MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 3

$217\times 217+183\times 183=$ ?

0%
$79698$
0%
$80578$
0%
$80698$
0%
$81268$

Explanation

${(217)}^{2}+{(183)}^{2}={(200+17)}^{2}+{(200-17)}^{2}$

$=2\times [{(200)}^{2}+{(17)}^{2}]$ [Ref:

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

$=2[40000+289]$

$=2\times 40289$

$=80578$

$(a + b)^{2} = (a + b)\times$ ____

0%
$ab$
0%
$2ab$
0%
$(a + b)$
0%
$(a - b)$

Explanation

We know that

$x^{2}=x\times x$

Then

$(a+b)^{2}=(a+b)\times (a+b)$

Hence, the answer is (C)

The value of

$(2.3 a^{5}b^{2})\times (1.2a^{2}b^{4})$ , when

$a = 1$ and

$b = 0.5$ is ________.

0%
$0.011625$
0%
$0.043125$
0%
$0.031825$
0%
$0.041525$

Explanation

Given expression is

$(2.3a^5b^2)\times (1.2a^2b^4)$

$= 2.76 \times a^{7}b^{6}$

Also given

$a=1, b=0.5$

Subtituting the values, we get

$2.76 \times 1^{7}\times (0.5)^{6}$

$= 2.76 \times (0.015625)$

$=0 .043125$

Find the product.

$\left ( \dfrac{2}{3}xy \right ) \times \left ( \dfrac{-9}{10}x^2 y^2 \right )$

0%
$-\dfrac{1}{5}x^6 y^-3$
0%
$-\dfrac{2}{7}x^2 y^7$
0%
$-\dfrac{3}{5}x^3 y^3$
0%
$-\dfrac{2}{5}y^3 x^3$

Explanation

$\left ( \frac{2}{3}xy \right ) \times \left ( \frac{-9}{10}x^2 y^2 \right )$

$=-\frac{3}{5}x^3 y^3$

Obtain the product of:

$a, a^2, a^3$

0%
$a^5$
0%
$a^4$
0%
$a^6$
0%
$6^a$

Explanation

${\textbf{Step -1 : Perform the product of }}\mathbf{a,{a^2},{a^3}.}$

$a \times {a^2} \times {a^3}$

$= {a^1} \times {a^2} \times {a^3}.$

$= {a^{\left( {1 + 2 + 3} \right).}}$

${\textbf{Step -2 : Calculate by adding the powers}}{\textbf{.}}$

$= {a^{\left( {1 + 2 + 3} \right).}}$

$= {a^6}.$

${\textbf{Hence , the product of }}\mathbf{a,{a^2},{a^3}{\text{ is }}{a^6}}\textbf {and correct answer is option C}$

Find the product of

$\left ( -\dfrac{10}{3}pq^3 \right ) \times \left ( \dfrac{6}{5}p^3 q \right )$

0%
$p^4q^4$
0%
$- 4p^4q^4$
0%
$4p^2q^4$
0%
$4q^4p^5$

Explanation

$\left ( -\frac{10}{3}pq^3 \right ) \times \left ( \frac{6}{5}p^3 q \right )$

$=-2pq^3\times 2p^3q$

$=-4p^4q^4$

Find the product of

$(p^2 q^2)$ and

$(2p + q)$ .

0%
$2{ p }^{ 4 }{ q }^{ 2 }+{ 2p }^{ 2 }{ q }^{ 2 }$
0%
$2{ p }^{ 4 }{ q }^{ 2 }+{ p }^{ 2 }{ q }^{ 2 }$
0%
$2{ p }^{ 2 }{ q }^{ 2 }+{ p }^{ 2 }{ q }^{ 2 }$
0%
$2{ p }^{ 3 }{ q }^{ 2 }+{ p }^{ 2 }{ q }^{ 3 }$

Explanation

$\begin{aligned}{}{p^2}{q^2}\left( {2p + q} \right)& = {p^2}{q^2} \times 2p + {p^2}{q^2} \times q\\ &= 2{p^3}q^2 + {p^2}{q^3}\end{aligned}$

Hence, the product of

$p^2q^2$ and

$2p+q$ is equal to

$2{p^3}q^2 + {p^2}{q^3}.$

Simplify the following:

$(\sqrt{3}-\sqrt{2})^{2}$ is equal to

$5-2\sqrt{6}$ .
If true then enter

$1$ and if false then enter

$0$ .

