MCQ Questions

Free PDF Download of CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Arithmetic ProgressionsMCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 5 Arithmetic Progressions

1. The n^th term of an A.P. is given by a_n = 3 + 4n. The common difference is
(a) 7
(b) 3
(c) 4
(d) 1

Answer/Explanation

Answer: c
Explaination:Reason: We have an = 3 + 4n
∴ a_n+1 = 3 + 4(n + 1) = 7 + 4n
∴ d = a_n+1 – a_n
= (7 + 4n) – (3 + 4n)
= 7 – 3
= 4

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2. If p, q, r and s are in A.P. then r – q is
(a) s – p
(b) s – q
(c) s – r
(d) none of these

Answer/Explanation

Answer: c
Explaination:Reason: Since p, q, r, s are in A.P.
∴ (q – p) = (r – q) = (s – r) = d (common difference)

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3. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
(a) 2, 4, 6
(b) 1, 5, 3
(c) 2, 8, 4
(d) 2, 3, 4

Answer/Explanation

Answer: d
Explaination:Reason: Let three numbers be a – d, a, a + d
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2

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4. The (n – 1)^th term of an A.P. is given by 7,12,17, 22,… is
(a) 5n + 2
(b) 5n + 3
(c) 5n – 5
(d) 5n – 3

Answer/Explanation

Answer: d
Explaination:Reason: Here a = 7, d = 12-7 = 5
∴ a_n-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

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5. The n^th term of an A.P. 5, 2, -1, -4, -7 … is
(a) 2n + 5
(b) 2n – 5
(c) 8 – 3n
(d) 3n – 8

Answer/Explanation

Answer: c
Explaination:Reason: Here a = 5, d = 2 – 5 = -3
a_n = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n

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6. The 10^th term from the end of the A.P. -5, -10, -15,…, -1000 is
(a) -955
(b) -945
(c) -950
(d) -965

Answer/Explanation

Answer: a
Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10^th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955

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7. Find the sum of 12 terms of an A.P. whose nth term is given by a_n = 3n + 4
(a) 262
(b) 272
(c) 282
(d) 292

Answer/Explanation

Answer: a
Explaination:Reason: Here a_n = 3n + 4
∴ a₁ = 7, a₂ – 10, a₃ = 13
∴ a= 7, d = 10 – 7 = 3
∴ S₁₂ = \(\frac{12}{2}\)[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282

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8. The sum of all two digit odd numbers is
(a) 2575
(b) 2475
(c) 2524
(d) 2425

Answer/Explanation

Answer: b
Explaination:Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

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9. The sum of first n odd natural numbers is
(a) 2n²
(b) 2n + 1
(c) 2n – 1
(d) n²

Answer/Explanation

Answer: d
Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) × 2n = n²Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

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10. If (p + q)^th term of an A.P. is m and (p – q)^tn term is n, then pth term is

Answer/Explanation

Answer: d
Explaination:Reason: Let a is first term and d is common difference
∴ a_{p + q} = m
a_{p – q} = n
⇒ a + (p + q – 1)d = m = …(i)
⇒ a + (p – q – 1)d = m = …(ii)
On adding (i) and (if), we get
2a + (2p – 2)d = m + n
⇒ a + (p -1)d = \(\frac{m+n}{2}\) …[Dividing by 2
∴ a_n = \(\frac{m+n}{2}\)

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11. If a, b, c are in A.P. then \(\frac{a-b}{b-c}\) is equal to

Answer/Explanation

Answer: a
Explaination:Reason: Since a, b, c are in A.P.
∴ b – a = c – b
⇒ \(\frac{b-a}{c-b}\) = 1
⇒ \(\frac{a-b}{b-c}\) = 1

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12. The number of multiples lie between n and n² which are divisible by n is
(a) n + 1
(b) n
(c) n – 1
(d) n – 2

Answer/Explanation

Answer: d
Explaination:Reason: Multiples of n from 1 to n² are n × 1, n × 2, n × 3, …, m× n
∴ There are n numbers
Thus, the number of mutiples of n which lie between n and n² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

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13. If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is
(a) 0
(b) 1
(c) -1.
(d) 2

Answer/Explanation

Answer: a
Explaination:Reason: Let common difference of A.P. be x
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
Given equation n-4b + 6c-4d + c
= a – 4(a + x) + 6(A + 2r) – 4(n + 3x) + (o + 4.v)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

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14. The next term of the sequence

Answer/Explanation

Answer: a
Explaination:

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15. n^th term of the sequence a, a + d, a + 2d,… is
(a) a + nd
(b) a – (n – 1)d
(c) a + (n – 1)d
(d) n + nd

Answer/Explanation

Answer: a
Explaination:Reason: an = a + (n – 1)d

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16. The 10th term from the end of the A.P. 4, 9,14, …, 254 is
(a) 209
(b) 205
(c) 214
(d) 213

Answer/Explanation

Answer: a
Explaination:Reason: Here l – 254, d = 9-4 = 5
∴ 10^th term from the end = l – (10 – 1 )d = 254 -9d = 254 = 9(5) = 254 – 45 = 209

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17. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
(a) 0
(b) 2
(c) 4
(d) 6

Answer/Explanation

Answer: d
Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 – 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 18
⇒ x = 6

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18. The sum of all odd integers between 2 and 100 divisible by 3 is
(a) 17
(b) 867
(c) 876
(d) 786

Answer/Explanation

Answer: b
Explaination:Reason: The numbers are 3, 9,15, 21, …, 99
Here a = 3, d = 6 and a_n = 99
∴ a_n = a + (n – 1 )d
⇒ 99 = 3 + (n – 1) x 6
⇒ 99 = 3 + 6n – 6
⇒ 6n = 102
⇒ n = 17
Required Sum = \(\frac{n}{2}\)[a + a_n] = \(\frac{17}{2}\)[3 + 99] = \(\frac{17}{2}\) × 102 = 867

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19. If the numbers a, b, c, d, e form an A.P., then the value of a – 4b + 6c – 4d + e is
(a) 0
(b) 1
(c) -1
(d) 2

