Mahesh

Practice the AP 10th Class Maths Bits with Answers Chapter 6 Progressions on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 6th Lesson Progressions with Answers

Question 1.
Which term of Answer:P., 18, 15, 12, ………….. equal to ‘0’ ?
Answer:
7
Explanation :
a = 18, d = 15 – 18 = -3
a_n = 0 ⇒ a + (n – 1)d = 0
18 + (n-1)(-3) = 0
(n-1)(-3) = -18
n – 1 = 6 ⇒ n = 7

Question 2.
Find the 21st term of an Answer:P., whose first two terms are – 3 and 4.
Answer:
137
Explanation :
a₁ = -3, a₂ = 4
⇒ d = 4 – (-3) = 7
a₂₁ = a + 20d
= (-3) + 20(7)
= -3 + 140 = 137

Question 3.
Which term of G.P., 3, 3\(\sqrt{3}\) , 9,
equals to 243 ?
Answer:
9
Explanation :

Question 4.
If a, b, c are in G.P., then find b’.
Answer:
\(\sqrt{\mathrm{ac}}\)

Question 5.
Find the sum of 10 terms of the progression log 2 + log 4 + log 8 + log 16 + ………………..
Answer:
55 log 2

Question 6.
Find nth term of a progression a, ar, ar² ……………
Answer:
ar^{n – 1}

Question 7.
Find the sum of first 100 natural num-bers.
Answer:
5050

Question 8.
Find the common difference of Answer:P. log₂ 2, log₂ 4, log₂ 8.
Answer:
1
Explanation :
log₂2,log₂2² , log₂2³
log₂2, 2 log₂2, 3 log₂2
1, 2, 3, ….
⇒ d = 1

Question 9.
In a GP, a₁ = 20 and a₄ = 540, then ‘ find ‘r’.
Answer:
3
Explanation :
a = 20, a₄ = a – r³ = 540
⇒ 20.r³ = 540
⇒ r³ = \(\frac{540}{20}\) = 27 ⇒ r³ = 3³ ⇒ r = 3

Question 10.
In an Answer:P., if a = 1, a_n = 20 and S_n = 399, then find ‘n’.
Answer:
38
Explanation :

Question 11.
Find the common difference of the AP x – y, x, x + y.
Answer:
y

Question 12.
Find the common difference of the AP 1,-1, -3.
Answer:
– 2

Question 13.
If k, 2k + 1, 2k + 3 are three consecutive terms in Answer:P., then find the value of’k’.
Answer:
1
Explanation :
k + 1 – k = 2k + 3 – (2k + 1)
⇒ k + 1 = 2 ⇒ k = 1

Question 14.
Find the common difference in the AP 2a – b, 4a – 3b, 6a – 5b.
Answer:
2a – 2b.

Question 15.
Which term of the arithmetic progres-sion 24,21,18, is the first negative term ?
Answer:
10th term
Explanation :
a = 24, d = 21-24 = -3
a_n = a + (n – 1)d = 0
⇒ 24 + (n- 1) (-3) = 0
⇒ 24 – 3n + 3 = 0
⇒ 27 – 3n = 0
⇒ n = 9
∴ First negative term is ’10’.

Question 16.
Find the next term in Answer:P. \(\sqrt{3}, \sqrt{12}, \sqrt{27}\)
Answer:
\(\sqrt{48}\)

Question 17.
Find the common difference of an arithmetic progression in which
a₂₅ – a₁₂ = -52
Answer:
-4
Explanation :
a + 24d – (a + 11d) = – 52
a_n + 24d – a – 11d = – 52
⇒ 13d = – 52 ⇒ d = \(-\frac{52}{13}\) = – 4.

Question 18.
Find the sum of first ‘n’ odd natural numbers.
Answer:
n²

Question 19.
Find the common difference of an Answer:P. for which a₁₈ – a₁₄ = 32.
Answer:
8

Question 20.
Write a formula for sum of first ‘n’ terms in an AP.
Answer:
S_n = \(\frac{n}{2}\) [2a + (n – 1 )d] (or)
S_n = \(\frac{n}{2}\)[a + l]

Question 21.
Which term of the G.P. \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \ldots\) is \(\frac{1}{2187}\) ?
Answer:
7th

Question 22.
If a_n = \(\frac{\mathbf{n}}{\mathbf{n}+\mathbf{1}}\) then find a₂₀₁₇
Answer:
\(\frac{2017}{2018}\)

Question 23.
In an arithmetic progression, 4th term is 11 and 7th term is 17, then find its common difference.
Answer:
2
Explanation :
a₄ = a + 3d = 11 and a₇ = a + 6d = 17

Question 24.
If x, x + 2, x + 6 are three consecutive terms in G.P. Find the value of’x’.
Answer:
2

Question 25.
The ’n^th’ term of an Answer:P. is a_n = 3 + 2n, then find the common difference.
Answer:
2

Question 26.
If a_n = \(\frac{n(n+3)}{n+2}\)then find a₁₇.
Answer:
\(\frac{340}{19}\)

Question 27.
In an AP a_n = \(\frac{5 n-3}{4}\), then find a₇.
Answer:
8
Explanation :
a₇ = \(\frac{5 \times 7-3}{4}=\frac{32}{4}\) = 8

Question 28.
Find the common difference of an arithmetic progression, whose 3rd term is 5 and 7^{th</sup? term is 9.
Answer:
1}

Question 29.
If 4, a, 9 are in G.P., then find ‘a’,
Answer:
±6

Question 30.
If \(-\frac{2}{7}\) x , \(-\frac{7}{2}\) are in Geometric Progression, then find the value of x.
Answer:
1

Question 31.
If (i) – 1.0, – 1.5, – 2.0, – 2.5,…… and
(ii)- 1,-3, -9, -27, ………
ate two progressions, then which of them is a geometric progression ?
A) (i) only
B) (ii) only
C) (i) and (ii) both
D) None of them
Answer:
B) (ii) only

Question 32.
In a G.P., a = 81, r = \(-\frac{1}{3}\), then find a₃.
Answer:
9
Explanation :
a₃ = Answer:r² = 81. (\(\left(\frac{-1}{3}\right)^{2}\)) ⇒ a₃ = 9

Question 33.
Write a G.P. with r = 2 and a = 7.
Answer:
7, 14, 28 ………………..

Question 34.
If a, b, c are in AP, then find ’b’.
Answer:
\(\frac{a+c}{2}\) = b.

Question 35.
3, \(\frac{3}{2}\) , \(\frac{3}{4}\), …………. then find ‘r’.
Answer:
r = \(\frac{t_{2}}{t_{1}}=\frac{\frac{3}{2}}{3}=\frac{3}{2} \times \frac{1}{3}=\frac{1}{2}\)

Question 36.
Find the sum of first 1000 positive integers.
Answer:
500500

Question 37.
In the AP – 9, – 14, – 19, – 24, …………….. then find the value of a₃₀ – a₂₀.
Answer:
-50
Explanation :
a = – 9, d = -14 + 9 = -5
a₃₀ = a + 29d
= – 9 + 29 (- 5)
= -9-145 = -154
a₂₀ = a + 19d
= -9 + 19 (-5)
= -9-95 = -104
∴ a₃₀ – a₂₀ = – 154 + 104 = – 50

Question 38.
If 4, x, 9 are in G.P., then find ‘x’.
Answer:
6

Question 39.
Find 15^th term of the AP x – 7, x – 2, x + 3, …………………….
Answer:
x + 63

Question 40.
\(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\) find a₇.
Answer:
\(\frac{1}{2187}\)

Question 41.
Write a formula 1² + 2² + 3² + … + n².
Answer:
Σn² = \(\frac{n(n+1)(2 n+1)}{6}\)

Question 42.
1 + \(\frac{1}{2}+\frac{1}{2^{2}}\) + ………….. then find ‘r’.
Answer:
\(\frac { 1 }{ 2 }\)

Question 43.
Find the common ratio of the G.P. 2, \(\sqrt{8}\), 4.
Answer:
\(\sqrt{2}\)

Question 44.
1,4,7,10, ……………… find ‘d’.
Answer:
3

Question 45.
If a_n = \(\frac{\mathbf{n}}{\mathbf{n}+\mathbf{2}}\), then find a₃.
Answer:
\(\frac{3}{5}\)

Question 46.
Write a_n in G.P.
Answer:
ar^n-1 = a_n

Question 47.
2, \(\frac{5}{2}\), 3, find S₂₅.
Answer:
S₂₅ = 200
Explanation :
a = 2, d = \(\frac{5}{2}\) – 2 = \(\frac{1}{2}\)
S₂₅ = \(\frac{n}{2}\)[2a + (n- 1) d]
= \(\frac{25}{2}\) [2(2) + 24 (1/2)]
= \(\frac{25}{2}\)[4 + 12]
= \(\frac{25}{2}\) x 16 = 25 x 8 = 200

Question 48.
Write G.M of a and b.
Answer:
\(\sqrt{\mathrm{ab}}\) = G.M.

Question 49.
In an Answer:P. a₁ = -4, a₆ = 6, then find a₂
Answer:
-2.

Question 50.
If a, b, c are in Answer:P., then find ‘b’.
Answer:
\(\frac{a+c}{2}\) = b.

Question 51.
Which term of the G.P. 2, 6, 18, 54,…. is 2 x 3¹⁰ ?
Answer:
11^th term
Explanation :
a = 2, r = 3, a_n = 2 x 3¹⁰
Answer:r^n-1 = 2x 3¹⁰
2 x 3^n-1 = 2 x 3¹⁰
⇒ n – 1 = 10 ⇒ n = 11

Question 52.
Find the sum of 15 terms of the Answer:P. 4, 7, 10, ………………
Answer:
375

Question 53.
Which term of the Answer:P. 100, 90, 80, …………… is zero ?
Answer:
11th term
Explanation :
a = 100, d = 90 – 100 = -10
a_n = a + (n – 1)d = 0
⇒ 100 + (n- 1)(- 10) = 0
⇒ (n – 1) (- 10) = – 100
⇒ n- 1 = 10
⇒ n = 11

Question 54.
Is the numbers, – 15, -11,-7, – 3, are in Answer:P. ? If so, find’d’.
Answer:
Yes, it is in Answer:P., with d = 4.

Question 55.
\(\frac{1}{\sqrt{2}},-2, \frac{8}{\sqrt{2}}\) find a₅
Answer:
\(32 \sqrt{2}\)

Question 56.
Find AM of 24 and 16.
Answer:
20

Question 57.
In the Answer:P. – 11,-9, – 7, find d’.
Answer:
2

Question 58.
Which term of Answer:P. 21, 18, 15, …………… is -81 ?
Answer:
35th term

Question 59.
Write a G.P. your own with r = – 2.
Answer:
5,- 10, 20,-40, ……………

Question 60.
Find the sum of first ’n’ natural num-bers.
Answer:
1 + 2 + 3 + 4 ……………+ n = Σ \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Question 61.
In AP a₁₂ = 37, d = 3, then find S₁₂.
Answer:
246
Explanation :
a₁₂ = 37, d = 3
⇒ a + 11d = 37
⇒ a + 11 x 3 = 37
⇒ a = 37 – 33 = 4
S₁₂ = \(\frac { 12 }{ 2 }\)[2 x 4 + 11 x 3]
= 6[8 + 33] = 6 x 41 = 246

Question 62.
If 3, x, 11 are in Answer:P., then find ‘x’.
Answer:
7 = x
Explanation :
x – 3 = 11- x ⇒ 2x = 14 ⇒ x = 7

Question 63.
In an AP 7a₇ = 11a₁₁, then find a₁₈.
Answer:
0 = a₁₈

Question 64.
Find number of terms of the Answer:P.
-5 + (-8) + (- 11) + + (-230).
Answer:
76
Explanation :
a = -5, d = -8 + 5 = -3, a_n ⇒ a + (n – 1) d = – 230
⇒ – 5 + (n – 1) (- 3) = – 230
⇒ (n – 1)(-3) = -225
⇒ n – 1 = \(\frac{-225}{-3}\) = 75
⇒ n = 75 + 1 ⇒ n = 76

Question 65.
– 8, – 6, – 4, find a₇.
Answer:
6 = a₇.

Question 66.
1, – 1, 1, – 1, 1, – 1, ………………upto 131 terms, then find S₁₃₁.
Answer:
1 = S₁₃₁.

Question 67.
ind the next term of the Answer:P. 51, 59, 67,75.
Answer:
83

Question 68.
a_n = 9 – 5n, find a₄.
Answer:
-11
Explanation :
Explanation :
a_n = 9 – 5n
⇒ a₄ = 9- 5×4 = 9-20 = -11

Question 69.
3, 6, 12, then find ‘r’.
Answer:
r = \(\frac{6}{3}=\frac{12}{6}\) = 2

Question 70.
Which term of Answer:P. 7 + 4 + 1 + is – 56 ?
Answer:
22

Question 71.
n – 1, n – 2, n – 3, ….. find a10.
Answer:
n – 10 = a₁₀

Question 72.
Write Answer:M. of M, P, C.
Answer:
\(\frac{M+P+C}{3}\)

Question 73.
In the formula a_n = 3.6, a = – 18.9, d = 2.5, then find ‘n’.
Answer:
10
Explanation :
a_n = 3.6, a = – 18.9, d = 2.5
⇒ a + (n – 1)d = 3.6
⇒ – 18.9 + (n – 1)(2.5) = 3.6
⇒ (n – 1)(2.5) = 3.6 + 18.9 = 22.5
⇒ n-1 = \(\frac{22.5}{2.5}\) = 9
⇒ n = 9 + 1 = 10

Question 74.
Write G.M. of a’ arid \(\frac{1}{a}\).
Answer:
G.M. = 1

Question 75.
In a series a_n = \(\frac{n(n+1)}{3}\), find a₂.
Answer:
a₂ = 2.

Question 76.
If a, b, c are in Answer:P., then find b – Answer:
Answer:
c – b

Question 77.
\(\frac{1}{4}, \frac{-1}{4}, \frac{-3}{4}, \frac{-5}{4}\) ………….. find ‘d’
Answer:
d = \(\frac{-1}{2}\)

Question 78.
Find 1 + 1 + 1 + + n terms.
Answer:
S_n = n

Question 79.
Find G.M. of x³ and \(\frac{1}{x^{3}}\)
Answer:
G.M. = 1

Question 80.
Write Answer:M. of x² + y² and x² – y².
Answer:
x²
Explanation :
Answer:M = \(\frac{x^{2}+y^{2}+x^{2}-y^{2}}{2}=\frac{2 x^{2}}{2}\) = x²

Question 81.
a, a², a³, …… then find r.
Answer:
r = a .

Question 82.
Reciprocals of terms of G.P. are in which progression ?
Answer:
G.P.

Question 83.
2, – 6, 18, – 54,………….find r.
Answer:
-3.

Question 84.
Find the value of – 5 + (- 8) + (- 11) + ………….. + (-230).
Answer:
-8930

Question 85.
In a G.P. 25, – 5,1, \(-\frac{1}{5}\),…. then find ‘r’.
Answer:
\(-\frac{1}{5}\) = r

Question 86.
If 2, x, 6 are in G.P., then find ‘x’.
Answer:
\(2 \sqrt{3}\) = x.

Question 87.
In a G.P. a₈ = 192, r = 2, then find a₁₂.
Answer:
3072 = a₁₂.

Question 88.
Which term of G.P., 2,8,32, is 512?
Answer:
5

Question 89.
1, 2, 3, ……….. find sum to ’10’ terms.
Answer:
S₁₀ = 55
Explanation :
a = 1,d = 1
S₁₀ = \(\frac{10}{2}\)[21 + 9 x 1]
= 5[2 + 9] = 5 x 11 = 55

Question 90.
\(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}\), …………. find a_n.
Answer:
\(\frac{5}{2^{n}}\) = a_n

Question 91.
4,16, , 256, ……….. then find
A,
64

Question 92.
In the Answer:P. 100, 103, 106, find d’.
Answer:
d = 3

Question 93.
Find the Answer:P. with first term is 8 and common difference is 2\(\frac { 1 }{ 2 }\).
Answer:
8, 10\(\frac { 1 }{ 2 }\), 13, …………….

Question 94.
How many terms of Answer:P. – 6, \(\frac { -11 }{ 2 }\), – 5, ……………. are needed to obtain a sum – 25 ?
Answer:
5 or 20 terms.
Explanation :
a = -6,d = \(\frac{-11}{2}\) + 6 = \(\frac{1}{2}\) S_n = -25
S_n = \(\frac{n}{2}\)[2a + (n – 1)d] = -25

⇒ n (n – 25) – 100
⇒ n2 – 25n + 100 = 0
⇒ n2 – 20n – 5n + 100 = 0
⇒ n (n – 20) – 5 (n – 20) = 0
⇒ (n – 5) (n – 20) = 0
∴ n = 5 or 20

Question 95.
(a + 3d), (a + d), (a – d), ……….. find the next term of the Answer:P.
Answer:
a – 3d

Question 96.
Find the 103^rd term of 1, -1,1,- 1, ….
Answer:
– 1

Question 97.
Find the sum of first 50 even numbers.
Answer:
2550
Explanation :
Sum of first 50 even numbers
= n(n + 1)
= 50(51) = 2550

Question 98.
Find the common ratio of the G.P.192, 36, 9, …………
Answer:
1/4

Question 99.
Find the 25^th term of
– 300, – 290, – 280,
Answer:
-60.

Question 100.
a, b, c are in AP, then write, \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in which progression ?
Answer:
Harmonic progression (H.P)

Question 101.
In Answer:P. a_p = q, a_q = p, then find a_{p + q}.
Answer:
0 = a_{p + q}

Question 102.
a, b, c are in Answer:P., then 3^a, 3^b, 3^c fire in which series ?
Answer:
In geometric Progression (G.P).

Question 103.
22, 32, 42, find a₇.
Answer:
a₇ = 82

Question 104.
a_n = 2ⁿ, then find a₅.
Answer:
32 = a₅.

Question 105.
Write G.M. of 5 and 125.
Answer:
25

Question 106.
Σn = 10, then Σn³.
Answer:
1000
Explanation :
(Σn)³ = (10)³ = 1000

Question 107.
Write G.M. of x, y, z.
Answer:
G.M. = \(\sqrt[3]{x y z}\)

Question 108.
Find the value of 16 + 11 + 6 + …. 23 terms.
Answer:
S₂₃ = – 897.

Question 109.
Write a formula to 1³ + 2³ + 3³ +….+ n³
Answer:
\(\frac{n^{2}(n+1)^{2}}{4}=\Sigma n^{3}\)

Question 110.
a_n = (n – 1) (n – 2), then find a₂.
Answer:
a₂ = 1

Question 111.
If a, b, c are in G.P., then find \(\frac{\mathbf{b}}{\mathbf{a}}\)
Answer:
\(\frac{b}{a}=\frac{c}{b}\)

Question 112.
-1, \(\frac{1}{4}, \frac{3}{2}\), ……….. find sum to 10 terms.
Answer:
46.25

Question 113.
Find the sum of first 40 positive integers which are divisible by 6.
Answer:
4920
Explanation :
S_n = 6 + 12 + 18 + 24 + …. + 240
= \(6\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]=6\left[\frac{40 \times 41}{2}\right]\)
= 6 x 20 x 41
= 4920

Question 114.
If a, b, c are in G.P., then find b2.
Answer:
ac = b²

Question 115.
Calculate the common ratio of the G.P. 4, 20, 100, 500, ……………
Answer:
r = 5

Question 116.
In the Answer:P., first term is 4 and common difference is – 1, then find Answer:P.
Answer:
4,3, 2, Answer:P.

Question 117.
Find the sum of first 20 odd numbers..
Answer:
400 = S₂₀.

Question 118.
How many numbers are divisible by ‘4’ lying between 101 and 250 ?
Answer:
37

Question 119.
If a₇ – a₃ = 32, then the common dif-ference of the Answer:P.
Answer:
8 = d

Question 120.
Which term of the Answer:P. 125, 120, 115, ………… is the first negative ?
Answer:
7th term.

Question 121.
If a₇ ÷ a₄ of a G.P is 343, then find the common ratio.
Answer:
r = 7.

Question 122.
If x, xy, xy², xy³, …. forms a G.P, then find its 15th term.
Answer:
xy¹⁴

Question 123.
Find the n^th term of a, a + d, a + 2d,…
Answer:
a + (n – 1) d = a_n

Question 124.
In a G.P. write a₆.
Answer:
ar⁵ = a₆.

Question 125.
Calculate the 16th term of 4, – 4, 4, – 4, ……..
Answer:
4 = a₆.

Question 126.
In Answer:P. aa₁₂. = 37, d = 3, then find ‘a’.
Answer:
4 = Answer:

Question 127.
3, – 3², 3³, find a₆.
Answer:
– 729 = a₆.

Question 128.
If there are’n’AM’s between’a’and h’, then find d.
Answer:
d = \(\frac{b-a}{n+1}\)

Question 129.
7, 10, 13, find a₅.
Answer:
19 = a₅.

