MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 1

Multiply :

$(a^2 + b)$ and

$(a + b^2)$

0%
$a^2 + a^2b^2 + ab - b^3$
0%
$a^3 + a^2b^2 + ab + b^3$
0%
$a^3 + a^2b^2 - ab + b^3$
0%
$a^2 + a^2b^3 + ab + b^3$

Explanation

$(a^2 + b) (a + b^2)$

$=a^2(a+b^2)+b(a+b^2)$

$=a^3+a^2b^2+ab+b^3$

Multiply

$(2x + 5)$ and

$(4x-3)$ .

0%
$4x^2 + 14x-15$
0%
$4x^2 + 14x+15$
0%
$8x^2 + 14x-15$
0%
$8x^2 + 14x+15$

Explanation

$(2x+5)\times (4x-3)$

$=2x(4x-3)+5(4x-3)$

$=8x^2-6x+20x-15$

$=8x^2+14x-15$

Multiply:

$a + b, 7a^2b^2$

0%
$a^3b^6 + a^2b^3$
0%
$7a^3b^2 + 7a^2b^3$
0%
$7a^5b^2 + 3a^2b^3$
0%
$a^3b^2 + 7a3b^3$

Explanation

$a + b, 7a^2b^2$

$=(a + b)\times (7a^2b^2)$

$=7a^3b^2+7a^2b^3$

Simplify:

$5x(2x+3y)$

0%
$10x^2+15xy$
0%
$10x^4+15y$
0%
$10y^2+5xy$
0%
$10x^2+15x$

Explanation

$5x(2x+3y)$

$=10x^2+15xy$

Evaluate using expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(9.4)^2$

0%
88.36
0%
88.46
0%
89.16
0%
89.56

Explanation

${ (a+b) }^{ 2 }\quad =\quad a^{ 2 }+b^{ 2 }+2ab\\ a=9,\quad b=0.4\\ =\quad 9^{ 2 }+0.4^{ 2 }+2*9*0.4\\ =\quad 88.36$

Obtain the product of:

$2, 4y, 8y^2, 16y^3$

0%
$1024^6$
0%
$1000y^6$
0%
$1024y^5$
0%
$1024y^6$

Explanation

$2, 4y, 8y^2, 16y^3$

$=2\times 4y\times 8y^2\times 16y^3$

$=1024y^6$

Simplify:

$(l + m)^2- 4lm$

0%
$(l -m)^2$
0%
$4l^2m^2$
0%
$(l+2m)^2$
0%
$(l-2m)^2$

Explanation

$(l+m)^2-4lm$

$=l^2+2lm+m^2-4lm$

$=l^2-2lm+m^2$

$=(l-m)^2$

Multiply

$(5-2x)$ and

$(3 + x)$

0%
$15-x+{2 x }^{ 2 }$
0%
$15-x-{ x }^{ 2 }$
0%
$15-x-{2 x }^{ 2 }$
0%
$15-x+{ x }^{ 2 }$

Explanation

$(5 -2x)\times (3 + x)$

$=5(3 + x)-2x(3 + x)$

$=15+5x-6x-2x^2$

$=15-x-2x^2$

Find the value of

$2x\times 3x^3y^2\times3y^3z^2$

0%
$18x^4y^5z^2$
0%
$12x^4y^5z^2$
0%
$18x^2y^6z^2$
0%
$18x^4y^5z^3$

Explanation

$2x\times 3x^3y^2\times3y^3z^2$

$=(2\times 3\times 3)(x\times x^3)(y^2\times y^3)(z^2)$

$=18x^4y^5z^2$ , use

$a^m\times a^n=a^{m+n}$

Find the product of the following pairs of monomials

$4$ &

$7p$

0%
$28p$
0%
$36p$
0%
$47p$
0%
$32p$

Explanation

$\begin{aligned}{}4 \times 7p &= \left( {4 \times 7} \right)p\\ &= 28p\end{aligned}$

Obtain the product of:

$xy, yz, zx$

0%
$x^3y^2z^2$
0%
$x^2y^2y^2$
0%
$x^2y^2z^2$
0%
$x^2y^2y^y$

Explanation

$xy, yz, zx$

$=xy\times yz\times zx$

$=x^2y^2z^2$

$(2x + 3y)^{2} = 4x^{2} + 9y^{2} + M$ , find M.

