Table of Contents
MCQ Questions for Class 8 Maths: Ch 9 Algebraic Expressions and Identities
1. Expressions consists of _____________ and _______________.
(a) variables, constants
(b) identities
(c) expressions
(d) none of these
► (a) variables, constants
2. Which of the following is an expression?
(a) 1/2
(b) 3
(c) 3x-2
(d) 2
► (c) 3x-2
3. The coefficient of x in the expression -7x +5 is
(a) 5
(b) -7
(c) 7
(d) 0
► (b) -7
4. The number of terms in the expression 2x2+3x+5 is
(a) 1
(b) 2
(c) 3
(d) 5
► (c) 3
5. Like terms in the expression 7x,5x2,7y, -5yx, -9x2, are
(a) 7x, -5yx
(b) 5x2, -5yx
(c) 5x2, -9x2
(d) 7x, 7y
► (c) 5x2, -9x2
6. Terms are added to form ___________.
(a) expressions
(b) terms
(c) identities
(d) none of these
► (a) expressions
7. Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y, –3xz + 5x – 2xy.
(a) 5xy + 3zx + 5x – 4y
(b) 5xy + 9yz +2zx + 5x – 4y
(c) 5xy + 9yz +3zx + 5x – 4y
(d) 5xy + 9yz +3zx + 4y
► (c) 5xy + 9yz +3zx + 5x – 4y
8. The expression x + y + z is in
(a) one variable
(b) no variable
(c) three variables
(d) two variables
► (c) three variables
9. Which of the following is a monomial ?
(a) 4x²
(b) a + 6
(c) a + 6 + c
(d) a + b + c + d
► (a) 4x²
10. How many terms are there in the expression 5 – 3xy ?
(a) 1
(b) 2
(c) 3
(d) 5
► (b) 2
11. How many terms are there in the expression 5xy + 9yz + 3zx + 5x – 4y ?
(a) 1
(b) 3
(c) 4
(d) 5
► (d) 5
12. If x = 3 is solution of x² + kx + 15, value of k is
(a) k = -8, x = 3
(b) k = 8, x = 5
(c) k = 6, x = 5
(d) None of above
► (a) k = -8, x = 3
13. The coefficient in the term -5x is
(a) 5
(b) -5
(c) 1
(d) 2
► (b) -5
14. n (4 + m) = 4n + ___
(a) 4m
(b) 4n
(c) 4mn
(d) nm
► (d) nm
15. The like terms of the following are
(a) x, 3x
(b) x, 2y
(c) 2y, 6xy
(d) 3x, 2y
► (a) x, 3x
16. Value of expression ‘a(a2+a +1)+5’ for ‘ a’ = 0 is
(a) a+5
(b) 1
(c) 6
(d) 5
► (d) 5
17. Which of the following is like term as 7xy?
(a) 9
(b) 9x
(c) 9y
(d) 9xy
► (d) 9xy
18. The area of triangle is’ xy’ where’ x’ is length and ‘y’ is breadth. If the length of rectangle is increased by 5 units and breadth is decreased by 3 units, the new area of rectangle will be
(a) (x-y)(x+3)
(b) (xy+15)
(c) (x+5)(y-3)
(d) (xy+5-3)
► (c) (x+5)(y-3)
19. The expression 7xy has the factors
(a) 7,x,y
(b) x,y
(c) 7,x
(d) 7,y
► (a) 7,x,y
20. Which of the following is a binomial?
(a) 4x + y + 2
(b) 2x + 7
(c) 3x + 4y – 6
(d) 3x
► (b) 2x + 7
21. When numbers/literals are added or subtracted, they are called _________.
(a) identities
(b) expressions
(c) variables
(d) terms
► (d) terms
22. The volume of rectangular box whose length, breadth and height is 2p,4q.8r respectively is
(a) 14pqr
(b) 2p+4q+8r
(c) 64pqr
(d) 64
► (c) 64pqr
Algebraic Expressions and Identities Class 8 Extra important Questions Short Answer Type
Question 11.
Simplify the following:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
Solution:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
= a2b2 – a2c2) + b2c2 – b2a2) + c2a2 – c2b2)
= 0
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
= x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3
= x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3
= 6x3 – x2y2 – 2xy3
Question 12.
Multiply (3x2 + 5y2) by (5x2 – 3y2)
Solution:
(3x2 + 5y2) × (5x2 – 3y2)
= 3x2(5x2 – 3y2) + 5y2(5x2 – 3y2)
= 15x4 – 9x2y2 + 25x2y2 – 15y4
= 15x4 + 16x2y2 – 15y4
Question 13.
Multiply (6x2 – 5x + 3) by (3x2 + 7x – 3)
Solution:
(6x2 – 5x + 3) × (3x2 + 7x – 3)
= 6x2(3x2 + 7x – 3) – 5x(3x2 + 7x – 3) + 3(3x2 + 7x – 3)
= 18x4 + 42x3 – 18x2 – 15x3 – 35x2 + 15x + 9x2 + 21x – 9
= 18x4 + 42x3 – 15x3 – 18x2 – 35x2 + 9x2 + 15x + 21x – 9
= 18x4 + 27x3 – 44x2 + 36x – 9
Question 14.
Simplify:
2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)
Solution:
2x2(x + 2) – 3x(x2 – 3) – 5x(x + 5)
= 2x3 + 4x2 – 3x3 + 9x – 5x2 – 25x
= 2x3 – 3x3 – 5x2 + 4x2 + 9x – 25x
= -x3 – x2 – 16x
Question 15.
Multiply x2 + 2y by x3 – 2xy + y3 and find the value of the product for x = 1 and y = -1.
Solution:
(x2 + 2y) × (x3 – 2xy + y3)
= x2(x3 – 2xy + y3) + 2y(x3 – 2xy + y3)
= x5 – 2x3y + x2y3 + 2x3y – 4xy2 + 2y4
= x5 + x2y3 – 4xy2 + 2y4
Put x = 1 and y = -1
= (1)5 + (1)2 (-1)3 – 4(1)(-1)2 + 2(-1)4
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2
Question 16.
Using suitable identity find:
(i) 482 (NCERT Exemplar)
(ii) 962
(iii) 2312 – 1312
(iv) 97 × 103
(v) 1812 – 192 = 162 × 200 (NCERT Exemplar)
Solution:
Question 17.
Solution:
Question 18.
Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2 (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a2 -b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq2
= RHS.
Hence Verified.
Question 19.
Find the value of \(\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }\), using a suitable identity. (NCERT Exemplar)
Solution:
Question 20.
Find the value of x, if 10000x = (9982)2 – (18)2 (NCERT Exemplar)
Solution:
RHS = (9982)2 – (18)2 = (9982 + 18)(9982 – 18)
[Since a2 -b2 = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964
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