Question 130.
Find AM of 5 and 95.
Answer:
50 = AM

Question 131.
f 4, x, 16 are in G.P., then find ‘x’.
Answer:
8
Explanation :
b² = ac ⇒ x = \(\sqrt{4 \times 16}=\sqrt{64}\) = 8

Question 132.
5, 1, – 3, – 7, find a₁₀.
Answer:
– 31 = a₁₀.

Question 133.
Write product of ‘n’ GM’s between a and b.
Answer:
(ab)^n/2 = G.M.

Question 134.
If a, b, c are in GP, then b is called
Answer:
Geometric mean.

Question 135.
How many 3-digit numbers are divisible by 7 ?
Answer:
128

Question 136.
If a = 3 and a₇ = 33, then find a₁₁.
Answer:
53 = a₁₁
Explanation :
a = 3,
a₇ = a + (n – 1)d = 33
⇒ 3 + (7 – 1) d = 33
⇒ 6d = 30 ⇒ d = 5
a₁₁ = a + 10d
= 3 + 10×5 = 3 + 50 = 53

Question 137.
Find the 10th term of the
AP:3, 11, 19, …………..
Answer:
75 = a₁₀

Question 138.
aa₂₈ – aa₂₃ = 15, then find the common difference of the Answer:P.
Answer:
3 = d

Question 139.
If x – 1, x + 3, 3x – 1 are in Answer:P., then find ‘x’.
Answer:
4 = x

Question 140.
Find the common ratio of the G.P.
3, 6, 12, 24, …………..
Answer:
2

Question 141.
Find the 8th term from the end of the Answer:P., 7, 10, 13, …………. 1814.
Answer:
163

Question 142.
\(\frac{\mathbf{b}+\mathbf{c}-\mathbf{a}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}-\mathbf{b}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}-\mathbf{c}}{\mathbf{c}}\) are in AP, then write \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in ,
Answer:
Arithmetic Progression.

Question 143.
Which term of the Answer:P, 24, 21, 18, …. is the first negative ?
Answer:
a₁₀ is first negative term.

Question 144.
Find AM of 10 and 20.
Answer:
15 = AM .

Question 145.
Write the next term of the
Answer:P. \(\sqrt{48}, \sqrt{75}, \sqrt{147}, \ldots \ldots \ldots\)
Answer:
\(\sqrt{192}\)

Question 146.
– 20, – 18, – 16, ………… which term of this Answer:P. is a first positive term ?
Answer:
12

Question 147.
Find the 17^th term of 1.1, 2.2, 3.3, 4.4 …………….
Answer:
18.7 = a₁₇

Question 148.
If \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) is the AM of ‘a’and ‘b’, then find ‘n’.
Answer:
n = 0.

Choose the correct answer satisfying the following statements.

Question 149.
Statement (A): Common difference of the AP : – 5, – 1, 3, 7, is 4.
Statement (B): Common difference of the a, a + d, a + 2d, is given by
d = 2^nd term – 1^st term.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
Common difference, d = – 1 – (- 5)
= 4
So, both A and B are correct and B explains Answer: Hence, (i) is the correct option.

Question 150.
Statement (A) : a_n – a_n-1 is not independent of n, then the given sequence is an AP.
Statement (B) : Common difference d = a_n – a_n-1 is constant or independent of n.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
We have, common difference of an Answer:P. d = a_n – a_n-1 is independent of ‘n’ or constant. So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 151.
Statement (A): The sum of the first ‘n’ terms of an AP is given by S_n = 3n² -4n. Then its nth term a_n = 6n – 7.
Statement (B) : n^th term of an AP, whose sum to ’n’ terms is S_n, is given
by a_n = S_n – S_n-1
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
n^th term of an AP be a_n = S_n – S_n-1
⇒ a_n = 3n² – 4n – 3(n – 1)² + 4 (n – 1)
⇒ a_n = 6n – 7
So, both A and B are correct and B explains Answer: Hence, (i) is the correct option.

Question 152.
Statement (A) : Common difference of an AP in which a₂₁ – a₇ = 84 is 14. Statement (B) : n^th term of an AP is given by a_n = a + (n – 1) d.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
We have, a_n= a + (n – 1)d
a₂₁ – a₇ = {a + (21 – 1)}d – {a + (7-l)d} = 84
⇒ a + 20d – a – 6d = 84
⇒ 14d = 84
⇒ d = 84/14 = 6
⇒ d = 6
So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 153.
Statement (A) : Three consecutive terms 2k + 1, 3k + 3 and 5k – 1 from an AP, then k is equal to 6.
Statement (B) : In an AP
a, a + d, a + 2d, the sum to n terms of the AP be S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d].
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
For 2k + 1, 3k + 3 and 5k- 1 to form an Answer:P.
(3k + 3)-(2k + 1) = (5k- l)-(3k + 3)
⇒ k + 2 = 2k-4
⇒ 2 + 4 = 2k – k = k
⇒ k = 6
So, both A and B are correct but B does not explain Answer:
Hence, (i) is the correct option.

Question 154.
Statement (A) : 10^th term from the end of AP : 100, 95, 90, 85,……………. 10 is 55.
Statement (B): The n^th term from the end of an AP having last term L and common difference d is L – (n – 1) d.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 155.
Statement (A) : If the sum of first ‘n’ terms of an AP is an² + bn, then its common difference is 2Answer:
Statement (B): In an AP with first term a and last term l, sum of n terms is
given by S_n = \(\frac{n}{2}\)(a + l).
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 156.
Statement (A) : The sum of all natural numbers between 100 and 1000 which are multiple of 7 is 70336.
Statement (B) : The 10^th term of an AP is 31 and 20^th term is 71. Then t₃₀ = 111
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 157.
Statement (A): 1, 2, 4, 8,……………. is a G.P., 4, 8, 16, 32 is a G.P. and 1 + 4, 2 + 8, 4 + 16, 8 + 32, …. is also a G.P.
Statement (B) : Let general term of a G.P. with common ratio ‘r’ be T_{k + 1} and general term of another G.P., with common ratio ‘r’ be Tk + v then the series
whose general term T_{k + 1} = T_{k + 1} + T_{k + 1} is also a G.P. with common ratio ‘r’.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
Let T_{k + 1} = ar^k and T_{k + 1} = br^k
Since T”_{k + 1} = ar^k + br^k = (a + b)r^k ,
∴ T”_{k + 1} is general term of a G.P.
Option (i) is correct.

Question 158.
Statement (A) : 1111 …………. 1 (upto 91 terms) is a prime number.
Statement (B) : If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}, \frac{a+b-c}{c}\) are in Answer:P., then \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are also in Answer:P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
Since 11 11 ………..,
= \(\frac{10^{91}-1}{10-1}\) = divisible by 9.
The given number is not prime. So, A is false, but B is true.
∴ Option (iii) is correct.

Question 159.
Statement (A) : Let the positive numbers a, b, c be in Answer:P, then \(\frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ac}}, \frac{1}{\mathrm{ab}}\) are also in Answer:P.
Statement (B): If each term of an Answer:P. is divided by abc, then the resulting sequence is also in Answer:P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 160.
Statement (A) : Let three distinct positive real numbers a, b, c are in G.P., then a², b², c² are in G.P.
Statement (B) : If we square each term of a G.P., then the resulting sequence is also in G.P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 161.
Statement (A) : The sum of the series with the n^th term, t_n = (9 – 5n) is 465, when number of terms n = 15.
Statement (B) : Given series is in Answer:P. and sum of ‘n’ terms of an Answer:P. is
S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n- 1) d]
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Read Che below passages and answer to the following questions.

Following given series are in Answer:P.
2, 4, 6, 8, …………..
3,6,9,12, ……….
First series contains 30 terms, while the second series contains 20 terms. Both of the above given series contains some terms, which are common to both of them.

Question 162.
Find the last term of both the above given Answer:P. are
Answer:
(60, 60)
Explanation :
For 2, 4, 6, 8, …………..
Last term, t30 = 2 + (30 – 1)2
= 2 + 2 (29) = 60
3,6,9,12,
Last term, t20 = 3 + (20 – 1)3 x = 3 + 57 = 60

Question 163.
Find the sum of both the above given Answer:P. are
Answer:
(930, 630).
Explanation :
For 2, 4, 6, 8,
S₃₀ = \(\frac { 30 }{ 2 }\) (2 + 60) = 930
For 3, 6, 9, 12,
S₂₀ = \(\frac { 20 }{ 2 }\) (3 + 60) = 630

Question 164.
Write number of terms identical to both the above given Answer:P. are
Answer:
10
Explanation :
Let mth term of the first series is common with the nth term of the second series.
t_m = t_n
2 + (m – 1)2 = 3 + (n – 1) 3
2 + 2m -2 = 3 + 3n – 3
2m = 3n
\(\frac{m}{3}=\frac{n}{2}=k \text { (let) }\)
m = 3k, n = 2k.
Hence, k = 1, 2, 3, , 10.
[ ∵ 1 ≤ m ≤ 30, 1 ≤ n ≤ 20
1 ≤ 3k ≤ 30, 1 ≤ 2k ≤ 20
\(\frac{1}{2}\) ≤ k ≤ 10, \(\frac{1}{2}\) ≤ k ≤ 10]
For each value of k, we get one identical term.
Thus, number of identical terms =10.

There are 25 trees at equal distances of 5 m in a line with a well, the distance of the well from the nearest tree being 10 m. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next.

Question 165.
Where the well located in the garden?
Answer:
Obviously the well must be on one side of the trees.

Question 166.
How much the distance between the trees ?
Answer:
5 m.

Question 167.
Which mathematical concept is used to find the total distance the gardener will cover in order to water all the trees?
Answer:
Arithmetic Progression.

Question 168.
Column – II give common difference for Answer:P. given column -1, match them cor-rectly.

Answer:
A – (iv), B – (iii).

Question 169.
Column – II give common difference for Answer:P. given column -1, match them cor. rectly.

Answer:
A – (ii), B – (i).

Question 170.
Column – II give n^th term for Answer:P. given column -I, match them correctly.

Answer:
A – (iv), B – (iii).

Question 171.
Column – II give n,h term for Answer:P. given column -I, match them correctly.

Answer:
A – (ii), B – (i).

Question 172.

Answer:
A – (i), B – (ii).

Question 173.

Answer:
A – (iv), B – (iii).

Question 174.
In which progression are the perimeters of triangles formed by joining the midpoints of sides of triangles succes-sively in the given figure.

Answer:
Geometric Progression.

Practice the AP 10th Class Maths Bits with Answers Chapter 7 Coordinate Geometry on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 7th Lesson Coordinate Geometry with Answers

Question 1.
Write the nearest point to origin,
i) (2, – 3)
ii) (5, 0)
iii) (0, – 5)
iv) (1, 3)
Answer:
(1,3)

Question 2.
The distance of a point (3, 4) from the origin is how many units ?
Answer:
5 units.

Question 3.
Write the formula to find the area of a triangle.
Answer:
Δ = \(\frac { 1 }{ 2 }\) bh and
Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Question 4.
Find the mid point of (2, 3) and (-2,3).
Answer:
(0, 3)

Question 5.
Find the distance to X – axis from the point (3, – 4).
Answer:
4 units

Question 6.
Find the centroid of the triangle formed by these points (0, 3); (3, 0) and (0, 0).
Answer:
(1, 1)

Question 7.
Where do the points lie on co-ordinate axis ?
(- 4, 0), (2, 0), (6, 0), (- 8, 0)
Answer:
On X-axis.

Question 8.
The graph of y = 5 represents.
Answer:
Parallel to X – axis.

Question 9.
Find sum of the distances from A(3, 4) to X – axis and from B(5, 7) to Y – axis.
Answer:
9 units.

Question 10.
Find the distance from origin to (2,3).
Answer:
\(\sqrt{13}\) units.

Question 11.
Find slope of the line passing through the points (0, sin 60°) and (cos 30°, 0).
Answer:
m = – 1

Question 12.
If the mid point of (x – y, 8) and (2, x + y) is (5, 10), then find (x, y).
Answer:
(10,2)

Question 13.
Where the point (0, 5) lies ?
Answer:
On Y – axis.

Question 14.
Find the area of a triangle whose verti-ces (points) Eire (0, 0), (3, 0) and (0, 4).
Answer:
6 sq. units.

Question 15.
Write the slope of Y – axis.
Answer:
Not defined.

Question 16.
Find the mid point of line segment joined by (4, 5) and (- 6, 3).
Answer:
(-1,4)

Question 17.
(x, y), (2,0), (3,2) and (1,2) are vertices of a parallelogram, then find (x, y).
Answer:
(0, 0)

Question 18.
Find centroid of a triangle, whose ver-tices are (- a, 0), (0, b) and (a, 0).
Answer:
(0, \(\frac { b }{ 3 }\))

Question 19.
Find the distance between two points A (a cos 0, 0), B (0, a sin 0).
Answer:
a units.

Question 20.
Find the distance between (0, 0), (x₁, y₁) points.
Answer:
\(\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}}\)

Question 21.
If A(log₂ 8, log₅ 25) and B(log₁₀ 10, log₁₀ 100), then find the mid-point of AB.
Answer:
(2,2)

Question 22.
Find the distance between (0, 7) and (- 7, 0).
Answer:
7\(\sqrt{2}\) units.

Question 23.
Find slope of the line passing through the points (- 1, 1) and (1, 1).
Answer:
0

Question 24.
Find the slope of the line passing through the points (4, 6) and (2, – 5).
Answer:
\(\frac { 11 }{ 2 }\)

Question 25.
In the given figure, find the area of ΔOAB.

Answer:
6 sq. units.

Question 26.
A line makes 45° with X – axis, then find its slope.
Answer:
m = tan θ = tan 45° = 1

Question 27.
If a line is passing through (2, 3) and (2, – 3), then write the nature of that line.
Answer:
The line is parallel to Y – axis and The slope of the line is not defined.

Question 28.
Find area of the triangle formed by the points A(0, 0), B(1, 0) and C(0, 1).
Answer:
\(\frac { 1 }{ 2 }\) sq. units.

Question 29.
Find the distance from X – axis to (- 4, 3) is units.
Answer:
3 units.

Question 30.
Find the area of the triangle BOA is …………… sq. units.

Answer:
3 sq. units.

Question 31.
Find the slope of the line that passes through the points P (x₁, y₁) and Q(x₂, y₂) and making an angle ‘θ’ with X – axis.
Answer:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Question 32.
The area of given triangle is 60 sq. units, then find x = …………units.

Answer:
12 units.

Question 33.
A line makes 45° with X – axis, then find its slope.
Answer:
1

Question 34.
Find the distance between the points (x₁, y₁) and (x₂, y₂) which are on the line parallel to Y – axis.
Answer:
|y – y₂| or |y₂ – y₁ |

Question 35.
If the co-ordinates of the vertices of a rectangle are (0, 0), (4, 0), (4, 3) and (0, 3), then find the length of its di¬agonal.
Answer:
5 units.

Question 36.
Find the distance from Y-axis to (4, 0) is ……………. units.
Answer:
4 units.

Question 37.
Draw the graph represented by y = x.

Question 38.
If origin is the centroid of a triangle, whose vertices are (3, 2), (- 6, y) and (3, – 2), then calculate ‘y’.
Answer:
y = 0

Question 39.
In a coordinate plane, if line segment AB is parallel to X – axis, then write about points A and B.
Answer:
X – coordinates of points A and B are equal.

Question 40.
Find the distance between the points (0, 7) and (- 7, 0).
Answer:
7\(\sqrt{2}\) units.

Question 41.
Find the distance of the point (- 8, 3) from the origin.
Answer:
\(\sqrt{73}\) units

Question 42.
If points (x, 0), (0, y) and (1, 1) are collinear, then find \(\frac{1}{x}+\frac{1}{y}\).
Answer:
1

Question 43.
Write a point on the X – axis is of the form.
Answer:
(x, 0)

Question 44.
Find the points (- 3, 0), (0, 5) and (3, 0) are the vertices of which type of triangle ?
Answer:
Isosceles triangle.

Question 45.
Find the area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b).
Answer:
0

Question 46.
Write a point on the Y – axis is of the form.
Answer:
(0, y)

Question 47.
The point which divides the line segment joining the points (3, 4) and (7, – 6) internally in the ratio 1 : 2 lies in the quadrant.
Answer:
Q₄

Question 48.
Find the distance between the points (- 2, 3) and (2, – 3).
Answer:
\(\sqrt{52}\) units.

Question 49.
AOBC is a rectangle whose three ver-tices are A(4, 0), B(0, 3) and O (0, 0), then find its diagonal ?
Answer:
5 units.

Question 50.
Find the distance of the point (-8, -7) from Y – axis.
Answer:
8 units.

Question 51.
A circle is drawn with origin as centre and passing through (2, 3), then find its radius.
Answer:
\(\sqrt{13}\) units.

Question 52.
Find the perimeter of a triangle whose vertices are A(12, 0), 0(0,.0) and B(0, 5).
Answer:
30 units.

Question 53.
Find the distance of the point (- 4, 3) from X – axis.
Answer:
3 units.

Question 54.
If the distance between the points (4, y) and (1, 0) is 5, then find ‘y’.
Answer:
y = ± 4.

Question 55.
Write the distance of (x, y) from X-axis.
Answer:
y units.

Question 56.
Find the distance of the point (- 9, 40) from the origin.
Answer:
41 units

Question 57.
If (0, 0), (a, 0) and (0, b) are collinear, then write the relation between ‘a’ and b’.
Answer:
ab = 0

Question 58.
Which ratio the centroid divides each median ?
Answer:
2:1

Question 59.
Find the value of ‘p’ if the distance be-tween (2, 3) and (p, 3) is 5. ,
Answer:
p = 7

Question 60.
Find the angle between X – axis and Y – axis.
Answer:
90°

Question 61. Find the distance between the points (a cos θ, 0) and (0, a sin θ).
Answer:
‘a’ units

Question 62.
(- 2, 8) belongs to which quadrant ?
Answer:
Q₂

Question 63.
Find the centroid of the triangle whose vertices are (2, – 3), (4, 6), (- 2, 8).
Answer:
(\(\frac{4}{3}, \frac{11}{3}\))

Question 64.
Guess shape of the closed figure formed by the points (- 2, 0), (2, 0), (2, 2), (0, 4) and (-2,-2).
Answer:
Pentagon

Question 65.
Find the midpoint of the line joining of (2, 3) and (- 2, 3).
Answer:
(0, 0)

Question 66.
If the centroid of the triangle formed with (a, b); (b, c) and (c, a) is O (0, 0), then the value of a³ + b³ + c³.
Answer:
3 abc

Question 67.
If the points (a, 2a), (3a, 3a) and (3,1) are collinear, then find k.
Answer:
k = \(\frac { -1 }{ 3 }\)

Question 68.
Write the coordinates of the midpoint joining P(x₁, y₁) and Q(x₂, y₂).
Answer:
\(\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)\)

Question 69.
Find the slope of line joining of (5, -1), (0, 8).
Answer:
\(-\frac{9}{5}\) =m

Question 70.
If the distance between the points (3, k) and (4, 1) is \(\sqrt{10}\), then find the value of k.
Answer:
4 (or)-2.

Question 71.
If (- 2, – 1), (a, 0), (4, b) and (1,2) are the vertices of a parallelogram, then find a’.
Answer:
a = 1

Question 72.
Write the slope of X – axis.
Answer:
0

Question 73.
P(2, 2), Q(- 4, 4) and R(5, – 8) are the vertices of a ΔPQR, then find length of median from ‘R’.
Answer:
\(\sqrt{157}\) units.

Question 74.
Find the value of ‘k’ if the distance be-tween (2, 8) and (2, k) is 3.
Answer:
k = 5.

Question 75.
Find the distance of a point (α, β) from the origin.
Answer:
\(\sqrt{\alpha^{2}+\beta^{2}}\)

Question 76.
If the points (1, 2), (-1, x) and (2, 3) are collinear, then find the value of x.
Answer:
x = 0.

Question 77.
If (- 2,8) and (6, – 4) are the end points of the diameter of a circle, then find the centre of the circle.
Answer:
(2, 2) = centre.

Question 78.
A(0, -1), B(2, 1) and C(0, 3) are the vertices of AABC, then find median
through ‘B’ has a length . units.
Answer:
2

Question 79.
Two vertices of a triangle are (3, 5) and (- 4, – 5). If the centroid of the triangle is (4, 3), find the third vertex.
Answer:
(13, 9).

Question 80.
If A, B, C are collinear, then find the area of AABC.
Answer:
Δ = 0

Question 81.
A circle drawn with origin as centre 13
passes through (\(\frac { 13 }{ 2 }\),0). Find the point which doesn’t lie in the interior of the circle.
Answer:
(-6,3)

Question 82.
Find area of triangle formed by (-4, 0), (0, 0) and (0, 5) is ……………… sq. units.
Answer:
10 sq.units.

Question 83.
Find the ratio in which the point (4, 8) divide the line segment joining the points (8, 6) and (0, 10).
Answer:
1:1

Question 84.
Write a formula to the coordinates of the point which divides the line join¬ing (x₁, y₁) and (x₂, y₂) in the ratio m: n internally.
Answer:
P = \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)

Question 82.
If A(4, 0), B(8, 0), then find \(\overline{\mathbf{A B}}\).
Answer:
4 units.

