MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 2

State whether True or False.

Multiply:

$-\dfrac{2}{3}a^7b^2$ and

$-\dfrac{9}{4}ab^5$ .

The answer is

$\dfrac{3}{2}a^8b^7$ .

0%
True
0%
False

Explanation

$-\frac{2}{3}a^7b^2 \times -\frac{9}{4}ab^5$

$=(-\frac{2}{3})(-\frac{9}{4})a^{7+1}b^{2+5}$

$=\frac{3}{2}a^8b^7$

Multiply:

$-5cd^2$ by

$-5cd^2$

0%
$25c^2d^5$
0%
$25c^3d^4$
0%
$25c^2d^3$
0%
$25c^2d^4$

Explanation

$-5cd^2 \times -5cd^2$

$=(-5)(-5)c^{1+1}d^{2+2}$

$=25c^2d^4$

Multiply:

$\frac{2}{3}ab$ by

$-\frac{1}{4}a^2b$

0%
$-\frac{1}{6}a^3b^3$
0%
$-\frac{1}{6}a^3b^2$
0%
$-\frac{1}{6}a^4b^2$
0%
$-\frac{1}{6}a^3b^4$

Explanation

$\frac{2}{3}ab \times -\frac{1}{4}a^2b$

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

$=-\frac{2}{12}a^3b^2$

State whether True or False.

Multiply:

$4a$ and

$6a+7$ .

The answer is

$24a^2+28a$ .

0%
True
0%
False

Explanation

$\\ 4a\times (6a+7)\\ =24{ a }^{ 2 }+28a\\ \\$

State whether True or False.

Multiply:

$2a^2-5a-4$ and

$-3a$ .

The anwser is

$-6a^3+15a^2+12a$ .

0%
True
0%
False

Explanation

$(2a^2-5a-4)\times (-3a)$

$=2a^2(-3a)-5a(-3a)-4(-3a)$

$=2(-3)a^{2+1}-(5)(-3)a^{1+1}-4(-3)a$

$=-6a^3+15a^2+12a$

Use identities to solve:

$(97)^{2}$

0%
$9,659$
0%
$9,409$
0%
$9,009$
0%
$9,209$

Explanation

$(97)^2$

$=(100-3)^2$
using,

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

$=(100)^2-2(100)3+(3)^2$

$=10000-600+9$

$=9409$

Simplify:

$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right)$

0%
$\displaystyle { 6x }^{ 2 }-{ 2y }^{ 2 }-{ 3x }^{ 2 }y$
0%
$\displaystyle { 6x }^{ 2 }$
0%
$\displaystyle { 3x }^{ 2 }y$
0%
$\displaystyle { -2y }^{ 2 }$

Explanation

We have to simplify

$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right)$

$\displaystyle ={ 3x }^{ 2 }-{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }-{ 3x }^{ 2 }-{ 2y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 }$

$\displaystyle =\left( { 3x }^{ 2 }-{ 3x }^{ 2 } \right) +\left( -{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 } \right) -{ 2y }^{ 2 }$

$=\displaystyle 0-{ 2y }^{ 2 }=-{ 2y }^{ 2 }$ .
Hence given expression simplified to

$-2y^2$

Use the identity

$(a+b)(a-b) = a^2-b^2$ to evaluate:

$33\times 27$ .

0%
$891$
0%
$881$
0%
$981$
0%
$841$

Explanation

We know,

$33 \times 27 = (30 + 3) \times (30 - 3)$ .

Applying the formula

$(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where

$a = 30 , b = 3$ ,

we get,

$33 \times 27 = (30 + 3) \times (30 - 3) = { 30 }^{ 2 }-{ 3 }^{ 2 } = 900 - 9 = 891$ .

Therefore, option

$A$ is correct.

Use the identity

$(a+b)(a-b) = a^2-b^2$ to evaluate:

$21\times 19$ .

0%
$389$
0%
$399$
0%
$289$
0%
$429$

Explanation

We know,

$21 \times 19 = (20+ 1) \times (20 - 1)$ .