0%
$1$
0%
$0$
0%
$-1$
0%
$-2$

Explanation

We know,

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

$a^{2} + b^{2} - 2{a}{b}$ .

Then,

$(\sqrt{3}-\sqrt{2})^{2}$
=

$(\sqrt{3})^{2} + (\sqrt{2})^{2} - 2\times\sqrt{3}\times\sqrt{2}$
=

$3 + 2 - 2\sqrt{6}$
=

$5 - 2\sqrt{6}$ .

Hence, the statement is correct.

Therefore, option

$A$ is correct.

Find the errors and correct the given expression.

$(z + 5)^2 = z^2 + 25$

0%
$(2z) = z^2 + 10z + 25$
0%
$(z + 5) = z^2 + 10z + 25$
0%
$(z + 5)^2 = z^2 + 10z + 25$
0%
$(z + 10)^2 = z + 10 - 5$

Explanation

LH.S=

$(z+5)^2=z^2+2(z)(5)+5^2 ..................... [(a+b)^2=a^2+2ab+b^2]$

$=z^2+10z+25 \neq$ R.H.S

The correct option is C.

Find the product of given expressions:

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

0%
$36a^2b^2c^2$
0%
$6a^2b^2c^2$
0%
$a^2b^2c^2$
0%
$2a^36b^2c^2$

Explanation

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

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

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

Use the identity

$(x + a) (x + b) = x^2 + (a + b) x + ab$ to find the given product:

$(2x + 5y)$

$(2x + 3y)$ .

0%
$4x^2 + 24xy - 15y^2$
0%
$4x^2 + 16xy + 25y^2$
0%
$4x^2 + 16xy + 15y^2$
0%
$x^2 + 16xy + 15y^2$

Explanation

Given,

$(2x + 5y) (2x+3y)$ .

We know,

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

On comparing, we get,

$x=2x ,a=5y ,b=3y$ .

Thus,

$(2x + 5y) (2x+3y)$

$=(2x)^2+(5y+3y)2x+(5y)\times(3y)$

$=4x^2+(8y)2x+15y^2$

$=4x^2+16xy+15y^2$ .

Hence, option

$C$ is correct.

Use the identity

$(x + a) (x + b) = x^2 + (a + b) x + ab$ to find the given product:

$(xyz +4) (xyz +2)$ .

0%
$x^2 y^2 z^2+ 6xyz$
0%
$x^2 y^2 z^2 + 8$
0%
$x^2 y^2 z^2 + 6xyz + 8$
0%
$x^2 y^2 z^2 - 6xyz + 8$

Explanation

Given,

$(xyz+ 4) (xyz+2)$ .

We know,

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

On comparing, we get,

$x=xyz ,a=4 ,b=2$ .

Thus,

$(xyz+4) (xyz+2)$

$=(xyz)^2+(4+2)xyz+4\times 2$

$=x^2y^2z^2+6xyz+8$ .

Therefore, option

$C$ is correct.

Simplify :

$(4m + 5n)^2 + (5m + 4n)^2$ .

0%
$41m^2 + 80mn + 41n^2$
0%
$4m^2 - 54mn + 41n^2$
0%
$41m + 80mn + 4n^2$
0%
$41m^2 + 10mn + 41n^2$

Explanation

Given,

$(4m + 5n)^2 + (5m + 4n)^2$ .

We know,

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

Then,

$(4m + 5n)^2 + (5m + 4n)^2$

$=[(4m)^2+2(4m)(5n)+(5n)^2]+[(5m)^2+2(5m)(4n)+(4n)^2]$

$=16m^2+40mn+25n^2+25m^2+40mn+16n^2$

$=41m^2+80mn+41n^2$ .