Question 83.
Find the slope of the line \(\frac{\mathbf{x}}{\mathbf{a}}+\frac{\mathbf{y}}{\mathbf{b}}\) = 1.
Answer:
m = \(\frac{-\mathrm{b}}{\mathrm{a}}\)

Question 87.
Find .the radius of the circle whose centre is (3, 2) and passes through (- 5, 6) is……………..units.
Answer:
4\(\sqrt{5}\) units.

Question 88.
In Heron’s formula ‘s’ represents.
Answer:
s = \(\frac{a+b+c}{2}\) = Semi perimeter

Question 89.
Slope of the line joining the points (2, 5) and (k, 3) is 2, then find k.
Answer:
k = 1.

Question 90.
If A(4, 5), B(7, 6), then find \(\overline{\mathbf{A B}}\).
Answer:
\(\sqrt{10}=\overline{\mathrm{AB}}\)

Question 91.
Write the distance of (x, y) from Y-axis.
Answer:
‘x’units.

Question 92.
A(2, 0), B(l, 2), C(l, 6), then find ∆ABC.
Answer:
∆ = 0, so the points are collinear.

Question 93.
Find the mid point of the line joining the points (1,1) and (0, 0).
Answer:
( \(\frac{1}{2}, \frac{1}{2}\) )

Question 94.
How much the slope of vertical line ?
Answer:
Not defined.

Question 95.
A(1, – 1), B(0, 6) and C(- 3, 0), then find G (centroid).
Answer:
G = (\(\frac{-2}{3}, \frac{5}{3}\))

Question 96.
A(a, b) and B(- a, – b), then find \(\overline{\mathbf{B A}}\).
Answer:
\(\overline{\mathrm{BA}}=2 \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\)

Question 97.
Find the centroid of the triangle formed with the line x + y = 6 with the coordinate axes.
Answer:
G .= (2, 2)

Question 98.
Find the area of triangle formed with (- 5,-1), (3,-5) and (5, 2).
Answer:
32 sq. units.

Question 99.
How much the slope of horizontal line?
Answer:
0 = m

Question 100.
Find the angle between the lines x = 2 and y = 3.
Answer:
θ = 90°.

Question 101.
Write the slope of the line y = mx.
Answer:
‘m’

Question 102.
The midpoint of the line joining the points (1, 2) and (1, p) is (1, – 1), then find p.
Answer:
p = – 4.

Question 103.
Name the point of concurrence of me-dians of a triangle is called
Answer:
Centroid

Question 104.
If AC = AB + BC, then the points A, B, C are called points.
Answer:
Collinear

Question 105.
ax + by + c = 0, represents a
Answer:
Straight line

Question 106.
If the points (k, k), (2, 3) and (4, – 1) are collinear, then find k.
Answer:
\(\frac{7}{3}\) = k

Question 107.
Write other name for x-coordinate of a points.
Answer:
Abscissa

Question 108.
Find the slope of the line joining the points A(-1.4, – 3.7) and B(-2.4, 1.3).
Answer:
m = – 5

Question 109.
If a < 0, then (- a, – a) belongs to which quadrant ?
Answer:
Q₁

Question 110.
If θ is the angle made by a line with X – axis, then find slope’m’.
Answer:
m = tan θ

Question 111.
Find the area of square formed with the vertices (0, – 1), (2, 1), (0, 3) and (-2, 1) taken in order as vertices.
Answer:
∆ = 8 sq. units.

Question 112.
Name the person who was introduced coordinate geometry.
Answer:
Rene Descartes

Question 113.
Find the coordinates of centroid of the triangle formed with the vertices (-1,3), (6, -3) and (-3, 6).

Question 114.
In quadrilateral ABCD, AB = BC = CD = AD and \(\overline{\mathbf{A C}} \neq \overline{\mathbf{B D}}\), then it is Answer:……………..type of quadrilateral.
Answer:
Rhombus

Question 115.
Write the slope of the line joining the points (2a, 3b) and (a, – b).
Answer:
m = \(\frac{4 \mathrm{~b}}{\mathrm{a}}\)

Question 116.
Write a formula to distance of (x, y) from origin.
Answer:
\(\sqrt{x^{2}+y^{2}}\)

Question 117.
In rhombus all sides are……………….
Answer:
Equal in length.

Question 118.
If the slope of a line passing through (- 2, 3) and (4, a) is \(\frac { -5 }{ 3 }\), then find Answer:
Answer:
a = – 7.

119.
A(2a, 4a), B(2a, 6a), C(2a+ \(\sqrt{3}\), 5), then write ΔABC is a type of tri¬
angle.
Answer:
Equilateral triangle.

Question 120.
How much each angle in equilateral triangle ?
Answer:
60° = each angle.

Question 121.
In the below figure, G is the centroid then AG : GD = ………………

Answer:
2:1

Question 122.
In the below figure AD : GD =

Answer:
3 : 1

Question 123.
Write the midpoint of a line segment divides it in the ratio.
im
Answer:
1 : 1

Question 124.
If the distance between the points (x₁, y₁) and (x₂, y₂) is |x₁ – x₂|, then they are parallel to ……………..
Answer:
x-axis.

Question 125.
Find slope of the line joining the points A(0,0), B(1/2,1/2)
Answer:
1 = m

Question 126.
Write the equation of X – axis.
Answer:
Y = 0

Question 127.
Write the point of concurrence of alti-tudes of a triangle is called ……………..
Answer:
Orthocentre

Question 128.
P(cos θ, – cos θ), Q (sin θ, sin θ), then find \(\overline{\mathbf{P Q}}\).
Answer:
\(\sqrt{2}\) units.

Question 129.
Diagonals in a parallelogram …………….. to each other.
Answer:
Bisect

Question 130.
Find slope of the line 3x – 2 = 0.
Answer:
Not defined = (\(\frac{0}{3}\))

Question 131.
If A(p, q), B(m, n) and C(p – m, q – n) are collinear, then find pn.
Answer:
qm

Question 132.
Write Y-axis can be represented as.
Answer:
X = 0

Question 133.
Write number of medians of a triangle.
Answer:
3

Question 134.
A(cot θ, 1), B(0, 0), then find \(\overline{\mathbf{B A}}\).
Answer:
cosec θ

Question 135.
If the point (4 – p) lie on X – axis, then find the value of p² + 2p – 1.
Answer:
– 1

Question 136.
A(t, 2t), B(- 2, 6), C(3, 1) and ΔABC = 5 sq.units, then find ‘t’.
Answer:
t = 2

Question 137.
y-intercept of the line x – 2y + 1 = 0 is …………..
Answer:
b = \(\frac { 1 }{ 2 }\)

Question 138.
In the below figure find ‘x’.

Answer:
-9 = x

Question 139.
In the below figure find ‘y’.

Answer:
3 = y

Question 140.
Where the X and Y axes will inter¬sects ?
Answer:
(0, 0)

Question 141.
If the point (a, 5) lies on Y – axis find the value of ‘a’.
Answer:
a = 0.

Question 142.
Write (3, 0), (8, 0), (1/2, 0) points lie on ………….
Answer:
X – axis.

Question 143.
Nature of the line that does not pass through origin and having a zero slope is
Answer:
Parallel to X – axis.

Question 144.
Y-intercept of the line y = mx + c is ….
Answer:
‘c’

Question 145.
In ΔABC, all the side are different, then it is called type of triangle.
Answer:
Scalene

Question 146.
A = (\(\frac{1}{2}, \frac{3}{2}\)) , B = (\(\frac{3}{2}, \frac{-1}{2}\)) then find \(\overline{\mathbf{B A}}\)
Answer:
\(\sqrt{5}\)

Question 147.
Find the area of below parallelogram, if ΔABC = 5 sq. units.

Answer:
10 sq. units

Question 148.
Find x-intercept of the line x – y +1 =0.
Answer:
– 1

Question 149.
In ΔPQR, PQ = QR, then it is called ……………… triangle.
Answer:
Isosceles

Question 150.
If (1, x) is at \(\sqrt{10}\) units from origin, then find the value of ‘x’.
Answer:
x = ± 3

Question 151.
A(1, – 1), B(2 1/2, 0), C(4, 1), then find area of ΔABC.
Answer:
Δ = 0.

Question 152.
Name the line joining the mid point of one side of a triangle from opposite vertex is called …………….
Answer:
Median

Question 153.
Find angle made by the line y = x with the positive direction of X – axis.
Answer:
45°.

Choose the correct answer satistying the following statements.

Question 154.
Statement (A): The point (0, 4) lies on Y – axis.
Statement (B) : The X co-ordinate of the point on Y – axis zero.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 155.
Statement (A): The value of y is 6, for which the distance between the points P(2, – 3) and Q(10, y) is 10.
Statement (B): Distance between two given points A(x₁, y₁) and B(x₂, y₂) is given by
AB = \(\sqrt{\left(x_{2}+x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 156.
Statement (A) : The point (- 1, 6) di¬vides the line segment joining the points (- 3, 10) and (6, – 8) in the ratio 2 : 7 internally.
Statement (B): Three points A, B and C are collinear if area of AABC = 0.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 157.
Statement (A) : Centroid of a triangle formed by the points (a, b), (b, c) and (c, a) is at origin, then a + b + c = 0.
Statement (B) : Centroid of a AABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by
\(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 158.
Statement (A): The area of the triangle with vertices (- 5,-1), (3, – 5), (5, 2) is 32 square units.
Statement (B): The point (x, y) divides the line segment joining the points A(xj, y2) and B(x2, y2) in the ratio k : 1 externally, then
\(x=\frac{k x_{2}+x_{1}}{k+1}, y=\frac{k y_{2}+y_{1}}{k+1}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 159.
Statement (A): The ratio in which the segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6) is 2 : 7.
Statement (B) : If A(x₁, y₁), B(x₂, y₂) are two points. Then the point C(x, y) such that C divides AB internally in the ratio k : 1 is given by
x = \(\frac{\mathrm{kx}_{2}+\mathrm{x}_{1}}{\mathrm{k}+1}, \mathrm{y}=\frac{\mathrm{ky}_{2}+\mathrm{y}_{1}}{\mathrm{k}+1}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 160.
Statement (A) : If three vertices of a parallelogram taken in order are (- 1, – 6), (2, – 5) and (7, 2), then its fourth vertex is (4, 1).
Statement (B) : Diagonals of a paral-lelogram bisect each other.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 161.
Statement (A) : The points (k -I- 1, 1), (2k + 1, 3) and (2k + 2, 2k) are col- linear, then k = 4.
Statement (B) : Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear if and only if
x₁(y₂ – y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) = 0
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 162.
Statement (A) : Let the vertices of a ΔABC are A(- 5, – 2), B(7, 6) and C(5, – 4), then coordinate of circum- centre is (1, 2).
Statement (B) : In a right angle tri¬angle, mid-point of hypotenuse is the circumcentre of the triangle.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 163.
Statement (A): If A(2a, 4a) and B(2a, 6a) are two vertices of a equilat¬eral triangle ABC, then the vertex C is given by (2a + a\(\sqrt{3}\) , 5a).
Statement (B): In equilaterahtriangle all the coordinates of three vertices can be rational.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 164.
Statement (A) : The equation of the straight line which passes through the point (2,-3) and the point of the inter-section of the lines x + y + 4 = 0 and 3x – y – 8 = 0 is 2x – y – 7 = 0.
Statement (B) : Product of slopes of two perpendicular straight lines is – 1.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Read Che below passages and answer to the following questions.

Let there be two points (4, 1) and (5,-2) in a two dimensional coordi¬nate system. A line which passes through the above give points and intersects the coordinate axes forms a triangle.

Question 165.
Write the equation of the line passing through the above given points.
Answer:
3x + y – 13 = 0.

Question 166.
Find the point of intersection of the above line with both the coordinate axes.
Answer:
(\(\frac { 13 }{ 3 }\),0) and (0, 13).

Question 167.
Find the area of the triangle so formed.
Answer:
\(\frac { 169 }{ 6 }\) sq. units.

In the diagram on a Lunar eclipse, if the position of Sun, Earth and Moon are shown by (- 4, 6) (k, – 2) and (5, – 6) respectively.

Question 168.
In Lunar eclipse what is the positions of Sun, Earth and Moon ?
Answer:
All are in same line, i.e., collinear.

Question 169.
To solve the above problem which mathematical concept is used ?
Answer:
Co-ordinate Geometry.

Question 170.
Which formula is used to find the value of k ?
Δ = \(\frac { 1 }{ 2 }\) |x₁(y₂ – y₃) + x₂(y₃-y₁) + x₃(y₁ – y₂) | = 0

Manowbhiram calculated the dis¬tance between T(5, 2) and R(-4, -1) to the nearest length is 9.5 units.

Question 171.
Do you agree with Manowbhiram ?
Answer:
Yes, I agree with him.

Question 172.
Which mathematical concept is used to you support Manowbhiram ?
Answer:
Co-ordinate Geometry (or) (Distance formula).

Question 173.
Column – II gives distance between pair of points given in column -I, match them correctly.

Answer:
A – (iv), B – (i)

Question 174.
Column – II gives distance between pair of points given in column -I, match them correctly.

Answer:
A – (ii), B – (iii)

Question 175.
Column – II gives the coordinates of the point ’p’ that divides the line segment join¬ing the points given in column -I, match them correctly.

Answer:
A – (iv), B – (ii)

Question 176.
Column – II gives the coordinates of the point ‘p’ that divides the line segment join¬ing the points given in column -I, match them correctly.

Answer:
A – (iii), B – (i).

Question 177.
Column – II gives the area of triangles whose vertices are given in column -1, match them correctly.

Answer:
A – (iv), B – (iii).

Question 178.
Column – II gives the area of triangles whose vertices are given in column -1, match them correctly.

Answer:
A – (ii), B – (i).

Question 179.

Answer:
A – (iv), B – (iii).

Question 180.

Answer:
A – (iii), B – (i).

Question 181.
Name the quadrilateral, which satis¬fies both the conditions given below.
Statement (A) : Diagonals are equal
Statement (B) : All sides are equal
a) Rhombus
b) Parallelogram
c) Rectangle
d) Square
Answer:
(d)

Question 182.
Find the area of the shaded triangle, in the figure given below.

Answer:
6 sq. units

Question 183.
What is the slope of the line joining the points (2, 0) and (- 2, 0).
Solution:

Question 184.
Name the point which is the point of intersection of medians of a triangle.
Answer:
Centroid of a triangle.

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles InText Questions

Check Your Progress [Page No. 66]

Question 1.
Find the complementary angles of
(i) 27°
Answer:
If the sum of any two angles is 90°, then the angles are called complementary angles.
A complementary angle of 27° is
(90 – 27) = 63°

(ii) 43°
Answer:
Complementary angle of 43° is
(90 – 43) = 47°

(iii) k°
Answer:
Complementary angle of k° is (90 – k)°

(iv) 2°
Answer:
Complementary angle of 2° is
(90 – 2) = 88°

Question 2.
Find the supplementary angles of
(i) 13°
Answer:
If the sum of any two angles is 180°, then the angles are called as supplementary angles.
Supplementary angle of 13° is
(180 – 13) = 167°

(ii) 97°
Answer:
Supplementary angle of 97° is
(180 – 97) = 83°

(iii) a°
Answer:
Supplementary angle of a° is
(180 – a)°

(iv) 46°
Answer:
Supplementary angle of 46° is
(180 – 46) = 134°

Question 3.
Find the conjugate angles of
(i) 74°
Answer:
If the sum of any two angles is 360°, then the angles are called as conjugate angles.
Conjugate angle of 74° is
(360 – 74) = 286°

(ii) 180°
Answer:
Conjugate angle of 180° is
(360- 180) = 180°

(iii) m°
Answer:
Conjugate angle of m° is (360 – m)°

(iv) 300°
Answer:
Conjugate angle of 300° is
(360.-300) = 60°

[Page No. 66]

Question 1.
Umesh said, “Two acute angles cannot form a pair of supplementary angles.” Do you agree ? Give reason.
Answer:
Yes, acute angle is always less than 90°. So, sum of two acute angles is always less than 180°.
Therefore, two acute angles cannot form a pair of supplementary angles (180°).

Question 2.
Lokesh said, “Each angle in any pair of complementary angles is always acute.” Do you agree? Justify your answer.
Answer:
Yes, sum of any two acute angles is 90°, then they are complementary angles. If they are not acute means they may . be right angle (90°) or obtuse angle (> 90°) or etc.

So, its impossible.
Therefore, each angle in any pair of complementary angles is always acute.

Let’s Think [Page No. 69]

Question 1.
In the figure, ∠AOB and ∠BPC are not adjacent angles. Why? Give reason.

Answer:
In the given figure, ∠AOB and ∠BPC are not adjacent angles. Because, they have no common vertex and no common arm.

Question 2.
In the figure, ∠AOB and ∠COD have common vertex O. But ∠AOB, ∠COD are not adjacent angles. Why? Give reason.

Answer:
In the given figure, ∠AOB and ∠COD are not adjacent angles. Because, they have common vertex. But they have no common arm.

Question 3.
In the figure, ∠POQ and ∠POR have common vertex O and common arm OP but ∠POQ and ∠POR are not adjacent angles. Why? Give reason.

Answer:
∠POQ and ∠POR have common vertex O and common arm OP. But they are not lie either side of the common arm. That’s why they are not adjacent angles.

Check Your Progress [Page No. 70]

Question 1.
In the adjacent figure \(\overrightarrow{\mathbf{P R}}\) is a straight line and O is a point on the line. \(\overrightarrow{\mathbf{O Q}}\) is a ray.

(i) If ∠QOR= 50°, then what is ∠POQ?
Answer:
Given ∠QOR= 50°
∠POQ, ∠QOR are linear pair.
⇒ ∠POQ + ∠QOR = 180°
⇒ ∠POQ + 50° = 180°
⇒ ∠POQ — 180° – 50° = 130°
.-. ∠POQ =130°

(ii) If ∠QOP = 102°, then what is ∠QOR?
. Sol. Given ∠QOP = 102°
∠QOP and ∠QOR are linear pair.
⇒ ∠QOP + ∠QOR = 180°
⇒ 102° + ∠QOR = 180°
⇒ 102°- 102° + ∠QOR = 180°- 102°
⇒ ∠QOR = 78°

Let’s Explore [Page No. 70]

Question 1.
A linear pair of angles must be adja-cent but adjacent angles need not be linear pair. Do you agree? Draw a figure to support your answer.
Answer:
Yes.

Question 2.
Mahesh said that the sum of two angles 30° and 150° is 180°, hence they are linear pair. Do you agree? Justify your answer.
Answer:
No, sum of two angles is 180°, then they are said to be supplementary angles.
If the two angles are on the same straight line and they are adjacent they are said to be linear pair.
So, the two angles 30° and 150° are need not be linear pair.

Let’s Think [Page No. 70]

Question 1.
In the adjacent figure, AB is a straight line, O is a point on AB. OC is a ray. Take a point D in the interior of ∠AOC, join OD.
Find ∠AOD + ∠DOC + ∠COB.

Answer:
Given ∠AOC and ∠COB are linear pair. But ∠AOC = ∠AOD + ∠DOC

⇒ ∠AOC + ∠COB = 180° (linear pair)
⇒ ∠AOD + ∠DOC + ZCOB = 180°

Question 2.
In the given figure, AG is a straight line. Find the value of ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6.

Answer:
Given ∠AOC and ∠COG are linear pair. ∠AOC + ∠COG = 180° (linear pair)
But ∠AOC = ∠AOB + ∠BOC
= ∠1 + ∠2
∠COG – ∠COD + ∠DOE + ∠EOF + ∠FOG – ∠3 + ∠4 + ∠5 + ∠6
=> (∠AOB + ∠BOC) + (∠COD + ∠DOE + ∠EOF + ∠FOG)
= 180°
=> ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6
= 180°
Therefore, the sum of angles at a point on the same side of the line is 180°.

Let’s DO Activity [Page No. 72]

Take a white paper. Draw 3 distinct pairs of intersecting lines on this paper. Measure the angles so formed and fill the table.

From the above table, we observe that “vertically opposite angles are equal.

Check Your Progress [Page No. 73]

In the figure three lines p, q and r inter-sect at a point O. Observe the angles in the figure. Write answers to the following.
Question 1.
What is the vertically opposite angle to ∠1?
Answer:
Vertically opposite angle to ∠1 is ∠4.

Question 2.
What is the vertically opposite angle to ∠6?
Answer:
Vertically opposite angle to ∠6 is ∠3.

Question 3.
If ∠2 = 50°, then what is ∠5?
Answer:
Vertically opposite angle of ∠2 is ∠5. So, ∠5 = ∠2 = 50° ∠5 = 50°

Let’s Think [Page No. 75]

Question 1.
In the figure, the line l intersects other two lines m and n at A and B respectively. Hence l is a transversal. Is there

Give reason.
Answer:
Yes.
1) The line m intersects other two lines / and n at two distinct points A and C respectively. Hence m is a transversal line.
2) The line n intersects other two lines / and m at two distinct points B and C respectively. Hence n is a transversal line.