Answer/Explanation

Answer: a
Explaination:Reason: Let x be the common difference of the given AP
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
∴ a – 4b + 6c – 4d + e = a – 4 (a + x) + 6(a + 2x) – 4(a + 3x) + (a + 4x)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

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20. If 7 times the 7^th term of an A.P. is equal to 11 times its 11^th term, then 18^th term is
(a) 18
(b) 9
(c) 77
(d) 0

Answer/Explanation

Answer: d
Explaination:Reason: We have 7a₇ = 11a₁₁
⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]
⇒ 7(a + 6d) = 11(a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 4a = -68d
⇒ a = -17d
∴ a₁₈ = a + (18 – 1)d = a + 17d = -17d + 17d = 0

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21. If p, q, r are in AP, then p³ + r³ – 8q³ is equal to
(a) 4pqr
(b) -6pqr
(c) 2pqr
(d) 8pqr

Answer/Explanation

Answer: b
Explaination:
∵ p, q, r are in AP.
∴ 2q = p + r
⇒ p + r – 2q = 0
∴ p³ + r³ + (-2p)³ = 3 × p × r × -2q
[Using ifa + 6 + c = 0 ⇒ a³ + b³ + c³ = 3 abc]
⇒ p³ + r³ – 8q³ = -6pqr.

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22. In an AP, if a = 3.5, d = 0, n = 101, then a will be [NCERT Exemplar Problems]
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5

Answer/Explanation

Answer: b
Explaination: (b) a₁₀₁ = 3.5 + 0(100) = 3.5

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23. The list of numbers -10, -6, -2, 2, … is [NCERT Exemplar Problems]
(а) an AP with d = -16
(b) an AP with d = 4
(c) an AP with d = -4
(d) not an AP

Answer/Explanation

Answer: b
Explaination: (b) An AP with d = 4.

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24. Two APs have the same common difference. . The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is [NCERT Exemplar Problems]
(a) -1
(b) -8
(c) 7
(d) -9

Answer/Explanation

Answer: c
Explaination:
a₄ – b₄ = (a₁ + 3d) – (b₁ + 3d)
= a₁ – b₁= – 1 – (-8) = 7

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25. In an AP, if d = -2, n = 5 and an = 0, the value of a is
(a) 10
(b) 5
(c) -8
(d) 8

Answer/Explanation

Answer: d
Explaination:
d = – 2, n = 5, a_n = 0
∵ a_n = 0
⇒ a + (n – 1)d=0
⇒ a + (5 – 1)(- 2) = 0
⇒ a = 8
Correct option is (d).

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26. If the common difference of an AP is 3, then a₂₀ – a₁₅ is
(a) 5
(b) 3
(c) 15
(d) 20

Answer/Explanation

Answer: c
Explaination:
Common difference, d = 3
a₂₀ – a₁₅ = (a + 19d) – (a+ 14d)
= 5d=5 × 3 = 15

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27. The next term of the AP √18, √50, √98, …….. is
(a) √146
(b) √128
(c) √162
(d) √200

Answer/Explanation

Answer: c
Explaination:
(c) √18, √50, √98, ….. = 3√2, 5√2, 7√2, ……
∴ Next term is 9√2 = √162

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28. The common difference of the AP

(a) p
(b) -p
(c) -1
(d) 1

Answer/Explanation

Answer: c
Explaination: (c) Common difference = a₂ – a₁

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29. If the n^th term of an AP is (2n +1), then the sum of its first three terms is
(a) 6n + 3
(b) 15
(c) 12
(d) 21

Answer/Explanation

Answer: b
Explaination:
a₁= 2 × 1 + 1 = 3,
a₂ = 2 × 2 + 1 = 5,
a₃ = 2 × 3 + l= 7
∴ Sum = 3 + 5 + 7 = 15

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30. An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is
(a) 16 m
(b) 47 m
(c) 31 m
(d) 52 m

Answer/Explanation

Answer: c
Explaination:
S₃₁ =\(\frac{31}{2}\) (2a + 30d)
a₁₆ = a + 15d = zw
⇒ S₃₁ = \(\frac{31}{2}\) × 2(a + 15d)
⇒ S₃₁ = 31m

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31. The first term of an AP of consecutive integers is p² + 1. The sum of 2p + 1 terms of this AP is
(a) (p + 1)²
(b) (2p + 1) (p + 1)²
(c) (p+1)³
(d) p³ + (p + 1)³

Answer/Explanation

Answer: d
Explaination:

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32. If the sum of first n terms of an AP is An + Bn² where A and B are constants, the common difference of AP will be
(a) A + B
(b) A – B
(c) 2A
(d) 2B

Answer/Explanation

Answer: d
Explaination:
S_n = An + Bn²
S₁ = A × 1 +B × 1²
= A + B
∵ S₁ = a₁
∴ a₁ = A + B … (1)
and S₂ = A × 2 + B × 2²
⇒ a₁ + a₂ = 2A + 4B
⇒ (A + B) + a₂ = 2A + 4B[Using (i)]
⇒ a₂ = A + 3B
∴ d = a₂ – a₁ = 2B

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33. If p – 1, p + 3, 3p – 1 are in AP, then p is equal to ______ .

Answer/Explanation

Answer:
Explaination:
∵ p – 1, p + 3 and 3p – 1 are in AP.
∴ 2(p + 3) = p – 1 + 3p – 1
⇒ 2p + 6 = 4/> -2.
⇒ -2p = -8
⇒ p = 4.

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34. Write down the first four terms of the sequences whose general terms are
(i) T_n = 2n + 3
(ii) T_n =3^{n + 1}
(iii) T₁ = 2, T_n = T_n-1+ 5, n ≥ 2

Answer/Explanation

Answer:
Explaination:
(i) T_n= 2n + 3
T₁ = 2 × 1 + 3 = 5,
T₂ = 2 × 2 + 3 = 7,
T₃ = 2 × 3 + 3 = 9,
T₄ = 2 × 4 + 3 = 11
∴ Ist four terms are 5, 7, 9 and 11.