0%
$12xy$
0%
$10xy$
0%
$12$
0%
$10$

Explanation

$(2x + 3y)^{2} = 4x^{2} + 9y^{2} + 12xy$ , .................

$(a+b)^2= a^2+2ab+b^2$

hence,

$M = 12xy$

Find the product of the following pairs of monomials.

$4p, 7pq$

0%
$p^2q$
0%
$28p^2q$
0%
$28pq$
0%
$28p^2$

Explanation

$4p, 7pq$
The product of the monomial pair is,

$=4\times p\times 7\times p\times q$

$=(4\times 7)\times(p\times p\times q)$

$=28p^2q$

Find the product of the following pair of monomials.

$4p$ &

$7p$

0%
$28p+2$
0%
$28$
0%
$28p^2$
0%
$28p$

Explanation

Given:

$4p, 7p$
The product of the monomial pair is:

$=4\times p\times 7\times p$

$=(4\times 7)\times(p\times p)$

$=28p^2$

Find the product of the following pairs of monomials.

$4p\ \&\ 0$

0%
$4p$
0%
$4$
0%
4 $+$ p
0%
$0$

Explanation

$4p, 0$
The product of the monomial pair is,

$=4\times p\times 0$

$=0$

Find the product of the following pairs of monomials.

$4p^3, 3p$

0%
$12p$
0%
$12p^4$
0%
$12p+4$
0%
$p^4$

Explanation

$4p^3, 3p$
The product of the monomial pair is,

$=4\times p\times p\times p\times 3\times p$

$=(4\times 3)\times(p\times p\times p\times p)$

$=12p^4$

Obtain the product of:

$m, mn, mnp$

0%
$m^2n^3p$
0%
$n^3m^2p$
0%
$m^3n^2p$
0%
$m^4n^2p$

Explanation

$m, mn, mnp$

$=m\times mn\times mnp$

$=m^3n^2p$

The coefficient of

$x^2$ in

$(2-3x^2) (x^2-5)$ is :

0%
$-17$
0%
$-10$
0%
$-3$
0%
$17$

Explanation

$(2-3x^2)(x^2-5)$

$=-3x^4+17x^2-10$

A Coefficient is a constant number multiplied with a variable.

Here as

$17$ is multiplied with

$x^2$ . so it becomes the coefficient.

Hence,

$D$ is the correct answer.

Multiply: $$4p, q + r$$

0%
$4r + 4pr$
0%
$4q + pr$
0%
$4p + pr$
0%
$4pq + 4pr$

Explanation

$4p, q + r$

$=4p\times( q + r)$

$=4pq+4pr$

Multiply:

$8ab^2$ by

$-4a^3b^4$ .

0%
$-32a^4b^7$
0%
$-32a^4b^6$
0%
$-32a^4b^5$
0%
$-32a^5b^6$

Explanation

$8ab^2\times (-4a^3b^4)$

$=8\times (-4)a^{1+3}b^{2+4}$

$=-32a^4b^6$

Multiply:

$ab, a b$

0%
$a^2b$
0%
$a^2b^2$
0%
$ab^3$
0%
$a^3b^3$

Explanation

$ab\times a b$

$={ a }^{ 1+1 }\times{ b }^{ 1+1 }$

$={ a }^{ 2 }{ b }^{ 2 }$

Find the product of

$x ,x^2, x^3 ,x^4$

0%
$x^{11}$
0%
$x^{8}$
0%
$x^{9}$
0%
$x^{10}$

Explanation

The exponent or product rule states that while multiplying two powers having the same base, we have to add the exponents.

