# Class 8 Maths MCQ – Identity

This set of Class 8 Maths Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “Identity”.

1. Select the term which is not an identity.
a) (a + b)2 = a2 + b2
b) (a + b)2 = a2 + b2 + 2ab
c) (a – b)2 = a2 + b2 – 2ab
d) (a + 1)(a + 2) = a2 + 3a + 2

Explanation: Identities are the special type of equations which are true for every value of the variables (a + 1)2 = a2 + b2 is only true for a = 1, b = 0; a = 0, b = 1.

2. Find the value of (2a + 1)2 using standard identity.
a) 4a2 + 2 + 2a
b) 4a2 + 2 – 4a
c) 4a2 – 1 + 4a
d) 4a2 + 1 + 4a

Explanation: Using the identity (a + b)2 = a2 + b2 + 2ab
⇒ (2a + 1)2 = (2a)2 + (1)2 + 2(2a)(1)
⇒ (2a + 1)2 = 4a2 + 1 + 4a.

3. Find the value of (m + 4n)2 using the standard identity.
a) m2 + 4n2 – 8mn
b) m2 + 16n2 + 4mn
c) m2 + 16n2 + 8mn
d) m2 + 4n2 + 8mn

Explanation: We know that (a + b)2 = a2 + b2 + 2ab.
⇒ (m + 4n)2 = (m)2 + (4n)2 + 2(m)(4n)
⇒ (m + 4n)2 = m2 + 16n2 + 8mn.

4. Calculate the value of (11x – 2y)2.
a) 121x2 + 4y2 – 22xy
b) 121x2 + 4y2 – 44xy
c) 121x2 + 4y2 + 44xy
d) 121x2 + 4y2 + 22xy

Explanation: The expression is of the form (a – b)2 where, a = 11x and b = 2y and (a – b)2 = a2 + b2 – 2ab.
⇒ (11x – 2y) = (11x)2 + (2y)2 – 2(11x)(2y)
⇒ (11x – 2y) = 121x2 + 4y2 – 44xy.

5. Evaluate the value of (2q – 3p)2.
a) 4q2 + 9p2 + 12pq
b) 4q2 + 9p2 – 12pq
c) 4q2 + 9p2 + 6pq
d) 4q2 + 9p2 – 6pq

Explanation: The expression is of the form (a – b)2 where, a = 2q and b = 3p and (a – b)2 = a2 + b2 – 2ab.
⇒ (2q – 3p)2 = (2q)2 + (3p)2 – 2(2q)(3p)
⇒ (2q – 3p)2 = 4q2 + 9p2 – 12pq.
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6. Using the standard identity, find the value of (3.9)2.
a) 15.11
b) 16.11
c) 15.21
d) 16.21

Explanation: 3.9 can be written as (4 – 0.1) and (a – b)2 = a2 + b2 – 2ab.
⇒ (4 – 0.1)2 = (4)2 + (0.1)2 – 2(4)(0.1)
⇒ (4 – 0.1)2 = 16 + 0.01 – 0.8 = 15.21.

7. Find the value of (12.4)2.
a) 152.77
b) 153.66
c) 153.76
d)152.76

Explanation: We can use the identity (a + b)2 = a2 + b2 + 2ab.
⇒ (12 + 0.4)2 = (12)2 + (0.4)2 + 2(12)(0.4)
⇒ (12 + 0.4)2 = 144 + 0.16 + 9.6 = 153.76.

8. Using the standard identity, find the value of (2m + 2)(2m + 3).
a) 4m2 + 10m + 5
b) 4m2 + 10m + 6
c) 4m2 + 10m – 6
d) 4m2 + 10m – 5

Explanation: The expression is of the form (x + a)(x + b). Now, (x + a)(x + b) = x2 + (a + b)x + ab.
⇒ (2m + 2)(2m + 3) = (2m)2 + (2 + 3)2m + (2)(3)
⇒ (2m + 2)(2m + 3) = 4m2 + 10m + 6.

9. Evaluate the value (x + 12)(x + 1).
a) x2 + 13x – 13
b) x2 + 13x + 13
c) x2 + 13x – 12
d) x2 + 13x + 12

Explanation: The expression is of the form (x + a)(x + b). Now, (x + a)(x + b) = x2 + (a + b)x + ab.
⇒ (x + 12)(x + 1) = x2 + (12 + 1)x + 12(1)
⇒ (x + 12)(x + 1) = x2 + 13x + 12.

10. Using the standard identity, find the value of 201 × 204.
a) 4104
b) 401004
c) 40104
d) 41004

Explanation: 201 and 204 can be written as (200 + 1) and (200 + 4) respectively. (x + a)(x + b) = x2 + (a + b)x + ab
⇒ (200 + 1)(200 + 4) = (200)2 + (1 + 4)200 + (1)(4)
⇒ (200 + 1)(200 + 4) = 40000 + 1000 + 4 = 41004.

11. Calculate the value of 904 × 902.
a) 815408
b) 8105408
c) 810548
d) 81548

Explanation: 904 and 902 can be written as (900 + 4) and (900 + 2) respectively. Also, (x + a)(x + b) = x2 + (a + b)x + ab.
⇒ (900 + 4)(900 + 2) = (900)2 + (4 + 2)900 + (2)4
⇒ (900 + 4)(900 + 2) = 810000 + 5400 + 8 = 815408.

12. Find the value of (10001 + 12)(10001 – 12).
a) 1000190857
b) 1000019857
c) 10019857
d) 100019857

Explanation: We know that (a + b)(a – b) = a2 – b2.
⇒ (10001)2 – (12)2 = 100020001 – 144
⇒ (10001)2 – (12)2 = 100019857.

13. Find the value of (98)2 – (2)2.
a) 9600
b) 960
c) 96000
d) 900

Explanation: a2 – b2 = (a + b)(a – b)
⇒ (98)2 – (2)2 = (98 + 2)(98 – 2)
⇒ (98)2 – (2)2 = 100(96) = 9600.

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

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