(ii) T_n = 3ⁿ⁺¹
⇒ T₁ = 3¹⁺¹ = 9,
T₂ = 3²⁺¹ = 27,
T₃= 3³⁺¹ = 81,
T₄ = 3⁴⁺¹ = 243
∴ Ist four terms are 9, 27, 81 and 243

(iii) T₁ = 2, T_n = T_n-1 + 5, n ≥ 2
⇒ T₂ = T_2-1 + 5
= T₁ + 5 = 2 + 5 = 7
T₃ = T_3-1 + 5
= T₂ + 5 = 7 + 5 = 12
and T₄=T_4-1 + 5
= T₃ + 5 = 12 + 5 = 17
∴ Ist four terms are 2, 7, 12 and 17.

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35. Find:
The 10th term of 10.0, 10.5, 11.0, 11.5, …..

Answer/Explanation

Answer:
Explaination:
a=10, d = 10.5 – 10 = 0.5
a₁₀ = a + 9d= 10 + 9 × 0.5 = 14.5

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36. In an A.P., if the common difference (d) = – 4 and the seventh term (a₇) is 4, then find the first term. [CBSE 2018]

Answer/Explanation

Answer:
Explaination:
Let a = first term,
Given, d = – 4, a₇ = 4
⇒ a + (7 – 1)d = 4
[ ∵ nth term of an AP = a_n = a + (n – 1)d]
⇒ a + 6d = 4
⇒ a + 6 × (-4) = 4
⇒ a – 24 = 4.
⇒ a = 4 + 24 = 28
∴ First term, a = 28

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37. Write the n^th term of the A.P. [Delhi 2017 (C)]

Answer/Explanation

Answer:
Explaination:

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38. Which term of the AP 21, 18, 15, … , is zero?

Answer/Explanation

Answer:
Explaination:
Here, a = 21, J= 18-21 = -3
Let a_n = 0
⇒ a + (n – 1 )d = 0
⇒ 21 +(n – 1)(-3) = 0
⇒ (n – 1)(- 3) = -21
⇒ n – 1 = \(\frac{-21}{-3}\)
⇒ n = 8
∴ 8th term is zero.

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39. For what value ofp, are 2p+ 1, 13, 5p – 3 three consecutive terms of an AP?

Answer/Explanation

Answer:
Explaination:
If terms are in AP, then
13 – (2p + 1) = (5p – 3) – 13
⇒ 13 – 2p – 1 = 5p – 3 – 13
⇒ 28 = 7p
⇒ p = 4.

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40. What is the common difference of an A.P. in which a₂₁ – a₇ = 84? [AI 2017]

Answer/Explanation

Answer:
Explaination:
Let ‘cf be the common difference of the AP whose first term is ‘a’
Now, a₂₁ – a₇ = 84
⇒ (a + 20d) – (a + 6d)= 84
⇒ 20d – 6d = 84
⇒ 14d = 84
⇒ d = 6

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41. The first term of an AP is p and its common difference is q. Find its 10th term.

Answer/Explanation

Answer:
Explaination:
∵ a = p and d = q
a_n = a + (n – 1)d
⇒ a₁₀ =p + 9q

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42. Which term of the AP 14, 11, 8, ….is -1?

Answer/Explanation

Answer:
Explaination:
Here, a = 14, d= 11 – 14 = – 3
Let a_n = – 1
⇒ a + (n – 1 )d = – 1
⇒ 14 + (n – 1)(- 3) = – 1
⇒ (n – 1)(- 3) = – 1 – 14
⇒ n – 1 = \(\frac{-15}{-3}\) = 5
⇒ n = 6
∴ 6th term of the AP is -1.

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43. Write the next two terms of the AP: 1,-1, -3, -5, …

Answer/Explanation

Answer:
Explaination:
a₁ = 1
a₁ = a₂ – a₁ = – 1 – 1 = – 2
a₅ = a₁ + 4d
= 1 + 4(-2)
= 1 – 8 = — 7
a₆ = a₅ + d = – 7 – 2 = – 9
Next two terms are – 7 and – 9.

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44. If a_n = \(\frac{n(n-3)}{n+4}\), then find 18th term of this sequence.

Answer/Explanation

Answer:
Explaination:

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45. If the first term of an AP is 2 and common difference is 4, then sum of its first 40 terms is ______ .

Answer/Explanation

Answer:
Explaination:
S₄₀(2 × 2 + 39 × 4)
= 40 × (80)= 3200

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46. Three numbers in an AP have sum 24. Its middle term is ____ .

Answer/Explanation

Answer:
Explaination:
Let numbers be a – d, a and a + d.
⇒ a – d+a + a + d= 24
⇒ 3a = 24
⇒ a = 8
∴ Middle term = 8.

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47. The value of the expression 1 – 6 + 2 – 7 + 3 – 8 + ….. to 100 terms is ______ .

Answer/Explanation

Answer:
Explaination:
1 – 6 + 2 – 7 + 3 – 8 + … 100 terms
= (1 + 2 + 3 +… up to 50 terms) + (-6 -7 – 8 – … up to 50 terms)
= \(\frac{50}{2}\)[2 × 1 + 49 × 1] + [\(\frac{50}{2}\){2 × (-6) + 49 × (-1)}]
= 25 × 51 + 25 x (-12 – 49) = 25(51 – 61)
= -250.

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48. If the sum of first m terms of an AP is 2m² + 3m, then what is its second term?

Answer/Explanation

Answer:
Explaination:
S_{m = 2m² + 3m
∴ S1 = 2 × 1² + 3 × 1 = 5 = a1
and S2 = 2 × 2² + 3 × 2 = 14 …(i)
⇒ a1 + a2 = 14
⇒ 5 + a2 = 14
⇒ a2 = 9}

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49. If the sum of first p terms of an AP is ap² + bp, find its common difference.

Answer/Explanation

Answer:
Explaination:
S_{p = ap² + bp
S1= a × 1² + b × 1
= a + b = a1
and S2 = a × 2² + b × 2 = 4a + 2b …(i)
⇒ a1 + a2 = 4a + 2b
⇒ a + b + a2 = 4a + 2b
⇒ a2 = 3a + b [Using eq. (i)]
Now, d = a2 – a1
= (3a + b)-(a + b) = 2a.}

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50. If sum of first n terms of an AP is 2n² + 5n. Then find S₂₀.