$\Rightarrow x\times x^2\times x^3\times x^4$

$\Rightarrow x^{1+2+3+4}$

$\Rightarrow x^{10}$

The product of

$4x^2 - 7x + 19$ and

$(- 2x)$ when

$x= 0$ is:

0%
$19$
0%
$-38$
0%
$17$
0%
$0$

Explanation

Any real number when multiplied by 0 gives 0.
Here at

$x=0$ ,

$-2x = 0$ and hence the result of multiplication will also be zero.

Simplify:

$5{ x }^{ 2 }(x+5)+25$

0%
$\\ 5{ x }^{ 3 }+25{ x }+25$
0%
$5{ x }^{ 3 }+50x$
0%
$5{ x }^{ 3 }+25{ x }^{ 2 }+25$
0%
$5{ x }^{ 3 }+25x$

Explanation

$5{ x }^{ 2 }(x+5)+25\\ =5{ x }^{ 3 }+25{ x }^{ 2 }+25$

Multiply:

$a^2 -9, 4a$

0%
$4a^3 - 36a$
0%
$3a^3 - 36a$
0%
$3a^3 +36a$
0%
$4a^3 + 36a$

Explanation

$a^2- 9, 4a$

$=a^2- 9\times 4a$

$=4a^3-36a$

State whether True or False.

Multiply:

$-\dfrac{3}{2}x^5y^3$ and

$\dfrac{4}{9}a^2x^3y$ .

The answer is

$-\dfrac{2}{3}a^2x^8y^4$ .

0%
True
0%
False

Explanation

$-\frac{3}{2}x^5y^3 \times \frac{4}{9}a^2x^3y$

$=(-\frac{3}{2})(\frac{4}{9})ax^{5+3}y^(3+1)$

$=-\frac{2}{3}a^2x^8y^4$

State whether True or False.

Multiply:

$abx, -3a^2x$ and

$7b^2x^3$ .

The value is

$-21a^3b^3x^5$ .

0%
True
0%
False

Explanation

$(abx)\times (3a^2x)\times (7b^2x^3)$

$=(-3)(7)a^{1+2}b^{1+2}x^{1+1+3}$

$=-21a^3b^3x^5$

State whether True or False.

On multiplying

$a^2, ab$ and

$b^2$ , we get the answer as

$a^3b^3$ .

0%
True
0%
False

Explanation

The given terms are

$a^2,\ ab$ and

$b^2$

On multiplying the given terms, we get:

$a^2 \times ab \times b^2$
Using the laws of exponent, we know that,

$a^m\times a^n=a^{m+n}$

$\therefore \ a^2 \times ab \times b^2=a^{2+1}b^{1+2}$

$\Rightarrow a^3b^3$

$\therefore \ a^2 \times ab \times b^2=a^3b^3$

Hence, the given statement is true.

State whether True or False.

Multiply:

$-3bx, -5xy$ and

$-7b^2x^3$ .

The answer is

$-105b^3x^5y$ .

0%
True
0%
False

Explanation

$(3bx)\times (5xy)\times (7b^2x^3)$

$=(-3)(-5)(-7)b^{1+2}x^{1+1+3}y^{1}$

$=-105b^3x^5y$

State whether True or False.

Multiply:

$2a^3-3a^2b$ and

$-\dfrac{1}{2}ab^2$ .

The value is

$-a^4b^2+\dfrac{3}{2}a^3b^3$ .

0%
True
0%
False

Explanation

As given

$2a^3-3a^2b$ and

$-\frac{1}{2}ab^2$

$\Rightarrow 2a^{3}\times-\frac{1}{2}ab^{2} - 3a^{2}b\times-\frac{1}{2}ab^{2}$

$\Rightarrow -a^{3}\times ab^{2} + \frac{3}{2}a^{2}b\times ab^{2}$

$\Rightarrow -a^{3+1}b^{2}+\frac{3}{2}a^{2+1}b^{1+2}$

$\Rightarrow -a^{4}b^{2}+\frac{3}{2}a^{3}b^{3}$

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 1 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 1 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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