Question 2.
How many transversals can be drawn for the given pair of lines?
Answer:
One and only one transversal line can be drawn for the given pair of lines.

Check Your Progress [Page No. 76]

Observe the figures (i) and (ii) then fill the table.


Answer:

Check Your Progress [Page No. 78]

In the figure, p ∥ q and t is a transversal. Observe the angles formed.

Question 1.
If ∠1 = 100°, then what is ∠5?
Answer:
In the given figure ∠5 = Z1 (corresponding angles)
Given ∠1 = 100°
So, ∠5 = ∠1 = 100°
∴ ∠5 – 100°

Question 2.
If ∠8 = 80°, then what is ∠4?
Answer:
In the given figure ∠4 = ∠8 (corresponding angles)
Given ∠8 = 80°
So, ∠4 — ∠8 = 80°
∴ ∠4 = 80°

Question 3.
If ∠3 = 145°, then what is ∠7?
Answer:
In the given figure ∠7 = ∠3 (corresponding angles)
Given ∠3 = 145°
So, ∠7 = ∠3 = 145°
∴ ∠7 – 145°

Question 4.
If ∠6 = 30°, then what is ∠2?
Answer:
In the given figure ∠2 = ∠6 (corresponding angles)

Given ∠6 = 30°
So, ∠2 = ∠6 = 30°
∴ ∠2 = 30°

Let’s Think [Page No. 78]

Question 1.
What is the relation between alternate exterior angles formed by a transversal on parallel lines?
Answer:

Alternate exterior angles are equal. That is ∠1 = ∠7 and ∠2 = ∠8

Check Your Progress [Page No. 80]

In the figure, m∥ n and l is a transversal.

Question 1.
If ∠3 = 116°, then what is ∠5?
Answer:
In the figure ∠5 = ∠3 (Alternate interior angles)
Given ∠3 = 116°
So, ∠5 = ∠3 = 116°
∴ ∠5 = 116°

Question 2.
If ∠4 = 51°, then what is ∠5?
Answer:
In the figure the interior angles in the same side of transversal are supple-mentary’.
So, ∠5 + ∠4 = 180° we know ∠4 = 51°
∠5 + 51° = 180°
∠5 + 51° – 51° = 180° – 51°
=> ∠5 = 129°

Question 3.
If ∠1 = 123° then what is ∠7?
Answer:
In the given figure
∠7 = ∠1 (Alternative exterior angles) Given ∠1 = 123°
So, ∠7 = ∠1 = 123°
∴ ∠7 = 123° .

Question 4.
If ∠2 = 66° then what is ∠7?
Answer:
In the given figure,
sum of the exterior angles are the same side of transversal are supplementary. So, ∠2 + ∠7 = 180°
=> 66° + ∠7 = 180° (Given ∠2 = 66°)
=> 66° + ∠7 – 66° = 180° – 66°
=> ∠7 = 114°
∴ ∠7 = 114°

Let’s Think [Page No. 80]

Question 1.
What is the relation between co-exterior angles, when a transversal cuts a pair of parallel lines?

Answer:
Co-exterior angles are supplementary. That is ∠2 + ∠7 = 180° and ∠1 + ∠8 = 180°

Let’s Do Activity [Page N0. 80]

Take a white paper and draw a pair of non-parallel lines p and q and a transversal shown in the fIgure 1. Measure the corresponding angles and fill the table. Measure the pair óf corresponding angles and fill the table.

Check Your Progress [Page No. 81].

From the figure, state which property that is used in each of the following.

Question 1.
If ∠3 = ∠5, then p ∥ q.
Answer:
Given ∠3 = ∠5
Alternative interior angles are equal.

Question 2.
If ∠3 + ∠6 = 180°, then p ∥ q.
Answer:
Given ∠3 + ∠6 = 180°
Interior angles on the same Side of transversal are supplementary.

Question 3.
If ∠3 = ∠8, then p∥q.
Answer:
Given ∠3 = = ∠8 .
∠3 and ∠8 are not corresponding angles and not alternate interior angles.
So, ∠3 ≠ ∠8.

Let’s Explore [Page No. 81]

Question 1.
When a transversal intersects two lines and a pair of alternate exterior angles are equal, what can you say about the two lines?
Answer:
If a pair of alternate exterior angles are equal, then the two lines are parallel to each other.

Question 2.
When a transversal intersects two lines and a pair of co-exterior angles are supplementary, what can you say about the two-lines?
Answer:
If a pair of co-exterior angles are supplementary’ then the two lines are parallel to each other.

Examples:

Question 1.
In the given figure, ∠B and ∠E are complementary angles. Find the value of x.

Answer:
From the figure,
∠B = x + 10°and ∠E =35°
Since ∠B and ∠E are complementary angles,
∠B + ∠E =90°
⇒ x + 10° + 35° = 90°
⇒ x + 45° = 90°
⇒ x = 90°- 45°
x = 45°

Question 2.
If the ratio of supplementary angles is 4 : 5, then find the two angles.
Answer:
Given ratio of supplementary angles = 4:5
Sum of the parts in the ratio = 4 + 5 = 9
Sum of the supplementary angles = 180°
First angle = \(\frac{4}{9}\) × 180° = 80°
Second angle = \(\frac{5}{9}\) × 180 °= 100°

Question 3.
Find the linear pair of angles which are equal to each other?
Answer:
Let the equal linear pair of angles are x° and x°.
⇒ x° + x° = 180° .
⇒ 2x° = 180°
⇒ x° = \(\frac{180^{\circ}}{2}\)
∴ x° = 90°
Hence, each angle = 90°

Question 4.
In the given figure, PS is a straight line, find x°.

Answer:
From the given figure, ∠POQ = 60°
∠QOR = x°
∠ROS = 47°
But ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + x° + 47° = 180°
⇒ x° + 107° = 180°
⇒ x° = 180°- 107°
∴ x° = 73°

Question 5.
Observe the figure, then find x, y and z.

Answer:
From the figure, x = 110° (vertically opposite angles are equal)
y + 110° = 180°
y = 180°- 110° = 70°
z = y ⇒ z = 70°
Hence x = 110°, y = 70° and z = 70°

Question 6.
In the given figure AB ∥ CD and AE is transversal. If ∠BAC =120°, then find x and y.

Answer:
In the given figure, AB ∥ CD and AE is transversal.
∠BAC = 120
∠ACD = x
∠DCE = y
∠BAC = ∠DCE (correpsonding angles are equal)
y = 120°
x + y = 180 ° (Linear pair of angles are supplementary)
x + 120° – 180°
x = 180°- 120°
∴ x = 60°
Hence x = 60°, y =120°.

Question 7.
In the given figure, BA ∥ CD and BC is transversal. Find x.

Answer:
In the given figure, BA ∥ CD and BC is transversal.
∠C = x + 35° and ∠B = 60°
∠C = ∠B (∵ alternate interior angles are equal)
x + 35° = 60° .
x – 60° – 35°
∴ x = 25°

Question 8.
In the figure \(\overrightarrow{\mathbf{M N}} \| \overrightarrow{\mathbf{K L}}\) and \(\overline{\mathrm{MK}}\) is transversal. Find x.

Answer:
From the figure, \(\overrightarrow{\mathbf{M N}} \| \overrightarrow{\mathbf{K L}}\) and \(\overline{\mathrm{MK}}\) is transversal.
∠M = 2x and ∠K = x + 30°
∠M + ∠K = 180° (Q co-interior angles are supplementary)
⇒ 2x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30°
⇒ 3x = 150° ⇒ x = \(\frac{150^{\circ}}{3}\)
∴ x = 50°

Question 9.
In the figure \(\overline{\mathrm{AB}} \| \overline{\mathrm{DE}}\) and C is a point in between them. Observe the figure, then find x, y and ∠BCD.

Answer:
In the figure \(\overline{\mathrm{AB}} \| \overline{\mathrm{DE}}\) and C is a point in between them.
Draw a parallel line CF to \(\overline{\mathrm{AB}}\) through C.
\(\overline{\mathrm{AB}} \| \overline{\mathrm{CF}}\) and \(\overline{\mathrm{BC}}\) is a transversal,
x + 103° = 180°
x = 180°- 103°
x = 77°

From the figure,
\(\overline{\mathrm{DC}} \| \overline{\mathrm{CF}}\) and \(\overline{\mathrm{CD}}\) is a transversal,
y + 103° = 180°
y = 180°- 103°
y = 77°
and ∠BCD = x + y = 77° + 77° = 154°

Question 10.
In the figure transversal p intersects two lines m and n. Observe the figure, check whether m ∥ n or not.

Answer:
In the given figure, it is given that each angle in the pair of corresponding angles is 45°. So they are equal. Since a pair of corresponding angles are equal the lines are parallel. Hence, m ∥ n.

Practice Questions [Page No. 87]
Indicate the group (a, b, c, d, e, f) to which given below belongs to

Question 1.
State, District, Mandai
Answer:

Group: b

Question 2.
Boys, Girls, Artistists
Answer:

Group: c

Question 3.
Hours, Days, Minutes
Answer:

Group: b

Question 4.
Women, Teacher, Doctor
Answer:

Group: c

Question 5.
Food, Curd, Spoon
Answer:

Group: f

Question 6.
Humans, Dancer. Player
Answer:

Group: f

Question 7.
Building, Brick, Bridge
Answer:

Group: c

Question 8.
Tree, Branch, Leaf
Answer:

Group: b

Question 9.
Gold. Silver, Jewellary
Answer:

Group: f

Question 10.
Bulbs, Mixtures, Electricals
Answer:

Group: f

Question 11.
Women, Illiteracy, Men
Answer:

Group: c

Question 12.
Medicine, Tablets. Syrup
Answer:

Group: f

Question 13.
Carrots, Oranges, Vegetables
Answer:

Group: e

Question 14.
Female, Mothers. Sisters
Answer:

Group: f

Question 15.
Table. Furniture, Chair
Answer:

Group: f

Question 16.
Fruits, Mango. Onions
Answer:

Group: e

Question 17.
School, Teacher, Students
Answer:

Group: f

Question 18.
Rivers. Oceans. Springs
Answer:

Group: d

Question 19.
India. Andhra Pradesh, Visakhapatnam
Answer:

Group: b

Question 20.
Animals. Cows. Horses
Answer:

Group: f

Question 21.
Fish. Tiger, snakes
Answer:

Group: a

Question 22.
Flowers. Jasmine. Banana
Answer:

Group: e

Question 23.
Authors, teachers, Men
Answer:

Group: c

Question 24.
Dog. Fish, Parrot
Answer:

Group: a

Question 25.
Rose, Flower, Apple
Answer:

Group: e

Question 26.
School, Benches. Class Room
Answer:

Group: b

Question 27.
Pen, Stationary, Powder
Answer:

Group: e

Question 28.
Crow, Pigeon. Bird
Answer:

Group: f

Question 29.
Mammals, Elephants, Dinosaurs
Answer:

Group: f

Question 30.
Writers. Teachers, Researchers
Answer:

Group: d

REASONING [Practice Questions]

Question 1.
Musician, Instrumentalist, Violinist.
Answer:

Group: b

Question 2.
Officer, Woman, Doctor
Answer:

Group: c

Question 3.
Girls, Students, 7” class girls
Answer:

Group: d

Question 4.
Pencil, Stationary, Toothpaste
Answer:

Group: e

Question 5.
Food, Curd, Fruits
Answer:

Group: f

Question 6.
Bird Pigeon, Cow
Answer:

Group: e

Question 7.
Notes, Pad, Pen
Answer:

Group: a

Question 8.
Banana, Guava, Apple
Answer:

Group: a

Question 9.
Tomato, Food, Vegetables
Answer:

Group: b

Question 10.
Cow, Dog Pet
Answer:

Group: c

Practice the AP 10th Class Maths Bits with Answers Chapter 4 Pair of Linear Equations in Two Variables on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 4th Lesson Pair of Linear Equations in Two Variables with Answers

Question 1.
Pair of equations 4x + 6y = 7 and 2x + 3y = 8. How many solutions have ?
Answer:
No solution.

Question 2.
Find the point where the line 2x – 3y = 8 intersects X – axis ?
Answer:
(4,0)
Explanation:stitute y = 0 in 2x – 3y = 8
⇒ 2x – 3-0 = 8 ⇒ x = 4
∴ The point on X – axis = (4, 0).

Question 3.
Find the solution for the equations \(\sqrt{3} x+\sqrt{5 y}=0\) and \(\sqrt{7} \mathrm{x}+\sqrt{11} \mathrm{y}=0\)
Answer:
x = 0, y = 0.

Question 4.
Find the value of ‘x’ in the equation 3x- (x-4) = 3x + 1.
Answer:
3
Explanation:
3x – x + 4 = 3x + 1
⇒ 3x – 3x — x = 1 – 4
⇒ x = 3

Question 5.
The pair of equations a₁x + b₁y + C₁ = 0 and a₂x + b₂y + c₂ = 0 are consistent, then write the condition for that.
Answer:
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{~b}_{2}}\) and \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{~b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)

Question 6.
The graph y = ax + b is a straight line, find the point x where it intersects x – axis ?
Answer:
(\(-\frac{b}{a}\), 0)

Question 7.
Find the value of ‘k’ for which the sys-tem of equations kx – y = 2 and
6x – 2y = 3 has no solution.
Answer:
3
Explanation:
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{~b}_{2}}\) when pair of equations has no solution.
\(\frac{\mathrm{k}}{6}=\frac{1}{2} \Rightarrow \mathrm{k}=\frac{6}{2}\) = 3

Question 8.
Find the point of intersection of x + y = 6 and x – y = 4.
Answer:
(5,1)
Explanation:
x + y = 6 and
x-y = 4 ⇒ x = 4 + y
4 + y + y = 6 ⇒ 2y = 2 ⇒ y = 1
x = 6 – y = 6 – 1 ⇒ x = 5 .
∴ Point of intersection (x, y) = (5, 1).

Question 9.
If pair of equations 6x + 2y – 9 = 0 and kx + y – 7 = 0 has no solution, then find ‘k’.
Answer:
3
Explanation:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \Rightarrow \frac{6}{k}=\frac{2}{1} \Rightarrow \frac{6}{2}\) = k ⇒ k = 3

Question 10.
If the pair of equations 2x+3y+k = 0, 6x + 9y + 3 = 0 having infinite solu-tions, find the value of ‘k’.
Answer:
1
Explanation:
Infinite solutions, so

Question 11.
A pair of linear equations in two variables are 2x – y = 4 and 4x – 2y = 6. The pair of equations are
Answer:
Inconsistent.
Explanation:
\(\frac{2}{4}=\frac{1}{2} \neq \frac{4}{6} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ Pair of equations are inconsistent.

Question 12.
How many solutions to the pair of equations y = 0 and y = – 3 ?
Answer:
No solution.

Question 13.
If 7x – 8y = 9, then find ‘y’.
Answer:
\(\frac{7 x-9}{8}\)

Question 14.
Write the standard form of a linear equation.
Answer:
ax + by + c = 0

Question 15.
For which value of ‘k’ will the follow-ing pair of linear equations have no solution 3x + y = 1;
(2k- l)x + (k – l)y = 2k – 1 ?
Answer:
2
Explanation:
No, solution, so \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\)
⇒ \(\frac{3}{2 k-1}=\frac{1}{k-1}\)
⇒ 3(k – 1) = 1 (2k – 1)
⇒ 3k – 3 = 2k – 1 ⇒ 3k – 2k = 3 – 1 ⇒ k = 2

Question 16.
The lines 3x + 8y – 13 = 0 and – 6x – 16y + 23 = 0 are type of lines.
Answer:
Parallel
Explanation:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{3}{-6}=\frac{8}{-16} \neq \frac{-13}{23}\)
∴ Parallel lines.

Question 17.
2x + 3y = 1, 3x – y = 7, then find (x, y).
Answer:
(2,-1)

Question 18.
Where the line x – y = 8 intersects X – axis ?
Answer:
(8,0)

Question 19.
x + \(\frac { 6 }{ y }\) = 6 and 3x – \(\frac { 8 }{ y }\) = 5, then find ‘y’.
Answer:
2

Question 20.
Write the nature of the graph of the line y = 5x.
Answer:
The graph of the line passes through the origin.
Explanation:
y – 5x ⇒ y = mx passes through the origin.

Question 21.
Pair of linear equations px + 3y – (p – 3) = 0, 12x + py – p = 0 has infinitely many solutions, then find p’.
Answer:
± 6.
Explanation:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \Rightarrow \frac{p}{12}=\frac{3}{p}=\frac{(p-3)}{p}\)
p² = 36 ⇒ p = \(\sqrt{36}\) = ± 6

Question 22.
In which quadrant (2, 0) belongs ?
Answer:
Q₁

Question 23.
x + y = 10, x – y = – 4, then find ‘x’.
Answer:
3
Explanation:
x – y = -4 ⇒ x = y – 4 = y – 4 + y= 10
⇒ 2y – 4 = 10 ⇒ 2y = 14 ⇒ y = 7,
x – 7 = -4 ⇒ x = 7 – 4 = 3
∴ x = 3

Question 24.
Find the solution of the equations \(\sqrt{2} x+\sqrt{3} y=0\) and \(\sqrt{3} x-\sqrt{8} y=0\).
Answer:
x = 0, y = 0 (or) (0, 0)

Question 25.
If 3x + 4y + 2 = 0and9x + 12y + k = 0 represent coincident lines, then find the value of ‘k’.
Answer:
x = 0, y = 0 (or) (0.0)
Explanation:

Question 26.
If ax + b = 0, then find ‘x’. b
Answer:
\(-\frac{b}{a}\)

Question 27.
If x + y = 7, x – y = 1, then find 2x.
Answer:
8
Explanation:
x – y = 1 ⇒ x = 1 + y and
x + y = 7 ⇒ 1 + y + y = 7
⇒ 1 + 2y = 7 ⇒ 2y = 6 ⇒ y = 3,
x = 1 + 3 = 4
∴ 2x = 2(4) = 8

Question 28.
If x – y = 0; 2x – y = 2, then find the value of ‘y’.
Answer:
2

Question 29.
How many solutions to the pair of lin-ear equations 3x + 4y + 5 = 0 and 12x + 16y +15 = 0 have ?
Answer:
No solution. They are parallel lines.

Question 30.
Find the solution to \(\frac{a^{2}}{x}-\frac{b^{2}}{y}=0\) ; \(\) = a + b, x ≠ 0, y ≠ 0
Answer:
(a², b²)

Question 31.
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then find the value of ‘k’.
Answer:
k = \(\frac{15}{4}\)

Question 32.
The lines represented by 5x + 3y – 7 = 0 and 6y + 10x – 14 = 0 are type …………… of lines.
Answer:
Coincident
Explanation:
\(\frac{5}{10}=\frac{3}{6}=\frac{7}{14} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
So given pair of equations are coinci-dent lines.

Question 33.
Find slope of the line x = 2y.
Answer:
Slope (m) = \(\frac{1}{2}\)
Explanation:
x = 2y ⇒ y = \(\frac { 1 }{ 2 }\) x ⇒ m = \(\frac { 1 }{ 2 }\)

Question 34.
If x = 1 and y = \(\frac{1}{2}\), then find x – y.
Answer:
\(\frac{3}{2}\)
Explanation:
x – y = 1 – \(\frac { -1 }{ 2 }\) = 1 + \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\)

Question 35.
Find the value of x if y = \(\frac{3}{4}\) x and 5x + 8y = 33.
Answer:
x = 3

Question 36.
Write the slope of X – axis.
Answer:
y = 0

Question 37.
Find the value of y in – 5x + 10y =100 at x = 0.
Answer:
y = 10

Question 38.
2u + 3v = 2 and 4u – 6v = 0, then find ‘v’.
Answer:
\(\frac{1}{3}\)

Question 39.
If – x + y = – 10, then write ‘x’ as sub-ject.
Answer:
x = y + 10
Explanation:
-x + y = -10 ⇒ x = y + 10

Question 40.
Write shape of the graph of 3x-y = -1.
Answer:
Straight line.

Question 41.
The pair of linear equations
px + 2y = 5 and 3x + y = 1 has unique solution, then find value of’p’.
Answer:
p ≠ 6.
Explanation:
\(\frac{p}{3} \neq \frac{2}{1}\) ⇒ p ≠ 6

Question 42.
The lines represented by 5x + 7y – 14 = 0 and 10x + 3y-8 = 0 are……………….lines.
Answer:
Consistent.

Question 43.
3x – 5y = – 1 and – y + x = – 1, then find (x, y).
Answer:
(-2,-1).

Question 44.
How many solutions to the pair of lin-ear equations – 3x + 4y = 7 and
\(\frac { 9 }{ 2 }\) x – 6y + \(\frac { 21 }{ 2 }\) = 0 ?
Answer:
Infinitely many solutions.
Explanation:

So infinitely, many solutions.

Question 45.
For which value of ‘p’ the lines repre-sented by 8x + 2py = 2 and 2x + 5y + 1 = 0 are parallel ?
Answer:
10
Explanation:
\(\frac{8}{2}=\frac{2 p}{5}\) ⇒ P = \(\frac{8 \times 5}{2 \times 2}\) ⇒ P = 10

Question 46.
If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\), then the lines are …………………. type of lilies.
Answer:
Coincident lines (or) dependent lines.