Answer/Explanation

Answer:
Explaination:
S_{n = 2n² + 5n
S20 = 2(20)² + 5 × 20
= 2 × 400 + 100 = 900}

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Free PDF Download of CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Pair of Linear Equations in Two Variables MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 3 Pair of Linear Equations in Two Variables

1. A pair of linear equations a₁x + b₁y + c₁ = 0; a₂x + b₂y + c₂ = 0 is said to be inconsistent, if

Answer

Answer: b

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2. Graphically, the pair of equations 7x – y = 5; 21x – 3y = 10 represents two lines which are
(a) intersecting at one point
(b) parallel
(c) intersecting at two points
(d) coincident

Answer

Answer: b

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3. The pair of equations 3x – 5y = 7 and – 6x + 10y = 7 have
(a) a unique solution
(b) infinitely many solutions
(c) no solution
(d) two solutions

Answer

Answer: c

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4. If a pair of linear equations is consistent, then the lines will be
(a) always coincident
(b) parallel
(c) always intersecting
(d) intersecting or coincident

Answer

Answer: d

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5. The pair of equations x = 0 and x = 5 has
(a) no solution
(b) unique/one solution
(c) two solutions
(d) infinitely many solutions

Answer

Answer: c

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6. The pair of equation x = – 4 and y = – 5 graphically represents lines which are
(a) intersecting at (- 5, – 4)
(b) intersecting at (- 4, – 5)
(c) intersecting at (5, 4)
(d) intersecting at (4, 5)

Answer

Answer: b

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7. For what value of k, do the equations 2x – 3y + 10 = 0 and 3x + ky + 15 = 0 represent coincident lines

Answer

Answer: a

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8. If the lines given by 2x + ky = 1 and 3x – 5y = 7 are parallel, then the value of k is

Answer

Answer: a

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9. One equation of a pair of dependent linear equations is 2x + 5y = 3. The second equation will be
(a) 2x + 5y = 6
(b) 3x + 5y = 3
(c) -10x – 25y + 15 = 0
(d) 10x + 25y = 15

Answer

Answer: c

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10. If x = a, y = b is the solution of the equations x + y = 5 and 2x – 3y = 4, then the values of a and b are respectively
(a) 6, -1
(b) 2, 3
(c) 1, 4
(d) 19/5,  6/5

Answer

Answer: d

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11. The graph of x = -2 is a line parallel to the
(a) x-axis
(b) y-axis
(c) both x- and y-axis
(d) none of these

Answer

Answer: b

---

12. The graph of y = 4x is a line
(a) parallel to x-axis
(b) parallel to y-axis
(c) perpendicular to y-axis
(d) passing through the origin

Answer

Answer: d

---

13. The graph of y = 5 is a line parallel to the
(a) x-axis
(b) y-axis
(c) both axis
(d) none of these

Answer

Answer: a

---

14. Two equations in two variables taken together are called
(a) linear equations
(b) quadratic equations
(c) simultaneous equations
(d) none of these

Answer

Answer: c

---

15. If am bl then the system of equations ax + by = c, lx + my = n, has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these

Answer

Answer: a

---

16. If in the equation x + 2y = 10, the value of y is 6, then the value of x will be
(a) -2
(b) 2
(c) 4
(d) 5

Answer

Answer: a

---

17. The graph of the equation 2x + 3y = 5 is a
(a) vertical line
(b) straight line
(c) horizontal line
(d) none of these

Answer

Answer: b

---

18. The value of k, for which equations 3x + 5y = 0 and kx + lOy = 0 has a non-zero solution is
(a) 6
(b) 0
(c) 2
(d) 5

Answer

Answer: a

---

19. The value of k, for which the system of equations x + (k + l)y = 5 and (k + l)x + 9y = 8k – 1 has infinitely many solutions is
(a) 2
(b) 3
(c) 4
(d) 5

Answer

Answer: a

---

20. The value of k for which the equations (3k + l)x + 3y = 2; (k2 + l)x + (k – 2)y = 5 has no solution, then k is equal to
(a) 2
(b) 3
(c) 1
(d) -1

Answer

Answer: d

---

21. The pair of equations x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)

Answer

Answer: d

---

22. Asha has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively
(a) 35 and 15
(b) 15 and 35
(c) 35 and 20
(d) 25 and 25

Answer

Answer: d

---

23. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24

Answer

Answer: c

---

24. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is
(a) 27
(b) 72
(c) 45
(d) 36

Answer

Answer: d

---

25. The pair of linear equations 2x + 3y = 5 and 4x + 6y= 10 is
(a) inconsistent
(b) consistent
(c) dependent consistent
(d) none of these

Answer/Explanation

Answer: c
Explaination:

---

26. The pair of equations y = 0 and v = -7 has [NCERT Exemplar Problems]
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

Answer/Explanation

Answer: d
Explaination: (d) No solution (parallel lines)

---

27. The pair of equations x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)

Answer/Explanation

Answer: d
Explaination: (d) Intersecting at (a, b)

---

28. Match the Column:

(a) 1 – A, 2 – B, 3 – C
(b) 1 – B, 2 – C, 3 – A
(c) 1 – C, 2 – B, 3 – A
(d) 1 – A, 2 – C, 3 – B

Answer/Explanation

Answer: d
Explaination:

---

29. The pair of equations x = 4 and y = 3 graphically represents lines which are
(a) parallel
(b) intersecting at (3, 4)
(c) coincident
(d) intersecting at (4, 3)

Answer/Explanation

Answer: d
Explaination:
(d) When lines x = 4, y = 3 will intersect, then x coordinate = 4, y-coordinate = 3

---

30. a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where a₁, b₁, c₁, a₂, b₂, c₂ are all real numbers and a₁²+ b₁² ≠ 0, a₂² + b₂² ≠ 0, is called a
(a) family of two different straight lines
(b) family of two coincident lines
(c) pair of linear equations in two variables
(d) none of these

Answer/Explanation

Answer: c
Explaination:
(c) Definition of pair of linear equations in two variables.

