AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 7 పాక్షిక భిన్నాలు Exercise 7(b) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Exercise 7(b)

అభ్యాసం – 7 (బి)

క్రింది భిన్నాలను పాక్షిక భిన్నాలుగా విడగొట్టండి.

ప్రశ్న 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) [May, Mar. ’11]
సాధన:
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\) అనుకుందాం.
2x2 + 3x + 4 = A(x2 + 2) + (Bx + C) (x – 1) …….(1)
x = 1 వ్రాస్తే, 2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
(1) లో x2 గుణకాలను పోల్చగా
2 = A + B
⇒ B = 2 – A
⇒ B = 2 – 3
⇒ B = -1
(1) లో స్థిరపదాలను పోల్చగా
4 = 2A – C
⇒ C = 2A – 4
= 6 – 4
= 2
∴ A = 3, B = -1, C = 2
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(b)

ప్రశ్న 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
సాధన:
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{A}{2+x}+\frac{B x+C}{1-x+x^2}\) అనుకుందాం.
3x – 1 = A(1 – x + x2) (Bx + C) (2 + x) …….(1)
x = -2 వ్రాస్తే, -7 = A(1 + 2 + 4)
⇒ -7 = 7A
⇒ A = -1
(1) లో x2 గుణకాలను పోల్చగా
0 = A + B
⇒ B = -A = 1
స్థిరపదాలను పోల్చగా
-1 = A + 2C
⇒ 2C = -1 – A
⇒ 2C = -1 + 1
⇒ 2C = 0
⇒ C = 0
∴ A = -1, B = 1, C = 0
\(\frac{3 x-1}{\left(1-x+x^2\right)(2+x)}=-\frac{1}{2+x}+\frac{x}{1-x+x^2}\)

ప్రశ్న 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
సాధన:
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\) అనుకుందాం.
x2 – 3 = A(x2 + 1) + (Bx + C) (x + 2) …..(1)
x = -2 వ్రాస్తే, 4 – 3 = A(4 + 1)
⇒ 1 = 5A
⇒ A = \(\frac{1}{5}\)
(1) లో x2 గుణకాలను పోల్చగా
1 = A + B
⇒ B = 1 – A
⇒ B = 1 – \(\frac{1}{5}\)
⇒ B = \(\frac{4}{5}\)
(1) లో స్థిరపదాలను పోల్చగా
-3 = A + 2C
⇒ 2C = -3 – A
⇒ 2C = -3 – \(\frac{1}{5}\)
⇒ 2C = \(-\frac{16}{5}\)
⇒ C = \(-\frac{8}{5}\)
∴ A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(-\frac{8}{5}\)
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(b)

ప్రశ్న 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
సాధన:
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{\left(x^2+x+1\right)^2}\) అనుకుందాం.
x2 + 1 = (Ax + B) (x2 + x + 1) + (Cx + D) ……(1)
(1) లో x3 గుణకాలను పోల్చగా, A = 0
(1) లో x2 గుణకాలను పోల్చగా, A + B = 1 ⇒ B = 1
(1) లో x గుణకాలను పోల్చగా, A + B + C = 0
⇒ 1 + C = 0
⇒ C = -1
(1) లో స్థిరపదాలను పోల్చగా, B + D = 1
⇒ D = 1 – B
= 1 – 1
= 0
∴ A = 0, B = 1, C = -1, D = 0
∴ Ax + B = 1, Cx + D = -x
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\)

ప్రశ్న 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(b) Q5
∴ x3 + x2 + 1 = A(x – 1) (x2 + x + 1) + B(x2 + x + 1) + (Cx + D) (x – 1)2 ……(2)
x = 1 ను (2) లో వ్రాయగా
1 + 1 + 1 = A(0) + B(1 + 1 + 1) + (C(1) + D) (0)
⇒ 3B = 3
⇒ B = 1
(2) లో x3 గుణకాలను పోల్చగా
1 = A + C ….(3)
(2) లో x2 గుణకాలను పోల్చగా
1 = A(1 – 1) + B(1) + C(-2) + D(1)
⇒ 1 = B – 2C + D
⇒ 1 = 1 – 2C + D
⇒ 2C = D ……..(4)
x = 0 ను (2) లో వ్రాయగా
1 = A(-1) (1) + B(1) + D(-1)2
⇒ A + B + D = 1
⇒ -A + 1 + D = 1
⇒ A = D ……..(5)
(3), (4), (5) ల నుండి
1 = D + \(\frac{D}{2}\)
⇒ \(\frac{3D}{2}\) = 1
⇒ D = \(\frac{2}{3}\)
(5) నుండి A = \(\frac{2}{3}\)
(4) నుండి C = \(\frac{\mathrm{D}}{2}=\frac{\left(\frac{2}{3}\right)}{2}=\frac{1}{3}\)
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(b) Q5.1

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 7 పాక్షిక భిన్నాలు Exercise 7(a) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Exercise 7(a)

అభ్యాసం – 7(ఎ)

I. క్రింది భిన్నాలను పాక్షిక భిన్నాలుగా విడగొట్టండి.

ప్రశ్న 1.
\(\frac{2 x+3}{(x+1)(x-3)}\)
సాధన.
\(\frac{2 x+3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}\) అనుకుందాం.
∴ 2x + 3 = A(x – 3) + B(x + 1) …..(1)
(1) లో x = -1 వ్రాస్తే, 1 = A(-4) ⇒ A = \(-\frac{1}{4}\)
(1) లో x = 3 వ్రాస్తే, 9 = B(4) ⇒ B = \(\frac{9}{4}\)
\(\frac{2 x+3}{(x+1)(x-3)}=\frac{-1}{4(x+1)}+\frac{9}{4(x-3)}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

ప్రశ్న 2.
\(\frac{5 x+6}{(2+x)(1-x)}\)
సాధన:
\(\frac{5 x+6}{(2+x)(1-x)}=\frac{A}{2+x}+\frac{B}{1-x}\) అనుకుందాం.
5x + 6 = A(1 – x) + B(2 + x) …..(1)
(1) లో x = -2 వ్రాస్తే, -10 + 6 = A(1 + 2) ⇒ A = \(-\frac{4}{3}\)
(1) లో x = 1 వ్రాస్తే, 5 + 6 = B(2 + 1) ⇒ B = \(\frac{11}{3}\)
∴ \(\frac{5 x+6}{(2+x)(1-x)}=-\frac{4}{3(2+x)}+\frac{11}{3(1-x)}\)

II.

ప్రశ్న 1.
\(\frac{3 x+7}{x^2-3 x+2}\)
సాధన:
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\) అనుకుందాం.
3x + 7 = A(x – 2) + B(x – 1) ……(1)
(1) లో x = 1 వ్రాస్తే, 10 = -A ⇒ A = -10
(1) లో x = 2 వ్రాస్తే, 13 = B
∴ \(\frac{3 x+7}{x^2-3 x+2}=\frac{-10}{x-1}+\frac{13}{x-2}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

ప్రశ్న 2.
\(\frac{x+4}{\left(x^2-4\right)(x+1)}\) [Mar. ’14]
సాధన:
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-2}\)
x + 4 = A(x2 – 4) + B(x + 1)(x – 2) + C(x + 1)(x + 2) …….(1)
(1) లో x = -1 వ్రాస్తే, 3 = A(1 – 4)
⇒ 3 = -3A
⇒ A = -1
(1) లో x = -2 వ్రాస్తే,
2 = B(-2 + 1) (-2 – 2)
⇒ 2 = 4B
⇒ B = \(\frac{1}{2}\)
(1) లో x = 2 వ్రాస్తే,
6 = C(2 + 1) (2 + 2)
⇒ 6 = 12C
⇒ C = \(\frac{1}{2}\)
∴ \(\frac{x+4}{\left(x^2-4\right)(x+1)}=-\frac{1}{x+1}+\frac{1}{2(x+2)}+\frac{1}{2(x-2)}\)

ప్రశ్న 3.
\(\frac{2 x^2+2 x+1}{x^3+x^2}\)
సాధన:
\(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
2x2 + 2x + 1 = Ax(x + 1) + B(x + 1) + Cx2 ……(1)
(1) లో x = 0 వ్రాస్తే, 1 = B
(1) లో x = -1 వ్రాస్తే, 2 – 2 + 1 = C(1) ⇒ C = 1
ఇరువైపులా x2 గుణకాలు పోల్చగా
2 = A + C
⇒ A = 2 – C
= 2 – 1
= 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\)

ప్రశ్న 4.
\(\frac{2 x+3}{(x-1)^3}\)
సాధన:
\(\frac{2 x+3}{(x-1)^3}\)
x – 1 = y అనుకుంటే x = y + 1
⇒ \(\frac{2 x+3}{(x-1)^3}=\frac{2(y+1)+3}{y^3}=\frac{2 y+5}{y^3}\)
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) II Q4

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

ప్రశ్న 5.
\(\frac{x^2-2 x-6}{(x-2)^3}\)
సాధన:
x – 2 = y అనుకొనుము
⇒ x = y + 2
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) II Q5

III.

ప్రశ్న 1.
\(\frac{x^2-x+1}{(x+1)(x-1)^2}\)
సాధన:
\(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\) అనుకుందాం.
x2 – x + 1 = A(x – 1)2 + B(x + 1)(x – 1) + C(x + 1) …….(1)
x = -1 వ్రాస్తే, 1 + 1 + 1 = A(4) ⇒ A = \(\frac{3}{4}\)
x = 1 వ్రాస్తే, 1 – 1 + 1 = C(2) ⇒ C = +\(\frac{1}{2}\)
1 లో x2 గుణకాలను పోల్చగా
A + B = 1
⇒ B = 1 – A
⇒ B = 1 – \(\frac{3}{4}\)
⇒ B = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1}{2(x-1)^2}\)

ప్రశ్న 2.
\(\frac{9}{(x-1)(x+2)^2}\)
సాధన:
\(\frac{9}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\) అనుకొందాం.
9 = A(x + 2)2 + B(x – 1) (x + 2) + C(x – 1) ……(1)
x = 1 వ్రాస్తే, 9 = 9A ⇒ A = 1
x = -2 వ్రాస్తే, 9 = -3C ⇒ C = -3
(1) లో x2 గుణకాలను పోల్చగా
A + B = 0
⇒ B = -A = -1
∴ \(\frac{9}{(x-1)(x+2)^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{(x+2)^2}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

ప్రశ్న 3.
\(\frac{1}{(1-2 x)^2(1-3 x)}\)
సాధన:
\(\frac{1}{(1-2 x)^2(1-3 x)}=\frac{A}{1-3 x}+\frac{B}{1-2 x}+\frac{C}{(1-2 x)^2}\) అనుకుందాం.
1 = A(1 – 2x)2 + B(1 – 3x) (1 – 2x) + C(1 – 3x) ……..(1)
x = \(\frac{1}{3}\) వ్రాస్తే, 1 = A\(\left(1-\frac{2}{3}\right)^2\)
⇒ 1 = \(\frac{A}{9}\)
⇒ A = 9
x = \(\frac{1}{2}\) వ్రాస్తే, 1 = C(1 – \(\frac{3}{2}\))
⇒ 1 = \(-\frac{C}{2}\)
⇒ C = -2
(1) లో x2 గుణకాలను పోల్చగా
0 = 4A + 6B
6B = -4A – 36
B = -6
∴ \(\frac{1}{(1-2 x)^2(1-3 x)}=\frac{9}{1-3 x}-\frac{6}{1-2 x}-\frac{2}{(1-2 x)^2}\)

ప్రశ్న 4.
\(\frac{1}{x^3(x+a)}\)
సాధన:
\(\frac{1}{x^3(x+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+a}\) అనుకుందాం.
= \(\frac{A \cdot x^2(x+a)+B(x)(x+a)+C(x+a)+D x^3}{x^3(x+a)}\)
∴ 1 = A(x2) (x + a) + Bx(x + a) + C(x + a) + Dx3 ……..(1)
x = 0 ను (1) లో వ్రాస్తే, 1 = A(0) + B(0) + C(0 + a) + D(0)
⇒ 1 = C (a)
⇒ C = \(\frac{1}{a}\)
x = -a ను (1) లో వ్రాస్తే, 1 = A(0) + B(0) + C(0) + D(-a)3
⇒ 1 = D(-a3)
⇒ D = \(-\frac{1}{a^3}\)
(1) లో x3 గుణకాలను పోల్చగా
0 = A + D
⇒ A = -D
⇒ A = \(\frac{1}{a^3}\)
(1) లో x2 గుణకాలను పోల్చగా
0 = Aa + B
⇒ B = -aA
⇒ B = \(-a\left(\frac{1}{a^3}\right)\)
⇒ B = \(-\frac{1}{a^2}\)
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) III Q4

ప్రశ్న 5.
\(\frac{x^2+5 x+7}{(x-3)^3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) III Q5

AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a)

ప్రశ్న 6.
\(\frac{3 x^3-8 x^2+10}{(x-1)^4}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) III Q6
AP Inter 2nd Year Maths 2A Solutions Chapter 7 పాక్షిక భిన్నాలు Ex 7(a) III Q6.1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(c) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(c)

అభ్యాసం – 6(సి)

ప్రశ్న 1.
క్రింది సమాసాల విలువలను 4 దశాంశాలకు సవరించి కనుక్కోండి.
(i) \(\sqrt[5]{242}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(i)

(ii) \(\sqrt[7]{127}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(ii)

(iii) \(\sqrt[5]{32.16}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(iii)

(iv) √199
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(iv)

(v) \(\sqrt[3]{1002}-\sqrt[3]{998}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(v)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(v).1

(vi) \((1.02)^{3 / 2}-(0.98)^{3 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q1(vi)
= 2[0.0299995]
= 0.0599990
≈ 0.059999
∴ \((1.02)^{3 / 2}-(0.98)^{3 / 2}\) = 0.059999

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c)

ప్రశ్న 2.
x2 ఆపై x ఘాతాలు ఉపేక్షించేంతగా |x| స్వల్పమైతే క్రింది సమాసాల ఉజ్జాయింపు విలువలను కనుక్కోండి.
(i) \(\frac{(4+3 x)^{1 / 2}}{(3-2 x)^2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(i)

(ii) \(\frac{\left(1-\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{1 / 5}}{(3-x)^3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(ii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(ii).1

(iii) \(\sqrt{4-x}\left(3-\frac{x}{2}\right)^{-1}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(iii)

(iv) \(\frac{\sqrt{4+x}+\sqrt[3]{8+x}}{(1+2 x)+(1-2 x)^{-1 / 3}}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(iv)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(iv).1

(v) \(\frac{(8+3 x)^{2 / 3}}{(2+3 x) \sqrt{4-5 x}}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q2(v)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c)

ప్రశ్న 3.
s, t లు ధన వాస్తవసంఖ్యలు, s తో పోల్చినపుడు t విలువ చాలా తక్కువ అయితే \(\left(\frac{s}{s+t}\right)^{1 / 3}-\left(\frac{s}{s-t}\right)^{1 / 3}\) యొక్క ఉజ్జాయింపు విలువ కనుక్కోండి.
సాధన:
s తో పోల్చినపుడు విలువ చాలా తక్కువ కనుక \(\frac{t}{s}\) అత్యల్పం
∴ |\(\frac{t}{s}\)| < 1
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q3

ప్రశ్న 4.
p, q లు ధన వాస్తవ సంఖ్యలు, q తో సరి పోలిస్తే p విలువ చాలా తక్కువ అయితే \(\left(\frac{q}{q+p}\right)^{1 / 2}+\left(\frac{q}{q-p}\right)^{1 / 2}\) యొక్క ఉజ్జాయింపు విలువ కనుక్కోండి.
సాధన:
q తో సరిపోలిస్తే p విలువ చాలా తక్కువ కనుక \(\frac{p}{q}\) అత్యల్పం
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q4
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q4.1

ప్రశ్న 5.
x4, ఆపై x ఘాతాలు ఉపేక్షిస్తే \(\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}\) యొక్క ఉజ్జాయింపు విలువ కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q5

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c)

ప్రశ్న 6.
3√3 విలువను \(\frac{2}{3}\) యొక్క ఆరోహణ ఘాతాలలో వ్రాయండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(c) Q6

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(b) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(b)

అభ్యాసం – 6(బి)

I.

ప్రశ్న 1.
క్రింది సమాసాలకు ద్విపద విస్తరణ వ్యవస్థితంచే x ల సమితులు కనుక్కోండి. [T.S. Mar. ’16, Mar. ’11]
(i) \((2+3 x)^{-2 / 3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q1(i)

(ii) \((5+x)^{3 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q1(ii)

(iii) (7 + 3x)-5
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q1(iii)

(iv) \(\left(4-\frac{x}{3}\right)^{-1 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q1(iv)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 2.
క్రింది విస్తరణలో సూచించిన పదాలు కనుక్కోండి.
(i) \(\left(1+\frac{x}{2}\right)^{-5}\) లో 6వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q2(i)

(ii) \(\left(1-\frac{x^2}{3}\right)^{-4}\) విస్తరణలో 7వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q2(ii)

(iii) \((3-4 x)^{-2 / 3}\) విస్తరణలో 10వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q2(iii)

(iv) \(\left(7+\frac{8 y}{3}\right)^{7 / 4}\) విస్తరణలో 5వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q2(iv)

ప్రశ్న 3.
క్రింది విస్తరణలలో మొదటి 3 పదాలు వ్రాయండి.
(i) \((3+5 x)^{-7 / 3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q3(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q3(i).1

(ii) (1 + 4x)-4
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q3(ii)

(iii) \((8-5 x)^{2 / 3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q3(iii)

(iv) \((2-7 x)^{-3 / 4}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q3(iv)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 4.
క్రింది విస్తరణలో సాధారణ పదం ((r + 1)వ పదం) కనుక్కోండి.
(i) \((4+5 x)^{-3 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q4(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q4(i).1

(ii) \(\left(1-\frac{5 x}{3}\right)^{-3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q4(ii)

(iii) \(\left(1+\frac{4 x}{5}\right)^{5 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q4(iii)

(iv) \(\left(3-\frac{5 x}{4}\right)^{-1 / 2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) I Q4(iv)

II.

ప్రశ్న 1.
\(\frac{1+2 x}{(1-2 x)^2}\) విస్తరణలో x10 గుణకం కనుక్కోండి.
సాధన:
\(\frac{1+2 x}{(1-2 x)^2}\) = (1 + 2x) (1 – 2x)-2
= (1 + 2x) [1 + 2(2x) + 3(2x)2 + 4(2x)3 + 5(2x)4 + 6(2x)5 + 7(2x)6 + 8(2x)7 + 9(2x)8 + 10(2x)9 + 11(2x)10 + …….. + (r + 1) . (2x)r +……]
∴ \(\frac{1+2 x}{(1-2 x)^2}\) లో x10 గుణకం = (11) (2)10 + 10 (2) (29)
= 210 (11 + 10)
= 21 × 210

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 2.
\((1-4 x)^{-3 / 5}\) విస్తరణలో x4 గుణకం కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q2
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q2.1

ప్రశ్న 3.
(i) \(\frac{(1-3 x)^2}{(3-x)^{3 / 2}}\) విస్తరణలో x5 గుణకాన్ని కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q3(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q3(i).1

(ii) \(\frac{(1+x)^2}{\left(1-\frac{2}{3} x\right)^3}\) విస్తరణలో x8 గుణకాన్ని కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q3(ii)

(iii) \(\frac{(2+3 x)^3}{(1-3 x)^4}\) విస్తరణలో x7 గుణకాన్ని కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q3(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q3(iii).1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 4.
\(\frac{\left(1+3 x^2\right)^{3 / 2}}{(3+4 x)^{1 / 3}}\) విస్తరణలో x3 గుణకాన్ని కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) II Q4

III.

