of body tissues through beans, dais, meat, eggs, rice, wheat, etc.
    3. Vitamins: that provide protection and ensure the working of many vital systems of the body through foods such as fruits, leafy vegetables, sprouts, unpolished rice, etc.
    4. Minerals are required in small quantities for many important functions such as iron and blood formation. This is provided through green leafy vegetables, ragi, etc.

Question 10.
Read the following paragraph and write your comments.
It is also important to ensure that soil and other natural resources are not damaged or depleted in the process. Some scientists and people working in the field of agriculture report that the way rice and wheat are cultivated in India – by intensive and unscientific application of chemical fertilizers and insecticides – have led to a continuous but unsustainable Increase in the yield levels. In fact, these methods have led to soil degradation and depletion of ground water resources. If this continues, we may soon come to a situation that yields start falling rather than going up.
Answer:

  1. The given paragraph says that improper usage of chemical fertilizers and insecticides causes for damage of cultivable soil and depletion of ground water levels.
  2. If these methods are followed continuously there will be no ground water resources.
    My comments on this paragraph are as follows:
  3. The population is growing day by day.
  4. Cultivable land is not increasing in that proportion.
  5. The food grain production is to be increased.
  6. A part of green revolution it has become compulsory for the farmers to use fertilizers and pesticides to grow more crops.
  7. Instead of using chemical fertilizers, the farmers should use go for organic farming.
  8. Compost is to be encouraged. Multiple cropping is also one alternate.
  9. Cultivable land is to be expanded by changing the wastelands as farming lands.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 11.
Read the following paragraph and interpret.

If a country is to able to produce food grains for the whole of its population this would be considered as the first requirement. How do we measure whether if there is food for all ? Whether this food reaches families would be examined later. We are at first estimating what is available. This means that per person (or per capita) availability of foodgrains in the country should be sufficient and also increased over the years. Is the increase in foodgrain availability really happening?

Answer:

  1. According to the paragraph every country should produce the foodgrains how much it needs.
  2. Every country should examine two things.
  3. The first one is how much foodgrains are produced in the country and the second one is how it is reaching to the people.
  4. Many countries don’t concentrate on this sensitive issue.
  5. It is very primary thing to look into the matter of food grain production and its availability.
  6. When the production is less than the need or demand, automatically the prices go up.
  7. Government should encourage the farmers to produce more foodgrains.
  8. There should be control on exports also.

Question 12.
Read the following paragraph and write your opinion on it.

It Is said that there Is a change In consumption pattern with people demanding more fruits, vegetables, milk, meat, poultry and fisheries. This Is a good sign for the consumers as well as producers. Consumers need a diverse food basket and a balanced diet. Farmers producing food grains can go In for crop diversification In order to increase their Incomes.

Answer:

  1. The paragraph says that the people demand different types of food like vegetarian and non-vegetarian food.
  2. Taking this type of food is a good sign.
  3. By utilizing this diverse food, the people maintain good health.
  4. Where Is a lot of gap between the production of foodgrains and its availability.
  5. The pattern mentioned in this paragraph is good for health but the problem is its availability.
  6. The access of food is very less when it is compared to the Nutritionists suggestions.
  7. In Agriculture, there is a shift from food crops to commercial crops.
  8. Many farmers are concentrating on commercial crops.
  9. Consequently, there will be scarcity of foodgrains.
  10. Food grains are to be grown to what extent it is necessary.

Question 13.
Read the following paragraph and write your comments.

The national average calorie levels In both rural and urban areas are below the needed calorie requirements. Also, the consumption of calories has gone down between 1983 and 2004. This Is shocking since as we have seen our economy Is growing at a rapid rate. Production of goods and services has Increased many times.

Answer:

  1. This paragraph tells us that the people in rural and urban areas are not taking sufficient calories intake in their diet.
  2. It is surprising that the economy is growing but calorie intake Is decreasing.
  3. Production is increasing but consumption Is decreasing.
  4. According to the reports of nutritionists it is clearly understood that the people are not consuming as per requirements.
  5. Irrespective of caste, religion, region and gender the rich people take more than requirement but the problem is with the poor people.
  6. Government should take new steps to help the poor people.
  7. The PDS functioning should be made correct.
  8. Quality commodities are to be provided at cheaper prices.

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security

Question 14.
Write a letter to the Tahsildar about irregular functioning of ration shop in your area.

Tadepalli,
Date : xx xx xxxx

To
The Tasildar,
Mandal Revenue Office,
Tadepalli

Sub: PDS – irregular functioning of ration shop in our locality – Request for take action to regularize the functioning of the shop -reg.

Respected sir,

I am from Prakashnagar, Tadepalli. I would like to bring the following to your notice and favourable action. The ration shop under the Public Distribution System running in our locality is not functioning well.

In our locality the ration shop is always kept closed. Many cardholders come eager to get their commodities. As the shop is closed they return scolding the government and the dealer.

They are going to open market and purchase their commodities there. They lose the wages for that day also. Very often the shop is open but not all the commodities are given. Sometimes they are collecting more prices. Kerosene is being sold in open market.

Hence I request you to take necessary action to regulate the functioning of the ration shop and make it useful to the rural poor people. By doing so, the real purpose of PDS is served. I hope you take immediate and favourable action.

Thanking you sir,

Yours truly,
………………….
………………….

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security 3

AP SSC 10th Class Social Studies Important Questions Chapter 11 Food Security