---

31. Match the Column:

(a) 1 – A, 2 – B, 3 – C
(b) 1 – B, 2 – A, 3 – C
(c) 1 – B, 2 – C, 3 – A
(d) 1 – C, 2 – A, 3 – B

Answer/Explanation

Answer: c
Explaination:

---

32. A pair of linear equations which has a unique solution x = 2, y = -3 is [NCERT Exemplar Problems]
(a) x + y = —1; 2x – 3y = -5
(b) 2x + 5y = -11; 4x + 10y = -22
(c) 2x – y = 1 ; 3x + 2y = 0
(d) x-4y- 14 = 0; 5x-y- 13 = 0

Answer/Explanation

Answer: b
Explaination:
(b) 2x + 5y = – 11; 4x + 10y = – 22

---

33. The father’s age is six times of his son’s age. Four years hence, the age of the father will be four times of his son’s age. The present ages, in years, of the son and the father are, respectively. [NCERT Exemplar Problems]
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24

Answer/Explanation

Answer: c
Explaination:
Let F and S be the ages of father and sos respectively.
ATQ F = 6S …(i)
F + 4 = 4(S + 4)
⇒ F + 4 = 4S + 16 …(ii)
Substituting F = 6S in (ii), we get
6S + 4 = 4S + 16
2S = 12 .
⇒ S = 6; F = 36

---

34. If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, then the respective values of a and b are
[CBSE 2018(C)]
(a) 3, 5
(b) 5, 3
(c) 3, 1
(d) -1, -3

Answer/Explanation

Answer: c
Explaination:

---

35. The difference between a two digit number and the number obtained by interchanging the digits is 27. What is the difference betw een the two digits of the number?
(a) 9
(b) 6
(c) 12
(d) 3

Answer/Explanation

Answer: d
Explaination:
Let number is 10x + y (x, y are the digits).
Number obtained by interchanging the digits = 10y+x
∴ 10x + y – (10y + x) = 27
9(x -y) = 27
∴ x – y = 3

---

36. The pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines if
(a) a = b
(b) 3a = 2b
(c) 2 a = 3 b
(d) ab = 6

Answer/Explanation

Answer: c
Explaination:

---

37. The equations ax + by + c = 0 and dx + ey + c = 0 represent the same straight line if
(a) ad = be
(b) ac = bd
(c) be = ad
(d) ab = de

Answer/Explanation

Answer: d
Explaination:

---

38. Using the following equations:
\(\frac{4}{x}\) + 6y = 10
\(\frac{1}{x}\) – 6y = 5
find the value of p if p = 5x.
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: a
Explaination:

---

39. Find the number of solutions of the following pair of linear equations:
x + 2y — 8 = 0
2x + 4 v =16

Answer/Explanation

Answer:
Explaination:

∴ Given pair of linear equations has infinite many solutions.

---

40. Write whether the following pair of linear equations is consistent or not. x + y = 14, x – y = 4

Answer/Explanation

Answer:
Explaination:

⇒ The equations have unique solution.
∴ Pair of linear equations is consistent.

---

41. Given the linear equation 3x + 4y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines.

Answer/Explanation

Answer:
Explaination:
Lines are parallel when \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ One of the linear equation in two variables can be 6x + 8y + k = 0 where k is constant not equal to -16.

---

42. Find the value of k for which the given pair of linear equations has no common solution.
4x + Ay = 5
8x + 12y = 10

Answer/Explanation

Answer:
Explaination:

---

43. Find the value of p for which the given pair of linear equations has infinite many solutions.
4x + 3y – 4=0
px + 9y + 12 = 0

Answer/Explanation

Answer:
Explaination:

---

44. The larger of two supplementary angles exceeds the smaller by 20 degrees. Find the angles.

Answer/Explanation

Answer: c
Explaination:
Let larger angle = x° and smaller angle =y° Now x -y = 20 andx +y = 180
∴ Algebraic representation is x – y = 20, x+y= 180

---

45. On solving the following pair of linear equations: 2x – y = 2; 5x + 2y = 14 by substitution method, values of x and y are _____ and _____ respectively.

Answer/Explanation

Answer:
Explaination:
2x – y = 2 …(i)
5x + 2y = 14 …(ii)
Substituting y = 2x – 2 in (ii), we get 5x + 2(2x – 2)=14
9x – 4 = 14
⇒ x= \(\frac{18}{9}\) = 2
⇒ x = 2, y = 2

---

46. Solve the following pair of linear equations by substitution method 2x – 3y +15 = 0; 3x – 5 = 0.

Answer/Explanation

Answer: c
Explaination:

---

47. There are two positive numbers such
that sum of twice the first and thrice the second is 39, while the sum of thrice the first and twice the second is 36. The larger of the two is _____ .

Answer/Explanation

Answer: c
Explaination:

---

48. Solve for x and y by the method of elimination:
4x -3 y = 1; 5x – 7y = 2.

Answer/Explanation

Answer:
Explaination:
Given equations are
4x – 3y = 1 …(i)
5x – 7y = -2 …(ii)
For making coefficient of y equal in both the equations multiplying equation (/) with 7, we get
7 × (4x – 3y) = 7 × 1
⇒ 28x – 21y = 7 …(iii)
Multiplying equation (ii) with 3, we get
3 × (5x – 7y) = 3 × -2
⇒ 15x – 21y = -6 …(iv)
Subtracting equation (iv) from (iii), we get

when x – 1, equation (i) becomes
4 × – 3y = 1
⇒ -3y = -3
⇒ y = 1
∴ x= 1, y= 1

---

49. If the pair of equations x + y = √2 and x sin θ + y cos θ = 1 has infinitely many solutions, then θ = _____ .

Answer/Explanation

Answer:
Explaination:

---

50. Find the value of k so that the following system of equation has infinite solutions: 3x – y – 5 = 0, 6x – 2y + k = 0

Answer/Explanation

Answer:
Explaination:

---

51. For which values of p, does the pair of equations given below has unique solution? 4x + py + 8 = 0 and 2x + 2y + 2 = 0

Answer/Explanation

Answer:
Explaination:
For unique solution, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{4}{2} \neq \frac{p}{2}\)
⇒ p ≠ 4

---

52. The equations \(\frac{x y}{x+y}=\frac{1}{9}\) and \(\frac{x y}{x-y}=\frac{1}{4}\) are equivalent to the equations:

Answer/Explanation

Answer: d
Explaination:

---

53. On solving the following equations: \(\frac{5}{x}\) + 6y = 3; \(\frac{-5}{x}\) + 3y = 6, values of x and y are ____ and ____ respectively.