ప్రశ్న 1.
క్రింది అనంతశ్రేణుల మొత్తాలు కనుక్కోండి.
(i) \(1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\ldots \ldots \ldots\)
సాధన:
దత్తశ్రేణి S = \(1+\frac{1}{1} \cdot \frac{1}{3}+\frac{1.3}{1.2}\left(\frac{1}{3}\right)^2+\frac{1.3 \cdot 5}{1.2 .3}\left(\frac{1}{3}\right)^3\) + ……..
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q1(i)

(ii) \(1-\frac{4}{5}+\frac{4.7}{5.10}-\frac{4.7 .10}{5.10 .15}+\ldots \ldots\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q1(ii)

(iii) \(\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5 .7}{4.8 .12}+\ldots\) (Mar. ’11)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q1(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q1(iii).1

(iv) \(\frac{3}{4.8}-\frac{3.5}{4.8 .12}+\frac{3.5 .7}{4.8 .12 .16}-\ldots \ldots\) [T.S. Mar. ’16]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q1(iv)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 2.
t = \(\frac{4}{5}+\frac{4.6}{5.10}+\frac{4.6 .8}{5.10 .15}+\) …….∞ అయితే, 9t = 16 అని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q2

ప్రశ్న 3.
x = \(\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\frac{1.3 .5 .7}{3.6 .9 .12}+\ldots \ldots\) అయితే 9x2 + 24x = 11 అని చూపండి. [T.S. Mar. ’16]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q3
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q3.1
3x = 3√3 – 4
⇒ 3x + 4 = 3√3
ఇరువైపుల వర్గం చేయగా
(3x + 4)2 = (3√3)2
⇒ 9x2 + 24x + 16 = 27
⇒ 9x2 + 24x = 11

ప్రశ్న 4.
x = \(\frac{5}{(2 !) \cdot 3}+\frac{5.7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots\) అయితే x2 + 4x విలువ కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q4
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q4.1

ప్రశ్న 5.
క్రింది అనంత శ్రేణి మొత్తం కనుక్కోండి. [A.P. Mar. ’16, Mar. ’05]
\(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1.3}{1.2} \cdot \frac{1}{10^4}+\frac{1.3 .5}{1.2 .3} \cdot \frac{1}{10^6}+\ldots\right)\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q5

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b)

ప్రశ్న 6.
x ఒక శూన్యేతర అకరణీయ సంఖ్య అయితే \(1+\frac{x}{2}+\frac{x(x-1)}{2.4}+\frac{x(x-1)(x-2)}{2.4 .6}+\ldots \ldots\) \(=1+\frac{x}{3}+\frac{x(x+1)}{3.6}+\frac{x(x+1)(x+2)}{3.6 .9}+\ldots\) అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q6
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(b) III Q6.1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(a) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Exercise 6(a)

అభ్యాసం – 6(ఎ)

I.

ప్రశ్న 1.
ద్విపద సిద్ధాంతాన్ని ఉపయోగించి క్రింది సమాసాలను విస్తరించి వ్రాయండి.
(i) (4x + 5y)7
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q1(i)

(ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q1(ii)

(iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q1(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q1(iii).1

(iv) (3 + x – x2)4
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q1(iv)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 2.
క్రింది పదాలు వ్రాసి సూక్ష్మీకరించండి.
(i) \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) లో 6వ పదం [May ’13]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q2(i)

(ii) (3x – 4y)10 లో 7వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q2(ii)

(iii) \(\left(\frac{3 p}{4}-5 q\right)^{14}\) లో 10వ పదం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q2(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q2(iii).1

(iv) \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) లో rవ పదం (1 ≤ r ≤ 9)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q2(iv)

ప్రశ్న 3.
క్రింది సమాసాల ద్విపద విస్తరణలలో పదాల సంఖ్యను కనుక్కోండి.
(i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
సాధన:
(x + a)n విస్తరణలో n పదాల సంఖ్య = (n + 1)
∴ \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) ద్విపద విస్తరణలో పదాల సంఖ్య = 9 + 1 = 10

(ii) (3p + 4q)14
సాధన:
(3p + 4q)14 ద్విపద విస్తరణలో పదాల సంఖ్య = 14 + 1 = 15

(iii) (2x + 3y + z)7 [Mar. ’07; Mar. ’14, ’13]
సాధన:
(a + b + c)n విస్తరణలో పదాల సంఖ్య = \(\frac{(n+1)(n+2)}{2}\)
n ధన పూర్ణాంకం కనుక
(2x + 3y + z)7 లో పదాల సంఖ్య = \(\frac{(7+1)(7+2)}{2}\)
= \(\frac{8 \times 9}{2}\)
= 36

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 4.
(4x – 7y)49 + (4x + 7y)49 విస్తరణలో శూన్యేతర గుణకాలు కలిగిన పదాలు ఎన్ని?
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q4

ప్రశ్న 5.
(1 + x)39 విస్తరణలో చివరి 20 గుణకాల మొత్తం కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q5

ప్రశ్న 6.
(1 + x)2n, (1 + x)2n-1 విస్తరణలలో A, B లు వరుసగా xn గుణకాలు అయితే \(\frac{A}{B}\) విలువ కనుక్కోండి.
సాధన:
ఇచ్చిన దత్తాంశం ప్రకారం (1 + x)2n మరియు (1 + x)2n-1 విస్తరణలో xn గుణకాలు A మరియు B అనుకొందాం.
∴ A = 2nCn
B = 2n-1Cn
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) I Q6

II.

ప్రశ్న 1.
క్రింది గుణకాలను కనుక్కోండి.
(i) \(\left(3 x-\frac{4}{x}\right)^{10}\) లో x-6 గుణకం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q1(i)

(ii) \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) లో x11 గుణకం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q1(ii)

(iii) \(\left(7 x^3-\frac{2}{x^2}\right)^9\) లో x2 గుణకం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q1(iii)

(iv) \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\) లో x-7 గుణకం
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q1(iv)

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 2.
క్రింది సమాసాల ద్విపద విస్తరణలలో x లేని పదం (స్థిర పదం) కనుక్కోండి.
(i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q2(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q2(i).1

(ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q2(ii)

(iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q2(iii)

(iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q2(iv)

ప్రశ్న 3.
క్రింది సమాసాల ద్విపద విస్తరణలలో మధ్యపదం (పదాలు) కనుక్కోండి.
(i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)
సాధన:
(x + a)n విస్తరణలో n సరిసంఖ్య అయిన మధ్యపదం \(T_{\left(\frac{n+1}{2}\right)}\), n బేసిసంఖ్య అయిన రెండు పదాలు \(\frac{T_{n+1}}{2}\), \(\frac{T_{n+3}}{2}\) మధ్యపదాలు అవుతాయి.
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(i)

(ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(ii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(ii).1

(iii) (4x2 + 5x3)17
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(iii)

(iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(iv)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q3(iv).1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 4.
క్రింది సమాసాల ద్విపద విస్తరణలలో సంఖ్యాపరంగా గరిష్ఠ పదం (పదాలు) కనుక్కోండి.
(i) (4 + 3x)15, x = \(\frac{7}{2}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(i).1

(ii) (3x + 5y)12, x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(ii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(ii).1

(iii) (4a – 6b)13, a = 3, b = 5
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(iii).1
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(iii).2

(iv) (3 + 7x)n, x = \(\frac{4}{5}\), n = 15
సాధన:
(3 + 7x)n = \(\left[3\left(1+\frac{7}{3} x\right)\right]^n\)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(iv)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q4(iv).1

ప్రశ్న 5.
క్రింది వాటిని నిరూపించండి.
(i) 2.C0 + 5.C1 + 8.C2 + ….. + (3n+2).Cn = (3n + 4) . 2n-1
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(i)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(i).1

(ii) n ఒక సరిధన పూర్ణంకమైతే C0 – 4 . C1 + 7 . C2 – 10 . C3 + …… = 0
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(ii)

(iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(iii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(iii).1

(iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3+\ldots \ldots\) \(+\frac{3^n}{n+1} \cdot C_n=\frac{4^{n+1}-1}{3(n+1)}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(iv)

(v) C0 + 2 . C1 + 4 . C2 + 8 . C3 + ….. + 2n . Cn = 3n [May ’07]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(v)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q5(v).1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 6.
క్రింది మొత్తాలను కనుక్కోండి.
(i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \cdot \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \cdot \frac{{ }^{15} C_3}{{ }^{15} C_2}+\ldots \ldots\) \(+15 \cdot \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q6(i)

(ii) C0 . C3 + C1 . C4 + C2 . C5 + ….. + Cn-3 . Cn
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q6(ii)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q6(ii).1

(iii) 22 . C0 + 32 . C1 + 42 . C2 + …… + (n+2)2 . Cn
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q6(iii)

(iv) 3C0 + 6C1 + 12C2 + …… + 3 . 2n . Cn
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q6(iv)

ప్రశ్న 7.
ద్విపద సిద్ధాంతాన్ని ఉపయోగించి ప్రతి ధన పూర్ణాంకం nకు 50n – 49n – 1 ను 492 భాగిస్తుందని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q7
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q7.1

ప్రశ్న 8.
ద్విపద సిద్ధాంతాన్ని ఉపయోగించి ప్రతి ధన పూర్ణాంకం nకు 54n + 52n – 1 ను 676 భాగిస్తుందని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q8

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 9.
(1 + x + x2)n = a0 + a1x + a2x2 + …… + a2n x2n, అయితే
(i) a0 + a1 + a2 + ……. + a2n = 3n
(ii) a0 + a2 + a4 + …… + a2n = \(\frac{3^n+1}{2}\)
(iii) a1 + a3 + a5 + …….. + a2n-1 = \(\frac{3^n-1}{2}\)
(iv) a0 + a3 + a6 + a9 + …… = 3n-1 అని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q9
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q9.1
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q9.2

ప్రశ్న 10.
(1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ……. + b21 x21 అయితే
(i) b0 + b2 + b4 + …… + b20
(ii) b1 + b3 + b5 + ….. + b21 విలువలు కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q10
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q10.1

ప్రశ్న 11.
\(\left(2+\frac{8 x}{3}\right)^n\) విస్తరణలో x11, x12 గుణకాలు సమానమైతే n విలువ కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q11

ప్రశ్న 12.
22013 ను 17 తో భాగించగా వచ్చే శేషాన్ని కనుక్కోండి.
సాధన:
24 = 16
24 ను 17 తో భాగించగా వచ్చే శేషం – 1
22013 = (24)503 . 21
∴ 22013 ను 17 తో భాగించగా వచ్చే శేషం (-1)503 . 2
= (-1) . 2
= -2

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 13.
(1 + x)21 ద్విపద విస్తరణలో (2r + 4), (3r + 4) పదాల గుణకాలు సమానమయితే విలువ కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) II Q13

III.

ప్రశ్న 1.
(1 + x)n విస్తరణలో x9, x10, x11 పదాల గుణకాలు అంకశ్రేఢిలో ఉంటే, n2 – 41n + 398 = 0 అని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q1

ప్రశ్న 2.
(1 + x)n విస్తరణలో 3 వరస గుణకాలు 36, 84, 126 అయితే n విలువ కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q2

ప్రశ్న 3.
(a + x)n విస్తరణలో 2, 3, 4 పదాల గుణకాలు వరుసగా 40, 70, 1080 అయితే a, x, n విలువలు కనుక్కోండి. [T.S. Mar ’16, May ’06]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q3
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q3.1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 4.
(1 + x)n విస్తరణలో r, (r + 1), (r + 2) పదాల గుణకాలు అంకశ్రేఢిలో ఉంటే n2 – (4r + 1)n + 4r2 – 2 = 0 అని చూపండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q4

ప్రశ్న 5.
\(\left(2 x^3-\frac{3}{x^2}\right)^{14}\) విస్తరణలో x32, x-18 గుణకాల మొత్తం కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q5
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q5.1

ప్రశ్న 6.
(x + a)n ద్విపద విస్తరణలో బేసిపదాల మొత్తం P, సరిపదాల మొత్తం Q అయితే (i) P2 – Q2 = (x2 – a2)n (ii) 4PQ = (x + a)2n – (x – a)2n అని నిరూపించండి. [A.P. Mar. ’16]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q6
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q6.1

ప్రశ్న 7.
(1 + x)n ద్విపద విస్తరణలో 4 వరుస పదాల గుణకాలు వరుసగా a1, a2, a3, a4 అయితే \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\) అని చూపండి. [Mar. ’11; May ’07]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q7
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q7.1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 8.
\(\left({ }^{2 n} C_0\right)^2-\left({ }^{2 n} C_1\right)^2+\left({ }^{2 n} C_2\right)^2-\left({ }^{2 n} C_3\right)^2+\) ……. \(+\left({ }^{2 n} C_{2 n}\right)^2=(-1)^{22 n} C_n\) అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q8

ప్రశ్న 9.
(C0 + C1) (C1 + C2) (C2 + C3) …. (Cn-1 + Cn) = \(\frac{(n+1)^n}{n !}\) · C0 . C1 . C2 …… Cn అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q9
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q9.1

ప్రశ్న 10.
\((1+3 x)^n\left(1+\frac{1}{3 x}\right)^n\) విస్తరణలో x లేని పదం (స్థిర పదం) కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q10

ప్రశ్న 11.
(1 + x)2n ద్విపద విస్తరణలోని మధ్యపదం \(\frac{1.3 .5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\) అనిచూపండి. [May ’06]
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q11
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q11.1

ప్రశ్న 12.
(1 + 3x – 2x2)10 = a0 + a1x + a2x2 + …. + a20x20 అయితే
(i) a0 + a1 + a2 + …… + a20 = 210
(ii) a0 – a1 + a2 – a3 + ……… + a20 = 410 అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q12

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 13.
(3√3 + 5)2n+1 = x, f = x – [x] అయితే (ఇక్కడ [x] అనేది x పూర్ణాంక భాగాన్ని సూచిస్తుంది.), x . f విలువ కనుక్కోండి.
సాధన:
(3√3 + 5)2n+1 = x
f = x – [x] ⇒ 0 < f < 1
F = (3√3 – 5)2n+1 అనుకోండి.
5 < 3√3 < 6
⇒ 0 < 3√3 – 5 < 1
⇒ 0 < (3√3 – 5)2n+1 < 1
⇒ 0 < F < 1
⇒ 0 > -F > -1
⇒ -1 < -F < 0
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q13

ప్రశ్న 14.
R, n లు ధన పూర్ణాంకాలు, n బేసి పూర్ణాంకం, 0 < F < 1, (5√5 + 11)n = R + F అయితే
(i) R ఒక సరి పూర్ణాంకం
(ii) (R + F) . F = 4n అని చూపండి.
సాధన:
(i) R, n లు ధన పూర్ణాంకాలు
0 < F < 1, (5√5 + 11)n = R + F
(5√5 – 11)n = f అనుకోండి.
121 < 125 < 144 కనుక
ఇప్పుడు 11 < 5√5 < 12
⇒ 0 < 5√5 – 11 < 1
⇒ 0 < (5√5 – 11)n < 1
⇒ 0 < f < 1
⇒ 0 > -f > -1
∴ -1 < -f < 0
R + F – f = (5√5 + 11)n – (5√5 – 11)n
= \(\left[{ }^n C_0(5 \sqrt{5})^n+{ }^n C_1(5 \sqrt{5})^{n-1}(11)\right.\)
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q14

ప్రశ్న 15.
I, n లు ధన పూర్ణాంకాలు 0 < f < 1, (7 + 4√3)n = I + f అయితే
(i) I ఒక బేసి పూర్ణాంకం
(ii) (I + f) (I – f) = 1 అని చూపండి.
సాధన:
I, n లు ధన పూర్ణాంకాలు
(7 + 4√3)n = I + f, 0 < f < 1
7 – 4√3 = F అనుకోండి.
∴ 36 < 48 < 49
ఇప్పుడు 6 < 4√3 < 7
⇒ -6 > 4√3 > -7
⇒ -7< -4√3 < -6
⇒ 0 < (7 – 4√3)n < 1
∴ 0 < F < 1
I + f + F = (7 + 4√3)n + (7 – 4√3)n
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q15
= \(2\left[{ }^n C_0 7^n+{ }^n C_2 7^{n-2}(4 \sqrt{3})^2+\ldots . .\right]\)
= 2k, k పూర్ణాంకం
∴ I + f + F సరిపూర్ణాంకం.
⇒ f + F పూర్ణాంకం, (I పూర్ణాంకం కనుక)
కానీ 0 < f < 1, 0 < F < 1
⇒ 0 < f + F < 2
∴ f + F = 1 …….(1)
⇒ I + 1 సరి పూర్ణాంకం.
∴ I బేసి పూర్ణాంకం.
(I + f) (I – f) = (I + f) F, (1) నుండి
= (7 + 4√3)n (7 – 4√3)n
= [(7 + 4√3) (7 – 4√3)]n
= (49 – 48)n
= 1

AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a)

ప్రశ్న 16.
n ధన పూర్ణాంకం అయితే \(\sum_{r=1}^n r^3\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{(n)(n+1)^2(n+2)}{12}\) అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 6 ద్విపద సిద్ధాంతం Ex 6(a) III Q16

ప్రశ్న 17.
\(\left(5^{1 / 6}+2^{1 / 8}\right)^{100}\) విస్తరణలో కరణీయ పదాల సంఖ్యను కనుకోండి.
సాధన:
సాధారణ పదం
\(T_{r+1}={ }^{100} C_r\left(5^{1 / 6}\right)^{100-r}\left(2^{1 / 8}\right)^r\) = \({ }^{100} C_r 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)
0 ≤ r ≤ 100 లో r = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100 అయితే \(\frac{100-r}{6}\) ఒక పూర్ణాంకం
0 ≤ r ≤ 100 లో r = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\(\frac{r}{8}\) ఒక పూర్ణాంకం
0 ≤ r ≤ 100 అయితే r = 16, 40, 64, 88 అయితే \(\frac{100-r}{6}, \frac{r}{8}\) లు రెండూ పూర్ణాంకాలు
∴ \(\left(5^{1 / 6}+2^{1 / 8}\right)^r\) విస్తరణలో అకరణీయ పదాలు = 4
∴ \(\left(5^{1 / 6}+2^{1 / 8}\right)^r\) విస్తరణలో కరణీయ పదాల సంఖ్య = 101 – 4 = 97

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers

Question 1.
By comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\), \(\frac{c_{1}}{c_{2}}\) K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Answer:
Given: 5x – 4y + 8 = 0
7x + 6y – 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\); \(\frac{b_{1}}{b_{2}}\) = \(\frac{-4}{6}\); \(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the given pair of linear equations represents a pair of intersecting lines.

b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Answer:
Given : 9x + 3y + 12 = 0
18x + 6y + 24= 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
The lines are coincident.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

c) 6x – 3y + 10 = 0
2x – y + 9 = 0
Answer:
Given: 6x – 3y + 10 = 0
2x – y + 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = \(\frac{3}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = \(\frac{3}{1}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)
Here \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
∴ The lines are parallel.

Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
a) 3x + 2y = 8
2x – 3y = 1
Answer:
Given equaions are 3x + 2y = 8 and 2x – 3y = 1
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);
\(\frac{b_{2}}{b_{-3}}\) = \(\frac{-4}{6}\);
\(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the linear equations are consistent.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 1
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 2
The lines intersect at (2, 1), so the solution is (2, 1).

b) 2x – 3y = 8
4x – 6y = 9
Answer:
Given: 2x – 3y = 8 and 4x – 6y = 9
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{9}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Lines are inconsistent and have no solution.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 3
Lines are parallel.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 4a
The lines are parallel and no solution exists.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

c) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
9x – 10y = 12
Answer:
Given pair of equations \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 and 9x – 10y = 12
Now take \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 ⇒ \(\frac{9x+10y}{6}\) = 7 ⇒ 9x + 10y = 42
and 9x – 10y =12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{9}\) = \(\frac{1}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{10}{-10}\) = \(\frac{1}{-1}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-42}{-12}\) = \(\frac{7}{2}\)
Since \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\) they are intersecting lines and hence consistent pair of linear equations.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 5
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 6
Solution: The unique solution of given pair of equations is (3.1, 1.4)

d) 5x – 3y = 11
-10x + 6y = -22
Answer:
Given pair of equations 5x – 3y = 11 and -10x + 6y = -22
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{-10}\) = \(\frac{-1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{11}{-22}\) = \(\frac{-1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
∴ The lines are consistent.
∴ The given linear equations represent coincident lines.
Thus they have infinitely many solutions.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 7
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 8

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

e) \(\frac{4}{3}\)x + 2y = 8
2x + 3y = 12
Answer:
Given pair of equations \(\frac{4}{3}\)x + 2y = 8 ⇒ \(\frac{4x+6y}{3}\) = 8 ⇒ 4x + 6y = 24 ⇒ 2x + 3y = 12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\) = 2;
\(\frac{b_{1}}{b_{2}}\) = \(\frac{6}{3}\) = 2;
\(\frac{c_{1}}{c_{2}}\) = \(\frac{24}{12}\) = 2
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent.
∴ The given equations have infinitely many solutions.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 9
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 10

f) x + y = 5
2x + 2y = 10
Answer:
Given pair of equations x + y = 5 and 2x + 2y = 10
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent and have infinitely many solutions.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 11
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 12

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

g) x – y = 8
3x – 3y = 16
Answer:
Given pair of equations x – y = 8 and 3x – 3y = 16
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{16}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 13
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 14

h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
Answer:
Given pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{-2}\) = \(\frac{-1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-6}{-4}\) = \(\frac{3}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
The equations are consistent.
∴ They intersect at one point giving only one solution.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 15
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 16
The solution is x = 2 and y = 2

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
Answer:
Given pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-2}{-4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-2}{-5}\) = \(\frac{2}{5}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 17
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 18

Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.”
Help her friend to find how many pants and skirts Neha bought.
Answer:
Let the number of pants = x and the number of skirts = y
By problem y = 2x – 2 ⇒ 2x – y = 2
y = 4x – 4 ⇒ 4x – y = 4
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 19
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 20
The two lines are intersecting at the point (1,0)
∴ x = 1; y = 0 is the required solution of the pair of linear equations.
i.e., pants =1
She did not buy any skirt.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Answer:
Let the number of boys be x.
Then the number of girls = x + 4
By problem, x + x + 4 = 10
∴ 2x + 4 = 10
2x = 10-4
x = \(\frac{6}{2}\) = 3
∴ Boys = 3 Girls = 3 + 4 = 7 (or)
Boys = x, Girls = y
By problem x + y = 10 (total)
and y = x + 4 (girls)
⇒ x + y = 10 and x – y = – 4
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 21
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 22
∴ Number of boys = 3 and the number of girls = 7

Question 5.
5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Answer:
Let the cost of each pencil be Rs. x
and the cost of each pen be Rs. y.
By problem 5x + 7y = 50
7x + 5y = 46
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 23
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 24
The lines are intersecting at the point (3, 5).
x = 3 and y = 5 is the solution of given equations.
∴ Cost of one pencil = Rs. 3 and pen = Rs. 5

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 6.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer:
Let the width of the garden = x cm
then its length = x + 4 cm
Half the perimeter = \(\frac{1}{2}\) × 2(7+ b) = l + b
By problem, x + x + 4 = 36
2x + 4 = 36
2x = 36 – 4 = 32
∴ x = 16 and x + 4 = 16 + 4 = 20
i.e., length = 20 cm and breadth = 16 cm.
(or)
Let the breadth be x and length = y
then x + y = 36 ⇒ x + y = 36
y = x + 4 ⇒ x – y = -4
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 25
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 26
The two lines intersect at the point (16, 20)
i.e., length = 20 cm and the breadth = 16 cm.

Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Answer:
i) Given: 2x + 3y – 8 = 0
The lines are intersecting lines.
Let the other linear equation be ax + by + c = 0
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\); we have to choose appropriate values satisfying the condition above.
Thus the other equation may be 3x + 5y – 6 =0

ii) Parallel line \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0
4x + 6y – 10 = 0

iii) Coincident lines \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 8.
The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle = x units
breadth = y units Area = l . b = xy sq. units
By problem, (x – 5) (y + 2) = xy – 80 and          (x + 10) (y – 5) = xy + 50
⇒ xy + 2x – 5y – 10 = xy – 80 and                    xy – 5x + 10y – 50 = xy + 50
⇒ 2x – 5y = xy – 80 – xy + 10 and                   -5x + 10y = xy + 50 – xy + 50
⇒ 2x – 5y = – 70 and                                       -5x + 10y = 100
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 27
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 28
The two lines intersect at the point (40, 30)
∴ The solution is x = 40 and y = 30
i.e., length = 40 units; breadth = 30 units.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 9.
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Answer:
Let the number of benches = x say and the number of students = y
By problem
y = 3x + 1 ⇒ 3x – y + 1 = 0
and y = 4(x – 1) ⇒ 4x – y – 4 = 0
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 29
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 30
The two lines intersect at (5, 16)
∴ The solution of the equation is x = 5 and y = 16
i.e., Number of benches = 5 and the number of students = 16

AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2

10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use π = 3.14)
Answer:
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 1
Diameter of the base of the cone d = 6 cm.
∴ Radius of the base of the cone
r = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height of the cone = h = 4 cm
Slant height of the cone l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
∴ C.S.A of the cone = πrl
= \(\frac{22}{7}\) × 3 × 5
= \(\frac{330}{7}\) cm2
Radius of the hemisphere = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
C.S.A. of the hemisphere = 2πr2
= 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{396}{7}\)
Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
= \(\frac{330}{7}\) + \(\frac{396}{7}\)
= \(\frac{726}{7}\) ≃ 103.71 cm2.

AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 2.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use π = 3.14]
Answer:
Total surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 2
Cone:
Radius (r) = 8 cm
Height (h) = 6 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{8^{2}+6^{2}}\)
= \(\sqrt{64+36}\)
= √100
= 10 cm
C.S.A. = πrl
= \(\frac{22}{7}\) × 8 × 10
= \(\frac{1760}{7}\) cm2
Cylinder:
Radius (r) = 8 cm;
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm2
Hemisphere:
Radius (r) = 8 cm
C.S.A. = 2πr2
= 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm2
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A. = \(\frac{8096}{7}\) = 1156.57 cm2.

AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 3.
A medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.
Answer:
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 3
Surface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder
i) Now for Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5 mm
C.S.A of each hemisphere = 2πr2
C.S.A of two hemispheres
= 2 × 2πr2 = 4πr2
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × \(\frac{5}{2}\)
= \(\frac{550}{7}\)
= 78.57 mm2.

ii) Now for Cylinder:
Length of capsule = AB =14 mm
Then height (length) cylinder part = 14 – 2(2.5)
h = 14 – 5 = 9 mm
Radius of cylinder part (r) = \(\frac{5}{2}\)
Now C.S.A of cylinder part = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × 9
= \(\frac{900}{7}\)
= 141.428 mm2
Now total surface area of capsule
= 78.57 + 141.43 = 220 mm2

Question 4.
Two cubes each of volume 64 cm3 are joined end to end together. Find the surface area of the resulting cuboid.
Answer:
Given, volume of the cube.
V = a3 = 64 cm3
∴ a3 = 4 × 4 × 4 = 43 , Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a = 2 × 4 = 8 cm,
breadth = a = 4 cm.
height = a = 4 cm is formed.
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 4
∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm2
∴ The surface area of resulting cuboid is 160 cm2.

AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m2.
Answer:
Total surface area of the tank = 2 × C.S.A. of hemisphere + C.S.A. of cylinder.
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 5
Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
C.S.A. of hemisphere = 2πr2
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7
= 3.08 m2.
2 × C.S.A. = 2 × 3.08 m2 = 6.16 m2
Cylinder:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
Height (h) = 8 m
C.S.A. of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8
= 35.2 m2
∴ Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m2
Cost of painting its surface area @ Rs. 20 per sq.m, is
= 41.36 × 20 = Rs. 827.2.

Question 6.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.
Answer:
Let the length of the edge of the cube = a units
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 6
T.S.A. of the given solid = 5 × Area of each surface + Area of hemisphere
Square surface:
Side = a units
Area = a2 sq. units
5 × square surface = 5a2 sq. units
Hemisphere:
Diameter = a units;
Radius = \(\frac{a}{2}\)
C.S.A. = 2πr2
= 2π\(\left(\frac{a}{2}\right)^{2}\)
= 2π\(\frac{a^{2}}{4}\) = \(\frac{\pi \mathrm{a}^{2}}{2}\) sq. units
Total surface area = 5a2 + \(\frac{\pi \mathrm{a}^{2}}{2}\) = a2\(\left(5+\frac{\pi}{2}\right)\) sq. units.

AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 7.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.
AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 7
Answer:
Surface area of the given solid = C.S.A. of the cylinder + 2 × C.S.A. of hemisphere.
If we take base = radius
Cylinder:
Radius (r) = 3.5 cm
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 10
= 220 cm2
Hemisphere:
Radius (r) = 3.5 cm
C.S.A. = 2πr2
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= 77 cm2
2 × C.S.A. = 2 × 77 = 154 cm2
∴ T.S.A. = 220 + 154 = 374 cm2.

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.2

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.2

Question 1.
Find mode of the following data.
(i) 2, 3, 7, 5, 3, 2, 6, 7, 1,2.
Answer:
Given data : 2, 3, 7, 5, 3, 2, 6, 7, 1,2.
By arranging the numbers with same values together
1, 2, 2, 2, 3, 3, 5, 6, 7, 7.
As 2 occurs more frequently than other observations in the data.
∴ Mode = 2

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.2

(ii) K, A, B, C, B, C, D, K, B, D, B, K, A, K.
Answer:
Given data : K, A, B, C, B, C, D, K, B, D, B, K, A, K
By arranging the letters in the alphabetical order of same type together.
A, A, B, B, B, B, C, C, D, D, K, K, K, K.
As B and K occurs most frequently than other observations in the data.
∴ Mode = B and K.

(iii) First ten natural numbers.
Answer:
First 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
In the given observations there is no repeated number.
So, the given data has no mode.

(iv) 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8.
Answer:
Given data : 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8.
In the given observations, data is repeated an equal number of times. .
So, the given data has no mode.

Question 2.
20 students were participated in ‘SWATCH BHARAT ABHIYAN’ campaign. The number of days each student participated were 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4 and 5. Find mode of the data.
Answer:
Given data 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4, 5.
By arranging the numbers with same value together
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4; 4, 5, 5, 5.
As 2 occurs more frequently than other observations in the data.
∴ Mode = 2.

Question 3.
The number of goals scored by a 3, 2, 4, 6, 1, 3, 2, 4, 1 and 6. Find the mode of data.
Answer:
Given data: 3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
By arranging the numbers with same values together.
1, 1, 2, 2, 3, 3, 4, 4, 6, 6.
In the given observation, data is repeated an equal number of times.
So, the given data has no mode.

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.2

Question 4.
Find the mode of letters in the adjacent figure. Verify whether it is Unimodal or Bimodal Data.
AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.2 1
Answer:
In the figure data is: S, A, H, S, A, M, S, T, M, T, H, % A, T, S, M, H, M, A, S, T, M, A, T, S, T, H, M.
By arranging the letters, of same type together.
A, A, A, A, A, H, H, H, H, M, M, M, M/M, M, S, S, S, S, S, S, T, T, T, T, T, T, T.
As T occurs most frequently in the data.
∴ Mode = T
Data having only one mode is known as unimodal data.
So, given data is unimodal data.

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Exercise 5(e) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Exercise 5(e)

అభ్యాసం – 5(ఇ)

I.

ప్రశ్న 1.
nC4 = 210, అయితే n విలువ ఎంత?
సాధన:
సూచన: nCr = \(\frac{n !}{(n-r) ! r !}\) = \(\frac{n \cdot(n-1)(n-2) \ldots \ldots / n-(r-1)]}{1.2 .3 \ldots \ldots \ldots . .}\)
nC4 = 210
⇒ \(\frac{n(n-1)(n-2)(n-3)}{1.2 .3 .4}\) = 10 × 21
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 21 × 1 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 7 × 3 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 9 × 8 × 7
∴ n = 10

ప్రశ్న 2.
12Cr = 495, అయితే r విలువ కనుక్కోండి.
సాధన:
సూచన: nCr = nCn-r
12Cr = 495
= 5 × 99
= 11 × 9 × 5
= \(\frac{12 \times 11 \times 9 \times 5 \times 2}{12 \times 2}\)
= \(\frac{12 \times 11 \times 10 \times 9}{1.2 .3 .4}\)
= 12C4 లేదా 12C8
∴ r = 4 లేదా 8

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 3.
10 . nC2 = 3 . n+1C3, అయితే n విలువ ఎంత?
సాధన:
10 . nC2 = 3 . n+1C3
⇒ 10 × \(\frac{n(n-1)}{1.2}=\frac{3(n+1) n(n-1)}{1.2 .3}\)
⇒ 10 = n + 1
⇒ n = 9

ప్రశ్న 4.
nPr = 5040, nCr = 210 అయితే n, r విలువలను కనుక్కోండి. [A.P. Mar. ’16]
సాధన:
సూచన: nPr = r! nCr మరియు nPr = n(n – 1) (n – 2)…. [n – (r – 1)]
nPr = 5040, nCr = 210
r! = \(\frac{{ }^n P_r}{{ }^n C_r}=\frac{5040}{210}=\frac{504}{21}\) = 24 = 4!
∴ r = 4
nPr = 5040
nP4 = 5040
= 10 × 504
= 10 × 9 × 56
= 10 × 9 × 8 × 7
= 10P4
∴ n = 10
∴ n = 10, r = 4

ప్రశ్న 5.
nC4 = nC6, అయితే n ఎంత?
సాధన:
సూచన: nCr = nCs ⇒ r = s or r + s = n
nC4 = nC6
∴ n = 4 + 6 = 10

ప్రశ్న 6.
15C2r-1 = 15C2r+4 అయితే r విలువ కనుక్కోండి. [Mar. ’14, ’05]
సాధన:
15C2r-1 = 15C2r+4
⇒ 2r – 1 = 2r + 4 లేదా (2r – 1) + (2r + 4) = 15
⇒ 4r + 3 = 15
⇒ 4r = 12
⇒ r = 3
∴ 2r – 1 = 2r + 4
⇒ -1 = 4 ఇది అసాధ్యం
∴ r = 3

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 7.
17C2t+1 = 17C3t-5, అయితే t విలువ ఎంత?
సాధన:
17C2t+1 = 17C3t-5
⇒ 2t + 1 = 3t – 5 లేదా (2t + 1) + (3t – 5) = 17
⇒ 1 + 5 = t లేదా 5t = 21
⇒ t = 6 లేదా t = \(\frac{21}{5}\) ఇది పూర్ణాంకము కాదు
∴ t = 6

ప్రశ్న 8.
12Cr+1 = 12C3r-5, అయితే r విలువ కనుక్కోండి. [T.S. Mar. ’16, Mar. ’08]
సాధన:
12Cr+1 = 12C3r-5
⇒ r + 1 = 3r – 5 లేదా (r + 1) + (3r – 5) = 12
⇒ 1 + 5 = 2r లేదా 4r – 4 = 12
⇒ 2r = 6 లేదా 4r = 16
⇒ r = 3 లేదా r = 4
∴ r = 3 లేదా 4

ప్రశ్న 9.
9C3 + 9C5 = 10Cr, అయితే r విలువ కనుక్కోండి?
సాధన:
సూచన: nCr = nCn-r
10Cr = 9C3 + 9C5
nCr + nCr-1 = (n+1)Cr
9C6 + 9C5 = 10C6 లేదా 10C4
∴ r = 4 లేదా 6

ప్రశ్న 10.
ఆరుగురు పురుషులు ముగ్గురు స్త్రీల నుంచి అయిదుగురు సభ్యులున్న కమిటీలు ఎన్ని ఏర్పరచవచ్చు?
సాధన:
వ్యక్తుల సంఖ్య = 6 + 3 = 9
ఈ 9 మంది నుండి 5 గురు సభ్యులున్న కమిటీ ఏర్పరచే విధానాలు = 9C5
= 9C4
= \(\frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4}\)
= 126

ప్రశ్న 11.
పై ప్రశ్నలో కనీసం ఇద్దరు స్త్రీలు ఉండే కమిటీలు ఎన్ని?
సాధన:
కమిటీలో కనీసం ఇద్దరు స్త్రీలు ఉండేటట్లుగా కమిటీలను ఈక్రింది విధంగా ఎన్నుకోవచ్చు.
(i) ముగ్గురు పురుషులు, ఇద్దరు స్త్రీలు
ముగ్గురు పురుషులు, ఇద్దరు స్త్రీలను ఎన్నుకొనే విధానాల సంఖ్య = 6C3 × 3C2
= 20 × 3
= 60
(ii) ఇద్దరు పురుషులు, ముగ్గురు స్త్రీలు
ఇద్దరు పురుషులు, ముగ్గురు స్త్రీలను ఎన్నుకొనే విధానాల సంఖ్య = 6C2 × 3C2
= 15 × 1
= 15
∴ కనీసం ఇద్దరు స్త్రీలు ఉండేటట్లుగా కమిటీలను ఎన్ను కొనే విధానాల సంఖ్య = 60 + 15 = 75

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 12.
nC5 = nC6, అయితే 13Cn విలువ ఎంత? [Mar. ’13]
సాధన:
nC5 = nC6
⇒ n = 6 + 5 = 11
13Cn = 13C11
= 13C2
= \(\frac{13 \times 12}{1 \times 2}\)
= 78

II.

ప్రశ్న 1.
3 ≤ r ≤ n కు (n-3)Cr + 3 (n-3)Cr-1 + 3 (n-3)Cr-2 + (n-3)Cr-3 = nCr అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) II Q1

ప్రశ్న 2.
10C5 + 2 . 10C4 + 10C3 విలువ ఎంత?
సాధన:
సూచన: nCr + nCr-1 = (n+1)Cr
10C5 + 2 . 10C4 + 10C3
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) II Q2

ప్రశ్న 3.
సూక్ష్మీకరించండి 34C5 + \(\sum_{r=0}^4(38-r) C_4\)
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) II Q3

ప్రశ్న 4.
ఒక తరగతిలో 30 మంది విద్యార్థులున్నారు. వారిలో ప్రతి విద్యార్థి మిగిలిన విద్యార్థులందరితో ఒక చదరంగం ఆటను ఆడితే మొత్తం ఎన్ని చదరంగం ఆటలు వారు ఆడినట్లు?
సాధన:
తరగతిలోని విద్యార్థుల సంఖ్య = 30
ప్రతి విద్యార్థి మిగిలిన విద్యార్థులందరితో ఒక్కో చదరంగం ఆటను ఆడతాడు.
కనుక మొత్తం చదరంగం ఆటల సంఖ్య = 30C2
= \(\frac{30 \times 29}{1 \times 2}\)
= 435

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 5.
ఏడుగురు బాలికలు, ఆరుగురు బాలురు నుంచి ముగ్గురు బాలురు, ముగ్గురు బాలికలు ఉండే కమిటీలను ఎన్ని రకాలుగా ఏర్పరచవచ్చు?
సాధన:
ఏడుగురు బాలికలు, ఆరుగురు బాలురు నుండి ముగ్గురు బాలురు, ముగ్గురు బాలికలు ఉండే కమిటీల సంఖ్య = 7C3 × 6C3
= 35 × 20
= 700

ప్రశ్న 6.
10 మంది వ్యక్తుల నుంచి నిర్దేశించిన ఒక వ్యక్తి ఉండేలా ఆరుగురు సభ్యుల కమిటీలు ఎన్ని ఏర్పరచవచ్చు?
సాధన:
నిర్దేశించిన వ్యక్తి కమిటీలో ఉండి, మిగిలిన 9 మంది నుండి 5 గురు వ్యక్తులను ఎన్నుకొనే విధాల సంఖ్య = 9C5
∴ 10 మంది వ్యక్తుల నుంచి నిర్దేశించిన వ్యక్తి ఉండేలా ఆరుగురు సభ్యుల కమిటీలు ఎన్నుకొనే విధాల సంఖ్య = 9C5
= \(\frac{9 \times 8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4 \times 5}\)
= 126

ప్రశ్న 7.
ఇచ్చిన 9 పుస్తకాల నుంచి నిర్దేశించిన ఒక పుస్తకం లేకుండా 5 పుస్తకాలను ఎన్ని రకాలుగా ఎంచుకోవచ్చు?
సాధన:
9 పుస్తకాలనుంచి నిర్దేశించిన ఒక పుస్తకం లేకుండా 5 పుస్తకాలు ఎన్నుకోవాలి. అంటే నిర్దేశించిన ఆ పుస్తకం తీసివేసి, మిగిలిన 8 పుస్తకాల నుండి 5 పుస్తకాలు ఎంచుకోవాలి. ఈ పనిని 8C5 విధాలుగా చేయవచ్చు.
కనుక కావలసిన సంయోగాల సంఖ్య = 8C5
= 8C3
= \(\frac{8 \times 7 \times 6}{1 \times 2 \times 3}\)
= 56

ప్రశ్న 8.
EQUATION పదంలోని అక్షరాల నుంచి 3 అచ్చులు, 2 హల్లులు ఎన్ని రకాలుగా ఎంచుకోవచ్చు? [May ’11, Mar. ’07]
సాధన:
EQUATION అనే పదంలో {E, U, A, I, O} అను 5 అచ్చుల {Q, T, N} అను 3 హల్లులు కలవు.
అందులో 3 అచ్చులు, 2. హల్లులు ఎన్నుకొనే విధాలు = 5C3 × 3C2
= 10 × 3
= 30

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 9.
12 భుజాలున్న ఒక బహుభుజి కర్ణాల సంఖ్య కనుక్కోండి.
సాధన:
ఇచ్చట n = 12
n భుజాలున్న బహుభుజి కర్ణాల సంఖ్య = \(\frac{n(n-3)}{2}\)
= \(\frac{12 \times 9}{2}\)
= 54

ప్రశ్న 10.
ఒక వరుసలో ఉన్న n వ్యక్తుల నుంచి పక్క పక్కనే ఉన్న ఇద్దరు వ్యక్తులను ఎన్ని రకాలుగా ఎంచుకోవచ్చు?
సాధన:
ఒక వరుసలో ఉన్న n వ్యక్తుల నుంచి, పక్కపక్కనే ఉన్న ఇద్దరు వ్యక్తులను ఎంచుకొనే విధాల సంఖ్య = n – 1

ప్రశ్న 11.
4 సరూప నాణేలను 5 గురు బాలురకు ఎవరికైనా ఎన్నైనా ఇచ్చే పద్ధతిలో ఎన్ని రకాలుగా పంచవచ్చు?
సాధన:
4 సరూప నాణేలను ఈ క్రింది విభిన్న సమూహాలుగా విభజించవచ్చు.
(i) ఒక సమూహంలో 4 నాణేలు
(ii) రెండు సమూహాలలో వరుసగా 1, 3 నాణేలు
(iii) రెండు సమూహాలలో వరుసగా 2, 2 నాణేలు
(iv) రెండు సమూహాలలో వరుసగా 3, 1 నాణేలు
(v) మూడు సమూహాలలో వరుసగా 1, 1, 2 నాణేలు
(vi) మూడు సమూహాలలో వరుసగా 1, 2, 1 నాణేలు
(vii) మూడు సమూహాలలో వరుసగా 2, 1, 1 నాణేలు
(viii) నాలుగు సమూహాలలో వరుసగా 1, 1, 1, 1 నాణేలు
ఈ సమూహాలను 5 గురు బాలురకు పంచే విధాల సంఖ్య
= \({ }^5 C_1+2 \times{ }^5 C_2+{ }^5 C_2+{ }^5 C_3 \times \frac{3 !}{2 !}+{ }^5 C_4\)
= 5 + 20 + 10 + 30 + 5
= 70

III.