Therefore, option

$A$ is correct.

Simplify:

$(m^2 - n^2m)^2 + 2m^3n^2$ .

0%
$m^4 - n^4m^2$
0%
$m^4 + n^4m^2$
0%
$m^4 + n^4$
0%
$m^4 + n^4m^2 + 2m^3n^2$

Explanation

Given,

$(m^2 - n^2m)^2 + 2m^3n^2$ .

We know,

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

Then,

$(m^2 - n^2m)^2 + 2m^3n^2$

$=(m^2)^2-2(m^2)(n^2m)+(n^2m)^2+2m^3n^2$

$=m^4-2m^3n^2+n^4m^2+2m^3n^2$

$=m^4+n^4m^2$ .

Therefore, option

$B$ is correct.

Use the identity

$(x + a) (x + b) = x^2 + (a + b) x + ab$ to find the given product:

$(4x + 5)$

$(4x-1)$ .

0%
$16x^2 + 16x - 5$
0%
$16x^2 - 16x - 5$
0%
$16x^2 + 16x + 5$
0%
$16x^2 + 12x - 5$

Explanation

Given,

$(4x + 5) (4x- 1)$ .

We know,

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

On comparing, we get,

$x=4x ,a=5 ,b=-1$ .

Thus,

$=(4x+ 5) (4x -1)$

$=(4x)^2+[(5)+(-1)]4x+(5)\times(- 1)$

$=16x^2+(4)4x+(-5)$

$=16x^2+16x-5$ .

Hence, option

$A$ is correct.

Use the identity

$(x + a) (x + b) = x^2 + (a + b) x + ab$ to find the given product:

$(4x-5)$

$(4x-1)$ .

0%
$16x^2-24x + 5$
0%
$16x^2+ 24x + 1$
0%
$16x^2+ 20x + 4$
0%
$16x^2+ 20x + 5$

Explanation

Given,

$(4x- 5) (4x -1)$ .

We know,

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

On comparing, we get,

$x=4x ,a=-5 ,b=-1$ .

Thus,

$=(4x- 5) (4x -1)$

$=(4x)^2+[(-5)+(-1)]4x+(-5)\times(- 1)$

$=16x^2+(-6)4x+(5)$

$=16x^2-24x+5$ .

Hence, option

$A$ is correct.

The product of

$4a^{2}, -6b^{2}$ and

$3a^{2}b^{2}$ is

0%
$a^{2}b^{2}$
0%
$13a^{4}b^{4}$
0%
$-72a^{4}b^{4}$
0%
$a^{4}b^{4}$

Use the identity

$(x + a) (x + b) = x^2 + (a + b) x + ab$ to find the given product:

$(2a^2 + 9)$

$(2a^2 + 5)$ .

0%
$4a^4 + 28a^2 - 5$
0%
$4a^4 + 28a^2 + 45$
0%
$4a^3 + 28a^2 + 45$
0%
$a^4 + 28a^2 + 45$

Explanation

Given,

$(2a^2 + 9) (2a^2+5)$ .

We know,

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

On comparing, we get,

$x=2a^2 ,a=9 ,b=5$ .

Thus,

$(2a^2 + 9) (2a^2+5)$

$=(2a^2)^2+(9+5)2a^2+(9)\times(5)$

$=4a^2+(14)2a^2+45$

$=4a^4+28a^2+45$ .

Therefore, option

$B$ is correct.

$\displaystyle \left( x+8 \right) \left( x-8 \right)$ is equal to

0%
$\displaystyle { x }^{ 2 }-16x+64$
0%
$\displaystyle { x }^{ 2 }-64$
0%
$\displaystyle { x }^{ 2 }+64$
0%
$\displaystyle { x }^{ 2 }+16x-64$

Explanation

Required multiplication is

$=\displaystyle \left( x+8 \right) \left( x-8 \right)$

$\displaystyle =x\left( x-8 \right) +8\left( x-8 \right)$

$\displaystyle ={ x }^{ 2 }-8x+8x-64$

$\displaystyle ={ x }^{ 2 }-64$

Simplify:

$(2.5p -1.5q)^2- (1.5p-2.5q)^2$ .