Answer/Explanation

Answer:
Explaination:

---

54. Solve forx andy : x + \(\frac{6}{y}\) = 6, 3x – \(\frac{8}{y}\) = 5

Answer/Explanation

Answer:
Explaination:

---

We hope the given MCQ Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

Free PDF Download of CBSE Class 10 Maths Chapter 2 Polynomials Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Polynomials MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 2 Polynomials

1. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5

Answer

Answer: b

---

2. Given that two of the zeroes of the cubic poly-nomial ax³ + bx² + cx + d are 0, the third zero is

Answer

Answer: a

---

3. If one of the zeroes of the quadratic polynomial (k – 1) x² + kx + 1 is – 3, then the value of k is

Answer

Answer: a

---

4. A quadratic polynomial, whose zeroes are -3 and 4, is
(a) x²- x + 12
(b) x² + x + 12
(c) \(\frac{x^{2}}{2}-\frac{x}{2}-6\)
(d) 2x² + 2x – 24

Answer

Answer: c

---

5. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a – 0, b = -6

Answer

Answer: d

---

6. The number of polynomials having zeroes as -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3

Answer

Answer: d

---

7. Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is

Answer

Answer: b

---

8. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is -1, then the product of the
other two zeroes is
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1

Answer

Answer: a

---

9. The zeroes of the quadratic polynomial x2 + 99x + 127 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal

Answer

Answer: b

---

10. The zeroes of the quadratic polynomial x² + kx + k, k? 0,
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal

Answer

Answer: a

---

11. If the zeroes of the quadratic polynomial ax² + bx + c, c # 0 are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign

Answer

Answer: c

---

12. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.

Answer

Answer: a

---

13. Which of the following is not the graph of quadratic polynomial?

Answer

Answer: d

---

14. The number of polynomials having zeroes as 4 and 7 is
(a) 2
(b) 3
(c) 4
(d) more than 4

Answer

Answer: d

---

15. A quadratic polynomial, whose zeores are -4 and -5, is
(a) x²-9x + 20
(b) x² + 9x + 20
(c) x²-9x- 20
(d) x² + 9x- 20

Answer

Answer: b

---

16. The zeroes of the quadratic polynomial x² + 1750x + 175000 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: a

---

17. The zeroes of the quadratic polynomial x² – 15x + 50 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: c

---

18. The zeroes of the quadratic polynomial 3x² – 48 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: b

---

19. The zeroes of the quadratic polynomial x² – 18x + 81 are
(a) both negative
(b) one positive and one negative
(c) both positive and unequal
(d) both equal and positive

Answer

Answer: d

---

20. The zeroes of the quadratic polynomial x² + px + p, p ≠ 0 are
(a) both equal
(b) both cannot be positive
(c) both unequal
(d) both cannot be negative

Answer

Answer: b

---

21. If one of the zeroes of the quadratic polynomial (p – l)x² + px + 1 is -3, then the value of p is

Answer

Answer: b

---

22. If the zeroes of the quadratic polynomial Ax² + Bx + C, C # 0 are equal, then
(a) A and B have the same sign
(b) A and C have the same sign
(c) B and C have the same sign
(d) A and C have opposite signs

Answer

Answer: b

---

23. If x³ + 1 is divided by x² + 5, then the possible degree of quotient is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: b

---

24. If x³ + 11 is divided by x² – 3, then the possible degree of remainder is
(a) 0
(b) 1
(c) 2
(d) less than 2

Answer

Answer: d

---

25. If x⁴ + 3x² + 7 is divided by 3x + 5, then the possible degrees of quotient and remainder are:
(a) 3, 0
(b) 4, 1
(c) 3, 1
(d) 4, 0

Answer

Answer: a

---

26. If x⁵ + 2x⁴ + x + 6 is divided by g(x), and quotient is x² + 5x + 7, then the possible degree of g(x) is:
(a) 4
(b) 2
(c) 3
(d) 5

Answer

Answer: c

---

27. If x⁵ + 2x⁴ + x + 6 is divided by g(x) and quo-tient is x² + 5x + 7, then the possible degree of remainder is:
(a) less than 1
(b) less than 2
(c) less than 3
(d) less than 4

Answer

Answer: c

---

28. What is the number of zeroes that a linear poly-nomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: b

---

29. What is the number(s) of zeroes that a quadratic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: c

---

30. What is the number(s) of zeores that a cubic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: d

---

31. If one of the zeroes of the cubic polynomial x³ + px² + qx + r is -1, then the product of the other two zeroes is
(a) p + q + 1
(b) p-q- 1
(c) q – p + 1
(d) q – p – 1

Answer

Answer: c

---

32. If one zero of the quadratic polynomial x² + 3x + b is 2, then the value of b is
(a) 10
(b) -8
(c) 9
(d) -10

Answer

Answer: d

---

33. If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is:
(a) 2
(b) -2
(c) 4
(d) -4

Answer

Answer: b

---

34. If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as
(a) value of p(x)
(b) zero of p(x)
(c) constant term of p{x)
(d) none of these

Answer/Explanation

Answer: b
Explaination:
Zero of p(x)
Let p(x) = ax + b
Put x = k
p(k) = ak + b = 0
∴ k is zero of p(x).