Applying the formula

$(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where

$a = 20 , b = 1$ ,

we get,

$21 \times 19 = (20+ 1) \times (20 - 1) = { 20 }^{ 2 }-{ 1 }^{ 2 } = 400 - 1 = 399$ .
Therefore, option

$B$ is correct.

Solve for

$x$ :

$(502)^{2}$

0%
$2,52,004$
0%
$2,62,104$
0%
$2,22,004$
0%
$2,52,864$

Explanation

$(502)^2$

$=(500+2)^2$
using,

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

$=(500)^2+2(500)(2)+(2)^2$

$=250000+2000+4$

$=252004$

Use identities to evaluate:

$(101)^{2}$

0%
$11,601$
0%
$12,761$
0%
$10,111$
0%
$10,201$

Explanation

$(101)^2$

$=(100+1)^2$
using,

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

$=(100)^2+2(100)(1)+1^2$

$=10000+200+1$

$=10201$

$49x^2-b=\left (7x+\dfrac {1}{2}\right )\left (7x-\dfrac {1}{2}\right )$ , then the value of

$b$ is :

0%
$0$
0%
$\dfrac {1}{\sqrt 2}$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{2}$

Explanation

$(7x+\frac12)(7x-\frac {1}{2})=49x^2-b$

$\Rightarrow 49x^2-\frac14=49x^2-b$ (

$\because (a-b)(a+b)=a^2-b^2$ )

$\Rightarrow b=\frac14$
Option C is correct.

Evaluate using the expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(92)^2$

0%
8444
0%
8464
0%
8474
0%
8414

Explanation

Given:

${92}^{2} = {(90 + 2)}^{2}$

It is in the form of

${(a+b)}^{2}$ , where

$a = 90, b = 2$ .

Applying the formula

${ (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab$ we get,

${92}^{2} = {(90 + 2)}^{2}$

$= {90}^{2} + { 2 }^{ 2 } + 2\times 90 \times 2$

$= 8100 + 4 + 360 = 8464$

Evaluate using expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(188)^2$

0%
35444
0%
35484
0%
35344
0%
35384

Explanation

${188}^{2} = {(200 - 12)}^{2}$

It is the form of

${(a - b)}^{2}$ , where

$a = 200, b = 12$

Applying the formula

${ (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$

${188}^{2} = {(200 - 12)}^{2} = {200}^{2} + { 12 }^{ 2 } - 2\times 200 \times 12 = 40000 + 144 - 4800 = 35344$

Obtain the product of

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

0%
$1045y^5$
0%
$1024y^6$
0%
$1634y^3$
0%
$102y^5$

Explanation

Product of

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

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

$=2\times 4 \times 8\times 16 \times y^{(1+2+3)}$

$[\because a^m \times a^n =a^{m+n} ]$

$=1024y^6$

The product of

$\displaystyle\frac{2}{3}xy$ and

$\dfrac{3}{2}xz$ is equal to

0%
$\displaystyle\frac{1}{6}$ xyz
0%
$x^2yz$
0%
$6x^2yz$
0%
none of these

Explanation

$\dfrac{2}{3} xy \times \dfrac{3}{2}xz$ =

$x^2yz$

Obtain the product of

$a,-a^2,a^3$

0%
$a^5$
0%
$-a^6$
0%
$a^3$
0%
$-a^5$

Explanation

$a,a^2,a^3$

$=(a)\times (-a^2)\times (a^3)$

$=-a^6$

Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

$xy, 2x^2y, 2xy^2$

0%
$16x^3y^4$
0%
$4x^2y^3$
0%
$4x^4y^4$
0%
$4x^2y^2$

Explanation

Given:

$xy, 2x^2y, 2xy^2$

Volume

$=l\times b\times h$

$=(xy\times 2x^2y\times 2xy^2)$

$=4x^4y^4$

Find the product of

$4p, 0$

0%
4
0%
$1$
0%
$4p$
0%
$0$

Explanation

$4p\times(0)$

$=0$

Obtain the volume of rectangular boxes with the given length, breadth and height respectively.