0%
$4p^2 - 4q^2$
0%
$6.25p^2 -4q^2$
0%
$4p^2 - 2.25q^2$
0%
$6.25p^2 - 2.25q^2$

Explanation

Given,

$(2.5p - 1.5q)^2 - (1.5p - 2.5q)^2$ .

We know,

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

Then,

$(2.5p - 1.5q)^2 - (1.5p - 2.5q)^2$

$=[(2.5p)^2-2(2.5p)(1.5q)+(1.5q)^2]-[(1.5p)^2-2(1.5p)(2.5q)+(2.5q)^2]$

$=6.25p^2-7.5pq+2.25q^2-2.25p^2+7.5pq-6.25q^2$

$=6.25p^2+2.25q^2-2.25p^2-6.25q^2$

$=6.25p^2-2.25p^2+2.25q^2-6.25q^2$

$=4p^2-4q^2$ .

Therefore, option

$A$ is correct.

Simplify:

$(ab + bc)^2 - 2ab^2c$ .

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

Explanation

Given,

$(ab + bc)^2 - 2ab^2c$ .

We know,

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

Then,

$(ab + bc)^2 - 2ab^2c$

$=(ab)^2+2(ab)(bc)+(bc)^2-2ab^2c$

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

$=a^2b^2+b^2c^2$ .

Therefore, option

$C$ is correct.

Find the product of:

$\left ( \cfrac{2}{5}x^{2}y^{2}z^{5} \right ) \left ( -\cfrac{15}{8}x^{5}y^{3}z^{3} \right ) \left ( \sqrt{3}x^{7}y^{2}z^{3} \right )$

Answer:

$-\cfrac{3\sqrt{3}}{4}x^{14} y^{7} z^{11}$

0%
True
0%
False

Explanation

$\cfrac { 2 }{ 5 } x^{ 2 }{ y }^{ 2 }{ z }^{ 5 }\times \cfrac { -15 }{ 8 } { x }^{ 5 }{ y }^{ 3 }{ z }^{ 3 }\times \sqrt { 3 } { x }^{ 7 }{ y }^{ 2 }{ z }^{ 3 }$

$=x^{ 2+5+7 }{ y }^{ 2+3+2 }{ z }^{ 5+3+3 }\times \cfrac { -15 }{ 8 } \sqrt { 3 } \times \cfrac { 2 }{ 5 }$

$=x^{ 14 }{ y }^{ 7 }{ z }^{ 11 }\times \cfrac { -3 }{ 4 } \sqrt { 3 }$

So, the correct answer is option

$A$ .

The value of the product

$(4a^{2}+3b)(9b^{2}+4a)$ at

$a = 1,$ b

$= -2$ is

0%
$60$
0%
$-80$
0%
$70$
0%
$-50$

Explanation

$a = 1,$

$b = -2$ we have

$4a^2 + 3b= 4 + 3(-2) = -2$ and

$9b^2 + 4a = 9(-2)^2 + 4(1) = 9 \times 4 + 4 = 40$
Thus product =

$-2 \times 40 = -80$

Simplify :

$\displaystyle \left(\frac{3}{2}x - 0.45y\right)^2$

0%
$2.25x^2-1.35xy+0.2015y^2$
0%
$2.15x^2-1.35xy+0.2025y^2$
0%
$2.25x^2-1.35xy+0.2025y^2$
0%
$2.25x^2-1.25xy+0.2025y^2$

Explanation

$\displaystyle \left(\frac{3}{2}x - 0.45y\right)^2$

$=\displaystyle \left(1.5x - 0.45y\right)^2$

$=\displaystyle (1.5x)^2 + (0.45y)^2- 2(1.5x)(0.45y)^2$

$= 2.25x^2 +0.2025 y^2 -1.35xy$

Use the identity

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

$103\times 97$ .