---

35. If one of the zeroes of the quadratic polynomial (k -1)x² + kx + 1 the value of k is [NCERT Exemplar Problems]

Answer/Explanation

Answer: c
Explaination:
(k – 1)² + kx + 1
One zero is – 3, so it must satisfy the equation and make it zero.
∴ (k- 1) (- 3)² + k(-3) + 1 =0
⇒ 9k – 9 – Ik + 1 = 0
⇒ 6k – 8 = 0
⇒ k = \(\frac{8}{6}\) = \(\frac{4}{3}\)

---

36. If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then [NCERT Exemplar Problem, CBSE 2011]
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6

Answer/Explanation

Answer: d
Explaination:
x² + (a + 1)x + b
∵ x = 2 is a zero
and x = – 3 is another zero
∴ (2)² + (a + 1)² + 6 = 0
and (- 3)² + (a + 1) (- 3) + b = 0
⇒ 4 + 2cr + 2 + & = 0
and 9 – 3a – 3 + b = 0
⇒ 2a + b = – 6 …(i)
and – 3a + b = – 6 …(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = – 6.

---

37. Which of the following is not the graph of a quadratic polynomial?[NCERT Exemplar Problems]

Answer/Explanation

Answer: a
Explaination:
∵ It crosses the x-axis in three points.

---

38. Zeroes of a polynomial can be determined graphically. No. of zeroes of a polynomial is equal to no. of points where the graph of polynomial
(a) intersects y-axis
(b) intersects x-axis
(c) intersects y-axis or intersects x-axis
(d) none of these

Answer/Explanation

Answer: b
Explaination:
(b) Intersects x-axis.

---

39. If graph of a polynomial does not intersects the x-axis but intersects y-axis in one point, then no. of zeroes of the polynomial is equal to
(a) 0
(b) 1
(c) 0 or 1
(d) none of these

Answer/Explanation

Answer: a
Explaination:
(a) Zero

---

40. A polynomial of degree n has
(a) only 1 zero
(b) at least n zeroes
(c) atmost n zeroes
(d) more than n zeroes

Answer/Explanation

Answer: c
Explaination:
Maximum number of zeroes of a
polynomial = degree of the polynomial.

---

41. If p(x) = axr + bx + c, then –\(\frac{b}{a}\) is equal to
(a) 0
(b) 1
(c) product of zeroes
(d) sum of zeroes

Answer/Explanation

Answer: d
Explaination:
(d) Sum of zeroes = –\(\frac{b}{a}\)

---

42. If p(x) = ax² + bx + c one zero is and a + b + c = 0, then one zero is
(a) \(\frac{-b}{a}\)
(b) \(\frac{c}{a}\)
(c) \(\frac{b}{c}\)
(d) none of these

Answer/Explanation

Answer: b
Explaination:
p(1) = 0; a(1)² + b(1) + c = 0
⇒ a + b + c = 0
∴ one zero (α) = 1
αβ = product of zeroes = \(\frac{c}{a}\)
⇒ 1.β = \(\frac{c}{a}\)
⇒ β = \(\frac{c}{a}\)
∴ zeroes are 1 and \(\frac{c}{a}\)

---

43. If p{x) = ax2 + bx + one of the zeroes is c and a + c = b, then

Answer/Explanation

Answer: c
Explaination:
p(-1) = 0; a(-1)² + b(-1) + c = 0
⇒ a – b + C = 0,
∴ One zero (a) = -1
αα = product of zeroes = \(\frac{c}{a}\)
⇒ (-1).β = \(\frac{c}{a}\)
⇒ β = \(\frac{-c}{a}\)

---

44. The number of polynomials having zeroes as -2 and 5 is [NCERT Exemplar Problems]
(a) 1
(b) 2
(c) 3
(d) more than 3

Answer/Explanation

Answer: d
Explaination:
∵ x² – 3x – 10, 2x² – 6x – 20,
\(\frac{1}{2}\)x² – \(\frac{3}{2}\)x – 5, 3x² – 9x – 30 etc.,
have zeroes – 2 and 5.

---

45. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is [NCERT Exemplar Problems]

Answer/Explanation

Answer: b
Explaination:
∵ αβ + βγ + γα = \(\frac{c}{a}\)
Let, α = 0
So, 0 + βγ + 0 = \(\frac{c}{a}\)
⇒ βγ= \(\frac{c}{a}\)

---

46. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.

Answer/Explanation

Answer: a
Explaination:
f(x) = x² + ax + b
Given: zeroes are α and – α
Sum of zeroes = α – α = 0
∴ f(x) = x² + b, which is not linear
Product of zeroes = α(-α) = – α² = \(\frac{b}{1}\)
⇒ -α² = b
It is possible when, b < 0.
Hence, it has no linear term and the constant term is negative.

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47. If 4x² – 6x – m is divisible by x – 3, the value of m is exact divisor of
(a) 9
(b) 45
(c) 20
(d) 36

Answer/Explanation

Answer: a
Explaination:
Here p(3) = 0
⇒ 4(3)² – 6 × 3-m = 0
⇒ 36 – 18 – m = 0
⇒ m=18
∴ Value of m is exactly divisible by 9.

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48. Which one of the following statements is correct
(a) if x⁶ + 1 is divided by x + 1, then the remainder is -2.
(b) if x⁶ + 1 is divided by x – 1, then the remainder is 2.
(c) if x⁶ + 1 is divided by x + 1, then the remainder is 1.
(d) if x⁶ + 1 is divided by x – 1, then the remainder is -1.

Answer/Explanation

Answer: b
Explaination:
p(x) = x⁶ + 1
when divided by x – 1, then remainder = p(1)
∴ p(1) = 1⁶ + 1 = 2

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49. Consider the following statements
(i) x – 2 is a factor of x³ – 3x² + 4x – 4.
(ii) x + 1 is a factor of 2x³ + 4x + 6.
(iii) x – 1 is a factor of x⁵ + x⁴ – x³ + x² -x + 1.
In these statements
(a) 1 and 2 are correct
(b) 1, 2 and 3 are correct
(c) 2 and 3 are correct
(d) 1 and 3 are correct

Answer/Explanation

Answer: a
Explaination:
x – 2 is a factor of x³ – 3x² + 4x – 4
∵ remainder is zero Similarly x + 1 is a factor of 2x³ + 4x + 6
but x – 1 is not a factor of x⁵ + x⁴ – x³ + x² – x + 1
∵ remainder is not zero
∴ Statements 1 and 2 are correct.