ప్రశ్న 1.
\(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots \ldots(4 n-1)}{\{1.3 .5 \ldots \ldots(2 n-1)\}^2}\) అని నిరూపించండి.
సాధన:
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) III Q1

ప్రశ్న 2.
ఒక సమితి A లో 12 మూలకాలున్నాయి. ఆ సమితిలో
(i) 4 మూలకాలున్న ఉపసమితులెన్ని?
(ii) కనీసం 3 మూలకాలున్న ఉపసమితులెన్ని?
(iii) 3 లేదా అంతకంటే తక్కువ మూలకాలున్న ఉపసమితులెన్ని? [May ’07]
సాధన:
సమితి A లో వున్న మూలకాల సంఖ్య = 12
(i) 4 మూలకాలున్న ఉపసమితుల సంఖ్య = 12C4
= \(\frac{2 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}\)
= 495
(ii) కనీసం 3 మూలకాలున్న ఉపసమితులు.
పై లెక్క ప్రకారం కనీసం రెండు మూలకాలున్న ఉపసమితులు = 12C0 + 12C1 + 12C2
= 1 + 12 + 66
= 79
A సమితికున్న మొత్తం ఉపసమితుల సంఖ్య = 212
∴ కనీసం 3 మూలకాలున్న ఉపసమితులు = 212 – (79)
= 4096 – 79
= 4017
(iii) 3 లేదా అంతకంటే తక్కువ మూలకాలున్న ఉప సమితులు సున్నా మూలకాలు
(i.e.,) మూలకాలు లేని ఉపసమితిల సంఖ్య = 12C0 = 1
ఒకే ఒక మూలకము వున్న ఉపసమితులు = 12C1 = 12
రెండు మూలకములు వున్న ఉపసమితులు = 12C2
= \(\frac{12 \times 11}{1 \times 2}\)
= 66
మూడు మూలకములు వున్న ఉపసమితులు = 12C3
= \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3}\)
= 220
∴ కనీసం 3 మూలకాలున్న ఉపసమితులు = 1 + 12 + 66 + 220 = 299

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 3.
ఏడుగురు బాట్స్మెన్, ఆరుగురు బౌలర్లు నుంచి కనీసం అయిదుగురు బౌలర్లు ఉన్న పదకొండు మంది క్రికెట్ టీమును ఎన్ని రకాలుగా ఏర్పరచవచ్చు?
సాధన:
కనీసం 5 గురు బౌలర్లు ఉన్న పదకొండు మంది క్రికెట్ టీమును క్రింద చూపిన విధాలుగా ఎంచుకోవచ్చు.
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) III Q3
∴ కోరిన విధంగా క్రికెట్ టీముని ఎంచుకొనే విధానాలు = 42 + 21 = 63

ప్రశ్న 4.
5 అచ్చులు, 6 హల్లులు నుంచి 3 అచ్చులు, 3 హల్లులు ఉండేలా ఎన్ని 6 అక్షరాల పదాలు ఏర్పరచవచ్చు.
సాధన:
అచ్చుల సంఖ్య = 5
హల్లుల సంఖ్య = 6
5 అచ్చుల నుండి 3 అచ్చులు ఎన్నుకొనే విధానాల సంఖ్య = 5P3
6 హల్లులు నుండి 3 హల్లులు ఎన్నుకొనే విధానాల సంఖ్య = 6P3
ఈ 6 అక్షరాలను వాటిలో వాటిని మార్చి వ్రాయగల పదాల సంఖ్య 6!
∴ 6 అక్షరాల పదాలలో 3 అచ్చులు 3 హల్లులు ఉండేలా ఎన్నుకోగల పదాల సంఖ్య = 5C3 × 6C3 × 6!

ప్రశ్న 5.
ఒక రైలు మార్గంలో 8 స్టేషన్లు ఉన్నాయి. వీటిలో 3 స్టేషన్లలో రైలు ఆపాలి. ఆ మూడు స్టేషన్లలో ఏ రెండూ పక్కపక్కన లేకుండా ఎన్ని రకాలుగా ఎంచుకోవచ్చు? [Mar. ’08]
సాధన:
మొదటి రైలు ఆగే స్టేషన్ ముందు గల స్టేషన్ల సంఖ్య x1 అనుకోండి. ఇట్లే మొదట, రెండు రైలు ఆగే మధ్య x2 స్టేషన్లు, రెండు, మూడు రైలు ఆగే స్టేషన్ల మధ్య x3, స్టేషన్లు మరియు మూడవసారి రైలు ఆగిన తరువాత x4 స్టేషన్లు ఉన్నాయి అనుకోండి.
అప్పుడు x1 ≥ 0, x2 ≥ 1, x3 ≥ 1, x4 ≥ 0 మరియు x1 + x2 + x3 + x4 = n – 3
ఈ సమీకరణానికి గల సాధనల సంఖ్య 6C3
∴ ఏ రెండు స్టేషన్లు పక్క పక్కన లేకుండా 8 స్టేషన్లలో 3 స్టేషన్లు ఎంచుకొనే విధానాలు = 6C3
= \(\frac{6 \times 5 \times 4}{1 \times 2 \times 3}\)
= 20

ప్రశ్న 6.
ఆరుగురు భారతీయులు, అయిదుగురు అమెరికా దేశస్థుల నుంచి అయిదుగురు సభ్యులున్న కమిటీని, ఆ కమిటీలో భారతీయుల సంఖ్య పెద్దదిగా ఉండేలా ఎన్ని రకాలుగా ఎంచుకోవచ్చు? [Mar. ’13, ’08]
సాధన:
కమిటీలో భారతీయుల సంఖ్య పెద్దదిగా ఉండేటట్లు కమిటీని ఎన్నుకొనే విధాలు ఈ క్రింద ఇవ్వబడినవి.
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) III Q6
∴ కమిటీని కోరిన విధంగా ఎంచుకొనే విధానాల సంఖ్య = 200 + 75 + 6 = 281

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 7.
ఒక ప్రశ్నాపత్రంలో A, B, C అనే మూడు భాగాలలో వరుసగా 3, 4, 5 ప్రశ్నలున్నాయి. ఒక్కో భాగం నుంచి కనీసం ఒక ప్రశ్న ఉండే విధంగా మొత్తం 6 ప్రశ్నలు ఎన్ని రకాలుగా ఎంచుకోవచ్చు?
సాధన:
మొదటి పద్దతి
ఒక్కొక్క భాగం నుంచి ఒక ప్రశ్న ఉండే విధంగా 6 ప్రశ్నలు ఎన్నుకొనే విధాలు ఈ క్రింది ఇవ్వబడినవి.
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) III Q7
రెండవ పద్ధతి
ఒక్కో భాగం నుంచి కనీసం ఒక ప్రశ్న ఉండే విధంగా మొత్తం 6 ప్రశ్నలు ఎంచుకొనే విధాలు = మొత్తం 12 ప్రశ్నల నుండి 6 ప్రశ్నలు ఎంచుకొనే విధాలు – C భాగం నుండి కాక మిగిలిన రెండు భాగాల నుండి 6 ప్రశ్నలు ఎన్నుకోవడం – B నుండి కాక మిగిలిన రెండు భాగాల నుండి 6 ప్రశ్నలు ఎన్నుకోవటం – A నుండి కాక మిగిలిన రెండు భాగాల నుండి 6 ప్రశ్నలు ఎన్నుకోవడం
= 12C67C68C69C6
= 805

ప్రశ్న 8.
12 విభిన్నమైన వస్తువులను ఎన్నిరకాలుగా
(i) 4 సమభాగాలుగా చేయవచ్చు
(ii) నలుగురు వ్యక్తులకు సమానంగా పంచవచ్చు?
సాధన:
(i) 12 విభిన్న వస్తువులను 4 సమభాగాలుగా విభజించే విధానాలు = \(\frac{(12) !}{(3)^4 4 !}\)
(ii) నలుగురు వ్యక్తులకు సమానంగా 12 విభిన్న వస్తువులు పంచే విధాల సంఖ్య = \(\frac{(12) !}{(3 !)^4}\)

AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e)

ప్రశ్న 9.
ఒక తరగతిలో నలుగురు బాలురు, ఆ బాలికలున్నారు. ప్రతీ ఆదివారం వారిలో కనీసం ముగ్గురు బాలురు ఉండేలా 5 గురు ఉన్న సమూహం విహారయాత్రకు వెళ్తారు. ప్రతీ ఆదివారం వేర్వేరు సమూహాలు విహారయాత్రకు వెళ్తాయి. వారి తరగతి ఉపాధ్యాయిని విహార యాత్రకు వచ్చిన ప్రతీ అమ్మాయికి ప్రతిసారి ఒక్కో బొమ్మ ఇవ్వగా వచ్చే మొత్తం బొమ్మల సంఖ్య 85 ఐతే g విలువ కనుక్కోండి.
సాధన:
బాలుర సంఖ్య = 4
బాలికల సంఖ్య = g
కనీసం ముగ్గురు బాలురు ఉండేలా ఈ క్రింది పట్టికలో తెలిపిన విధంగా ఎన్నుకొనవచ్చును.
AP Inter 2nd Year Maths 2A Solutions Chapter 5 ప్రస్తారాలు-సంయోగాలు Ex 5(e) III Q9
G1 లో బాలికల సంఖ్య = [4C3 × 9C2] × 2 ఎందువలననగా ప్రతిసారి ఇద్దరు బాలికలు ఉంటారు.
G2 లో బాలికల సంఖ్య = [4C3 × 9C2] × 1 ఎందువలననగా ప్రతిసారి ఒక బాలిక ఉంటుంది.
మొత్తం బాలికలకు ఇచ్చిన బొమ్మల సంఖ్య = 85
⇒ [4C3 × 9C2] × 2 + [4C3 × 9C2] × 1 = 85
⇒ 4 . \(\frac{g(g-1)}{2}\) × 2 + 1 . g . 1 = 85
⇒ 4g2 – 4g + g – 85 = 0
⇒ 4g2 – 3g – 85 = 0
⇒ 4g2 – 20g + 17g – 85 = 0
⇒ 4g(g – 5) + 17(g – 5) = 0
⇒ (g – 5) (4g + 17) = 0
g ≠ \(\frac{17}{5}\) కనుక
∴ g = 5
∴ బాలికల సంఖ్య = 5

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

AP State Board Syllabus AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000.

AP State Syllabus SSC 10th Class Social Studies Important Questions 19th Lesson Emerging Political Trends 1977 to 2000

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 1 Mark Important Questions and Answers

Question 1.
Expand the term AIADMK.
Answer:
All India Anna Dravida Munnetra Kazagam.

Question 2.
Give any two examples for Regional Political parties.
Answer:
TDP, YSRCP, JANA SENA, TRS, AIADMK, DMK, etc.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 3.
Which welfare schemes initiated by N.T. Rama Rao are still continuing with some changes in Andhra Pradesh?
Answer:

  1. Mid-day meal scheme in government schools.
  2. Sale of rice at subsidy rates to the poor.

Question 4.
Identify at least any two states presently ruled by regional parties in India on the given Indian political map.
Answer:
AP SSC 10th Class Social Studies Important Questions Chapter 18 Independent India (The First 30 years – 1947-77) 9

Question 5.
What was the contribution of Telecom revolution?
Answer:
The contribution of Telecom Revolution:
A network of telephonic communication in the country using satellite technology increased.

Question 6.
Mention any two initiations of N.T. Rama Rao.
Answer:

  1. Sale of rice at Rs. 2/- kg
  2. Mid day meal scheme in government schools.
  3. Liquor prohibition

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 7.
Write about the 73rd amendment of the constitution.
Answer:
73rd amendment: The 73rd constitutional amendment created institutions of local self government at the village level and so Gram Panchayat, Mandal Parishad and Zilla Parishad are formed.

Observe the table given below and answer the questions 8 & 9.
Results of Telangana State Assembly and Parliament Elections – 2014

S.No.Name of the PartyAssembly Seats wonParliament Seats won
1.T.R.S.6311
2.Congress Party212
3.T.D.P.202
4.Others152
Total11917

Question 8.
Name the two parties that secured more than 15 Assembly seats.
Answer:
Parties that secured more than 15 Assembly seats.

  1. TRS
  2. Congress Party
  3. T.D.P

Question 9.
Why did TRS secure more seats in 2014 elections?
Answer:
TRS secured more seats in 2014 elections because it played a key role in Telangana agitation.

Question 10.
What is meant by the Coalition government?
Answer:
During the time of General Election to the Assembly and Lok Sabha, no party gain the majority to form the government at the centre or state at that time. Two or more than two political parties come together to form a single government.
(OR)
A number of national and regional parties had to come together to form governments at the centre.

Question 11.
Name some non-political movements.
Answer:
Environmental movements, the feminist movement, civil liberties movement, literacy movements.

Question 12.
Which became powerful motors of social change?
Answer:
A number of non-political movements emerged and became powerful motors of social change.

Question 13.
Which parties decided to merge together and form the Janata Party?
Answer:
The Congress, Swatantra Party, Bharatiya Jan Sangh, the Bharatiya Lok Dal and the Socia¬list Party decided to merge together and form the Janata Party.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 14.
Who supported the Janata Party?
Answer:
The DMK, the SAD and the CPI (M) chose to maintain their separate identities but supported the Janata Party in a common front against the Congress.

Question 15.
Who played an important role in bringing together all the anti-Congress and anti-Emergency parties?
Answer:
Senior leaders like Jayaprakash Narayan and Acharya JB Kriplani played an important role in bringing together all the anti-Congress and anti-Emergency parties to fight the elections.

Question 16.
What was the argument of the Janata Party regarding the dismiss of nine state governments?
Answer:
The Janata Party argued that the Congress party had lost its mandate to rule in the States as it had been defeated.

Question 17.
Which created a bad state in A.P.?
Answer:
In Andhra Pradesh, the frequent change of Chief Ministers by the central Congress leadership and the imposition of leaders from above created a bad taste.

Question 18.
Who moved to Assom and Bengal?
Answer:
The Bangladeshis moved to Assom and Bengal.

Question 19.
Name some communities of Assom.
Answer:
Bodos, Khasis, Mizos and Karbis.

Question 20.
Who was Bhindtanwale and what was his demand?
Answer:
Bhindranwale, the leader of the group of militant Sikhs began to preach separatism and also demanded the formation of a Sikh State- Khalistan.

Question 21.
What did the militants try?
Answer:
The militants tried to impose an orthodox life code on all Sikhs and even non-Sikhs of Punjab.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 22.
Who made a declaration in April 1986?
Answer:
In April 1986, an assembly at the Akal Takht, made a declaration of an independent state of Khalistan.

Question 23.
Where were the militants engaged in?
Answer:
The militants were also engaged in large scale kidnapping and extortion to raise funds for their work.

Question 24.
How were the methods used by the govern¬ment for the suppression of militancy in Punjab?
Answer:
The Government used very harsh methods for the suppression of militancy in Punjab, many of which were seen as a violation of. Constitutional rights of citizens.

Question 25.
What did Rajiv Gandhi begin?
Answer:
Rajiv Gandhi began a peace initiative in Punjab, Assam and Mizoram and also in the neighbouring country of Sri Lanka.

Question 26.
What is called the telecom revolution?
Answer:
Rajiv Gandhi initiated what is called the ‘telecom revolution’ in India which speeded up and spread the network of telephonic communication in the country using satellite technology.

Question 27.
What had been under dispute for some time regarding Babri Masjid?
Answer:
Some sections of the Hindus had begun a campaign for building a temple for Lord Rama in Ayodhya in the place of Babri Masjid.

Question 28.
What is the speciality of Elections held in 1989?
Answer:
The issue of corruption in administration and in political circles became the main plank of the election campaign for non-Congress political forces in the next elections held in 1989.

Question 29.
What is Policy Paralysis?
Answer:
Policy Paralysis means the coalition could not implement any policy which called for serious change for fear of withdrawal of support by one or the other partners.

Question 30.
Which was the first coalition to be re-elected?
Answer:
The UPA was the first coalition to be re-elected.

Question 31.
Who led the Left Front Government in West Bengal in 1977?
Answer:
Jyoti Basu of CPM led the Left Front Government in West Bengal in 1977.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 32.
On what did the Operation Barga depend?
Answer:
Operation Barga depended heavily on collective action by the share croppers and Panchayati Raj Institutions thus avoiding bureaucratic delays and domination of the landowning classes.

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 2 Marks Important Questions and Answers

Question 1.
Read the following paragraph and answer the questions.

The Government used very harsh methods for the suppression of militancy in Punjab, many of which were seen as a violation of the constitutional rights of citizens. Many observers felt that such violations of constitutional rights and human rights were justified.as the constitutional machinery was on the edge of collapse due to militant activity.
Express your views on the information given above.

Answer:
There was a threat to the integration of the Indian nation due to the militancy in Punjab. If the government had not taken such actions, the map of India would be different today. So I think the government was correct.

Question 2.
Read the given data to answer the questions.
AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 1

A) Which were the parties that participated in the governments of the National Front and United Front and supported the government from the outside?
Answer:
To National Front: CPM, CPI, and BJP.
To United Front: CPM.

B) Mention the name of the party that participated in the above three governments.
Answer:
J.K.N.C.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 3.
Based on the information given below, answer the following questions.

End of Emergency and formation of Janata party government under Morarji Desai and Charan Singh1977
Formation of Congress government led by Indira Gandhi1980
Formation of TDP1982
Operation Blue Star and assassination of Indira Gandhi1984
Rajiv Gandhi Accords with H.S. Longowal on Punjab and AASU on Assam.1985

a) Name the first non-Congress party which formed the government at the centre.
Answer:
Janata Party is the first non-Congress party which formed the government at the centre.

b) Who is the founder of Telugu Desam Party?
Answer:
Nandamuri Taraka Rama Rao (NTR) is the founder of Telugu Desam Party.