Question 47.
4m – 2n = 2 and 6m – 5n = 9, then find ‘n’.
Answer:
n = – 3.

Question 48.
If 99x + lOly = 499,101x+ 99y = 510, then find ‘x’.
Answer:
x = 3

Question 49.
Find the solution to x – y = 1 and 2x – 2y = 7.
Answer:
No solution (or) not possible.
Explanation:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{2}=\frac{1}{2}\),
so no solution.

Question 50.
141x + 93y = 189, 93x + 141y = 45, then find y’.
Answer:
y = -1

Question 51.
\(\frac{120}{x}+\frac{12}{x}\) = 11, then find ‘x’. x x
Answer:
x = 12.

Question 52.
Find the solution to 2x – 2y – 2 = 0, 4x – 4y – 5 = 0.
Answer:
No solution (or) not possible.

Question 53.
500x + 240y = 8, 130x + 240y = \(\frac{43}{10}\)
then find the value of ‘x’.
Answer:
\(\frac{1}{100}\)

Question 54.
Find the value of ‘y’ when \(\frac{x+y}{x y}\) = 2 and \(\frac{x-y}{x y}\) = 6.
Answer:
y = \(\frac { 1 }{ 4 }\)

Question 55.
Where the two lines 2x + y – 6 = 0 and 4x – 2y – 4 = 0 intersect, find that intersecting point.
Answer:
(2,2)

Question 56.
\(\frac{x+3}{2}-y=2, \frac{x-3}{2}+2 y=4 \frac{1}{2}\) then find ‘x’
Answer:
\(\frac { 14 }{ 3 }\)

Question 57.
How much the angle between any two parallel lines ?
Answer:
0°

Question 58.
Find the values of ’k’ for which the pair of linear equations 3x – 2y = 7, and 6x + ky + 11 = 0 has a unique solution.
Answer:
All numbers expect “’-“4r are the solution.

Question 59.
Find slope of the line y = x.
Answer:
m = 1

Question 60.
If x = 1, then find the value of’y’ satis- 5 3
fying the equation \(\frac{5}{x}+\frac{3}{y}\) = 6 ;
Answer:
y = 3.
Explanation:
\(\frac{5}{1}+\frac{3}{y}\) = 6 ⇒ \(\frac{3}{y}\) = 6 – 5 = 1 ⇒ y = 3

Question 61.
Write the point (- 3, – 8) is in the…………………quadrant.
Answer:
Q₃

Question 62.
Write the point (7, -5) is in the quadrant.
Answer:
Q₄

Question 63.
Find slope of the line ax + by + c = 0.
Answer:
m = \(\frac{-a}{b}\)

Question 64.
Write the nature of the line x = 2020.
Answer:
Slope not defined and it parallel to Y – axis.

Question 65.
Write the nature of the line x = 7.
Answer:
It is parallel to Y – axis.

Question 66.
Write nature of the graph of a pair of linear equations in two variables is represented by
Answer:
Straight lines.

Question 67.
Find the number of solutions to the pair of equations 6x – 7y + 8 = 0 and 12x – 14y +16 = 0.
Answer:
Infinitely many solutions.

Question 68.
If 2^x + 3^y = 17 and 2^{x + 2} – 3^{y + 1} = 5, then find y’.
Answer:
y = 2.
Explanation:
2^x + 3^y = 17
⇒ a + b = 17
⇒ 3a + 3b = 51
2^x . 2² – 3^y . 3¹ = 5
⇒ 4a – 3b = 5 ……………….. (ii)
Solving equations (1) & (2)
7a = 56 ⇒ a = 8,
b = 17-a = 17-8 = 9
3^y = b = 3² ⇒ y = 2

Question 69.
If \(\frac{2}{x}+\frac{3}{y}\) = 13 and \(\frac{5}{x}-\frac{4}{y}\) = -2
find the solution.
Answer:
( \(\frac{1}{2}, \frac{1}{3}\) )

Question 70.
If 5x + py + 8 = 0 and 10x + 15y + 12 = 0 has no solution, then find the value of
p’.
Answer:
P = 7\(\frac{1}{2}\)

Question 71.
The larger of two supplementary angles exceeds the smaller by 38°. Find them.
Answer:
71° and 109°.
Explanation:
x, 180 – x =⇒ y = 180 – x
x + 38 = 180 – x
⇒ 2x = 180-38 = 142
x = \(\) = 71°
⇒ y = 180-71 = 109°

Question 72.
Write the number of solutions to 4x + 6y – 7 = 0 and 8x + 5y – 8 = 0.
Answer:
Only one solution.

Question 73.
Find the number of solutions to the pair of equation llx – 7y = 6 and 4x + 9y = 8.
Answer:
Only one solution.

Question 74.
\(\frac{2}{x}+\frac{3}{y}\) = 2, \(\frac{12}{x}-\frac{9}{y}\) = 3, then find ’x’.
Answer:
x = 2.

Question 75.
Write the pair of equations 4x – 2y + 6 = 0 and 2x – y + 8 = 0 has ……………. solutions.
Answer:
No solution.

Question 76.
Sita has pencils and pens which are together 40 in number. If she has 5 less pencils and 5 more pens the number of pens become four times the number of pencils. Represent this situation in a linear equation form.
Answer:
x + y = 40

Question 77.
Which type of equations
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are …
Answer:
Pair of linear equations.

Question 78.
If ax + by = c and px + qy = r has unique solution, then write the condition.
Answer:
\(\frac{a}{b}=\frac{p}{q}\)

Question 79.
If the pair of equations kx + 14y + 8 = 0 and 3x + 7y + 6 = 0 has a unique so-lution, then find ’k’.
k ≠ 6

Question 80.
\(\frac{\mathbf{a x}}{\mathbf{b}}-\frac{\mathbf{b y}}{\mathbf{a}}\) = a + b, ax – by = 2ab, then find ‘x’.
Answer:
x = 3b

Question 81.
The pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has a unique solution, then write the condition.
Answer:
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

Question 82.
Write the type of pair of lines
3x – 2y + 6 = 0, 6x – 4y + 8 = 0 are represents
Answer:
Parallel lines.

Question 83.
The age of a daughter is one third the age of her father. If the present age of father is ‘x’ years, then write the age of the daughter after 18 years in linear form.
Answer:
y = \(\frac{\mathrm{x}}{3}\) + 18

Question 84.
How many solutions \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{\mathbf{2}}}=\frac{\mathbf{b}_{1}}{\mathbf{b}_{\mathbf{2}}}=\frac{\mathbf{c}_{1}}{\mathbf{c}_{\mathbf{2}}}\) will have ?
Answer:
Infinitely many solutions.

Question 85.
\(\frac{x+1}{2}+\frac{y-1}{3}=9, \frac{x-1}{3}+\frac{y+1}{2}=8\) then find x’.
Answer:
x = 13.

Question 86.
If the pair of equations 2x + y = 7 and 6x – py – 21 = 0 has infinite number of solutions, then find p.
Answer:
p = – 3
Explanation:
Infinite solutions,
so \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \Rightarrow \frac{2}{6}=\frac{1}{-p}\)
⇒ \(\frac{1}{3}=\frac{-1}{\mathrm{p}}\)
⇒ p = – 3

Question 87.
The ratio of incomes of two persons is 11:7 and the ratio of their expendi-tures is 9 : 5. If each of them manages to save ₹400 per month, then find the monthly income of first person.
Answer:
₹ 2200
Explanation:
11x – 9y = 400 and 7x – 5y = 400
On solving these equations income of first person was ₹ 2200.

Question 88.
Where the two lines 2x – y = 1, x + 2y = 13 will intersect each other ?
Answer:
(3,5)

Question 89.
Find the lines x – y = 1; 2x + y = 8 where they intersects at each other ?
Answer:
(3,2)
Explanation:
x = y + 1, 2(y + 1) + y = 8
⇒ 2y + 2 + y = 8
⇒ 3y = 6 ⇒ y = 2
⇒ x – 2 = 1 ⇒ x = 3

Question 90.
Write a line parallel to the line x + 2y + 1 = 0.
Answer:
2x + 4y + 1 = 0

Question 91.
If the equations (2m – l)x + 3y-5=0, 3x + (n – 1 )y – 2 — 0 has infinite number of solutions, then find ‘n’.
Answer:
n = \(\frac{11}{5}\).

Question 92.
Find where the line 2x + y = 7 inter-sects X – axis ?
Answer:
(\(\frac{7}{2}\), 0)

Question 93.
If \(\frac{5}{x-1}+\frac{1}{y-2}=2, \frac{6}{x-1}+\frac{-3}{y-2}=1\) then find ‘x’.
Answer:
x = 4

Question 94.
For what value of It, 2x + 3y = 4 and (k + 2)x + 6y = 3k + 2 will have infi-nitely many solutions ?
Answer:
k = 2
Explanation:
Infinitely many solutions, so

k + 2 = 4
k = 4 – 2 = 2

Question 95.
A fraction becomes \(\frac { 9 }{ 11 }\) if ‘2’ is added to both numerator and denominator. If ‘3’ is added to both numerator and denominator it becomes \(\frac { 5 }{ 6 }\), then find
the fraction.
Answer:
\(\frac { 7 }{ 9 }\) = fraction.
Explanation:
\(\frac{x+2}{y+2}=\frac{9}{11}\)
⇒ 11x + 22 = 9y + 18
⇒ 11 x – 9y = -4
\(\frac{x+3}{y+3}=\frac{5}{6}\)
⇒ 6x + 18 = 5y + 15
⇒ 6x – 5y = – 3
On solving these equations x = 7 and y = 9
∴ Fraction = \(\frac{x}{y}=\frac{7}{9}\)

Question 96.
Find the value of It’ for which the sys-tem of equations kx + 3y = 1, 12x + ky = 2 has no solution.
Answer:
k = – 6

Question 97.
The age of a father 8 years ago was 5 times that of his son 8 years. Hence, his age will be 8 years more than twice the age of his son. Then find the present age of father.
Answer:
48 years.

Question 98.
In the above problem, find the age of son.
Answer:
16 years.

Question 99.
If ad ≠ be, then find the pair of linear equations ax + by = p and cx + dy =q has how many solutions ?
Answer:
2 solutions.

Choose the correct answer satisfying the following statements.

Question 100.
Statement (A): Pair of linear equations 9x + 3y 4- 12 = 0, 18x 4- 6y + 24 = 0 have infinitely many solutions.
Statement (B): Pair of linear equations a₁x + b₁y + c₁ = 0 , a₂x + b₂y + c₂ = 0 have infinitely many solutions, if
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
From the given equations, we have

So, both A and B are correct and B explains A. Hence, (i) is the correct option.

Question 101.
Statement (A) : For k = 6, the system of linear equations x + 2y + 3 = 0 and 3x + ky + 6 = 0 is inconsistent. Statement (B) : The system of linear equations a₁x + b₁y + c₁ = 0,
a₂x + b₂y + c₂ = 0 is inconsistent if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
For inconsistent solution we have bs
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
So, A is correct but B is incorrect.
Hence, (ii) is the correct option.

Question 102.
Statement (A): The value of q = ±2, if x = 3, y = 1 is the solution of the line 2x + y – q² – 3 = 0.
Statement (B): The solution of the line will satisfy the equation of the line.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
As x = 3, y = 1 is the solution of
2x + y – q² – 3 = 0
⇒ 2 x 3 + 1 – q² – 3 = 0
⇒ 4 – q² = 0 ⇒ q² = 4 ⇒ q = ±2
So, both A and B are correct and B explains A.
Hence, (i) is the correct option.

Question 103.
Statement (A): The value of k for which the system of equations kx – y = 2, 6x – 2y = 3 has a unique solution is 3.
Statement (B) : The system of linear equations a₁x + b₁y + c₁ = 0,
a₂x + b₂y + c₂ = 0 has a unique solution if \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
Given system of linear equations has a
unique solution, if \(\frac{\mathrm{k}}{6} \neq \frac{1}{2}\)
⇒ \(\frac{\mathrm{k}}{6} \neq \frac{1}{2}\) ⇒ k ≠ 3
So, A is incorrect and B is corrrect.
Hence, (iii) is the correct option.

Question 104.
Statement (A) : The lines 2x – 5y = 7 and 6x – 15y = 8 are parallel lines.
Statement (B) : The system of linear equations a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 have infinitely many
solutions if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\).
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel,
if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
So, both A and B are correct.
Hence, (i) is the correct option.

Question 105.
Statement (A) : The system of equations 2x + y + 3 = 0 and 2x 4- y – 3=0 has no solution.
Statement (B) : The system of equations a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 has a unique solution when \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 106.
Statement (A) : If the system of equations 2x + 3y = 7 and 2ax + (a + b)y = 28 has infinitely many solutions, then 2a – b = 0.
Statement (B) : The system of equations x – 5y = 3 and 2x- lOy = 5 has a unique solution.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 107.
Statement (A): If the pair of lines are coincident, then we say that pair of lin-ear equations is consistent and it has a unique solution.
Statement (B) : If the pair of lines are parallel, then the pair of linear equations has no solution and is called inconsistent pair of equations.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 108.
Statement (A) : 3x + 4y – 5 = 0 and 6x -I- ky + 9 = 0 represent parallel lines if k = 8.
Statement (B): a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 represent parallel lines if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
In statement (A), given lines represent parallel lines, if
\(\frac{3}{6}=\frac{4}{\mathrm{k}} \neq \frac{5}{9} \Rightarrow \mathrm{k}=\frac{6 \times 4}{3}=8\)
∴ Statement (A) is true.
∴ Statement (B) is also true.
∴ Since reason is the correct explana¬tion for statement (A).
Option (i) is true.

Question 109.
Statement (A) : x + y – 4 = 0 and 2x + ky – 3 = 0 has no solution if k = 2.
Statement (B): a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 are consistent if \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
For (A), given equation has no solution, if
\(\frac{1}{2}=\frac{1}{\mathrm{k}} \neq \frac{-4}{-3}, \text { i.e., } \frac{4}{3}\)
⇒ k = 2 [ \(\frac{1}{2} \neq \frac{4}{3}\) holds]
∴ (A) is true.
Since (B) does not give result of (A), so option (i) is true.

Question 110.
Statement (A): If the system of equa-tions 2x + 3y = 7 and 2ax + (a + b)y = 28 has infinitely many solutions, then 2a – b = 0.
Statement (B) : The system of equations 3x – 5y = 9 and 6x- lOy = 8 has a unique solution. –
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
For (A), given system of equations has infinitely many solutions, if
\(\frac{2}{2 a}=\frac{3}{a+b}=\frac{-7}{-28}, \text { i.e., }\)
⇒ \(\frac{1}{a}=\frac{3}{a+b}=\frac{1}{4}\)
⇒ 3a = a + b ⇒ 2a – b = 0
Also clearly a = 4 and
a + b = 12 ⇒ b = 8
∴ 2a – b = 8 – 8 = 0
∴ (A) is true.
But (B) is false ∵ \(\frac{3}{6}=\frac{-5}{-10}\)
[∵ 3(-10) = (-5) (6) = -30]
For unique solution, if
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
∴ Option (ii) is true.

Question 111.
Statement (A): If kx – y – 2 = 0 and 6x – 2y – 3 = 0 are inconsistent, then k = 3.
Statement (B): a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 are inconsistent if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
Statement (A) is true.

⇒ k = 3.
Statement (B) is also true.
∴ Option (i) is true.

Question 112.
Statement (A): 3x – 4y = 7 and
6x – 8y = k have infinite number of solutions if k = 14.
Statement (B): a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 have a unique solution if \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 113.
Statement (A) : The linear equations x – 2y – 3 = 0 and 3x + 4y – 20 = 0 have exactly one solution.
Statement (B) : The linear-equations 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 have a unique solution.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 114.
Statement (A) : kx + 2y = 5 and 3x + y = 1 have a unique solution if
k = 6.
Statement (B) : x + 2y = 3 and 5x + ky + 7 = 0 have a unique solution k ≠ 1.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iv)

Read the below passages and answer to the following questions.

If we have two simultaneous equations ax + by = c and bx + ay — d, l (c + d c-d^
then x = \(\frac{1}{2}\left(\frac{\mathrm{c}+\mathrm{d}}{\mathrm{a}+\mathrm{b}}+\frac{\mathrm{c}-\mathrm{d}}{\mathrm{a}-\mathrm{b}}\right)\) and y = \(\frac{1}{2}\left(\frac{c+d}{a+b}-\frac{c-d}{a-b}\right)\)

Question 115.
Find the solution of 217x + 131y = 913 and 131x + 217y = 827.
Answer:
x = 3, y = 2
Explanation:
We have 217x + 131y = 913
131x + 217y = 827

Question 116.
Find the solution of 37x + 41y = 70 and 4lx + 37y = 86.
Answer:
x = 3, y = – 1
Explanation:
We have,
37x + 41y = 70
41x + 37y = 86

Question 117.
Find the solution of x + 2y = \(\frac { 3 }{ 2 }\) and 2x + y = \(\frac { 3 }{ 2 }\)
Answer:
x = \(\frac { 1 }{ 2 }\), y = \(\frac { 1 }{ 2 }\)
A system of linear equations is given as follows :
a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0
Explanation:
We have

Question 118.
Write the condition for two lines to have a unique solution.
Answer:
\(\frac{a_{1}}{a_{2}} \neq \frac{c_{1}}{c_{2}}\)

Question 119.
Write the condition for two lines to have infinitely many solutions.
Answer:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

Question 120.
Write the condition both lines are par-allel only.
Answer:
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
6 pencils and 4 notebooks together cost ₹ 90 whereas. 8 pencils and 3 notebooks together cost ₹ 85.

Question 121.
Create an equation to first situation.
Answer:
6x + 4y = 90.

Question 122.
Create an equation to second situation.
Answer:
8x + 3y = 85

Question 123.
Which mathematical concept is used to find the cost of notebook and pencil ?
Answer:
Pair of linear equations.
A boat goes 30 km upstream and 44 km downstream in 10 hrs. In 13 hrs it can go 40 km upstream and 55 km downstream.

Question 124.
Prepare an equation to first condition.
Answer:
\(\frac{30}{x-y}+\frac{44}{x+y}=10\)

Question 125.
Prepare an equation to second condition.
Answer:
\(\frac{40}{x-y}+\frac{55}{x+y}=13\)

Write the correct option to match the column -I and column – II, which gave value of ‘x1 and ‘y’ for pair of equation given in column -I.

Question 126.

Answer:
A – (iv), B – (ii)

Question 127.

A – (iii), B – (i)

Question 128.

A – (ii), B – (iv)

Question 129.

A – (i), B – (iii)

Answer Questions 130 and 131 based on the data given below.
“The cost of 1 kg potatoes and 2kg to-matoes was ₹ 30 on a certain day. After two days the cost of 2 kg potatoes and 4 kg tomatoes was found to be ₹ 66”.

Question 130.
Write a pair of linear equations in two variables x and y from the datAnswer:
Solution:
x + 2y = 30, 2x + 4y = 66 (or) x + 2y = 33

Question 131.
Which system of linear equations in two variables does the data represent ?
Answer:
Parallel lines, inconsistent, no solution.

Question 132.
For what value of ‘k’ is the pair of linear equations x + 2y = 7 and3x – ky = 21 has infinitely many solutions ?
Answer:
Given equations are x + 2y = 7 and 3x – ky = 21 has infinitely many solutions, so

Question 133.
What is the value of ‘x’ in 4x – 7y = 9 ify = 3?
Answer:
Given , 4x – 7y = 9
If y = 3, then 4x – 7(3) = 9
=> 4x-21 = 9
=> 4x = 30
x = \(\frac{30}{4}=\frac{15}{2}\)

Question 134.
Lahari bought two pens and five pencils spending Rs. 30. Express this information as a linear equation in variables x and y.
Answer:
Let the cost of each pen be ₹ x and the cost of each pencil be ₹ y.
By problem, 2x + 5y = 30

Practice the AP 10th Class Maths Bits with Answers Chapter 2 Sets on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 2nd Lesson Sets with Answers

Question 1.
Which type of set of human being that reside on moon is ……………….
Answer:
null set

Question 2.
Write the number of subsets of the null set Φ.
Answer:
1

Question 3.
Ifn(A) = 8, n(B) = 3, n(A ∩ B) = 2, then find n(A ∪ B).
Answer:
9
Explanation:
n(A∪B) = n (A) + n (B) – n (A ∩ B)
= 8 + 3-2 = 9

Question 4.
The number of subsets of a set is 16, then find the set has ………… elements.
Answer:
4
Explanation:
2ⁿ = 16 = 2⁴
⇒ no. of elements in the set = 4

Question 5.
Write the number of subsets of the set A = {l, 2,3, 4}.
Answer:
16
Explanation:
n (A) = 4, no. of subsets = 2ⁿ = 2⁴ = 16

Question 6.
If A⊂ B, n(A) = l2and n(13) = 20, then find the value of n (B – A).
Answer:
8
Explanation:
A ⊂ B, son (B – A) = 20- 12 .= 8

Question 7.
Roster form of (x: x is a prime number and a divisor of 6).
Answer:
{2,3}

Question 8.
Write an example for finite set in your own.
Answer:
{x/x∈N and x² = 9}

Question 9.
If A⊂B,n(A) = 4 and n(B) = 6,then find n(A∪ B).
Answer:
6
Explanation:
A ⊂ B, so n (A ∪ B) = n (B) = 6

Question 10.
If A⊂B, then A∩B is
Answer:
A
Explanation:
A⊂B, so A∩B = A

Question 11.
If the union of two sets is one of the set itself, write the relation between the two sets.
Answer:
One set is a subset of other set.