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50. If f(x) = 5x – 10 is divided by x – √2, then the remainder will be
(a) non zero rational number
(b) an irrational number
(c) 0
(d) \(f\left(\frac{1}{\sqrt{2}}\right)\)

Answer/Explanation

Answer: b
Explaination:
Remainder = f(√2) = 5 × √2 – 10
= an irrational number

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51. Zeroes of p(z) = z² – 27 are ______ and ______ .

Answer/Explanation

Answer:
Explaination:
For zeroes z² – 27 = 0
⇒ z² = 27
⇒ z = ±√27
⇒ z = ± 3√3

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52. Verify that x = 3 is a zero of the polynomial. p(x) = 2x³ – 5x² – 4x + 3

Answer/Explanation

Answer:
Explaination:
Here p(x) = 2x³ – 5x² – 4x + 3
∴ p{3) = 2(3)³ – 5 x (3)² -4 x 3 + 3
= 54 – 45- 12 + 3 = 0
∵ P(3) = 0
∴ x = 3 is a zero of p(x)

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53. The graph of y =f(x) is given below. How many zeroes are there of f(x)?

Answer/Explanation

Answer:
Explaination:
Graph of y = f(x) intersect x-axis in one point only.
Therefore number of zeroes of f(x) is one.

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54. The graph of y = f(x) is given, how many zeroes are there of f(x)?

Answer/Explanation

Answer:
Explaination:
∵ Graph y =f(x) does not intersect x-axis.
∴ f(x) has no zeroes.

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55. The graph of y = f(x) is given below, for some polynomial f(x). Find the number of zeroes of f(x).

Answer/Explanation

Answer:
Explaination:
∵ Graph of f(x) intersects x-axis at three different points.
∴ Number of zeroes of f(x) = 3.

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56. The graph of x = p(y) is given below, for some polynomial p(y). Find the number of zeroes of p(y).

Answer/Explanation

Answer:
Explaination:
∵ Graph of p(y) intersects y-axis in four different points.
∴ Numbers of zeroes = 4

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57.
Graph of the polynomial p(x) =px² + 4x – 4 is given as above. Find the value of p.

Answer/Explanation

Answer:
Explaination:
Graph of p(x) touches the x-axis at (2, 0)
∴ x = 2 is a zero of the p(x)
⇒ p(2) = 0
⇒ p(2)² + 4 × 2 – 4 = 0
4p + 4 = 0
⇒ p = -1

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58. If the product of the zeroes of x2 – 3kx + 2k1 – 1 is 7, then values of k are _____ and _____ .

Answer/Explanation

Answer:
Explaination:
Product of zeroes = 7
⇒ 2k² – 1 = 7
⇒ 2k² = 8
⇒ k² = 4
⇒ k = ± 2

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59. If zeroes ofp(x) = 2x² -Ix + k are reciprocal of each other, then value of k is _____ .

Answer/Explanation

Answer:
Explaination:
∵ Zeroes are reciprocal of each other
∴ Product of zeroes = 1
⇒ \(\frac{k}{2}\) = 1
⇒ k = 2

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60. Find the product of the zeroes of – 2x² + kx + 6.

Answer/Explanation

Answer:
Explaination:
Here a = – 2, b = k, c = 6
Product of zeroes = \(\frac{c}{a}\)
i. e., α × β = \(\frac{6}{-2}\) = -3

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61. Find the sum of the zeroes of the given quadratic polynomial -3x² + k.

Answer/Explanation

Answer:
Explaination:
Since polynomial is -3x² + 0x + k
∴ a = -3, b = 0, c = k
and sum of zeroes = \(\frac{-b}{a}\)
i. e., α + β = \(\frac{-b}{a}\)
⇒ α + β = \(\frac{0}{-3}\) = 0

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62. If one zero of the polynomial x² -4x+ 1 is 2 + √3, write the other zero.

Answer/Explanation

Answer:
Explaination:
Let other zero be α ,

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63. Write the polynomial, the product and sum of whose zeroes are –\(\frac{9}{2}\) and –\(\frac{3}{2}\) respectively.

Answer/Explanation

Answer:
Explaination:

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64. The value of m, in order that x² – mx – 2 is the quotient where x³ + 3x² – 4 is divided by x + 2 is ____ .

Answer/Explanation

Answer:
Explaination:

Since quotient = x² – mx – 2
∴ x² – mx – 2 = x² + x – 2
On comparison of coefficient of x, we get m = – 1.

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65. If one factor of x³ + 7kx² – 4kx + 12 is (x + 3), then the value of k is ______ .

Answer/Explanation

Answer:
Explaination:
p(x) = x³ + Ike – 4kx + 12
p(-3) = (-3)³ + 7k(-3)² – 4k (- 3) + 12
∵ x + 3 is a factor
∴ p(-3) = 0
⇒ -27 + 63k + 12k+ 12 = 0
⇒ 75k – 15 = 0
⇒ k = \(\frac{1}{2}\)

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66. A polynomial of degree five is divided by a quadratic polynomial. If it leaves a remainder, then find the degree of remainder.

Answer/Explanation

Answer:
Explaination:
Degree of remainder is always less than divisor.
∴ Degree of remainder will be less than 2.
Hence, degree is 1 or 0.

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67. Check whether 3x – 7 is a factor of polynomial 6x³ + x² – 26x – 25?

Answer/Explanation

Answer:
Explaination:

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68. If x³ + x² – ax + b is divisible by x² – x, write the values of a and b.

Answer/Explanation

Answer:
Explaination:
∵ p(x) = x3+xl-ax + b is divisible by x² -x.
x (x – 1) is a factor of p(x).
⇒ x = 0 and x = 1 are zeroes of p(x).
⇒ p(0) = 0
⇒ (0)³ + (0)² – a × 0 + b = 0
⇒ b = 0 and p(1) = 0
⇒ 1³ + (1)² – a × 1 + b = 0
⇒ 2 – a + 0 — 0
⇒ a = 2

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