Question 4.
Which are the newest states of India, when they created?
Answer:

StateYear of formation
1. Uttaranchal / Uttarkhand2000
2. Jharkand2000
3. Chattisghar / Chattisghad2000
4. Telangana2014

Question 5.
Read the table and answer the given equations.

Assassination of Rajiv Gandhi and government led by Congress party with P.V. Narsimha Rao as P.M.1991
Economic liberalization1990
Demolition of Babri Masjid1992
National Front Government with Deve Gowda and I.K. Gujral as P.M.s1996
NDA government led by A.B. Vajpayee1998

a) Which party won in 1996 elections and formed government?
Answer:
National Front.

b) Name the Coalition Governments mentioned in the above table.
Answer:
National Front and NDA Governments.

Question 6.
Write about people’s welfare schemes started by present Governments.
Answer:

  1. Supply of rice at the cost of Rs. 1 per Kg to the white ration cardholders.
  2. Pensions for the old age people and widows.
  3. Free textbooks, uniforms and Midday meal scheme in government schools.
  4. Housing schemes for the poor people.
  5. Health scheme for the poor people.
  6. Fees reimbursement to the poor for higher education, etc.

Question 7.
Read the following text and answer the questions given below.

The Congress returned to power in 1980. The Congress immediately paid back the Janatb in the same coin by dismissing the Janata and non-Congress governments in nine States. The Congress was victorious in all the States except Tamil Nadu and West Bengal.

A) Which party ruled before 1980s?
Answer:
Janata Party.

B) In which two states, the Congress party was defeated?
Answer:
Tamilnadu and West Bengal.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 8.
Prepare a table by classifying the given political parties into National and Regional Parties. “BJP, YSRCP, TDP, CPM, CPI, DMK, Congress-1, AGP”.

National partiesRegional parties

Answer:

S.No.National PartiesRegional Parties
1.Bharatiya janata PartyDMK
2.Congress -1TDP
3.CPIAGP
4.CPMYSRCP

Question 9.
Based on the information given below, answer the following questions.

Election and formation of Janata Dal government with VP Singh and Chandrasekhar1989
Decision to implement Mandal Commission recommendation1989
Ram Janmabhoomi Rath Yatra1990
Assassination of Rajiv Gandhi and government led by Congress party with P.V. Narsimha Rao as P.M.1991
Economic Liberalization1990
Demolition of Babri Masjid1992
National Front Government with Deve Gowda and IK Gujral as PMs1996
NDA government led by AB Vajpayee1998

i) Who was the Prime Minister at the time of demolition of Babri Masjid?
Answer:
P.V. Narasimha Rao.

ii) Give two examples of the Coalition government.
Answer:

  1. Janata Dal government.
  2. National Front government.
  3. National Democratic Alliance (NDA).

Question 10.
Sometimes coalition governments cause ‘Policy Paralysis’. Do you agree with this statement?
Write your opinion.
Answer:
Yes. I agree with this statement. The coalition could not implement any policy which called for serious change for fear of withdrawal of support by one or the other partners.

Question 11.
“Coalition Governments cause political instability.” Comment.
Answer:

  1. Sometimes no single party wins a majority of seats to form a government of its own. In such the situation, a number of political parties come together and form coalition governments.
  2. A common agreement between these parties has to be arrived at, but this is not so easy.
  3. Different parties put pressure on the government for their different interests.
  4. The government cannot implement any policy for fear of withdrawal of support by one or the other partners. The governments become instable.
  5. This is called policy paralise which is frequent in the coalition government.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 12.
What are the situations that paved to strengthen the regional parties in present days?
Answer:
The situations that paved to strengthen the regional parties

  1. Regional aspirations – regional movements.
  2. Intermediate castes strengthening – gaining political power.
  3. To gain political power.
  4. Defections and corruption.

Question 13.
Write the main reasons for Assam movement.
Answer:

  1. Demand for autonomy.
  2. Protest against the domination of Bangladesh.
  3. Migration from Bangladesh.
  4. Fear of losing their cultural roots.
  5. Trade and other establishments were in the hands of outsiders.
  6. No preference in employment for locals.

Question 14.
Observe the following table and analyse it.
Table: Seat share of various Political parties in 2014 (lok Sabha)

S.No.Political pattyWon Seats
1Bharatiya Janata Party (BJP)282
2Indian National Congress (INC)45
3Telugu Desam Party (TDP)16
4Telangana Rashtra Samithi (TRS)11
5Left parties [CPI + CPI (M)]10

Answer:

  1. In 2014 General elections BjP got with 282 seats and form the largest party and form the government also.
  2. Indian National Congress got only 45 seats.
  3. Left parties CPI + CPI (M) joined together got 10 seats.
  4. The Regional parties like TDP 16 seats 8i TRS 11 seats gained In Lok’Sabha elections.

Question 15.
What are the important changes that occured in India between 1975-85?
Answer:
Many changes occurred In India between 1975-85. Some of them are:

  1. Emergency was declared by smt. Indira Gandhi as she was asked to quit her Prime Minister post by Allahabad high court.
  2. Janatha Government came into power in 1979.
  3. Congress Party came to power in the elections after Janatha govt, failure.
  4. Non-political movements like environment movements, feminist movements, civil liberties movement and literacy movements came up.

Question 16.
At present, what is the necessity of coalition politics?
Answer:
In the present multiparty system in India it is impossible for any single party to win a majority of seats and form a government of its own but in 2019 elections BJP has won the election as single party. It went as coalition.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 17.
Read the following paragraph and comment on it.

In Andhra Pradesh, the frequent change of Chief Ministers by the central Congress leadership and the imposition of leaders from above created a bad taste. There was a feeling that the Andhra Pradesh leadership was not getting respect from the national Congress leadership. This was perceived as an insult to the pride of the Telugu people. N.T. Rama Rao(NTR), popular film actor, chose to take up this cause. He began the Telugu Desam Party (TDP) on his 60th birthday in 1982. He said that the TDP stood for the honour and self respect of the Telugu speaking people (Teluguvari atma gauravam). He argued that the state could not be treated as a lower office of the Congress party.

Answer:

  1. The Congress government frequently changed the Chief Ministers.
  2. The Congress was not giving respect to Andhra Pradesh leadership.
  3. The TDP was formed for the honour and self-respect of the Telugu speaking people.
  4. He introduced welfare schemes like midday meals to government schools, liquor prohibition and the sale of rice for Rs. 2/- per kg.
  5. These populist measures helped the TDP sweep the 1982 elections.
  6. TDP emerged as a strong regional party, and challenged the Congress domination.

Question 18.
What are the effects of changes of the Telecom Revolution on the Human lifestyles.
Answer:

  1. Telecom Revolution is the result of privatization of Telecommunications.
  2. Number of industries invested in telecommunications.
  3. “Mobiles” and Smart phones have created sensation.
  4. They reduced the distance between the buyers and sellers.
  5. Every family has a mobile in India.
  6. Telemarketing is a creative innovation.
  7. Smartphones have internet access and due to that internet facility is accessible to villagers through telephones.

Question 19.
What was Operation Blue Star?
Answer:

  1. Sikhs became militant in Punjab under Bhindranwale.
  2. People belonging to non-Sikhs were subjected to communal attack.
  3. Sikh separatist groups hid in the Golden Temple.
  4. Army had to intervene to vacate the campus.
  5. This was called ‘Operation Blue Star’.

Question 20.
What factors influenced central government to use armed forces to reduce tensions in Assam?
Answer:

  1. Three factors influenced the use of armed forces in the North Eastern Region.
  2. Firstly, it was a sensitive border area adjacent to China, Mynmar and Bangladesh.
  3. Secondly, rebel groups demanding separation from India, procured arms from outside.
  4. Thirdly, they indulged in large-scale ethnic violence against minority communities.
  5. The government thought this was the only way to bring about peace in the area.

Question 21.
Read the given information.
AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 2

Now answer the following questions.
a) Which party was included in “Governing parties” in all the above coalition governments?
Answer:
Jammu & Kashmir National Conference (JKNC)

b) Which party gave support to NDA government?
Answer:
TDP.

c) Which party gave support to National Front and United Front from outside?
Answer:
CPM.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 22.
What did the emergence of competitive alternatives ensured?
Answer:
The emergence of competitive alternatives ensured that Indian voters could always exercise a reasonable choice. This also allowed many different political viewpoints and sectional interests to become active in state level and national politics.

Question 23.
How was the rule of the first non-Congress government?
Answer:
The Janata Party had come to power promising a restoration of democracy and freedom from authoritarian rule. However, the disunity among the partners had a serious effect on the governance and its rule is most often remembered for internal squabbles and defections. The factional struggle in the party soon culminated in the fall of the government within three years leading to fresh elections in 1980.

Question 24.
What happened whenever there was any political instability?
Answer:
Whenever there was any political instability or natural calamity in the neighbouring country, thousands of people moved into the State creating huge discomfort for the locals. The local people felt that they would lose their cultural roots and soon be outnumbered by the ‘outsiders’.

Question 25.
What was there besides culture and demographics?
Answer:
Besides culture and demographics, there was also an economic dimension. Trade and other establishments were in the hands of non-Assamese communities. The major resources of the State, including tea and oil were again not benefitting the locals.

Question 26.
What was the dominant thrust of the movement?
Answer:
The dominant thrust of the movement was that Assam was being treated as an “internal colony” and this had to stop. The main demands were that the local people should be given greater preference in employment, the “outsiders” should be removed and the resources should be used for the benefit of the locals.

Question 27.
Which has led to violent attempts of ethnic cleansing in Assam?
Answer:
Too much emphasis on ethnic identities had a negative impact on other communities of Assam like the Bodos, Khasis, Mizos and Karbis. Many of them too demanded autonomous status. They began to assert themselves and wanted to drive out people of other communities from their areas.

Question 28.
What did Punjab claim?
Answer:
It laid claims to the new capital city of Chandigarh which remained a union territory directly administered by the Centre. Punjab also claimed more water from Bhakra Nangal dam and greater recruitment of Sikhs in the army.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 29.
Write about the resolution of Akali Dal.
Answer:
The Akali Dal had passed a set of resolutions in 1978 during the Janata Party rule in the centre, calling upon the central government to implement them. Its most significant demand was to amend the Constitution to give more powers to the states and ensure greater decentralisation of powers.

Question 30.
What happened after Rajiv Gandhi’s entrance?
Answer:
After Rajiv Gandhi became the Prime Minister, he held talks with SAD and entered into an agreement with Sant Langowal, the SAD president. Though fresh elections were held in Punjab and SAD won them, the peace was short-lived as Langowal was assassinated by the militants.

Question 31.
What did Rajiv Gandhi say in his speech?
Answer:
In a famous speech Rajiv Gandhi said that out of every Rupee spent on the poor barely 15 paise reaches them I It highlighted the fact that despite huge increases in development expenditure, the story of the poor remained the same.

Question 32.
Which factors influenced the central government to use armed forces to reduce tensions in Assam?
Answer:

  1. Three factors influenced the use of armed forces in the Northeastern region.
  2. Firstly, it was a sensitive border area adjacent to China, Myanmar and Bangladesh.
  3. Secondly, rebel groups demanding separation from India, procured arms from outside.
  4. Thirdly, they indulged in large scale ethnic violence against minority communities.
  5. The government thought this was the only way to bring about peace in the area.

Question 33.
What was meant by liberalization?
Answer:

  1. It meant a lot of things put together like the drastic reduction of government expenditure, reducing restrictions and taxes on imports, etc.
  2. It proved for reducing restrictions on foreign investments in India and allowed foreign countries to set up companies in India.
  3. It is required to the opening of many sectors of the economy to private investors.
  4. It brought in foreign goods and Indian businessmen were forced to compete with them.
  5. It had many positive and negative impacts on India.

Question 34.
“One of the greatest weakness was undoubtedly the low priority given to primary education and public health”. Comment on it.
Answer:

  1. The post-Independence era is marked with less priority to education and health.
  2. The optimum development of country depends mostly on the education and health levels of the population of it.
  3. It further forms part of Human Development Indicators also.
  4. So, I suggest more priority should be given to education and health now.

Question 35.
Read the given information.

In 1992 government led by P.V. Narasimha Rao passed an important amendment to the Constitution to provide local self-governments a Constitutional Status. The 73rd Constitutional Amendment created institutions of local self government at the village level while the 74th Constitutional Amendment did the same in towns and cities. These were path-breaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of universal adult franchise. One-third of the seats were to be reserved for women. Seats were also reserved for scheduled castes and tribes.

Answer the following questions.
a) Which constitutional amendment created institution of local self-government?
b) According to which amendment general elections were conducted in towns and cities?
c) How many seats are reserved for women in local bodies?
Answer:
a) 73rd Constitutional Amendment created institutions of local self-governments for villages.
b) According to 74th Constitutional Amendment general elections were conducted in towns/cities.
c) 1/3 seats were reserved for women in local self-government elections.

Question 36.
“Do you think that the reservations will promote the social development” ? Express your ideas.
Answer:

  1. Reservations will definitely promote social development.
  2. Scheduled castes and tribes were drowtodden and suffered in the social stature for centuries.
  3. To develop themselves and to question the injustice they meted out, reservations will of great help.
  4. Reservations both in education, jobs, and legislature help them.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 37.
Imagine and write the main reasons for the continuation of anti Hindi movement in Tamil Nadu till today.
Answer:

  1. DMK in Tamil Nadu believed the passage of Official Languages Act, 1963 was an attempt to first Hindi on the rest of the country.
  2. They started a statewide campaign protesting the imposition of Hindi.
  3. They organised strikes, dharnas, burning effigies, Hindi books as well as pages of constitution.
  4. Still there is same feeling in Tamil Nadu.

Question 38.
“Some people think that Social Welfare Schemes do not reach eligible persons”. Express your suggestions.
Answer:

  1. Despite all the attention to development in the country, much of it did not reach the real beneficiaries.
  2. Despite huge increases in development expenditure the story of the poor remained the same.
  3. The main reasons were political and beaurocratic corruption.
  4. Enlistment of various beneficiaries also plagued by officialdom and political pressures.

Question 39.
“India needed to adapt itself to the new technologies emerging in the world, especially computer and telecommunication technologies”. Comment.
Answer:

  1. Technologies like computer and communication technology are thursting the world.
  2. It is believed that we should also adopt them without fail.
  3. With initiatives of Rajiv Gandhi now called ‘Telecom Revolution1 was introduced in India.
  4. With the help of satellite technology communications spread widely and extensively.
  5. Everyone has access to mobile phones, the internet, email, facebook, Twitter, etc.

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 4 Marks Important Questions and Answers

Question 1.
Read the text given and answer the questions.

Panchayati Raj & 73rd, 74th Amendment

In 1992, Government led by P.V. Narasimha Rao passed an Important amendment to the Constitution to provide Local Self Governments a Constitutional status. The 73rd Constitutional Amendment created Institutions of local self-government at the village level, while the 74th Constitutional Amendment did the same in towns and cities. These were pathbreaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of Universal Adult Franchise. One-third of the seats were to be reserved for women. Seats were also reserved for scheduled castes and tribes. The concerns of the State governments were taken into account and it was left to the States to decide on what functions and powers were to be developed to their respective local self-governments. Consequently, the powers of local self-governments vary across the country.

i) What is Local Self Government?
Answer:
The Government that formed by the people at the village, town and city level to solve the local needs is Local Self Government.

ii) Which government recognised the Constitutional status of Local self Government?
Answer:
P.V. Narasimha Rao or Congress Government.

iii) What does the 73rd Constitutional Amendment say?
Answer:
Creation of Local Self government at the village level.

iv) 1/3 of seats were to be reserved for women in Local Self Governments. Comment.
Answer:
Women need political equality and they should Involve actively In the Local Governments.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 2.
Read the following paragraph and answer the given questions.

Liberalization measures brought in foreign goods and forced Indian business to compete with global manufacturers. It also led to the setting up of industries and business by foreign companies in India. However, it also meant a lot of hardship for the common people as the government was forced to cut subsidies to the people and as many factories closed down due to Influx of cheap foreign goods. This also led to privatization of many public amenities tike education, health, and transport and people had to pay high prices to private service providers.

Write your opinion on the consequences of liberalization.
(OR)
What are the consequences of economic Liberalization?
Answer:

  1. India was drawn into the world market.
  2. Liberalization paved the way to telecom revolution.
  3. Liberalization forced Indian business to compete with global manufactures.
  4. The government was forced to cut subsidies which results a great loss to people and local industries.
  5. It also led to the privatization of many public amenities like education, health and transport.
  6. It led to globalization.
  7. The policies of liberalisation have been of advantage particularly to well of sections only.

Question 3.
Explain the importance of regional parties in Democracy.
Answer:

  1. Multi-party system which includes national parties and regional parties strengthens the democracy.
  2. Regional parties reflect the spirit of the federalism.
  3. Regional parties have good understanding of the problems and needs of the respective states.
  4. They focus mainly on the development of their states.

Question 4.
Telecom revolution has brought several changes in human life nowadays. Explain them.
Answer:
Changes brought by the telecom revolution:

  1. Saves time
  2. Fast communication
  3. Online services
  4. Prosperous life
  5. Addiction
  6. Obesity
  7. Cost of living increased
  8. Affected human relations

Question 5.
Read the paragraph given below and interpret.

India was forced to open up and ‘liberalise’ its economy by allowing free flow of foreign capital and goods Into India. On the other hand, new social groups asserted themselves politically for the first time, and finally, religious nationalism and communal political mobilisation became Important features of our political life. All this put the Indian society into great turmoil, we are still coming to grips with these changes and adapting ourselves to them.

Answer:

  1. Liberalisation means relaxation of previous government restrictions usually in areas of social and economic policy.
  2. The twentieth century ended with India’s drawing into the world free market.
  3. India was forced to open up and liberalise its economy. It allowed free flow of foreign capital and goods into India.
  4. On the other hand, India seemed to have a thriving democracy in which voices of different sections of the population were making themselves heard and in which divisive and communal political mobilisation was threatening to destroy social peace.
  5. It had stood the test of time for over fifty years and had built a relatively stable economy and deeply rooted democratic politics.
  6. It still had not managed to solve the problem of acute poverty and gross inequality between castes, communities, regions and gender.

Question 6.
Observe the following table and write a paragraph analyzing it.
Summary of the 2014 -Indian General Elections

PartyAllianceVotes(%)Seats
BJPNDA31%282
INCUPA19.31%44

Answer:
The given table describes the summary of the 2014 general elections in India. In the given table two parties that is Bharatiya Janata Party and the Indian National Congress are compared. It is not only the party comparison but their alliances are also mentioned. The Bharatiya Janata Party alliance is National Democratic Alliance whereas the United Progressive Alliance is related to Indian National Congress. In these elections the NDAgot 31% of the votes whereas the UPAgot 19.31%. If we observe the seats, the BJP with its alliance won 282 whereas the INC won only 44. These elections are very crucial because the voter strongly rejected the pre-independence party which ruled India since 1947. For a long time it was a single largest party to win the seats in Lok sabha. The voters cleverly gave mandate to the Bharatiya Janata Party with the hopes that their future may be changed. The BJP announced the Prime Ministerial candidate, Narendra Modi in advance. He achieved and succeeded in Gujarat as Chief Minister. So the voters accepted him as Prime Minister also. They believed him. Congress lost faith of the people because of its failures. During Congress period there was a lot of corruption, scams and nepotism, etc. Many of the Congress members of Parliament were in court cases. Rajiv Gandhi himself declared that corruption is highly established in India. If the Bharatiya Janata Party with its alliance work for the development of the country, definitely they will win the next coming 2019 elections. So the party should keep this in mind and work in that direction.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 7.
Explain the effects along with the reasons for the emerging era of coalition politics.
Answer:
Reasons for the emergence of coalition era of politcs:

  1. Multi party system
  2. No single party securing required majority.
  3. Significance of regional parties increased.
  4. Congress party gradually lost people’s mandate after 1960s.

Effects:

  1. No political stability
  2. Isolating the ideologies
  3. Giving importance to party’s interest at the cost of nation’s interest.
  4. Coming to power in spite of securing less mandate.

Question 8.
Read the following paragraph and write your opinion.