Question 12.
The following venn diagram indicates

Answer:
A⊂B

Question 13.
If A From the venn diagram, find A ∪ B.

Answer:
{5, , 7, 8}

Question 14.
If A = {x : x is a letter in the word EX¬AMINATION}, then write its roster form.
Answer:
A = {e, x, m, i, n, a, t, o}

Question 15.
If A = {x : x is a letter in the word HEADMASTER}; then write its ros-ter form.
Answer:
A — {h, e, a, d, m, s, t, r}

Question 16.
If n (A) = 12 and n (A ∩ B) = 5, then find n (A – B).
Answer:
7
Explanation:
n (A – B) = n(A) – n(A∩B) = 12 – 5 = 7

Question 17.
The following venn diagram indicates

Answer:
A, B are disjoint sets.

Question 18.
The shaded region in the given figure shows.

Answer:
μ – B = B’

Question 19.
Write the relation between sets in the following venn diagram.

Answer:
A ∩ B = Φ

Question 20.
If A ={1,2, 3}, B = (3,4, 5), then find A Δ B.
Answer:
A Δ B = {1,2, 4, 5}
Explanation:
A Δ B = (A ∪ B) – (A ∩ B)
= {1, 2, 3, 4, 5}- {3} = {1,2, 4, 5}

Question 21.
(A’)’is equal to …………….
Answer:
A

Question 22.
An object of a set is called ……………..
Answer:
Element

Question 23.
2 is ………………. of set of natural numbers.
Answer:
An element

Question 24.
Number of elements in a singleton set is …………………..
Answer:
1

Question 25.
If A, B are disjoint sets such that n (A) = 4 and n (A ∪ B) = 7, then find n(B).
Answer:
3
Explanation:
n(A∪B) = n (A) + n (B) – n (A ∩ B)
⇒ 7 = 4 + n (B) – 0
⇒ n(B) = 7 – 4 = 3

Question 26.
‘O’ is to set of whole numbers.
Answer:
belong

Question 27.
n (A) = 4, then write n(p(A)).
Answer:
16
Explanation:
n(P(A)) = 2ⁿ = 2⁴ = 16

Question 28.
If A = {1, 2, 3} and B = {1, 2, 3, 4}, then we say A is a …………….. of B.
Answer:
Subset

Question 29.
Φ is equal to A.
Answer:
μ

Question 30.
A set is a ……………… of objects.
Answer:
Well defined collection.

Question 31.
{2, 4,6, 8, 10} is an example of which type of set ?
Answer:
Finite

Question 32.
If A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, then find A – B.
Answer:
{1,3}

Question 33.
A = {2, 4, 6, 8, 10}, then write its rule form.
Answer:
A = {x / x is an even number, x ≤ 10}

Question 34.
If B = {1, 7, 2, 0, 6}, then find n(B).
Answer:
5

Question 35.
A – (A -B) is equal to ……………..
Answer:
A ∩ B

Question 36.
The objects in the set are called ……………….. of the set.
Answer:
Elements

Question 37.
Let A, B are two sets such that n (A) = 5, n(B) = 7, then write the maximum number of elements in A ∪ B.
Answer:
12

Question 38.
Empty set is denoted by ………………..
Answer:
Φ

Question 39.
Write A Δ B.
Answer:
(A – B) ∪

(B – A) (or) (A∪B)-(A ∩ B)

Question 40.
A = {1, 2, 3}, B = {3, 4, 5}, then find A ∩ B.
Answer:
{3}

Question 41.
– 3 is of the set of whole numbers.
Answer:
not an element

Question 42.
If n(A ∪ B) = 8, n(A) = 6, n(B) = 4, then find n(A ∩ B).
Answer:
2

Question 43.
The number of elements in a set is called the…………….of the set.
Answer:
Cardinal number

Question 44.
A ∪ Φ is equal to …………………
Answer:
A

Question 45.
{x / x ≠ x} is which type of set ?
Answer:
Empty

Question 46.
B = {x/x ∈ N and x < 1000} is a ……………type of set.
Answer:
Finite

Question 47.
Write the symbol used for belongs to’.
Answer:
∈

Question 48.
n (Φ) is equal to ………………..
Answer:
0

Question 49.
Write (2, 6, 10} ∩ (8, 9, 11, 12, 13}.
Answer:
Φ

Question 50.
{x / x is a student of your school} is in which form ?
Answer:
Set Builder

Question 51.
Every set is ……………. of itself.
Answer:
Subset

Question 52.
A = {1, 2,7, 10}, then use symbol be-tween 7 and A.
Answer:
∈

Question 53.
If A = {1, 2, 3, 4}, then find the cardi-nality of set A.
Answer:
4

Question 54.
A≠B means, set A and B do not contains same elements. This statement is true (or) false.
Answer:
True

Question 55.
A = {1, 2, 3}, B = {12, 0, 5}, then find A-B.
Answer:
A

Question 56.
A = {x / x + 4 = 4}, then write Roster form of A.
Answer:
{0}

Question 57.
Which type of set has no elements in it ?
Answer:
Null set

Question 58.
If A ∪ B = A ∪ C and A ∩ B = A ∩ C, then write the relation between these sets.
Answer:
B = C

Question 59.
A set with only one element is known as ……………. set.
Answer:
singleton

Question 60.
The set of all real numbers is, which type of set ?
Answer:
Infinite set

Question 61.
Roster form of B = \(\left\{\frac{x}{x}+3=6\right\}\), B = ?
Answer:
{3}

Question 62.
‘μ’ is equal to
Answer:
Φ

Question 63.
If A ⊂ B and A ≠ B, then A’ is called the ………………. of B.
Answer:
Proper subset

Question 64.
{x / x is a natural number} is which type of set ?
Answer:
Infinite

Question 65.
Write the number of elements in the empty set.
Answer:
0

Question 66.
The null set is sometimes denoted as .
Answer:
{} = Φ

Question 67.
If in two sets A and B, every element of A is in B and every element of B is in A, then write it as
Answer:
A = B

Question 68.
Another name to Roster form is ……………….
Answer:
List

Question 69.
A’ – B’ is equal to
Answer:
B-A

Question 70.
If every element of A is also an element of B, then write this symboliically.
Answer:
A⊂B

Question 71.
A ⊂ B, then find A – B.
Answer:
Φ

Question 72.
“0 does not belong to the set of natural numbers”. Write the statement sym¬bolically.
Answer:
0 ∉ N

Question 73.
A = {1,2, 4}, B = {3, 5, 6}, then write the relation.
Answer:
A ∩ B = Φ

Question 74.
If A ⊂ B, then find A ∪ B.
Answer:
B

Question 75.
If B = {1,7, 2, 0,6}, then find n(B).
Answer:
5

Question 76.
Write Roster form of the set of natu¬ral numbers less than 6.
Answer:
(1, 2, 3, 4, 5}

Question 77.
If A ⊂ B, then find A-B.
Answer:
Φ

Question 78.
A ∪ Φ is equal to ……………
Answer:
A

Question 79.
A ∪ B = B ∪ A is called ……………. law.
Answer:
Commutative

Question 80.
If A = {1, 2, 2, 1, 3, 4, 3, 4}, then find n(A).
Answer:
4

Question 81.
Write cardinal number of null set.
Answer:
0

Question 82.
K = {x/x is a prime number less than 13}. Write list form of K.
Answer:
K= {2, 3, 5, 7, 11}

Question 83.
W – {0} is equal to …………………
Answer:
N

Question 84.
In the rule form, the slant bar stands for
Answer:
such that

Question 85.
A = {a, b, c}, B = {c, a, b}, then write the relation between A and B.
Answer:
A – B

Question 86.
Write the set formed from the letters of the word “SCHOOL “.
Answer:
{S, C, H, O, L}

Question 87.
A = {1, 2, 7}, B = {2, 1}, then write the relation between A and B.
Answer:
B⊂A

Question 88.
If A ⊂ B, B ⊂ C, then write the relation between A and C.
Answer:
A ⊂ C

Question 89.
If A ⊂ B, then find A∪(B – A).
Answer:
B

Question 90.
Write the set builder form of D = \(\left\{1, \frac{1}{2}, \frac{1}{3} ; \frac{1}{4}, \frac{1}{5}, \frac{1}{6}\right\}\)
Answer:
D = {x / x ∈ 1/n ,n ∈ N,n < 7}

Question 91.
In set builder form, the letter “X” denotes any………… that belongs to the set.
Answer:
Arbitrary element.

Question 92.
Write the Roster form of the set of multiples of 5 which lie between 25 and 50 is
Answer:
{30, 35, 40, 45}

Question 93.
Write the name of German mathemati¬cian who developed the theory of sets.
Answer:
George Cantor.

Question 94.
N∩W is equal to ………………..
Answer:
N

Question 95.
A = Φ, B = Φ, then find A∩B.
Answer:
Φ

Question 96.
Write the identity element under union of sets.
Answer:
Φ

Question 97.
A ∩ B = Φ, then find B ∩ A’.
Answer:
B

Question 98.
A = {all primes less than 20}
B = {all whole numbers less than 10}, then find A∩B.
Answer:
{2,3, 5, 7}

Question 99.
μ’ = Φ is called …………….. law.
Answer:
Complementary

Question 100.
A ∪ A = A is called……………law.
Answer:
Idempotent.

Question 101.
If A and B are disjoint sets, then write n (A ∪ B).
Answer:
n (A) + n (B)
Explanation:
n (A ∪ B) = n (A) + n (B)

Question 102.
If A = Φ, B = Φ, then find A∪B.
Answer:
Φ

Question 103.
n (A) = 3, then write the number of proper subsets of A.
Answer:
7
Explanation:
Proper subsets are 2ⁿ – 1 = 2³ – 1
= 8 – 1
= 7

Question 104.
A ∪ B = A ∩ B, then write the relation between A and B.
Answer:
A = B

Question 105.
n (A ∪ B) = 51, n (A) = 20, n (A ∩ B) = 13, then find n (B).
Answer:
44
Explanation:
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
⇒ 51 = 20 + n(B)- 13
⇒ 31 + 13 = n(B) = 44

Question 106.
A’ = B, then find A ∪ B.
Answer:
μ
Explanation:
A’= B ⇒ μ – A = B
⇒ A u B = μ

Question 107.
n (A) = 10, n (B) = 4, n (A ∩ B) = 2, then find n (A ∪ B).
Answer:
12

Question 108.
μ ∪ Φ is equal to
Answer:
μ

Question 109.
If A ∩ B = Φ then find n (A ∩ B).
Answer:
n (A) + n (B)

Question 110.
Write the identity element under intersection of sets.
Answer:
μ

Question 111.
A∪B = B, then write the relation between A and B.
Answer:
A⊂B

Question 112.
The given venn diagram represents.

Answer:
AΔB

Question 113.
Φ Δ Φ is equal to
Answer:
Φ

Question 114.
(A ∪ B)’ is equal to
Answer:
A’ ∩ B’
Explanation:
(A ∪ B)’ = A’∩B’

Question 115.
Draw the venn diagram of A – B.
Answer:

Question 116.
If the number of proper subsets of a given set is 31, then how many ele-ments the set contains ?
Answer:
5
Explanation:
2ⁿ – 1 = 31 ⇒ 2ⁿ = 32 = 2⁵
∴ no. of elements are 5.

Question 117.
Write the intersection of set of ratio-nal numbers and set of irrational num¬bers.
Answer:
Real numbers

Question 118.

This venn diagram represents
Answer:
A∩B

Question 119.
From the venn diagram, write the set A∪B.

Answer:
A ∪ B = {1, 2, 4, 5, 6, 7, 10}
Explanation:
A ∪ B = {1,2, 4, 5, 6, 7, 10}

If A = {x : x is a natural number}
B = {x : x is an even natural number}
C = {x : x is an odd natural number) and
D = {x : x is a prime number}

Question 120.
Find A∩B.
Answer:
A∩B = (1, 2, 3, 4, } ∩ (2, 4, 6, 8…………… }
= {2,4,6, 8, ….} = B{∵B⊂A}

Question 121.
Find A ∩C.
Answer:
A ∩ C = {1, 2, 3,4, …} ∩ {1, 3, 5, 7,…} = (1,3, 5, 7,…} = C(∵C⊂A}

Question 122.
Find A ∩D.
Answer:
A∩D = {1,2, 3, 4, …} ∩ {2, 3,5,7,…} = {2, 3, 5, 7,…} = D{∵ D⊂A)

By observing the below diagram and answer the following questions :

Question 123.
Find A∪B.
Answer:
A∪B = {2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} = {2,3,4,5,6,7,8,9,10}

Question 124.
Find A∩B.
Answer:
A∩B = {2, 3, 4, 5, 6} ∩ {7, 8, 9, 10}
= { } = Φ

Question 125.
Find A Δ B.
Answer:
A Δ B = (A ∪ B) – (A ∩ B) = A ∪ B . { ∵ A ∩ B = Φ}

By observing the below diagram and answer the following questions.

Question 126.
What do you observes in P and Q ?
Answer:
There are no common elements

Question 127.
Name the type of sets P and Q.
Answer:
P and Q are disjoint sets.

Question 128.
Write the relation between P and Q.
Answer:
P ∩ Q = Φ

Question 129.
Define disjoint sets.
Answer:
There is no common elements in any two sets such type of sets are called disjoint sets.

By observing the below information and answer the following questions.

D = The set of all letters in the word TRIGONOMETRY

Question 130.
Write the Roster form of set ‘D’.
Answer:
D = {T, R, I, G, O, N, M, E, Y}

Question 131.
Write the cardinal number of set D.
Answer:
n (D) = 9
By observing the below diagram and answer the following questions.

Question 132.
Find n(A).
Answer:
n (A) = 2

Question 133.
Find n(B).
Answer:
n (B) = 1

Question 134.
Find n(A ∩ B).
Answer:
n(A ∩ B) = Φ

Question 135.
Find n(A ∪ B).
Answer:
n(A ∪ B) = 3

Question 136.
Write the relation between n (A), n (B), n (A ∩ B) and n (A ∪ B).
Answer:
n (A) + n (B) = n(A∪B) + n(A∩B)
1 + 2 = 3 + 0 = 3

Write the correct matching options.

Question 140.
Roster form
A) {a, e, i, o, u} []
B) {2, 5, 10, 17} []

Choose the correct answer satistying the following statements.

Question 137.
Statement (A): If A = {1,2,3, 4, 5,6}, B = {7,8,9, 10, 11} and C = {6,8, 10, 12,14}, then A and B are disjoints sets.
Statement (B) : Two sets A and B are said to be disjoint, if A ∩ B = Φ
i) Both A and B are true
ii) A’ is true, ‘B’ is false
iii) A is false,’B’is true
iv) Both A and B are false
Answer:
i)

Question 138.
Statement (A) : The set of all rect-angles in contained in the set of all squares.
Statement (B) : The sets P = {a} and B = {{a}} are equal.
i) Both A and B are true
ii) A is true, ‘B’ is false
iii) A’ is false, ‘B’ is true
iv) Both A and 6 are false
Answer:
ii)

Question 139.
Statement (A) : For any two sets A and B, we have A – B = {x : x ∉ A and x∈B}
Statement (B) : For any two sets A and B, we have A-B = {x:x∈A and x∉B) andB-A = {x:x ∈ B and x ∉ A}
i) Both A and B are true
ii) A’ is true, ‘B’ is false
iii) A’ is false, ‘B’ is true
iv) Both A and B are false
Answer:
iii)

Write the correct matching options.

Question 140.
Roster form

Answer:
A – (i), B – (iv).

Question 141.

Answer:
A – (iii), B – (ii).

Question 142.

Answer:
A – (iv), B – (iii).

Question 143.

Answer:
A – (i), B – (ii).

Question 144.

Answer:
A – (ii), B – (iii).

Question 145.

Answer:
A – (i), B – (iv).

Question 146.
If A = {1,2,3} and Φ = { }, find A∩Φ
Answer:
Φ

Question 147.
Find n(A ∪ B) from the figure

Answer:
5

Question 148.
How many subsets does a set of three distinct elements have ?
Answer:
8 sub-sets

Question 149.
If A = {1,2,3} andB = {2,4,6}. What is n(A ∪ B) ?
Solution:
A = {1, 2, 3}, B = {2, 4, 6}
A∪B = {1, 2, 3}∪{2, 4,6} = {1,2,3,4,61
n(A ∪ B) = 5

Practice the AP 10th Class Maths Bits with Answers Chapter 5 Quadratic Equations on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 5th Lesson Quadratic Equations with Answers

Question 1.
If x² – px + q = 0(p,q∈R and p ≠ 0, q ≠ 0) has distinct real roots, then write
the condition.
Answer: p² > 4q.

Question 2.
If one root of 2x² + kx – 6 is 2., then find k.
Answer:
k = – 1
Explanation:
2(2)² + k(2) – 6 = 0
⇒ 8 + 2k – 6 = O
⇒ 2k + 2 = 0 ⇒ k = -1

Question 3.
If the equation x² + 5x + k = 0 has real and distinct roots, then find the value of ‘k’.
Answer:
k > 6.25
Explanation:
Real and distinct roots so,
b² – 4ac > 0
⇒ 25 – 4 . 1. k > 0
⇒ 25 > 4k = k > \(\frac{25}{4}\) > 6.25

Question 4.
Frame a quadratic equation, whose roots are 2 + \(\sqrt{3}\) and 2 – \(\sqrt{3}\) ?
Answer:
x² – 4x + 1 = 0
Explanation:
x² – (2 + \(\sqrt{3}\) +2 – \(\sqrt{3}\))x + (2 + \(\sqrt{3}\)) (2 – \(\sqrt{3}\))
⇒ x² – 4x + 1 = 0

Question 5.
In a quadratic equation ax² + bx + c = 0, if b² – 4ac > 0, then write the nature of the roots.
Answer:
Roots are real and distinct.

Question 6.
Create the quadratic equation, whose zeroes are \(\sqrt{2}\) and – \(\sqrt{2}\) ?
Answer:
x² – 2 = 0.
Explanation:
\(x^{2}-(\sqrt{2}-\sqrt{2}) x+(\sqrt{2})(-\sqrt{2})=0\)
⇒ x² – 2 = 0

Question 7.
For which positive value of x the qua-dratic equation 4x\(\sqrt{3}\) -9 = 0 satisfies ?
Answer:
\(\frac{3}{2}\)

Question 8.
If the roots of x² + 6x + 5 = 0 are a and P, then find the value of sum of the roots.
Answer:
-6
Explanation:
α + β = \(\frac{-b}{a}\) = — 6

Question 9.
Write the discriminant of 6x² – 5x + 1 = 0.
Answer:
D = 1
Explanation:
D = b² – 4ac = 25 – 4 . 6 . 1
⇒ 25 – 24 = 1 > 0 D = 1

Question 10.
Write the quadratic polynomial having \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 2 }\) as its zeroes.
Answer:
x² – \(\frac{5 x-1}{6}\) = 0 ⇒ 6x² – 5x + 1 = 0
Explanation:

Question 11.
If a number is 132 smaller than its square, then find the number.
Answer:
12
Explanation:
x + 132 = x²
⇒ x² – x – 132 = 0
By solve the equation, ∴ x = 12

Question 12.
Write the general form of a quadratic equation in variable ‘x’.
Answer:
ax² + bx + c = 0 (a ≠ 0).

Question 13.
Make the quadratic polynomial, whose zeroes are 2 and 3.
Answer:
x² – 5x + 6.

Question 14.
If α, β are the roots of x² – 10x + 9 = 0, thep find the value of | α – β |.
Answer:
8
Explanation:
x² – 9x – x + 9 = 0
⇒ x(x – 9) – 1 (x – 9) = 0
⇒ (x-9)(x- 1) = 0
x = 9 and 1, |α – β| = |9- 1| = 8

Question 15.
Write the discriminant of adjacent dia-gram indicates.

Answer:
b² – 4ac > 0.

Question 16.
If the roots of a quadratic equation px² + qx + r = 0 are imaginary, then write the condition of discriminant.
Answer:
q² < 4pr (or) q² – 4pr < 0.

Question 17.
Two angles are complementary. If the large angle is twice the measure of a smaller angle, then find the value of smaller angle.
Answer:
30°
Explanation:
x + y = 90°
⇒ x + 2x = 90°
⇒ 3x = 90° ⇒ x = 30°

Question 18.
Observe the following graphs.

Which as them are the graphs of qua-dratic polynomials ?
Answer:
(i) and (iv).

Question 19.
Write the possible number of roots to a quadratic equation.
Answer:
At a maximum of 2.