$2p, 4q, 8r$

0%
$64pqr$
0%
$16pqr$
0%
$32pqr$
0%
$8pr + r$

Explanation

Given :

$l,b,h$ of the rectangular box respectively are:

$2p, 4q, 8r$

Volume

$=l\times b\times h$

$=(2\times 4\times 8)pqr$

$=64pqr$

Obtain the product of

$a, 2b, 3c, 6abc$

0%
$36a^2b^2c^3$
0%
$36a^3b^2c^2$
0%
$36a^2b^2c^2$
0%
$36a^2b^3c^2$

Explanation

$a, 2b, 3c, 6abc$

$=(a)\times (2b)\times (3c)\times (6abc)$

$=36a^2b^2c^2$

Carry out the multiplication of the expressions of a given pair.

$ab,a -b$

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

Explanation

Given:

$ab,a -b$

$=ab\times (a-b)$

$=ab\times a-ab\times b$

$=a^2b-ab^2$

Simplify:

$x(3x+2)$

0%
$3x^2+2$
0%
$3x+2$
0%
$3x^2+2x$
0%
$3x^2+x$

Explanation

$x(3x+2)=3x^2+2$

$L.H.S=x(3x+2)$

$=3x^2+2x \neq R.H.S$
The correct statement is

$x(3x+2)=3x^2+2x$

Carry out the multiplication of the given expressions.

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

0%
$a^3b^2 - 7a^2b^3$
0%
$a^3b^2 + 7a^2b^3$
0%
$7a^3b^3 + 7a^2b^2$
0%
$7a^3b^2 + 7a^2b^3$

Explanation

Given:

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

Product

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

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

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

The value of

$(x^2y)(2x)(3y^3)$ is

0%
$12x^2y^4$
0%
$6x^3y^4$
0%
$3x^3y^4$
0%
$x^3y^4$

Explanation

$(x^2y)(2x)(3y^3)$
First:

$(2)(3) = 6$
Then:

$(x^3)(y^4)$
So,

$(x^2y)(2x)(3y^3)= 6x^3y^4$

$(a + b)^2-(a -b)^2$ will be equal to

$2ab$ :

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

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

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

$=4ab$ .

Hence, answer is

$4ab$ but not

$2ab$ .

Option

$B$ is correct.

Find the solution of

$(2x+6)^2$ .

0%
$x^2+24x+36$
0%
$4x^2+24x+6$
0%
$4x^2+24x+36$
0%
$4x^2-24x+36$

Explanation

$(2x+6)^2 = (2x+6)(2x+6)$
=

$2x.2x+2x.6+6.2x+6.6$
=

$4x^2+12x+12x+36$
=

$4x^2+24x+36$

What is the coefficient of

$x$ when

$-x + 6$ is multiplied by

$2x - 3$ ?

0%
$-15$
0%
$15$
0%
$-9$
0%
$9$

Explanation

We have,

$(-x+6)(2x-3)$

$=-x(2x-3)+6(2x-3)$

$=-2x^2+3x+12x-18$

$= -2x^2+15x-18$

Hence, coefficient of

$x$ is

$15$ .

The coefficient of the product of two monomials is not equal to the product of their coefficient.

0%
True
0%
False
0%
Sometimes
0%
Data insufficient

Explanation

The given statement is not true. We take an example.

$2x, 3y$ are two monomials with coefficients

$2$ and

$3$ respectively.
Now,

$(2x)(3y)=6xy$ , it is also monomial with coefficient

$6,$ which is a product of

$2$ and

$3.$

Hence, option

$B$ is correct.

$107\times 107+93\times93=$ ?

0%
$19578$
0%
$19418$
0%
$20098$
0%
$21908$
0%
None of these

Explanation

$107\times 107+93\times 93={107}^{2}+{93}^{2}$

$={(100+7)}^{2}+{(100-7)}^{2}$

$=2\times [{(100)}^{2}+{7}^{2}]$ ...................... [Ref:

${(a+b)}^{2}+{(a-b)}^{2}=2({a}^{2}+{b}^{2})$ ]

$=20098$

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 2 - MCQExams.com

0:0:2

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 2 - MCQExams.com

0:0:2

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.