The twentieth century closed with India which was drawn into the world market, India which seemed to have a thriving democracy in which voices of different sections of the population were making themselves heard and in which, divisive and communal political mobilisation were threatening to destroy social peace. It had stood the test of time for over fifty years and had built a relatively stable economy and deeply rooted democratic politics. It still had not managed to solve the problem of acute poverty and gross inequality between castes, communities, regions and gender.

Answer:
The given paragraph depicts about divisive and communal politics. These may destroy the social peace. After independence in India, stable government continued for 30, 40 years and unstability began. Main problem of solving poverty and inequalities with regard to caste, region is not yet solved.

My opinion is that the politics are only vote bank based. Sometimes the political leaders are there behind the communal riots. To throw out some Chief Minister of the same party, their party leaders encourage these riots. Caste based politics are shown at the time of tickets given to party candidates. Caste unions, and the caste group heads are distributed money to lure them to get their votes. Some constituencies are fixed for some religion because of their dominance in number. It is really a threat to democracy. Holy places of worship are also in some cases used to spread communal message. That destroys social peace.
My suggestion is that people should get awarness about this and act accordingly.

Question 9.
Observe the following table and analyse it.
The trend of Coalition Governments, 1989 – 2004

S.No.CoalitionDurationGoverning partiesSupporting parties
1.National Front1989 – 90JD, DMK, AGP, TDP, JKNCCPM, CPI, BJP
2.United Front1996 – 98JKNC, TDP, TMC, CPI, AGP, DMK, MGPCMP
3.National Democratic Alliance1998 – 2004JDU, SAD, TMC, AIADMK, JKNC, BJD, Shiva-SenaTDP

Answer:

  1. The given table is about the trend of Coalition Governments during the period of the years from 1989 to 2004.
  2. The details of three coalition governments and their duration, etc. are given in the table.
  3. During 1989-1990 Janata Dal-led National Front formed the government. The governing parties in this government were JD, DMK, AGP, TDP, JKNC. CPM, CPI and BJP supported this government.
  4. United Front formed the coalition government during 1996-1998. JKNC, TDP, TMC, CPI, AGP, DMK, MGP were the governing parties in this government. CPM supported this government.
  5. During 1998-2004 BJP-led National Democratic Alliance formed the government. The governing parties in this government were JDU, SAD, AIADMK, JKNC, TMC, BJD and Shiva Sena. TDP rendered support to the NDA government.
  6. The 1990s were years of very significant change in the post-Independence India.
  7. With the transformation to a competitive multi-party system, it became near impossible for any single party to win a majority of seats and form a government of its own.
  8. Since 1989, all governments that had formed at the national level have been either coalition or minority governments.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 10.
Explain about Assam movement in detail.
Answer:

Assam movement:

It is the struggle between Assamese and non-Assamese. These non-Assamese were none other than the people of Bangladesh. The youth of Assam formed All Assam Students Union (AASU) and was in the forefront of agitation. It led a number of strikes, agitations and marches to remove the so called outsiders. The problem of outsiders is not a cultural one but of economic issue. Every country or state wants to protect their cultural roots. The Assamese were most of them, Hindus and the outsiders were Muslims. The local people were afraid of their cultural roots.
Now they affect the trade and so the livelihoods of the locals had been in trouble. It is not only the problem of Assam, it happens at many states. Outsiders dominate a few areas of business and so the locals lose opportunities. In Assam the locals were not given priority or preference in employment. This was the demand of the Assamese. Gradually these demands led to communal polarisation as most of the outsiders are from Bangladesh Muslims. The movement between the Assamese and outsider Muslim led to form an idea of anti Indian stand.

Central Government took initiation and went on for talks for three years. An agreement was signed by the central government and the students union. In the next elections Assam Gana Parishad (an offshoot of AASU) came to power.

In conclusion, the formation of Bangladesh erstwhile Pakistan was taken place on the basis of religion. One’s religion can be given respect by all but it led to many disturbances. The Muslims, the outsiders of Assam occupied most of the areas of trade and business and there was distress and disappointment among the Assamese. The outsiders would have settled in Bangladesh only. They wanted their country to be separated and still they are coming to India illegally. Recently both the Prime Ministers of India and Bangladesh sat together and solved a few problems. If any problem arises, they should sit together and problems can be solved.

Question 11.
Prepare an album by collecting the photos of Prime Ministers of India and write their specialities.
Answer:
AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 6AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 7AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 8AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 9

Question 12.
You may notice how simple and genuine demands of the people of Punjab were hijacked by religious and anti-national extremists. What steps do you think would have prevented this?
Answer:

  1. The simple and genuine demands of Punjab were
    a) the contribution of state was ignored
    b) received unfair bargain when it was created
    c) capital remain UT
    d) more water from Bhakra Nangal and
    e) greater recruitment of Sikhs in the army.
  2. Akali Dal government was dismissed by Congress.
  3. A series of untoward incidents increased distance between Sikhs and the central government.
  4. Militant Sikhs demanded separate state.
  5. They occupied Golden Temple, then Congress used army to vacate.
  6. A fallout led to the assassination of Indira.
  7. Rioting in Delhi against Sikhs was followed.
  8. Later Langowal made an agreement with centre but was killed by militants.
  9. Militants engaged in extortion and kidnapping and lost faith of the people

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 13.
Understand the table and answer the following questions.
Some opposition parties of 1970’s

SI.No.Name of the partyPlace of ActivityIdeologies
1.BLD-Bharatiya Lok DalUttar PradeshSpecial attention to Indian farmers
2.Congress (O)Entire IndiaConservative section opposed policies of Indira Gandhi
3.CPI (M)
Communist Party of India (Marxist)
West Bengal, Tripura, Kerala, and remained statesRadical land reforms, trade unionism, socialist policies
4.DMK – Dravida Munnetra KazagamTamil Nadu1. Greater autonomy and powers to state
2. Fiercely opposed Hindi in the state
5.Jan SanghNorthern StatesA Hindu nationalist party
6.SAD – Shiromani Akali DalPunjab1. Great autonomy to states
2. Organised around Gurudwaras

a) Which political party fiercely opposes Hindi in the state ?
Answer:
DMK is the party which opposes Hindi in the state.

b) What is the place of activity for Jan Sangh?
Answer:
Jan Sangh is active in Northern States.

c) What is the ideology of CPI (M)?
Answer:
The ideologies of CPI (M) are radical land reforms, trade unionism and socialist policies.

d) Where is the political party which shows special attention to farmers, active ?
Answer:
The political party, which shows special attention to farmers is active in Uttar Pradesh.

e) Which party is of semireligious nature?
Answer:
SAD – Shiromani Akali Dal is of semireligious nature.

Question 14.
What were the implications of 1977 general elections?
Answer:

  1. It was a historical election for democracy.
  2. The Congress party was defeated at the national level for the first time.
  3. Janata Party became victorious and tried to consolidate itself.
  4. It dismissed nine Congress governments in states.
  5. It argued that Congress had lost its mandate to rule in the states as it had been defeated.
  6. Its stand somewhat proved correct by the results.
  7. Except Tamil Nadu and West Bengal, Janata Party came to power in states.
  8. The disunity among the partners had a serious effect on governance.
  9. The government fell within three years.
  10. It led to fresh elections in 1980.

Question 15.
Why was the public sympathy to Punjab militant Sikhs declined?
Answer:

  1. They formed armed attachments and engaged in terrorist activities.
  2. They clashed with police and other religious groups.
  3. Those who were not confirmed to militant approved behaviour were killed.
  4. There were civil casualities in derailing trains, exploding bombs, etc.
  5. They were engaged in kidnapping, extortion to raise funds.
  6. All this gradually alienated them from masses and even Sikhs.
  7. Over a period, public sympathy declined rapidly.
  8. Peace was finally returned to Punjab by the end of 1990s.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 16.
‘Coalition governments induce political instability’ – Elucidate.
Answer:

  1. Since 1989, all governments at national level were coalition/minority governments.
  2. A number of national and regional parties had come together.
  3. So political ideologies and programmes of all parties had to be accommodated.
  4. A common agreement had to be arrived at.
  5. No party could pursue extreme agendas.
  6. They needed to tone down their approaches.
  7. It caused considerable instability.
  8. Many coalitions did not last their full time.

Question 17.
How do political parties reap on communal polarisation? Provide an example.
Answer:

  1. The Hindus are led by Bharatiya Janata Party.
  2. In the year 1984 LokSabha elections they won only 2 seats.
  3. It made great strides when it took up the Ayodhya issue.
  4. It decided to campaign for the building of a temple at the site of mosque.
  5. It claimed that was the birthplace of Lord Rama.
  6. L.K. Advani in 1990, led a ‘Rath Yatra’ from Somanath to Ayodhya.
  7. This campaign was accompanied by intense communal polarisation.
  8. It caused a large number of communal conflicts.
  9. In 1991 General elections BJP’s strength went up to 120.
  10. It was then Rajiv was killed and sympathy wave followed the Congress, still, BJP withstood it.

Question 18.
What is meant by liberalisation?
Answer:

  1. It means a lot of things put together.
  2. It proposes drastic reduction of government expenditure.
  3. It asks for reducing restrictions and taxes on import of foreign goods.
  4. It provides for reducing restrictions on foreign investments in India.
  5. It is required to the opening of many sectors of the economy to private investors.
  6. It brought in foreign goods and forced Indian business to compete with them.
  7. It allowed foreign countries to set up companies in India.
  8. Common people suffered with cut of subsidies.
  9. Many factories were closed down due to influx of cheap foreign goods.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 19.
Study the timeline given below and answer the following questions.
AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 3a) Who was the Prime Minister that initiated peace agreements with Sri Lanka?
Answer:
Rajiv Gandhi initiated peace agreements with Sri Lanka.

b) Which government tried to implement Mandal Commissions report?
Answer:
Janata Dal government tried to implement Mandal Commissions report.

c) Name two important incidents occurred during the period of P.V. Narasimha Rao.
Answer:
Economic liberalization and the demolition of the Babri Masjid took place during the period of P.V. Narasimha Rao.

d) When was Indira Gandhi assassinated?
Answer:
Indira Gandhi was assassinated in 1984.

e) Who were the Prime Ministers of National Front Government?
Answer:
Deve Gowda and I.K. Gujral were the Prime Ministers of the National Front Government.

f) Who were the Prime Ministers of Janata Dal Government?
Answer:
V.P. Singh and Chandrasekhar were the Prime Ministers of Janata Dal Government.

g) Who led the Congress party after the assassination of Rajiv Gandhi?
Answer:
P.V. Narasimha Rao led the Congress party after the assassination of Rajiv Gandhi.

h) Who led the NDA government?
Answer:
A.B. Vajpayee led the NDA government.

i) When was the NDA Government formed?
Answer:
NDA formed the government in 1998.

Question 20.

Read the following information and answer the questions.

Some opposition parties of 1970s

BLD – Bharatiya Lok Dal – A party which was formed of socialists who called for special attention to Indian farmers, based mainly in Uttar Pradesh.

Congress (O) – The conservative section of the Congress which had opposed the policies of Indira Gandhi.

CPI (M) – Communist Party of India (Marxist)-a party with a national presence, which strove for radical land reforms, trade unionism and socialist policies.

DMK – Dravida Munnetra Kazagam – a party based mainly in Tamil Nadu which sought greater autonomy and powers for the state.

Jan Sangh – A Hindu nationalist party largely confined to the northern States.

SAD – Shiromani Akali Dal – a party based in Punjab catering specially to the Sikhs and organised around Gurudwaras. It therefore had a semi-religious character. It was also in favour of greater autonomy to the States.

a) Which party fought for autonomy in Tamil Nadu?
Answer:
Dravida Munnetra Kazagam fought for greater autonomy in Tamil Nadu.

b) Which party showed special attention to Indian farmers mainly in UP?
Answer:
Bharatiya Lok Dal showed special attention to farmers mainly in U.P.

c) Name the regional party of Punjab.
Answer:
Shiromani Akali Dal is the regional party of Punjab.

d) Name one Hindu nationalist party.
Answer:
“Jan Sangh” is one Hindu nationalistic party.

e) Which opposed the policies of Indira Gandhi?
Answer:
Congress (O) – The conservative section of the Congress opposed the policies of Indira Gandhi.

f) What was the main aim of SAD?
Answer:
It sought for greater autonomy to Punjab.

g) Which party was confined to North India only?
Answer:
Jan Sangh was confined to North India only.

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 21.

Read the following passage and interpret it.

Panchayati Raj & 73rd, 74th amendment

In 1992 government led by P.V. Narasimha Rao passed an important amendment to the Constitution to provide local self-governments a Constitutional Status. The 73rd Constitutional Amendment created institutions of local self government at the village level while the 74th Constitutional Amendment did the same in towns and cities. These were path-breaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of universal adult franchise.

One-third of the seats were to be reserved for women. Seats were also reserved for , scheduled castes and tribes. The concerns of the State governments were taken into j account and it was left to the States to decide on what functions and powers were to be devolved to their respective local self governments. Consequently, the powers of local self governments vary across the country.

Answer:

  1. In 1992 P.V. Narasimha Rao s government passed the important amendments of 73rd and 74th.
  2. The 73rd amendment created institutions of local self governments at the village levels.
  3. The 74th amendment created institutions of local self-governments at the town and city levels.
  4. They are path-breaking as the office bearers at the local level are elected on the basis of universal adult franchise.
  5. Seats are reserved for women and Scheduled Castes and Tribes too.
  6. Powers were devolved to their respective local self-governments.
  7. Hence we can say that these two amendments were path-breaking.

Question 22.
On the outline map of India locate the On the outline map of India locate the following.

  1. Andhra Pradesh
  2. Assom
  3. Punjab
  4. Tamil Nadu
  5. West Bengal
  6. Uttar Pradesh
  7. Nagaland
  8. Mizoram
  9. Bihar
  10. Gujarat
  11. Maharashtra
  12. Ayodhya

Answer:

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 4

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

Question 23.
Locate the following in the given map of World.

  1. Madagascar Island
  2. Nigeria
  3. Holland
  4. Amsterdam
  5. Brazil
  6. Jordan
  7. Israel
  8. Spain
  9. Palestine
  10. Bangladesh

Answer:
AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000 5

AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 8 Heredity Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 8th Lesson Heredity

10th Class Biology 8th Lesson Heredity Textbook Questions and Answers

Improve your learning

Question 1.
What are variations? How do they help organisms?
Answer:

  1. Differences in characters within very closely related groups of organisms are referred to as variations.
  2. Variations develop during reproduction in organisms.
  3. Variations are passed from parent to offspring through heredity.
  4. Beneficial variations are selected by the nature in evolution.
  5. Variations increase the survival chance of the organisms.
  6. These variations help the organisms to adapt to their environments.
  7. For example, green colour in the beetles is a variation that gave a survival advantage to the beetles as they cannot be seen by the crows.
  8. Some variations do not help organisms to survive. For example, colour variation occurs in red beetles and some blue beetles are produced instead of red beetles as they are eaten by crows easily.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 2.
One student (researcher) wants to cross pure tall plant (TT) with pure dwarf (tt) plant, what would be the Fj and F2 generations? Explain.
Answer:

  1. Pure tall plant has both the factors of the same type ‘TT’.
  2. Pure dwarf plant has both the factors of the same type ‘tt’.
  3. When a pure tall plant (TT) is crossed with pure dwarf plant (tt), all the offsprings in Fj generation are tall (Tt).
  4. So all the plants are heterozygous tall, as ‘T’ is the dominating factor.
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 1
  5. On self pollination of these F1 generation plants the new breed can have any combination of T and t like TT, Tt, Tt or tt.
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 2
  6. So in F2 generation 75% of plants are tall and 25% of plants are dwarf. Thus the phenotype ratio is 3 : 1.
  7. Among 75% of tall plants 25% are pure tall (TT) or homozygous tall, remaining 50% are heterozygous tall (Tt, tT).
  8. The remaining 25% dwarf plants are pure or homozygous dwarf (tt).
  9. So the genotype ratio is 1 : 2 : 1.

Question 3.
One experimenter cut the tails of parent rats, what could be the traits in offsprings? Do the daughter rats contain tails or not? Explain your argument.
Answer:

  1. If the tails of parent rats were cut, their offsprings will have normal tails.
  2. Daughter rats do not contain tails because the bodily changes are not inherited.
  3. So the change would not be passed to their offsprings.
  4. This was proved experimentally by Augustus Weisemann and rejected the theory ‘inheritance of acquired characters’ proposed by Lamarck.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 4.
In a mango garden a farmer saw one mango tree with full of mango fruits but with a lot of pests. He also saw another mango tree without pests but with few mangoes. But the farmer wants the mango tree with full of mango fruits and pests free. Is it possible to create new mango tree which the farmer wants? Can you explain how it is possible?
Answer:

  1. Yes, it is possible to create new mango tree which one the former wants with full of mango fruits and pests free.
  2. The former can cross two plants one with full of mangoes and pests and another plant with less mango fruits and without pests,
  3. In F1 generation he may get plants with full of mango fruits and without pests. Such plants are called hybrid plants.
  4. The F1 generation plants can be self pollinated and desired plants can be selected from the mixed population of F2 generation.
  5. The plant with desirable characters can be vegetatively propagated to get required number of plants.

Question 5.
ExplaIn monohybrid experiment with an example. Which law of inheritance can we understand? Explain.
Answer:

  1. We can understand the law of inheritance with an example of monohybrid cross between pure yellow pea seeds with pure green pea seeds.
  2. A pure breed (parental) yellow will have both the factors which denote them by ‘YY’ and pure breed (parental) green seed will have both the factors denote them by ‘yy’.
  3. During reproduction one factor (genes) from each parent is taken to form a new pair in the progeny (off spring).
  4. In F1 generation all pea plants are Yellow.
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 3
  5. F1 generation pea plants are self pollinated.
  6. In F2 generation 75% of the plants produced were Yellow seeds and the remaining 25% produced were green seed. It can be represented as
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 4
    In F2 generation the produced plants are YY, Yy; yY or yy.
    From the above example of monohybrid cross we can understand the following laws of inheritance.

1) When pure breed Yellow (YY) and green (yy) seeds were crossed, only Yellow seeds were expressed phenotypically in the F1 generation. It indicates that Yellow seed character is dominent over green seed characters.
This is “LAW OF DOMINANCE”.

2) When F1 plants are self pollinated each parent passes randomly selected allele (Y or y) of one of these factor to offsprings. This is segregation of alleles or genes during production of gametes.
This is “LAW OF SEGREGATION”.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 6.
What is the law of independent assortment? Explain with an example.
Answer:

  1. In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assorts independently of the other pairs. This is known as “Law of independent assortment”.
  2. If pea plants with two different pairs of characteristics (eg. : Round / yellow and green wrinkled) are breed with each other, the F1 progeny plants would have all round and yellow seeds.
  3. This implies that round and yellow seeds are dominant characters over green and wrinkled seeds.
  4. In F2 progeny there would be some plants with round and yellow seeds and some with green and wrinkled seeds.
  5. However, there would be some plants with mixed characters – yellow and wrinkled seeds and green and round seeds.
  6. This depicts that round/wrinkled trait and yellow / green trait are inherited independent of each other (law of independent assortment).
    The following punnet square explains this.
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 5
  7. The different combination of characters resulted from dihybrid cross.
    a) RRYY, RRYy, RrYy, RrYY, RRYy, RrYy, RrYy, RrYY, RrYy are having round and yellow seeds.
    b) RRyy, Rryy, Rryy have round and green seeds.
    c) rrYy, rrYy, rrYY have wrinkled and yellow seeds.
    d) rryy have wrinkled and green seeds.
  8. From the result, it can be concluded that the factors for each character or trait remains separate and maintain its identity in the gametes. This is known as “Law of independent assortment”.