Question 20.
If 1 is a common root of ax² + ax + 2 = 0 and x² + x + 6 = 0, then find a-b.
Answer:
2

Question 21.
Find the product of roots of quadratic equation ax² + bx + c = 0.
Answer:
\(\frac{\text { c }}{\text { a }}\)

Question 22.
Write the number of diagonals in a polygon, having ‘n’ sides.
Answer:
\(\frac{n(n-3)}{2}\)

Question 23.
Find the discriminant of quadratic equation 2x² + x – 4 = 0.
Answer:
33

Question 24.
A quadratic equation ax² + bx + c = 0 has two distinct real roots, then write the condition.
Answer:
b² – 4ac >0.

Question 25.
Draw the shape of quadratic equation which having distinct roots ?
Answer:

Question 26.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\) then find the number.
Answer:
2 or \(\frac { 1 }{ 2 }\)
Explanation:
x + \(\frac{1}{x}=\frac{5}{2}\)
⇒ \(\frac{x^{2}+1}{x}=\frac{5}{2}\)
⇒ 2x² + 2 = 5x
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x r x + 2 = 0
⇒ 2x(x – 2) – 1 (x – 2) – 0 ⇒ (x – 2) (2x – 1) = 0 1
∴ x = 2 or 1/2.

Question 27.
Find the roots of the equation 4x² – 4\(\sqrt{3}\) x + 3 = 0.
Answer:
\(-\frac{\sqrt{3}}{2}\)

Question 28.
Find the positive root of \(\sqrt{3 x^{2}+6}=9\)
Answer:
5
Explanation:
3x² + 6 = 81
⇒ 3x² = 81 – 6 = 75
⇒ x² = \(\) = 25 ⇒ x = 5

Question 29.
Find the roots of the quadratic equation (7x – 1) (2x + 3) = 0.
Answer:
\(\frac{1}{7}, \frac{-3}{2}\)

Question 30.
If the sum of the squares of two con-secutive odd numbers is 74, then find the smaller number.
Answer:
5 (or)-7
Explanation:
(2x + 1)² + (2x + 3)² – 74
⇒ 4x² + 4x + 1 + 4x² + 12x + 9 — 74′
⇒ 8x² + 16x + 10 = 74
⇒ 8x² + 16x – 64 = 0
⇒ 8(x² + 2x – 8) = 0
⇒ x² + 4x – 2x – 8 = 0
⇒ x(x + 4) – 2 (x + 4) = 0
⇒ x = – 4, 2
∴ x = – 4, then smaller number
= 2 . (-4) + 1 = -8 + 1 = -7
∴ x = 2, then smaller number
= 2 . (2) + 1 = 4 + 1 = 5

Question 31.
Write the standard form of a cubic polynomial.
Answer:
ax³ + bx² + cx + d = 0; (a ≠ 0).

Question 32.
Write the discriminant of 5x²– 3x – 2 = 0.
Answer:
49

Question 33.
Create the quadratic equation whose roots are – 2 and – 3.
Answer:
x² + 5x + 6 = 0

Question 34.
Find the roots of the quadratic equation \(\frac{x^{2}-8}{x^{2}+20}=\frac{1}{2}\)
Answer:
±6
Explanation:
2x² – 16 = x² + 20
⇒ x² – 36 ⇒ x = ±6.

Question 35.
Find the roots of the equation 3x² – 2\(\sqrt{6}\) x + 2 = 0.
Answer:
\(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

Question 36.
Find the roots of the quadratic equa- tion \(\left(x-\frac{1}{3}\right)^{2}\) = 9.
Answer:
\(\frac { 10 }{ 3 }\) (or) \(\frac { -8 }{ 3 }\).
Explanation:

Question 37.
On solving x² + 5 = – 6x, find the value of ‘x’
Answer:
– 1 or – 5.

Question 38.
Simplified form of \(\frac{\mathbf{x}}{\mathbf{x}-\mathbf{y}}-\frac{\mathbf{y}}{\mathbf{x}+\mathbf{y}}\)
Answer:
\(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}\)

Question 39.
Find the sum of roots of bx² + ax + c = 0.
Answer:
\(\frac{-a}{b}\)

Question 40.
Find the roots of 2x² – x + \(\frac { 1 }{ 8 }\) = 0.
Answer:
\(\frac{1}{4}, \frac{1}{4}\)

Question 41.
If x + \(\frac{1}{x}\) = 2, then find \(x^{2}+\frac{1}{x^{2}}\).
Answer:
2
Explanation:
x + \(\frac { 1 }{ x }\) = 2
⇒ \(x^{2}+\frac{1}{x^{2}}\) + 2 = 4 ⇒ x² + \(x^{2}+\frac{1}{x^{2}}\) = 2

Question 42.
If 3y² — 192, then find ‘y’.
Answer:
y = ± 8

Question 43.
How many diagonals has a pentagon?
Answer:
’9′

Question 44.
If α and β are the roots of the quadratic equation x² – 3x + 1 = 0, then find \(\left(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\right)\)
Answer:
7

Question 45.
If \(\mathbf{a}^{\mathbf{x}^{2}-4 \mathbf{x}+3}\) = 1, then find x (a # 0).
Answer:
1 or 3.

Question 46.
Find discriminant of the quadratic equation x + \(\frac { 1 }{ x }\) = 3.
Answer:
5

Question 47.
Create the quadratic equation with 2 < x < 3.
Answer:
x² – 5x + 6 < 0.
Explanation:
x² – (2 + 3)x + 2 . 3 < 0
⇒ x² – 5x + 6 < 0

Question 48.
p(x) = x² + 2x + 1, then find p(x²).
Answer:
x⁴ + 2x² + 1

Question 49.
x² – 7x – 60 = 0, then find ‘x’.
Answer:
12 and -5.

Question 50.
\(\frac{1}{a+3}+\frac{1}{a-3}+\frac{6}{9-a^{2}}\) is equal to ?
Answer:
\(\frac{2}{a+3}\)

Question 51.
Find the roots of \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\) = 0.
Answer:
\(\frac{-5}{\sqrt{2}} \text { or }-\sqrt{2}\)

Question 52.
Find the roots of a quadratic equation \((\sqrt{2} x+3)(5 x-\sqrt{3})=0\)
Answer:
\(\frac{-3}{\sqrt{2}}, \frac{\sqrt{3}}{5}\)

Question 53.
4x² + ky – 2 = 0 has no real roots, then find ‘k’.
Answer:
k < – \(\sqrt{32}\)

Question 54.
The sum of a number and its reciprocal is \(\frac { 50 }{ 7 }\), then find the number.
Answer:
7 (or) \(\frac { 1 }{ 7 }\)
Explanation:
x + \(\frac{1}{x}=\frac{50}{7}\)
⇒ \(\frac{x^{2}+1}{x}=\frac{50}{7}\)
⇒ 7x² + 7 — 50x
⇒ 7x² – 50x + 7 — 0
⇒ 7x² – 49x – x + 7 = 0
⇒ 7x (x – 7) – 1 (x – 7) – 0
⇒ (x – 7) (7x – 1) – 0
=+ x = 7 (or) 1/7

Question 55.
Find the roots of the quadratic equation \(\frac{9}{x^{2}-27}=\frac{25}{x^{2}-11}\)
Answer:
±6;

Question 56.
Write the nature of the roots of a qua-dratic equation 4x² – 12x + 9 = 0.
Answer:
Real and equal.

Question 57.
3x² + (- k)x + 8 = 0 has no real roots, then find k’.
Answer:
k < 4\(\sqrt{6}\)
Explanation:
No real roots. So D < 0,
(-k)² – 4 . 3 . 8 < 0
⇒ k² – 96 < 0
⇒ k² < 96
⇒ k < \(\sqrt{96}\)
⇒ k < \(4 \sqrt{6}\)

Question 58.
Find the discriminant of 3x² – 2x = \(\frac{-1}{3}\).
Answer:
D = 0

Question 59.
Find the product of the roots of 1 =x².
Answer:
-1

Question 60.
x(x + 4) = 12, then find ‘x’.
Answer:
– 6 or 2.

Question 61.
Form a quadratic equation from, x³ – 4x² – x + 1 = (x – 2)³.
Answer:
2x² – 13x + 9 = 0.
Explanation:
x³ – 4x² – x + 1 = x³ – 3 . x² . 2 + 3 . x . 2² – 2³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³ – 4x² – x + 1 = x³ + 6x² – 12x + 8 = 0
⇒ 2x² – 13x + 9 = 0

Question 62.
Find the product of the roots of x² + 7x = 0.
Answer:
0

Question 63.
\(\frac{2 a^{2}+a-1}{a+1}+\frac{3 a^{2}+5 a+2}{3 a+2}+\frac{4-a^{2}}{a+2}\) is equal to ?
Answer:
2 (a +1)

Question 64.
1 and \(\frac { 3 }{ 2 }\) are the roots of which qua-dratic equation ?
Answer:
2x² – 5x + 3 = 0.

Question 65.
If b² < 4ac, then draw the shape of graph.
Answer:

Question 66.
\(\sqrt{\mathbf{k}+\mathbf{1}}\) = 3, then find ‘k’.
Answer:
k = 8

Question 67.
\(\sqrt{x}=\sqrt{2 x-1}\), then find ‘x’.
Answer:
x = 1

Question 68.
If \(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}\) then find ‘x’
Answer:
3 or \(\frac { 4 }{ 3 }\)

Question 69.
Find the coefficient of ‘x’ in a pine qua-dratic equation.
Answer:
0

Question 70.
Write number of distinct line segments that can be formed out of n – points.
Answer:
\(\frac{n(n-1)}{2}\)

Question 71.
The product of two consecutive positive integers is 306, then find the largest number.
Answer:
18
Explanation:
x(x + 1) = 306 ⇒ x² + x – 306 = 0
by solving thik Q.E., x = 17
∴ Largest number = x + 1
= 17 + 1 = 18.

Question 72.
Write the nature of roots of 3x² + 13x – 2 = 0.
Answer:
Real and unequal.

Question 73.
If α and β are the roots of x² – 2x + 3 = 0, then find α² + β²
Answer:
α² + β² = – 2.

Question 74.
If (2x – 1) (2x + 3) = 0, then find ‘x’.
Answer:
\(\frac { 1 }{ 2 }\) or \(\frac { -3 }{ 2 }\)

Question 75.
Write the quadratic equation whose one root is 2 – \(\sqrt{3}\) .
Answer:
x² – 4x + 1 = 0

Question 76.
If b² – 4ac > 0, then write nature of the roots of the quadratic equation.
Answer:
Real and distinct.

Question 77.
Find product of the roots of ax² + bx + c = 0. c
Answer:
c/a

Question 78.
Write the nature of the roots of a qua-dratic equation 4x² + 5x + 1 = 0.
Answer:
Real and distinct.

Question 79.
Write the quadratic equation whose roots are – 1,6.
Answer:
x² – 5x – 6 = 0.

Question 80.
Create the quadratic equation whose roots are – 3 and – 4.
Answer:
x² + 7x 4- 12 = 0.

Question 81.
Find the roots of the quadratic equation (3x + 4)² – 49 = 0.
Answer:
1, \(\frac{-11}{3}\)

Question 82.
If x² – 2x + 1 = 0, then find x + \(\frac{1}{x}\).
Answer:
2

Question 83.
Write nature of the roots of 5x² – x + 1 = 0.
Answer:
Imaginary roots.

Question 84.
Write the nature of the roots of qua-dratic equation 3x² + x + 8 = 0.
Answer:
Imaginary roots.

Question 85.
Find product of the roots of the qua-dratic equation 3x² – 6x + 11 = 0.
Answer:
\(\frac{11}{3}\)

Question 86.
Form a quadratic equation whose roots are k and 1/k.
Answer:
x² – (\(\mathrm{k}+\frac{1}{\mathrm{k}}\))x + 1 = 0

Question 87.
If k² – 8kx + 16 = 0 has equal roots, then find the value of ‘k’.
Answer:
k = ± 1.
Explanation:
(-8k)² – 4(1) (16) = 0
⇒ 64k² = 64 ⇒ k² = 1 ⇒ k = ±1

Question 88.
If the roots of a quadratic equation ax² + bx + c = 0 are real and equal, then find ‘b²‘.
Answer:
4ac

Question 89.
3(x – 4)² – 5(x – 4) = 12, then find ‘x’.
Answer:
7 (or) 8/3.
Explanation:
3(x – 4)² – 5 (x – 4) = 12
3[x² + 16 – 8x] – 5x + 20 — 12
3x² + 48 – 24x – 5x + 20 – 12 — 0
⇒ 3x² – 29x + 56 = 0
⇒ 3x² – 21x – 8x + 56 — 0
⇒ 3x (x – 7) – 8 (x – 7) = 0
⇒ (x – 7) (3x – 8) = 0
⇒ x = 7 (or) x = \(\frac { 8 }{ 3 }\)

Question 90.
If a and pare the roots of x² + x + 1 = 0, then find α² + β².
Answer:
α² + β² = – 1.

Question 91.
\(\frac{1-\frac{1}{1+x}}{\frac{1}{1+x}}\) is equal to ?
Answer:
x

Question 92.
Find sum of the roots of a pure quadratic equation.
Answer:
0

Question 93.
\(\frac{\mathbf{x}}{\mathbf{a}-\mathbf{b}}=\frac{\mathbf{a}}{\mathbf{x}-\mathbf{b}}\) , then find ‘x’.
Answer:
b – a (or) – a

Question 94.
\(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\) x ≠ -4 x or 7 find x’.
Answer:
2 or 1

Question 95.
(1 – 5x) (9x +1) is equal to ?
Answer:
1 + 4x – 5x².

Question 96.
From the figure, find ’x’.

Answer:
± 10
Explanation:
By Pythagoras theorem,
x² = 6² + 8² = 64 + 36 = 100
x = \(\sqrt{100}\) = ± 10

Question 97.
Find the sum of the roots of the equation 3x² – 7x + 11 = 0.
Answer:
7/3

Question 98.
Find the roots of the quadratic equation \((\sqrt{5} x-3)(\sqrt{5} x-3)\) – 0.
Answer:
\(\frac{3}{\sqrt{5}}, \frac{3}{\sqrt{5}}\)

Question 99.
Write the nature of the roots of the quadratic equation \(\sqrt{3} x^{2}-2 x-\sqrt{3}\).
Answer:
Real and distinct.

Question 100.
If 5x² – kx + 11 = 0 has root x = 3, then find ’k’.
Answer:
k = \(\frac{56}{3}\)
Explanation:
5(3)² – k(3) + 11 = 0
⇒ 45 + 11 – 3k = 0
⇒ 56 – 3k = 0
⇒ 3k = 56 ⇒ k = \(\frac { 56 }{ 3 }\)

Question 101.
Find the value of ‘p’ for which 4x² – 2px + 7 = 0 has a real roots.
Answer:
p > 2\(\sqrt{7}\)

Question 102.
If one root of a quadratic equation is 7 – 7\(\sqrt{3}\) , then find the quadratic equation.
Answer:
x² – 14x + 46 = 0.

Question 103.
If b² – 4ac = 0, then write nature of the roots of the quadratic equation.
Answer:
Real and equal.

Question 104.
Find sum of the roots of ax² + bx + c = 0.
Answer:
\(-\frac{b}{a}\)

Question 105.
If the equation x² – kx + 1 = 0 has equal roots, then find the value of ‘k’.
Answer:
k = ± 2
Explanation:
b² – 4ac = (- k)² – 4 . 1 . 1 = 0
⇒ k² – 4 = 0
⇒ k² = 4 ⇒ k = \(\sqrt{4}\) = ± 2.

Question 106.
Find (he product of the roots of the qua-dratic equation \(\sqrt{2} \mathrm{x}^{2}-3 \mathrm{x}+5 \sqrt{2}\) = 0.
Answer:
5

Question 107.
Write the nature of roots of 3x² + 6x – 2 = 0.
Answer:
Real and distinct.

Question 108.
If the sum of the roots of the quadratic equation 3x² + (2k + 1)x – (k + 5) = 0 is equal to the product of the roots, then find the value of k.
Answer:
4
Explanation:
Sum of the roots = product of the roots
⇒ \(\frac{-(2 k+1)}{3}=\frac{-(k+5)}{3}\)
⇒ – 2k- 1 = -k – 5 ⇒ k = 4

Question 109.
Find the product of zeroes in the above equation.
Answer:
\(\frac{-11}{5}\)

Question 110.
Find the degree of any quadratic equation.
Answer:
2

Question 111.
In the quadratic equation
x² + x – 2 = 0, find the value of a + b + c.
Answer:
a + b + c = 0.

Question 112.
Find the value of \(\left(x+\frac{1}{x}\right)^{2}-\left(y+\frac{1}{y}\right)^{2}-\left(x y-\frac{1}{x y}\right) \cdot\left(\frac{x}{y}-\frac{y}{x}\right)\)
Answer:
0

Question 113.
Form a quadratic equation from x(2x + 3) = x² + 1.
Answer:
x² + 3x – 1 = 0.
Explanation:
2x² + 3x = x² + 1
⇒ x² + 3x – 1 = 0

Question 114.
(x – α) (x – β) = 0, then find the product.
Answer:
x² – (α + β)x + αβ = 0.

Question 115.
If α and β are die roots of x² – 5x + 6 = 0, then find the value of α – β.
Answer:
± 1.

Question 116.
For what values of m’ are the roots of the equation mx² + (m + 3)x + 4 = 0 are equal ?
Answer:
9 or 1.
Explanation:
(m + 3)² – 4 . m . 4 = 0
⇒ (m + 3)² – 16m = 0
⇒ m² + 9 + 6m- 16m = 0
⇒ m² – 10m + 9 = 0
⇒ m² – 9m – m + 9 = 0
⇒ m(m – 9) – 1 (m – 9) = 0
∴ m = 9 or 1

Question 117.
Find the roots of 2x² + x – 4 = 0.
Answer:
x = \(\frac{-1 \pm \sqrt{33}}{4}\)

Question 118.
If kx (x – 2) + 6 = 0 has equal roots, then find k’.
Answer:
k = 6.
Explanation:
kx² – 2kx + 6 = 0
⇒ (2k)² – 4 . k . 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k (k – 6) = 0 ⇒ k = 6

Question 119.
If ‘2’ is a root of x² + 5x + r = 0, then find ‘r’.
Answer:
r = -14

Question 120.
(α + β)² – 2αβ is sequal to ……………
Answer:
α² + β²

Question 121.
Find the value of \(\sqrt{\mathbf{a}+\sqrt{\mathbf{a}+\sqrt{\mathbf{a + \ldots \ldots \infty}}}}\)
Answer:
\(\frac{1+\sqrt{1+4 a}}{2}\)

Question 122.
If the sum of the roots of kx² – 3x + 1 = 0 is \(\frac{-4}{3}\) then find ‘k’.
Answer:
\(\frac{-9}{4}\)
Explanation:
\(\frac{3}{\mathrm{k}}=\frac{-4}{3} \Rightarrow \frac{3 \times 3}{-4}=\mathrm{k} \Rightarrow \mathrm{k}=\frac{-9}{4}\)

Question 123.
\(\frac{n(n+1)}{2}\) = 55, then find ‘n’
Answer:
10
Explanation:
⇒ n² + n = 110 = 0
⇒ n² + n – 110 = 0
⇒ n = 10

Question 124.
If ‘α’ is β root of ax² + bx + c = 0, then find aα² + bα + c.
Answer:
0

Question 125.
If α and β are the roots of the quadratic equation 2x² + 3x – 7 = 0, then find \(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\)
Answer:
\(\frac{-37}{14}\)

Question 126.
Find the sum of the roots of -7x + 3x² – 1 = 0.
Answer:
\(\frac{7}{3}\)

Question 127.
Find the roots of a quadratic equation \(\frac{\mathbf{x}}{\mathbf{p}}=\frac{\mathbf{p}}{\mathbf{x}}\)
Answer:
x = p
Explanation:
x² = p² ⇒ x = p

Question 128.
If (x – 3) (x + 3) = 16, then find the value of ‘x’.
Answer:
± 5.

Question 129.
Write the roots of a quadratic equation ax² + bx + c = 0.
Answer:
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Question 130.
Find the sum of the roots of the quadratic equation 5x² + 4\(\sqrt{3}\)x – 11 = 0.
Answer:
\(\frac{-4 \sqrt{3}}{5}\)

Question 131.
If one root of x² – (p – 1)x + 10 = 0 is 5, then find ‘p’.
Answer:
7
Explanation:
5² – (p – 1) 5 + 10 = 0
⇒ 25 + 10 – 5p + 5 = 0
⇒ 35 = 5p ⇒ p = 7

Question 132.
If one root of x2 – x + k = 0 is square of other, then find ‘k’.
Answer:
k = cube of one root
Explanation:
α = x, β = x²
Product of roots = αβ = \(\frac{\mathrm{c}}{\mathrm{a}}\)
⇒ x.x² = k ⇒ k = x³
k is cube of the first root.