0%
$8881$
0%
$9991$
0%
$9981$
0%
$9891$

Explanation

We know, $103 \times 97 = (100 + 3) \times (100 - 3)$ .

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 100 , b = 3$ ,

we get,
$103 \times 97 = (100 + 3) \times (100 - 3) = { 100 }^{ 2 }-{ 3 }^{ 2 } = 10000 - 9 = 9991$ .

Therefore, option $B$ is correct.

Evaluate (using formulae):

$\displaystyle \frac{2.43 \times 2.43 - 2 \times 2.43 \times 1.67 +1.67 \times 1.67}{2.43 - 1.67}$

0%
$0.76$
0%
$0.55$
0%
$1$
0%
$1.4$

Explanation

Given:

$\displaystyle \frac {2.43 \times 2.43 - 2 \times 2.43 \times 1.67 + 1.67 \times 1.67}{2.43-1.67}$

$=\displaystyle \frac {(2.43)^2 - 2 \times 2.43 \times 1.67 + (1.67)^2 }{0.76}$

$=\displaystyle \frac {(2.43-1.67)^2}{0.76}$ [

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

$=\displaystyle \frac {(0.76)^2}{0.76}$

$=0.76$

Use the identity

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

$9.8\times 10.2$ .

0%
$98.96$
0%
$99.86$
0%
$98.86$
0%
$99.96$

Explanation

We know, $9.8 \times 10.2 = (10 -0.2) \times (10+0.2)$ .

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 10 , b = 0.2$ ,

we get,
$9.8 \times 10.2 = (10 -0.2) \times (10+0.2)=10^2-(0.2)^2 \\ =100-0.04=99.96 .$

Therefore, option $D$ is correct.

State whether the statement is True or False.

Evaluate:

$(a+bc)(a-bc)(a^2+b^2c^2)$ is equal to

$a^4-b^4c^4$ .

0%
True
0%
False

Explanation

Given,

$(a+bc)(a-bc)(a^2+b^2c^2)$ .

We know,

$(a+b)(a-b)$

$=(a^{2}-b^{2})$ .

Multiply by using direct method, we get,

$(a+bc)(a-bc)(a^2+b^2c^2)$

$= [a^2-(bc)^2](a^2+b^2c^2)$

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

$= (a^2+b^2c^2)(a^2-b^2c^2)$ .

Now, by using formula

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

$= ((a^2)^2-(b^2c^2)^2)$

$= a^4-b^4c^4$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the statement is True or False:

On evaluating

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$ we get,

$49x^2-\dfrac{4}{9}y^2$ .

0%
True
0%
False

Explanation

Given,

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$ .

We know,

$(a+b)(a-b)$

$=(a^{2}-b^{2})$ .

Then, on multiplying, we get,

$(7x+\dfrac{2}{3}y)(7x-\dfrac{2}{3}y)$

$= (7x)^2-(\dfrac{2}{3}y)^2$

$= 49x^2-\dfrac{4}{9}y^2$ .

Hence, the statement is true.

Therefore, option

$A$ is correct.

Use the identity

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

$4.6\times 5.4$ .

0%
$24.84$
0%
$25.24$
0%
$24.94$
0%
$25.84$

Explanation

We are given the formula

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

We know,

$4.6 \times 5.4 = (5-0.4)(5+0.4)$ ,

where

$a=5$ and

$b=0.4$ .

Now, solve as shown below:

$4.6 \times 5.4 = (5-0.4)(5+0.4)$

$={ 5 }^{ 2 }-{ 0.4 }^{ 2 }=25-0.16=24.84$ .

Hence,

$4.6 \times 5.4 =24.84$ .

Therefore, option

$A$ is correct.

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 3 - MCQExams.com

0:0:4

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 3 - MCQExams.com

0:0:4

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.