Question 7.
How does sex determination take place in human?
(OR)
Explain sex determination in humans with the help of flow chart.
Answer:

  1. Each human cell contains 23 pairs (46) of chromosomes.
  2. Out of 23 pairs, 22 pairs of chromosomes are called autosomes.
  3. Remaining one pair is called allosomes or sex chromosomes.
  4. There are two types of sex chromosomes – one is ‘X’ and the other is ‘Y’.
  5. These two chromosomes determine the sex of an individual.
  6. Females have two ‘X’ chromosomes in their cells (XX).
  7. Males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 6
  8. All the gametes produced by women (ovum) will be with only X chromosomes.
  9. The gametes produced by man (sperm) will be of two types, one with X chromosomes and other with Y chromosomes.
  10. If the sperm carries X chromosome and fertilizes with the ovum, the resultant baby will have XX condition. So the baby will be a girl.
  11. If the sperm carries Y chromosome and fertilises with the ovum, the resultant baby will have XY condition. So the baby will be a boy.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 8.
Explain Darwin’s theory of ‘Natural selection’ with an example.
(OR)
What do you understand by the term Natural selection? Write Darwin’s theory of evolution.
Answer:

  1. Darwin proposed the theory of Natural selection.
  2. Nature only selects or decides which organism should survive or perish in nature.
  3. The organism with useful traits will survive and the organisms having harmful traits are going to perished or eliminated from its environment.
  4. For example, a group of twelve red beetles live in a bush of green leaves.
  5. They will grow their population by sexual reproduction.
  6. So they generate variations in their population. Let us assume crows eat the red beetles more the population of red beetles slowly reduced.
  7. Crows eat these red beetles and their population slowly reduces.
  8. During this time a colour variation arises by the sexual reproduction.
  9. So that there appears one beetle that is green in colour instead of red.
  10. Moreover this green colour beetle passes its colour to its offsprings; so that all its progeny are green.
  11. Crows cannot see the green coloured beetles on green leaves of the bushes and therefore crows cannot eat them.
  12. The crows can see the red beetles and eat them as a result, there are more and more green beetles than red ones which decrease in their number.
  13. The variation of green colour beetle gave a survival advantage to green beetles’ than red beetles. They were naturally selected.

Question 9.
What are variations? Explain with a suitable example.
Answer:

  1. Differences in characters within very closely related groups of organisms are referred to as variations.
  2. Often a new character in a group may lead to variations that are inherited.
  3. If we observe parents and offsprings, there will be some similar features in the offspring of the parents.
  4. At the same time we find differences between parents and offspring in their features.
  5. These differences are an example of variations.
  6. Variations are quite apparent among closely related groups of organisms.
  7. If we take roses as an another example, we observe number of varieties in them.
  8. But we can still find some characters similar to all plants.
  9. Thus rose plants have similar physical features, at the same time they have differences in characters like flower colour, number of petals, leaf size, stem, spines, etc.
  10. These differences in features are variations.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 10.
What variations generally have you observed in the species of cow?
Answer:
In the species of cow the following contrasting variations can be observed:

  1. White coloured – spotted
  2. Longhorns – short horns
  3. Height – dwarf
  4. Long-tail – short tail
  5. Elongated face – stunted face
  6. More milk giving – less milk giving, etc.

Question 11.
What are the characters that Mendel selected for his experiments on pea plants
(OR)
Write the seven pairs of contrasting characters in pea plant identified by Mendel and mention their traits.
Answer:
Mendel selected the following characters on pea plants for his experiment. They are:

CharacterDescription
1. Colour of the flower1. Purple or white.
2. Position of the flower2. Axial or terminal in position
3. Colour of the seed3. Either yellow or green.
4. Shape of the seed4. Either round or wrinkled.
5. Shape of the pod5. Inflated and constricted
6. Colour of the pod6. Yellow or green.
7. Length of the stem7. Tall and dwarf.

Question 12.
In what way Mendel used the word ‘Traits’? Explain with an example.
Answer:

  1. Trait is a separate variant of an organism.
  2. Mendel hypothesized that characters were carried as traits.
  3. An organism always carried a pair of factors for a character.
  4. He also hypothesized that distinguishing traits of the same character were present in the population of an organism.
  5. He assumed that the traits shown by the pea plants must be in the seeds that produce them.
  6. The seeds must have obtained these traits from the parent plants.
  7. The factors which are responsible for the character or trait of an organism, are now named as “genes”.
  8. By all these we can assume that Mendel used the word ‘traits’ for indicating the variant of an organism expressed by a pair of factors or genes.
  9. For example, height is a character of pea plant while the tallness is a trait expressed by a pair of factors either TT or Tt and dwarfness is another trait expressed by a pair of factors tt.

Question 13.
What are the differences that Mendel observed between parent and F2 generation?
Answer:
Mendel identified the following differences between parent and F2 generation.

ParentF2 Generation
1. They are pure breeds.1. They consist of mixed population.
2. They consist of homozygous alleles.2. They consist of homozygous alleles in some plants and heterozygous alleles in some other plants.
3. They have some fixed characteristic features.3. New combination of characters will appear.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 14.
Male is responsible for sex determination of baby – do you agree? If so write your answer with a flow chart.
Answer:

  1. Yes, I agree with the statement that male is responsible for sex determination of baby.
  2. There are two types of sex chromosomes in human beings, one is ‘X’ and other is ‘Y’.
    AP SSC 10th Class Biology Solutions Chapter 8 Heredity 7
  3. Females have two ‘X’ chromosomes in their cells (XX) whereas males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
  4. All the gametes produced by woman (ovum) will be with only X chromosomes.
  5. The gametes produced by man (sperm) will be of two types one with X chromosomes and other Y chromosomes.
  6. If the sperm carrying X chromosome fertilizes the ovum, the resultant baby will have XX condition. So the baby will be a girl.
  7. If the sperm carrying Y chromosome fertilizes the ovum, the resultant baby will have XY condition. So the baby will be a boy.
  8. So the gamete produced by the male is the deciding factor for sex determination of the baby.

Question 15.
Write a brief note on analogous organs.
(OR)
What are analogous organs?
Answer:

  1. The organs which are structurally different but functionally similar are known as ‘Analogous organs’.
  2. Wings of birds and bats is the example for analogous organs.
  3. The wings of bats are skin folds stretched mainly between elongated fingers.
  4. But the wings of birds are a feathery covering all along the arm.
  5. The designs of the two wings, their structure and components are different.
  6. They look similar because they have common use for flying, but their origins are not common.
  7. This makes the ‘analogous’ characteristics.
  8. This type of evolution is called convergent evolution.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 16.
How do scientists utilise information about fossils?
(OR)
“Fossils are valuable material that nature had preserved to know about ancient organisms.” Write the information you have collected about fossils.
Answer:

  1. Fossils are evidence of ancient life forms or ancient habitats which have been preserved by natural processes.
  2. The scientific study of fossils is called ‘Palaeontology’.
  3. Scientists utilise information about fossils to understand the evolutionary history of life.
  4. This information is also useful to study ecology and environmental history, such as ancient climates.
  5. This also helps to find out how old that certain layer of earth is.
  6. This information is also utilized as indicators of possible fossil fuel deposits which are of great interest to humanity.
  7. Thus scientists utilize the information on fossils to learn more about the earth’s past.

Question 17.
Mendel selected a pea plant for his experiments. Mention the reasons for the selection as these plants.
(OR)
Why did Mendel select the pea plant for his experiment? (OR)
Which characters in the pea plant are selected by Mendel, for his experiments?
What are the reasons for selecting pea plant by Mendel to conduct his experiments?
Answer:
Mendel chose the pea plant (Pisum sativam) for his breeding experiments for the following reasons.

  1. It is sexually reproducing.
  2. Flowers are bisexual.
  3. Predominantly self-pollinated.
  4. Predominantly self-fertilization.
  5. Well developed characters.
  6. Early hybridization.
  7. It is an annual plant.
  8. These plants have short maturity and can produce large number of seeds in a single generation.
  9. Pea plants have short life cycle.
  10. These plants can easy to grow either on the ground or in pots.

Question 18.
If the theory of inheritance of acquired characters proposed by Lamarck was true, how will the world be?
Answer:
If the theory of inheritance of acquired characters proposed by Lamarck was correct,

  1. All the organisms which lost some of their body parts should give birth to the offsprings without the lost parts.
  2. Rat which lost their tail should give birth to tail less rats.
  3. A handicapped who lost their legs in an accident should give birth to babies v without legs.
  4. A body builder’s children should be body builders.
  5. But all these are not happening because bodily changes won’t be passed to its offspring.

Question 19.
Collect information on the inherited traits in your family members and write a note on it.
Answer:

  1. My grandfather and father- had curling hair. I too have curling hair. So it’s an inherited trait in family.
  2. My mother and I both have long noses which appear similar. It’s another inherited trait.
  3. Eyes of my grandmother, my brother and mine are similar. It’s another inherited trait.
  4. Ear lobes of my father, brother and mine are similar. This is another inherited trait.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 20.
With the help of given information write your comment on evidences of evolution.

Mammals have fore limbs as do birds, reptiles and amphibians. The basic structure of the limbs is similar, though it has been modified to perform different functions.

Answer:

  1. The given information gives the evidences of evolution.
  2. Mammals. birds, reptiles and amphibians all these have forelimbs which have similar basic structure.
  3. But they are modified to perform different functions.
  4. This indicates that all the vertebrates have evolved from a common ancestor. These organs are called homologous organs. This type of evolution is called divergent evolution.
  5. In case of bat (mammal) and bird the designs of the two wings, their structure and components are different.
  6. They look similar because they have common use for flying, but their origins are not common.
  7. These organs which are structurally different but functionally similar are known as ‘Analogous organs’. This type of evolution is called “convergent evolution”.
  8. There are remarkable similarities in the embryos of above mentioned animals even in their limb formation. These are called embryological evidences.

Question 21.
Collect information about carbon dating method. Discuss with your physical science teacher.
(OR)
Write about the carbon dating method from the information collected by you.
Answer:

  1. Carbon dating is the method used to calculate the age of rocks, minerals or fossils.
  2. The breakdown of radioactive isotopes of certain elements such as carbon, uranium and potassium takes place at a known rate. So the age of rock or mineral containing isotopes can be calculated.
  3. Archaeologists use the exponential, radioactive decay of carbon 14 to estimate the death dates of organic material.
  4. The earth’s atmosphere contains various isotopes of carbon, roughly in constant proportions.
  5. These include the main stable isotope 12C and an unstable istope 14C.
  6. Through photosynthesis, plants absorb both forms from carbon dioxide in the atmosphere.
  7. When an organism dies, it contains the standard ratio of 14C to 12C.
  8. But as the 14C decays with no possibility of replenishment, the proportion of carbon 14 decreases at a known constant rate.
  9. The time taken for it to reduce by half is known as the half-life of 14C, which is 5730.
  10. The measurement of the remaining proportion of 14C in organic matter thus an estimation of its age.
  11. As the half life of carbon – 14 is 5,700 years, it is useful for dating objects up (o about 60,000 years old.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 22.
Draw a checker board, show the law of independent assortment with a flowchart and explain the ratio.
Answer:

AP SSC 10th Class Biology Solutions Chapter 8 Heredity 8

  1. The phenotypic ratio is 9 : 3 : 3 : 1. i.e., 9 round and yellow seeds 3 round and green seeds, 3 wrinkled and green seeds and 1 wrinkled and green seed.
  2. RRYY, RRYy, RrYy, RrYy, RRYy, RrYy, RrYy, RrYY and RrYy are round and yellow seeds.
  3. RRyy, Rryy, Rryy are round and green.
  4. rrYY, rrYy, rrYy are wrinkled and yellow.
  5. rryy are wrinkled and green.

From the above result, it can be concluded that factors for each character or trait remains separate and maintains its identity in the gametes. Thus in the inheritance of more than one pair of characters, the factors for each pair of characters assort independently of the other pairs. This is known as “Law of independent assortment”.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 23.
Explain the process to understand the monohybrid cross of Mendel experiment with a checker board.
Answer:
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 9

Question 24.
Prepare a chart showing the evolution of man through ages.
Answer:
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 10

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 25.
Nature selects only desirable characters. Prepare a cartoon.
Answer:
Nature selects only desirable characters
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 11

Question 26.
What is your understanding about survival of the fittest? Give some situations or examples that you observe in your surroundings.
Answer:

  1. Nature favours only useful variations.
  2. Each species tends to produce a large number of offspring.
  3. They compete with each other for food, space, mating and other species.
  4. In this struggle for existence only the fittest can survive.
  5. When cat tries to catch some rats, the rats which can run fast and hide in its hole can survive and which is slow can become prey for the cat.
  6. When we spray some insecticide on insects, most of them will die but few which can withstand that chemical will escape.
  7. When a pest attacks our garden plants, most of them may die but Which can withstand the pest can survive.
  8. When the dog tries to catch chickens, the chickens which will run fast and escape can survive but the slower ones will become food for the dog.

Question 27.
Write a monologue on the evolution of a human to perform a stage show on the theatre day in your school.
Answer:

  1. Hai, I am a human being. I am going to recall what had happened to me so far, how I had evolved, simply my journey from my origin to till now.
  2. Nearly 1.6 – 2.5 million years ago, during the gelasian pleistocene period, I used to wander in the forest. It is belived that, I evolved from apes.
  3. Between 1-1.8 million years ago, I gradually evolved into Homo erectus. I lived in this stage throughout most of the pleistocene. I used more diverse and sophisticated stone tools than my predecessors and it is belived that I travelled over oceans using rafts.
  4. Around 1,00,000 – 40,000 thousand years ago I evolved into Homo sapiens neanderthalensis. I was stronger than present in those days. I made advanced tools. I had language to communicate.
  5. Around 40 thousand years ago, I reached the present form of human being, the modern humans known as Homo sapiens. I learnt cultivation, construction of houses, cooking, etc. I had invented various things that help me to live comfortably.
  6. But my journey did not stop. It is still continuing. Let us see what may happen? Where can I reach? What changes may come in me? Hope for the best.
    Thank you.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Fill in the blanks.

  1. The process of acquiring change is called ———–.
  2. Mendel’s experiment explains about ———–.
  3. The four characters observed in the experiments on law of independent assortment are ———–.
  4. If we cross pollinate red flower plant with white flower we will get percent of ———– recessive trait plants.
  5. TT or YY, Tt or Yy are responsible for a ———– character.
  6. Female baby having 23 pairs of autosomes at the age of 18 years, has ———– pair autosomes and ———– of sex chromosomes.
  7. The population grows in ———– progression whereas food sources grow in ———– progression.
  8. A goat which walks properly can’t live for a long time. According to Darwin, this represents ———–.
  9. Forelimb of whale is for swimming whereas in horse it is used for ———–.
  10. The study of fossils is called ———–.

Answer:

  1. evolution
  2. heredity
  3. Round, wrinkled, yellow, green
  4. 100
  5. dominant
  6. 22, one pair
  7. geometrical, arithmetic
  8. survival of the fittest
  9. running
  10. palaeontology

Choose the correct answer.

  1. Which of the following is not a variation in rose plant?  [ ]
    A) Coloured petals
    B) Spines
    C) Tendrils
    D) Leaf margin
    Answer: C
  2. According to Mendel, alleles are  [ ]
    A) Pair of genes , Responsible for character
    B) gene
    C) Production of Gametes
    D) Recessive factors
    Answer: B
  3. Natural selection means  [ ]
    A) Nature selects desirable characters
    B) Nature rejects undesirable characters
    C) Nature reacts with an organism
    D) A, B
    Answer: A
  4. Palaeontologists deal with  [ ]
    A) Fossilised Embryological evidences
    B) Fossil evidences
    C) Fossilised Vestigial organ evidences
    D) All
    Answer: D

10th Class Biology 8th Lesson Heredity InText Questions and Answers

10th Class Biology Textbook Page No. 166

Question 1.
How does evolution take place?
Answer:

  1. Evolution takes place through the accumulation of new characters or variations in a species of organisms.
  2. Accumulation of variations occurs only when new characters are passed on from one generation to other and much more new characters are added to the pre-existing once.
  3. So this happens oVer a kirig period of time, sometimes several generations may pass.
  4. Hence it happens in a slow and steady manner.
  5. It is not just about change but producing something new and different.
  6. It is about the formation of new species and their adaptation to their environments.

10th Class Biology Textbook Page No. 168

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 2.
Is variation all about apparent differences? Or is it about some subtle differences as well that we most often overlook?
Answer:

  1. Variations are not always apparent differences.
  2. Sometimes these may be subtle differences that we most often overlook.
  3. When these subtle differences accumulate together they may become apparent.

10th Class Biology Textbook Page No. 171

Question 3.
How do parent plants pass on their traits to the seeds?
Answer:

  1. Every character or trait is controlled by a pair of factors called genes.
  2. At the time of sexual reproduction, one factor or each trait will pass to the gametes.
  3. By the fussion of male and female gametes zygote will form in which factors from both male and female parents get paired again.
  4. This zygote will develop into seed in the later stages.
  5. Thus parent plants pass on their traits to the seeds.

Question 4.
Will the seeds from tall plants always produce new tall plants?
Answer:

  1. No. Tall plants may or may not produce tall plants again.
  2. This is because tallness is a dominant character in most of the plants, especially in peas.
  3. So tall plant may be homozygous tall (TT) or heterozygous tall (Tt).
  4. If the parental plant is homozygous tall (pure breed), then they always produce new tall plants.
  5. If the parental plant is a heterozygous tall plant, then they produce the tall and dwarf plants in the ratio of 3 : 1.

10th Class Biology Textbook Page No. 175

Question 5.
What should be the percentage of each type of plants in F2 generation produced in dihybrid cross between pea plants with yellow, smooth seeds and green wrinkled seeds?
Answer:

  1. In F2 generation of dihybrid cross between pea plants with yellow, smooth seeds and green wrinkled seeds, new plants will produce with the following combination.
    i) Round and yellow
    ii) Round and green and
    iii) Wrinkled and green iv) Wrinkled and yellow
  2. They will produce in the ratio of 9 : 3 : 3 : 1 respectively.
  3. So 56 (56.25%) of plants should be with round and yellow seeds. 19 (18.75%) of plants should be with wrinkled and yellow seeds. 19 (18.75 %) of plants should be with round and green seeds and 6% (6. 25%) of plants should be with wrinkled and green seeds.

10th Class Biology Textbook Page No. 178

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 6.
What will happen if the sperm containing X chromosomes fertilizes the ovum?
Answer:

  1. If the sperm containing X chromosomes fertilizes the ovum which has X chromosome, the baby will have XX condition.
  2. So the baby will be a girl.

Question 7.
Who decides the sex of the baby – mother or father?
Answer:
Father decides the sex of the baby.

Question 8.
Is the sex also a character or trait? Does It follow Mendels’ law of dominance?
Answer:

  1. Yes, sex is also a character or a trait.
  2. It has two contrasting characters male and female.
  3. Male character is represented by a pair of allosomes ‘XY‘ (heterozygous).
  4. In this, we can consider Y as dominant and X as recessive.
  5. In this, recessive character is expressed only when it is homozygous recessive, i.e. female.
  6. Homozygous dominant is not existing as reproduction occurs between male (heterozygous dominent XY) and female (homozygous recessive XX) only.
  7. As X is not exhibiting its nature when Y is present along with it. it follows Mendel’s law of dominance.

Question 9.
Were all your traits similar to that of your parents?
Answer:

  1. No, all my traits are not similar to my parents.
  2. There are certain traits which differ from my parents.

10th Class Biology Textbook Page No. 185

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 10.
How does the evolution of organisms have taken place?
Answer:

  1. Variations which are beneficial are selected by nature and passed from parents to offspring through heredity.
  2. The same process happens with every new generation until the variation becomes common feature.
  3. As the environment changes, the organism within environment adopts and changes to the new living conditions.
  4. Over a long period of time, each species of organisms can acuumulate so many changes that it becomes a new species.
  5. Thus evolution of organisms took place from common pre-existing ancestors.

Question 11.
Are birds and bats more closely related to each other than to squirrels or lizards?
Answer:

  1. No, bats are mammals whereas birds belong to aves.
  2. Squirrels are mammals and lizards belong to reptiles.
  3. So bats and birds are not closely related to each other as they belong to two different groups.
  4. Both bats and birds have wings.

10th Class Biology Textbook Page No. 186

Question 12.
Do embryological evidences indicate that frogs have evolved from ancestors of fish?
Answer:
Yes the embryological evidences indicate that frogs have evolved from ancestors of fish.

Question 13.
Does the life history of every individual exhibit the structural features of their ancestors?
Answer:

  1. Yes. The life history of every individual exhibit the structural features of its ancestors.
  2. The resemblance is so close at an early stage.

10th Class Biology Textbook Page No. 189

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Question 14.
Think why did ancient human beings travelled from one place to other and how did they travelled?
Answer:

  1. Ancient human beings travelled from one place to other in search of better living conditions such as availability of food, water shelter and other facilities.
  2. They did not travel in a single line.
  3. They went forwards and backwards with groups, sometimes separating from each other.
  4. This travel is responsible for the formation of races.