Question 133.
If α, β are the roots of x2 – px + q = 0, then find α³ + β³.
Answer:
p³ – 3pq

Question 134.
x² + (x + 2)² = 290, then find ‘x’.,
Answer:
11 or – 13

Question 135.
Find the value of \(\sqrt{\mathbf{a} \sqrt{\mathbf{a} \sqrt{\mathbf{a}} \ldots \ldots \infty}}\)
Answer:
a

Question 136.
If \(\frac{-7}{3}\) is a root of 6x² – 13x – 63 = 0, then find other root.
Answer:
\(\frac{9}{2}\)

Question 137.
If b²2 – 4ac < 0, then write nature of the roots of the quadratic equation.
Answer:
Imaginary roots.

Choose the correct answer satistying the following statements.

Question 138.
Statement (A) : The equation x² + 3x + 1 = (x – 2)² is a quadratic equation.
Statement (B) : Any equation of the form ax2 + bx + c = 0 where a ± 0, is called a quadratic equation.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
We have, x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0, it is not of the form ax² + bx + c = 0
So, A is incorrect but B is correct.
Hence (iii) is the correct option.

Question 139.
Statement (A) : The roots of the qua-dratic equation x² + 2x + 2 = 0 are imaginary.
Statement (B) : If discriminant D = b² – 4ac < 0, then the roots of quadratic equation ax² + bx + c = 0 are imaginary.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
x² + 2x + 2 = 0
∴ Discriminant, D = b² – 4ac
= (2)² – 4 x 1 x 2
= 4 – 8 = -4 < 0
∴ Roots are imaginary.
So, both A and B are correct and B explains Answer: Hence, (i) is the correct option.

Question 140.
Statement (A) : The value of k = 2, if one root of the quadratic equation
6x² – x – k = 0 is \(\frac{2}{3}\)
Statement (B) : The quadratic equation ax² + bx + c = 0, a ≠ 0 has two roots.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
As one root is \(\frac{2}{3}\) ⇒ x = \(\frac{2}{3}\)

So, both A and B are correct but B does not explain Answer:
∴ Hence, (i) is the correct option.

Question 141.
Statement (A) : The equation 8x² + 3kx + 2 = 0 has equal roots, then the value of k is ± \(\frac{8}{3}\).
Statement (B) : The equation ax² + bx + c = 0 has equal roots if D = b² – 4ac = 0.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
8x² + 3kx + 2 = 0
∴ Discriminant, D = b² – 4ac
= (3k)² – 4 x 8 x 2
= 9k² – 64
For equal roots, D = 0
⇒ 9k² – 64 = 0
⇒ 9k² = 64
⇒ k² = \(\frac { 64 }{ 9 }\)
⇒ 9k² = ±\(\frac { 8 }{ 3 }\)
So, A and B both correct and B explains Answer: Hence, (i) is the correct option.

Question 142.
Statement (A) : The values of x are \(\frac{-a}{2}\), a for a quadratic equation 2x² + ax – a² = 0.
Statement (B) : For quadratic equation ax² + bx + c = 0.
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
2x² + ax – a² =0

So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 143.
Statement (A) : The equation (x – p) (x – r) + λ(x – q) (x – s) = 0, p < q < r < s, has non-real roots if λ > 0.
Statement (B) : The equation ax² + bx + c = 0, a, b,c ∈ R, has non-real roots if b² – 4ac < 0.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
Statement (A):
Let f(x) = (x – p) (x – r) + λ(x – q) (x- s)
f(p) = λ(p – q) (p – s)
f(q) = (q – p) (q – r)
f(s) = (s – p) (s – r)
f(r) = λ(r – q) (r – s)
If λ > 0, then f(p) > 0, f(q) < 0, f(r) < 0 and f(s) > 0.
⇒ f(x) = 0 has one real root between p and q and other real root between r and s.
Statement – B is obviously true. Option (iii) is correct.

Question 144.
Statement (A) : If roots of the equation x² – bx + c = 0 are two consecutive integers, then b² – 4c = 1.
Statement (B) : If a, b, c are odd integer, then the roots of the equation 4abc x² + (b² – 4ac)x – b = 0 are real and distinct.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
Statement (A) : Given equation , x² – bx + c = 0.
Let α, β be two roots such that |α – β| = 1. .
⇒ (α + β)² – 4αβ = 1.
⇒ b² – 4c = 1
Statement (B): Given equation
4abc x² + (b² – 4ac) x – b = 0
D = (b² – 4ac)² + 16 ab² c
D = (b² – 4ac)² > 0
Hence roots are real and unequal. Option (ii) is correct.

Question 145.
Statement (A) : If 1 ≤ a ≤ 2, then \(\sqrt{a+2 \sqrt{a-1}}+\sqrt{a-2 \sqrt{a-1}}=2\)
Statement (B) : If 1 ≤ a ≤ 2, then (a – 1) > 1.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
If 1 ≤ a ≤ 2 ⇒ 0 ≤ a- 1 ≤ 1

Statement – A is true.
Statement – B is false.
Option – (ii) is correct.

Question 146.
Statement (A): If one root is \(\sqrt{3}-\sqrt{2}\), then the equation of lowest degree with rational coefficients x⁴ – 10x² + 1 = 0.
Statement (B): For a polynomial equa-tion with rational coefficient irrational roots occurs in pairs.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
x = \(\sqrt{3}-\sqrt{2}\), x² = 5 – 2\(\sqrt{6}\)
(x² – 5)² = 24
x⁴ – 10x² + 25 = 24
x⁴ – 10x² + 1 = 0
For polynomial equation with rational coefficients irrational roots occurs in pairs.
Option (i) is correct.

Question 147.
Statement (A): Degree of the polynomial 5x² + 3x + 4 is 2.
Statement (B) : The degree of a poly-nomial of one variable is the highest value of the exponent of the variable.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Read the below passages and answer to the following questions.

Let us consider a quadratic equation x² + 3ax + 2a² = 0.
If the above equation has roots α,β and it is given that α² + β² = 5.

Question 148.
Find value of ‘a’.
Answer:
±1.
Explanation:
α + β = – 3a; αβ = 2a²
a² + p² = 5
⇒ (α + β)² – 2αβ = 5
⇒ (- 3a)² – 2(2a²) = 5
⇒ 9a² – 4a² = 5
⇒ 5a² = 5 ⇒ a = ± 1

Question 149.
Find value of ‘D’ for the above qua-dratic equation.
Answer:
D > 0.
Explanation:
D = (3a)² – 4(2a²)
= 9a² – 8a² = a² = 1 > 0

Question 150.
Find the product of roots.
Answer:
2
Explanation:
αβ = 2a² = 2(1) = 2

Let us consider a quadratic equation x² + λx + λ + 1.25 = O, where λ is a constant. The value of A such that the above quadratic equation has

Question 151.
Two distinct roots.
Answer:
λ > 5 or λ < – 1.
Explanation:
The equation has two distinct roots if b² – 4ac > 0.
∴ (λ – 5)(λ + 1) > 0
⇒ Either λ – 5 > 0 (or) λ + 1 > 0
⇒ λ > 5 (or) λ > -1
∴ λ > 5
⇒ λ – 5 <0 (or) λ + 1 < 0
⇒ λ < 5 (or) λ < – 1
∴ λ < -1 Hence the given equation has two dis-tinct roots for λ > 5 (or) λ < – 1

Question 152.
Two coincident roots.
Answer:
λ = 5 or λ = -1.
Explanation:
The equation has two coincident roots if b² – 4ac = 0
⇒ (λ – 5) (λ + 1) = 0
⇒ Either λ – 5 = 0 (or) λ = 5
⇒ λ + 1 = 0
⇒ λ = – 1
⇒ λ = 5 or – 1
Hence the given equation has coincident roots for λ = 5 or – 1.

The area of a rectangular plot is 528 m². The length of the plot is one more than twice its breadth.

Question 153.
Which mathematical concept is used to find area of above plot ?
Answer:
Quadratic equation.

Question 154.
Write the breadth and length of above given plot.
Answer:
Let breadth = x m, length = 2x + 1 m.

Question 155.
Write the equation of area of above given plot.
Answer:
Area = length x breadth
= x(2x + 1) – 2x² + x = 528 m².

The hypotenuse of a right triangle is 25 cm. We know that the difference in lengthof the other two sides is 5 cm.

Question 156.
Write the lengths of smaller and larger sides.
Answer:
Smaller side = x m
Larger side = (x + 5) cm.

Question 157.
Write the hypotenuse of the triangle.
Answer:
x² + (x + 5)² = (25)²
i.e., x² + 5x – 300 = 0

Question 158.
Which mathematical concept is used to find out the values of dimensions ?
Answer:
Quadratic equations.

Question 159.
Column -II give roots of quadratic equations given in column – I, match them correctly.

Answer:
A – (iv), B – (ii).

Question 160.
Column – II give roots of quadratic equations given in column -1, match them correctly.

Answer:
A – (i), B – (iii).

Question 161.
Write the correct matching.

Answer:
A – (ii), B – (iv).

Question 162.
Write die correct matching.

Answer:
A – (iii), B – (i).

Question 163.
Column – II give pair at two numbers for solution to problems given in column -I. Match them correctly.

Answer:
A – (iv), B – (ii).

Question 164.
Column – II give pair at two numbers for solution to problems given in column -I.
Match them correctly.

Answer:
A – (i), B – (iii).

Question 165.
D is the discriminait of the quadratic equation ax² + bx + e = 0.

Answer:
A – (ii), B – (i).

Question 166.
D Is the discriminant of the quadratic equation ax² + bx + c = O.

Answer:
A – (ii), B – (i).

Question 167.
Write a quadratic equation with roots 3 and 4.
Answer:
x² – 7x + 12 = 0

Question 168.
Draw the rough graph of the quadratic equation ax² + bx + c = 0, when b² – 4ac < 0.
Answer:

Practice the AP 10th Class Maths Bits with Answers Chapter 3 Polynomials on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 3rd Lesson Polynomials with Answers

Question 1.
Write the sum of zeroes of
bx² + ax + c.
Answer:
\(\frac{-a}{b}\)

Question 2.
Find product of the zeroes of
p(x) = (x-2).(x + 3)
Answer:
= -6
Explanation:
p(x) = x² + x – 6,
Product of zeroes = \(\frac{\mathrm{c}}{\mathrm{a}}\) = – 6.

Question 3.
5x – 3 represents which type of poly¬nomial ?
Answer:
Linear.

Question 4.
Find the value of ‘x’ which satisfies 2(x – 1) – (1 – x) = 2x + 3 ?
Answer:
6
Explanation:
2x – 2 – 1 + x = 2x + 3
⇒ 3x – 2x = 3 + 3 = 6
⇒ x = 6.

Question 5.
Write the degree of the polynomial
\(\sqrt{2}\)x² – 3x + 1.
Answer:
2

Question 6.
Write order of the polynomial 5x⁷ – 6x⁵ + 7x – 6.
Answer:
7

Question 7.
Find the product of zeroes of 2x² – 3x + 6.
Answer:
3
Explanation:
Product of zeroes = \(\frac{c}{a}=\frac{6}{2}\) = 3.

Question 8.
Find sum of zeroes of a polynomial x³ – 2x² + 3x – 4.
Answer:
2
Explanation:
Sum of zeroes = α + β + γ
= \(\frac{-b}{a}=\frac{-(-2)}{1}\) = 2

Question 9.
Find a quadratic polynomial, the sum of whose zeroes is zero and one zero is 4, is
Answer:
x² – 16 = 0
Explanation:
Sum of zeroes = \(\frac{-b}{a}\) = 0,
α + β = 0, β = 4
⇒ α = -4 x2 – (α + β)x + αβ = 0
⇒ x² – 0(x) -16 = 0
⇒ x² – 16 = 0.

Question 10.
If p(x) = x² – ax – 3 and p(2) = – 3, then find Answer:
Answer:
2
Explanation:
p(x) = (2)² – 2a – 3 = – 3
⇒ 1 — 2a = — 3
⇒ 4 – 2a = 0
⇒ 4 = 2a ⇒ a = 2

Question 11.
Write the zero value of polynomial px + q.
Answer:
\(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 12.
In a division, if divisor is x + 1, quo¬tient is x and remainder is 4, then find dividend.
Answer:
x (x + 1) + 4 = x² + x + 4
Explanation:
Dividend = Divisor x Quotient
+ Remainder
= (x + 1) x + 4 = x² + x + 4.

Question 13.
Find the zero value of linear polynomial ax – b.
Answer:
\(\frac{\mathrm{b}}{\mathrm{a}}\)

Question 14.
Find the sum of the zeroes of the poly-nomial x² + 5x + 6.
Answer:
– 5

Question 15.
4y² – 5y + 1 is an example for this type of polynomial.
Answer:
Quadratic polynomial.

Question 16.
Write the degree of the polynomial 5x⁷ – 6x⁵ + 7x – 4.
Answer:
7

Question 17.
How much the sum of zeroes of the polynomial 2x² – 8x + 6 ?
Answer:
4

Question 18.
f α,β are the zeroes of x² + x +1, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
-1
Explanation:
α + β = -1, αβ = 1
⇒ \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-1}{1}\) = -1

Question 19.
Write the number of zeroes of the poly¬nomial in the given graph.

Answer:
3

Question 20.
Write product of zeroes of the cubic polynomial 3x³ – 5x² – 11x – 3.
Answer:
1
Explanation:
Product of zeroes = αβγ
= \(\frac{-d}{a}=\frac{-(-3)}{3}=\frac{3}{3}\) = 1

Question 21.
The following is the graph of a poly¬nomial. Find the zeroes of the poly¬nomial from the given graph.

Answer:
– 2,1

Question 22.
Find the value of p(x) = 4x² + 3x + 1 at x = -1.
Answer:
2

Question 23.
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d and (a ≠ 0), then find αβγ

Question 24.
4x + 6y = 18 doesn’t pass through ori-gin, then its graph indicates
Answer:
A straight line

Question 25.
If ‘3’ is one of the zeroes of
p(x) = x² + kx – 9, then find the value of k.
Answer:
0
Explanation:
p(3) = 3² + 3k – 9 = 0
⇒ 3k = 0 ⇒ k = 0

Question 26.
When p(x) = x² – 8x + k leaves a re-mainder when it is divided by (x – 1), then find k.
Answer:
k > 7
Explanation:
p(1) = 1 – 8 + k = 0
⇒ k > 7

Question 27.
If α, β, γ are zeroes of x³ + 3x² – x + 2, then find the value of αβγ.
Answer:
-2

Question 28.
Make a quadratic polynomial having 2, 3 as zeroes.
Answer:
x² – 5x + 6 = 0
Explanation:
x² – (2 + 3)x + 2 . 3 = 0
⇒ x² – 5x + 6 = 0.

Question 29.
Write a quadratic polynomial, whose zeroes are \(\sqrt{2}\) and –\(\sqrt{2}\).
Answer:
x² – 2 = 0
Explanation:
x² – \((\sqrt{2}-\sqrt{2}) x+\sqrt{2}(-\sqrt{2})\) = 0
⇒ x² – 0(x) -2 = 0
⇒ x² – 2 = 0.

Question 30.
Find the coefficient of x7 in the poly¬nomial 7x¹⁷ – 17x¹¹ + 27x⁵ – 7.
Answer:
0

Question 31.
If α, β are the zeroes of the polyno¬mial x² – x – 6, then find α²β²
Answer:
36
Explanation:
αβ = \(\frac{\mathrm{c}}{\mathrm{a}}\) = – 6 ⇒ (αβ)² = (-6)² = 36

Question 32.
If ‘4’ is one of the zeroes of p(x) = x² + kx – 8, then the value of k.
Answer:
-2

Question 33.
If the polynomial p(x) = x³ – x² + 3x + k is divided by (x – 1), the remainder obtained is 3, then find the value of k.
Answer:
0
Explanation:
p(1) = 1-1 + 3 + k = 3 ⇒ k = 0

Question 34.
If one zero of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other, then find the value of k.
Answer:
5
Explanation:
α = x, β = 1/x
⇒ α . β = x . 1/x = k/5 ⇒ k = 5

Question 35.
Find the number of zeroes of the po!y: nomial, whose graph is given below.
Answer:

Question 36.
Number of zeroes that can be identi¬fied by the given figure.
Answer:

Question 37.
Observe the given rectangular figure, then write its area in polynomial func-tion.

Answer:
x² – 7x – 30 = 0
Explanation:
(10 – x)(x + 3) = 10x + 30 – x² – 3x = 0
⇒ x² – 7x – 30 = 0

Question 38.
f(x) = x + 3, then find zero of f(x).
Answer:
-3

Question 39.
One zero of the polynomial 2x² + 3x + k is 1/2, then find k.
Answer:
-2
Explanation:
im 19
α + β + γ = \(\frac{-b}{a}\) = -3
\(\sqrt{5}-\sqrt{5}+\gamma\) = -3
γ = -3

Question 40.
Find factors of x² + x(a + b) + ab.
Answer:
x + a and x + b

Question 41.
f(x) = 4x² + 4x – 3, then find f(\(\frac{-3}{2}\))
Answer:
0

Question 42.
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x³ + 3x² – 5x – 15, then find its third zero.
Answer:
-3
Explanation:

Question 43.
If \(\sqrt{3}\) and – \(\sqrt{3}\) are the zeroes of a polynomial p(x), then find p(x).
Answer:
x² – 3

Question 44.
If ‘m’ and ‘n’ are zeroes of the polyno-mial 3x² + 11x – 4, then find the value \(\frac{\mathbf{m}}{\mathbf{n}}+\frac{\mathbf{n}}{\mathbf{m}}\)
Answer:
\(\frac{-145}{12}\)
Explanation:
3x² + 11x – 4 = 3x² + 12x – x – 4 = 0
⇒(x + 4)(3x – 1)= 0
⇒x = -4, \(\frac { 1 }{ 2 }\)

Question 45.
In the product (x + 4) (x + 2). Write the constant term.
Answer:
8

Question 46.
Find the polynomial whose zeroes are – 5 and 4.
Answer:
x² + x – 20

Question 47.
Flnd product of zeroes of 3x² = 1
Answer:
\(-\frac{1}{3}\)

Question 48.
Find the sum of the zeroes of x² + 7x + 10.
Answer:
-7

Question 49.
If one root of the polynomial
fix) = 5x² + 13x + k is reciprocal of the other, then find ‘k’.
Answer:
5

Question 50.
f(x) = 3x – 2. then find zero of f(x).
Answer:
\(\frac{2}{3}\)

Question 51.
Write the zeroes from the below graph.

Answer:
-2, 0 and 2

Question 52.
Make a quadratic polynomial whose zeroes are 5 and -2.
Answer:
x² – 3x – 10.

Question 53.
If α, β are (1w zero ai polynomial
f(x) = x² – p(x + 1) – c then find (α + 1)(β + 1).
Answer:
1 – c
Explanation:
α + β = \(\frac{-b}{a}=\frac{+p}{1}\) = +p
αβ = \(\frac{c}{a}=\frac{-p-c}{1}\) = -(p+c)
(α + 1) (β + 1) = αβ + α + β + 1
= -p – c + p + 1
= 1 – c

Question 54.
Write number of constant polynomial x² + 7x – 7.
Answer:
1

Question 55.
Write the quadratic polynomial, whose sum and product of zeroes are 1 and – 2 respectively.
Answer:
x² – x – 2

Question 56.
Find the product of the zeroes of x³ + 4x² + x – 6.
Answer:
6

Question 57.
p(x) = \(\frac{x+1}{1-x}\) , then find p(0).
Answer:
1

Question 58.
Write the maximum number of zeroes that a polynomial of degree 3 can have.
Answer:
Three

Question 59.
If x + 2 is a factor of x² + ax + 2 b and a 4 b = 4, then find Answer:
Answer:
3
Explanation:
(- 2)² – 2a + 2b = 0
⇒ 4 – 2a + 2b = 0
⇒ -a + b = -2

⇒ b = 1,a = 4- b = 4 – 1 = 3.

Question 60.
The graph of ax + b represents which type of polynomial ?
Answer:
linear polynomial.

Question 61.
If ‘1’ is the zero of the quadratic poly- nomlal x² + kx – 5, then find the value ofk.
Answer:
4
Explanation:
1 + k – 5 = 0 ⇒ k = 4.

Question 62.
Find a quadratic polynomial, the sum of whose zeroes is W and product of zero is 3.
Answer:
x² – 9

Question 63.
If y = p(x) is represented by the given pqih, then find die number of zeroes.

Answer:
4

Question 64.
If α + β = 0, αβ = \(\sqrt{3}\), then write the quadratic polynomial.
Answer:
x² + \(\sqrt{3}\)

Question 65.
Find the degree of the polynomial ax⁴ + bx³ + cx² + dx + e.
Answer:
4.

Question 66.
If α, β, γ are the zeroes of the polyno- mid f(x) = x³ – px² + qx – r, then find
\(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\)
Answer:
\(\frac{\mathbf{p}}{\mathrm{r}}\)
Explanation:
\(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{\frac{p}{1}}{\frac{r}{1}}=\frac{p}{r}\)

Question 67.
If α and β are the zeroes of the poly¬nomial p(x) = 2x² + 5x + k satisfying the relation α² + β² + αβ = \(\frac{21}{4}\), then find k.
Answer:
2
Explanation:
α² + β² + αβ