10th Class Biology Textbook Page No. 183

Question 15.
In a forest there are two types of deer, in which one type of deer can run very fast. Whereas second type of deer can not run as fast as the first one. Lions, Tigers hunt deer for their food. Imagine which type of deer are going to survive in the forest and which type of deer population is going to be eliminated? And why?
Answer:

  1. Deer that can run fast can survive in the forest. Because they can escape easily from lions and tigers, when compared to second type.
  2. Deer that run slowly are going to be eliminated. Because they can be caught easily by its predators. So the survival chance will decrease.

10th Class Biology 8th Lesson Heredity Activities

Activity – 1

Think of your own family, what similarities do you share with your father and mother? Draw a table to represent the similarities of some characters like colour of eye (cornea), colour of hair, shape of nose, shape of face, type of earlobe (attached or free), inner thumb markings, etc. Write your characters in one column and that of your parents in the other columns.
Table – 1

CharactersIn meIn my Mother/FatherIn my Brother/SisterIn my grandma/grandpa

Answer:

S.No.CharactersIn meIn my Mother/FatherIn my Brother/SisterIn my grandma/grandpa
1.Colour of eyeBlackBlackBlackBlack
2.Colour of hairBlackBlackBlackBlack
3.Shape of noseLongLongShortShort
4.Shape of faceOvalRoundOvalOval
5.Type of earlobeFreeFreeFreeFree
6.Type of hairCurlingCurlingStraightStraight
7.Inner thumb markingConicalRoundRoundConical
8.Skin colourFairFairFairFair

1. Is there any character in you similar to that of your mother as well as your grandma?
Answer:
There are four characters in me similar to my mother as well as my grandma. They are

  1. Colour of eye
  2. Colour of hair
  3. Type of earlobe and
  4. Skin colour.

2. Is there any character in you similar only to that of your grandma?
Answer:
Two characters are similar in me and in my grandma. They are

  1. Shape of face and
  2. Inner thumb marking.

3. How do you think these characters may have been inherited by you from grandma?
Answer:
These characters are hereditary from parent to child.

4. Is there any character that is not present in grandma but present in your mother and you?
Answer:
Two characters are not present in grandma which are only present in me and my mother. They are

  1. Shape of nose
  2. Type of hair

5. Think where from your mother got that character?
Answer:
This character is the result of inherited traits transmitted from parent to progeny.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Activity – 2
Observe some of your friends and note their characters in the following table. Fill in yours as well.
Table – 2

Name of your friendColour of skinEarlobes Free / AttachedMarking on inner side of thumbLength of foreheadColour of eyes (Cornea)Any other features

Answer:

Name of your friendColour of skinEarlobes Free / AttachedMarking on inner side of thumbLength of foreheadColour of eyes (Cornea)Any other features
RaviBlackFreeRoundBroadBlueStraight hair, long nose, and face, etc.
GaneshBlackAttachedConicalNarrowBlackStraight hair, short nose, oval face, etc.
Vi jayFairFreeConicalBroadBlackCurling hair, short nose, round face, etc.
KarthikFairFreeRoundBroadBlackStraight hair, long nose, round face, etc.

1. Compare your characters to that of any one of your friend. How many characters did you find were similar among you and your friend?
Answer:
Only few characters such as black hair and black eye were similar among me and my friend.

2. Do you share more similar characters with your parents or with your friends?
Answer:
I share more similar characters with my parents than my friends.

3. Do you think that your differences from parents are same as differences from friends? Why / Why not?
Answer:
My differences from parents are not same as differences from friend. This is because the differences from parents are subtle as there is more genetic relation with parents but the differences from friends are apparent.

Activity – 3

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Observe seeds in a pea or bean pod. You may observe several parts to arrive at a generalisation.

1. Can you find two similar seeds there?
Answer:
No, all the seeds are not similar. They had certain variations.

2. What makes them vary? even though they are in the same pod. (Hint: You know that seeds are formed from ovules).
Answer:

  1. They vary from one another because they are produced from different ovules.
  2. Ovules of a plant are female gametes.
  3. These gametes carry different factors (genes) for different characters randomly.

3. Why variations are important? How are variations useful for an organism or a population?
Answer:

  1. Variations perhaps help a certain group of organisms in a community when conditions would otherwise be unfavourable for other groups.
  2. Desirable variations can be selected by nature.
  3. Desirable variations increase the chance of survival of an organism.
  4. Accumulation of variations after a long period leads to formation of new species.

Activity – 4

Let us do the following activity to understand the Mendelian principles of Heredity. Materials required :
a) 3 cm length and 1cm breadth chart pieces – 4
b) 2 cm length and 1cm breadth chart pieces – 4
c) Red buttons – 4
d) White buttons – 4
e) Chart, scale, sketch pen, pencil, 2 bags.

Method: Prepare a chart with 2×2 boxes along with number and symbol as shown in the figure.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity 12

Game 1: Monohybrid cross (starting with hybrid parents)
To start with take 1, 2 or 3, 4 . In case you start 1, 2 pick all the 16 long and short pieces and prepare such pairs in each of which you have a long and short piece.

Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4 short).One bag say ‘A’ represents male and the bag ‘B’ represents female. Now randomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with pairs of strips. You might have got the following combinations, two long strips, one long and one short strip, two short strips.

1. What is the number of long strip pairs?
Answer:
There are four long strip pairs.

2. What is the number of one long and one short pairs?
Answer:
There are eight, one long and one short strip pairs.

3. What is the number of short strips pairs?
Answer:
There are four short strip pairs.

4. What is the percentage of each type? Also find their ratios.
Answer:
The percentage of long strip pairs, one long and one short strip pairs and short strip pairs are 25%, 50% and 25% respectively and the ratio is 1 : 2 : 1.

5. What can you conclude from this game?
Answer:
From this game I have concluded that:

  1. Every individual possesses a pair of alleles, for any particular trait.
  2. Each parent passes a randomly selected copy (allele) of these to an offspring.
  3. The offspring then receives its own pair of alleles for that trait one each from both parents.
  4. If the long strip is considered as dominant 75% exhibit dominant and 25% exhibit recessive character. Thus the phenotype ratio is 3 : 1 in monohybrid cross.
  5. The genotype ratio is 1 : 2 : 1.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Activity – 5

Observe the below diagram showing variation in beetle population and its impact.
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 13Let us consider a group of twelve beetles. They live in bushes on green leaves. Their population will grow by sexual reproduction. So they were able to generate variations in population. Let us assume crows eat these red beetles. If the crows eat more Red beetles, their population is slowly reduced. Let us discuss the above three different situations in detail.
Answer:
Situation-1: In this situation, a colour variation arises during reproduction. So that there appears one beetle that is green in colour instead of red.
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 14Moreover, this green coloured beetle passes its colour to Its offspring (Progeny). So that all its progeny are green. Crows cannot see the green coloured beetles on green leaves of the bushes and therefore crows cannot eat them. But crows can see the red beetles and eat them. As a result there are more and more green beetles than red ones which decrease in their number.

The variation of colour in beetle ‘green’ gave a survival advantage to’green beetles’ than red beetles. In other words it was naturally selected. We can see that the ‘natural selection’ was exerted by the crows. The more crows there are, the more red beetles would be eaten and the more number of green beetles in the population would be. Thus the natural selection is directing evolution in the beetle population. It results in adaptation in the beetle population to fit in their environment better.

Let us think of another situation.
Situation-2: In this situation a colour variation occurs again in its progeny during reproduction, but now it results in ‘blue’ colour beetles instead of ‘red’colour beetle. This blue colour beetled can pass its colour to its progeny. So that all its progeny are blue.
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 15Crows can see blue coloured beetles on the green leaves of the bushes and the red ones as well. And therefore crows can eat both red and blue coloured beetles. In this case there is no survival advantage for blue coloured beetles as we have seen in case of green coloured beetles.

What happens initially in the population, there are a few blue beetles,but most are red. Imagine at this point an elephant comes by and stamps on the bushes where the beetles live. This kills most of the beetles. By chance the few beetles survived are mostly blue. Again the beetle population slowly increases. But in the beetle population most of them are in blue colour.

Thus sometimes accidents may also result in changes in certain characters of the population. Characters as we know are governed by genes. Thus there is change in the frequency of genes in small populations. This is known as “Genetic drift’, which provides diversity in the population.

Let us think of another situation :
Situation-3: In this case beetles population is increasing, but suddenly bushes were affected by a plant disease in which leaf material were destroyed or in which leaves are affected by this beetles got less food material.
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 16So beetles are poorly nourished. So the weight of beetles decrease but no changes take place in their genetic material (DNA). After a few years the plant disease are eliminated. Bushes are healthy with plenty of leaves.

What do you think will be condition of the beetles?
Answer:
The weight of beetles will increase once again as they get plenty of food material again.

AP SSC 10th Class Biology Solutions Chapter 8 Heredity

Activity – 6

Let us observe different stages of development of vertebrate embryos. Try to find out similarities and differences and discuss with your friends.
(OR)
What do you infer about the embroyological evidences of various organisms?
Answer:
AP SSC 10th Class Biology Solutions Chapter 8 Heredity 17

  1. There are remarkable similarities in the embryos of different animals from fish to man.
  2. The resemblance is so close at an early stage.
  3. Gradually the similarities are decreased when they become babies.
  4. The embryological evidences give us an idea that all the organisms have evolved from a common ancestors.

 

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

AP State Board Syllabus AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments.

AP State Syllabus SSC 10th Class Social Studies Important Questions 22nd Lesson Citizens and the Governments

10th Class Social 22nd Lesson Citizens and the Governments 1 Mark Important Questions and Answers

Question 1.
Write any two benefits of Lok Adalat in respect of entertaining disputes.
Answer:

  1. There is no court fee.
  2. Speedy trail of disputes
  3. Procedural flexibility
  4. The parties can directly interact with the judge.

Question 2.
When was the Right to Information Act passed by the Central Government?
Answer:
Right to Information Act was passed in 2005 by the Central Government.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 3.
How does the government department function?
Answer:
Every government department functions on the basis of certain rules and regulations.

Question 4.
What does Lok Adalat mean?
Answer:
Lok Adalat means people’s court.

Question 5.
Which two roles does RTI play?
Answer:
The first one is the role of the government departments and the second is that of the citizens.

Question 6.
Write about NALSA.
Answer:
Act and to frame the most effective and economical schemes for legal services.

Question 7.
What legal aid facilities are available under Legal Service Authority?
Answer:
Matrimonial disputes, maintenance cases, harassment cases by husband and in-laws, domestic violence cases, all types of civil cases, land disputes, compoundable criminal cases, etc.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 8.
When are the Lok Adalats constituted?
Answer:
The Lok Adalats are constituted in every state under the Legal Services Authority Act, 1987.

Question 9.
Write about District Legal Services Authority.
Answer:
District Legal Services Authority is constituted in every district to implement Legal Aid Programmes and Schemes in the district. The District Judge of the district is its Ex-Officio Chairman and the members are nominated by State Government.

Question 10.
Who is responsible officials in the information commission?
Answer:
National Legal Services Authority (NALSA) is the apex body constituted to lay down policies and principles for making legal services available under the provisions of the State Public Information Commissioner is the responsible official in the information commission along with Information Commissioners.

10th Class Social 22nd Lesson Citizens and the Governments 2 Marks Important Questions and Answers

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 1.
Read the following text and answer the following question.

The LSA Act was enacted to constitute legal services authority for providing free and competent legal services to the weaker sections of the society. 
To ensure that opportunities for securing justice were not denied to any citizen by reason of economic or other disabilities.
To organise Lok Adalats to ensure that the operation of the legal system promoted justice on Q basis of equal opportunity.
To develop an effective, alternative, innovative dispute resolution mechanism outside the courts.

Who are entitled to get free legal support?
(OR)
Who are eligible for benefit under Legal Service Authority?
Answer:
As per LSA Act and its objectives to give free legal support on an equality basis to economically backward. The following are entitled or eligible to get free legal support.

  1. A member of a SC or ST.
  2. A victim of trafficking in human beings or beggars.
  3. Women and children.
  4. A mentally ill or otherwise disabled person.
  5. A person who is a victim of a mass disaster, ethnic violence, etc.

Question 2.
Write about the Right to Information Act.
Answer:

  1. Right to Information Act was passed by the Central Government in 2005.
  2. The Act was passed as a result of people’s mobilization as well as in recognition of the provisions in the constitution.
  3. Two roles that need to be played for getting the benefits of RTI to the people.

Question 3.
Explain how RTI Act enriches Democracy.
Answer:

  1. It ensures the availability of all types of information to the citizen.
  2. Democracy requires an informed citizenry.
  3. Information should be transparent.
  4. This can help to control corruption and make the governments accountable.
    Thus RTI act enriches democracy.

Question 4.
What are the provisions there for people to seek information under RTI?
Answer:

  1. Any individual can get documents like government orders, reports, advice, logbooks, rules and regulations, attendance list, letters, etc.
  2. People seeking information have to pay a small amount to meet a copy of the document.
  3. If the person is below the poverty line he/she need not pay this amount.
  4. This information request can be sent through a handwritten letter or electronic mail.
  5. Information will be given in the official language of the state or in English or in Hindi.
  6. Individuals collecting the information can remain anonymous.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 5.
Read the following case study and give an explanation of how Swetha had benefited from RTI.

Gattu Swetha was a student of M.Sc at Kakatiya University in 2013. She completed her B.Sc. exams in March 2011. She got good marks in all subjects but in Chemistry she got 21 marks. On revaluation, she got only 9 marks. With RTI activists help she requested for a photocopy of her answer sheet. University refused to do so. She went to the State Chief Information Commissioner and later the university gave a photocopy and awarded 51 marks. Thus, she joined M.Sc.

Answer:
Swetha could not have accessed the photocopy of her answer sheet without the RTI enabling Act. She could not have got an entry into M.Sc. Earlier photostats of answer sheets were not given to students. They have only informed marks. Now students like Swetha can get a photocopy of their answer sheets even for competitive examinations that provide jobs. She did not stop even when the Public Information Officer of the University did not provide her with information. She went to the highest appellate authority in the state and was able to secure a photocopy. Thus, she made the best use of RTI enabling Act.

Question 6.
Read the following text and answer the following question.

The LSA through Lok Adalats settles long pending court cases in short time and without any expenses. Matrimonial disputes, maintenance cases, harassment cases by husband and in-laws, domestic violence cases, all types of civil cases, land disputes, compoundable criminal cases etc.

What kind of cases can be settled through Lok adalats?
Answer:
The following cases can be settled through Lok adalats.

  1. Matrimonial disputes.
  2. Maintenance cases.
  3. Harassment cases by husband and in-laws.
  4. Domestic violence cases.
  5. Long pending all civil cases, land disputes and compoundable criminal cases.

Question 7.
What exemptions are there for disclosure of information in the RTI Act 2005?
Answer:

  1. The law permits the government not to disclose certain information.
  2. Information that could affect the sovereignty and integrity of India.
  3. Information that could breach the privilege of Parliament and State Legislature.
  4. Information received in confidence from a foreign government. “
  5. Information that could endanger the life or physical safety of a person.
  6. Cabinet papers or records of Council of Ministers and Secretaries (before the final decision is taken)
  7. Most of our armed forces and security agencies are outside the purview of Information Commissions.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 8.
Provide an example where any citizen benefited from using RTI Act.
Answer:
T.AIIuga has 1 acre and 20 kuntas land in Vadali village of Mudinepalli mandal in Krishna district. After his death, his son Nallanna cultivated that land. But the untimely death of him left his kid unaware of land. Few years later his kid, Naresh recognized that their land was encroached. He requested the Tahsildar to survey his land with documents he had, which was of no use. Later, even he complained the same in grievance cell with the collector, but was of no use. With the suggestion of a representative of RTI Act, he applied to Information Officer. He has asked for the reason why his land was not surveyed. Later his land was identified, surveyed and shown to the applicant, Naresh.

Question 9.
When was RTI Act made? What was its constitutional prominence?
Answer:

  1. The Right to Information Act (RTI) was passed by the Parliament in 2005.
  2. Today the Right to Information is recognized within the Constitution under two Fundamental Rights as freedom of expression and the right to life.

Question 10.
Why do you think checking the information can help In improving accountability?
Answer:

  1. Information can be collected with the help of officials.
  2. Master rolls and paid amounts can be verified.
  3. This would bring out the corruption in practices.
  4. Officials are given the opportunity to defend and talk about the details of documents.
  5. When corruption was identified, cases were registered against concerned persons.
  6. Thus checking the information will make governments more accountable in their functioning.

Question 11.
What information shall be declared by every government office even without being asked by the people?
Answer:

  1. Under the RTI, it is also compulsory for every government office to declare certain information in public even without being asked by the people.
  2. One can identify them on the walls of these office buildings.
  3. They are the names, designations and phone numbers of Public Information Officer, Assistant Public Information Officer and First Appellate Authority.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments

Question 12.
What are the public authorities to follow the right to Information Act?
Answer:

  1. The institution created under the constitution.
  2. Made under the laws of Parliament or State legislature.
  3. Created through a notification of the appropriate government.
  4. It may include any body owned, controlled or substantially funded or Non-Governmental organizations substantially funded by the government either directly or indirectly through the funds.

10th Class Social 22nd Lesson Citizens and the Governments 4 Marks Important Questions and Answers

Question 1.
How are the Lok-adalats helping the common people? Explain.
Answer:

  1. There is no court fee.
  2. The procedural flexibility and speedy trial of disputes are provided.
  3. The parties to the dispute can directly interact with the judge.
  4. Free Legal advice is provided.
  5. Delay will be avoided in the settlement of disputes.

Question 2.
“Right to Information Act resembles the true spirit of Democracy”. Explain.
Answer:

  1. Democracy requires an informed citizenry.
  2. It needs transparency of information.
  3. Through R.T.I. Act, all the citizens can get such information.
  4. This Act can help to contain corruption.
  5. This also makes governments accountable to individual citizens also.
  6. Earlier various departments of the government responded only to the elected representatives.
  7. But today, they respond even to the common citizens.

Question 3.
Why is there a need for information to fight corruption?
Answer:

  1. Programmes that are undertaken to benefit the poor and remove poverty often do not reach the needy and the funds get diverted.
  2. A major reason for this is corruption.
  3. The reason is common people do not have proper information about the programmes and how they are being implemented.
  4. The people themselves had no way to verify how the money was spent by government.
  5. The information was not shared with the people.
  6. Yet, in a democracy, the money used for the welfare of people is theirs, so they have every right to know how this is being used.
  7. Earlier elected representatives checked the corruption.
  8. Now with the enactment of RTI, people can check the corruption.

Question 4.
Why do you think the information needed to be shared?
Answer:

  1. People believe that the information was crucial to their own welfare.
  2. Information is crucial to the human development and democratic rights.
  3. People can participate in government and ensure just development only if they have sufficient information in the form of official documents.
  4. Information will make governments more accountable in their functioning.
  5. Then only it is possible to monitor the functioning and check possibility of corrupt practice.
  6. In a situation where information has to be made public, arbitrary decisions by the elected representatives or the officials can be controlled.
  7. Information is crucial to the survival of the poor.

Question 5.
Collect the information of the movement which began for the information.
Answer:

  1. A group of people in Rajasthan organized themselves under the banner of Mazdoor Kisan Shakti Sanghatan (MKSS) and demanded information.
  2. There was no legal provision to take information from the government to the people.
  3. Initially, these documents were collected with the help of officials.
  4. People evaluated these documents through public meetings.
  5. Soon officials began to resist the idea of disclosing the information,
  6. This led to people’s movement, with rallies and marches for the next 3 years.
  7. People demanded that the information was crucial to their own welfare.

Question 6.
What exemptions are there for disclosure of information in RTI Act, 2005?
Answer:

  1. The law permits the government not to disclose certain information.
  2. Information that could affect the sovereignty and integrity of India.
  3. Information that could breach the privilege of Parliament and State Legislature.
  4. Information received in confidence from a foreign government.
  5. Information that could endanger the life or physical safety of a person.
  6. Cabinet papers or records of Council of Ministers and Secretaries (before the final decision is taken)
  7. Most of our armed forces and security agencies are outside the purview of Information Commissions.

AP SSC 10th Class Social Studies Important Questions Chapter 22 Citizens and the Governments