MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 5

Find the square of:

$391$

0%
154881
0%
153881
0%
152881
0%
151881

Explanation

${391}^{2} = {(400 - 9)}^{2}$

It is the form of ${(a - b)}^{2}$ , where $a = 400, b = 9$

Applying the formula ${ (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$
${391}^{2} = {(400 - 9)}^{2} = {400}^{2} + { 9 }^{ 2 } - 2\times 400 \times 9 = 1,60,000 + 81 - 7200 = 152881$

Find the square of:

$(3a + 7b)$ .

0%
$9a^{2}\, +\, 12ab\, +\, 22b^{2}$
0%
$9a^{2}\, +\, 2ab\, -\, 49b^{2}$
0%
$9a^{2}\, +\, 42ab\, +\, 49b^{2}$
0%
$9a^{2}\, -\, 42ab\, +\, 9b^{2}$

Explanation

Given,

$2a + b$ .

On squaring, we get,

$(2a+b)^2$ .

We know,

$(x+y)^2=x^2+y^2+2xy$ .

Then,

$(2a+b)^2$

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

$=4a^2+4ab+b^2$ .

Therefore, option

$D$ is correct.

Find the square of:

$2a + b$ .

0%
$4a^{2}\, -\, 4b\, +\, b^{3}$
0%
$a^{2}\, +\, 4ab\, +\, b^{3}$
0%
$a^{2}\, -\, 4b\, +\, b^{2}$
0%
$4a^{2}\, +\, 4ab\, +\, b^{2}$

Explanation

Given,

$2a + b$ .

On squaring, we get,

$(2a+b)^2$ .

We know,

$(x+y)^2=x^2+y^2+2xy$ .

Then,

$(2a+b)^2$

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

$=4a^2+4ab+b^2$ .

Therefore, option

$D$ is correct.

Find the square of the following number:

$998$

0%
$9,96,004$
0%
$9,15,004$
0%
$9,77,004$
0%
$9,83,004$

Explanation

$(998)^2=(1000-2)^2$

Using,

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

$(998)^2 = (1000)^2-2(1000)2+2^2$

$=1000000-4000+4$

$=996004$

Find the square of:

$607$

0%
368549
0%
368449
0%
368349
0%
368249

Explanation

$607$ can be written as $600+7$

$\therefore {607}^{2} = {(600 + 7)}^{2}$
It is the form of ${(a+b)}^{2}$ , where $a = 600, b = 7$
Applying the formula ${ (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab$
${607}^{2} = {(600 + 7)}^{2} = {600}^{2} + { 7 }^{ 2 } +2\times 600 \times 7 = 360000 + 49 + 8400 = 368449$

Multiply

$2x$ ,

$\dfrac{5y}{2}$ and

$z^2$

0%
$5x y z^2$
0%
$2x y z^2$
0%
$5x y^2 z$
0%
none of these

Explanation

$2x \times \dfrac{5y}{2} \times z^2 = 5xyz^2$

Evaluate:

$(4a\, +\, 3b)^{2}\, -\, (4a\, -\, 3b)^{2}\, +\, 48\, ab$ .

0%
$76 ab$
0%
$96 ab$
0%
$46 ab$
0%
$106 ab$

Explanation

Given, $(4a+3b)^{2} -(4a-3b)^{2}+48 ab$ .

We know, $(a+b)^2=a^2+2ab+b^2$

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

Then,

$(4a+3b)^{2} -(4a-3b)^{2}+48 ab$

$=((4a)^2+2(4a)(3b)+(3b)^2)-((4a)^2-2(4a)(3b)+(3b)^2)+48ab$

$=(4a)^2+2(4a)(3b)+(3b)^2-(4a)^2+2(4a)(3b)-(3b)^2+48ab$

$=24ab+24ab+48ab$

$=48ab+48ab$

$=96ab$ .

Therefore, option

$B$ is correct.

Evaluate:

$\displaystyle \left(\frac{2x}{7}\, -\, \frac{7y}{4}\right)^{2}$ .

0%
$\displaystyle \left(\frac{4x^{2}}{32}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$
0%
$\displaystyle \left(\frac{4x^{2}}{15}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$
0%
$\displaystyle \left(\frac{4x^{2}}{49}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$
0%
$\displaystyle \left(\frac{4x^{2}}{19}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$

Explanation

Given, $\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$ .

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

Then,

$\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$

$=(\dfrac{2x}{7})^2-2(\dfrac{2x}{7})(\dfrac{7y}{4})+(\dfrac{7y}{4})^2$

$=\dfrac{4x^{2}}{49}- xy+ \dfrac{49y^{2}}{16}$ .

Therefore, ption $C$ is correct.

Evaluate:

$\displaystyle\left(\dfrac{7}{8}{x}\, +\, \dfrac{4}{5}{y}\right )^{2}$ .

0%
$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$
0%
$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$
0%
$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$
0%
$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$

Explanation

Given, $\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$ .

We know, $(x+y)^2=x^2+2xy+y^2$ .

Then,

$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$

$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$

$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}}$ .

Therefore, option $D$ is correct.

The product of

$\displaystyle \left ( \frac{4p}{5}-3 \right )$ and

$\displaystyle \left ( \frac{5p}{8}-6 \right )$ is

0%
$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p-18$
0%
$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p-18$
0%
$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p+18$
0%
$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p+18$

Explanation

$\displaystyle \left ( \frac{4p}{5}-3 \right )\left ( \frac{5p}{8} -6\right )$

$\displaystyle =\frac{4p}{5}\times \left ( \frac{5p}{8}-6 \right )+(-3)\times \left ( \frac{5p}{8}-6 \right )$

$\displaystyle =\frac{p^{2}}{2}-\frac{24p}{5}-\frac{15p}{8}+18$

$\displaystyle =\frac{p^{2}}{2}-\frac{267}{40}p+18$

Simplify the following expression :

$x (y - z) - y (z - x) - z (x - y)$

0%
$2x (y - z)$
0%
$2y (z -x)$
0%
$2x ( z- y)$
0%
None

Explanation

$x(y-z) - y(z-x) - z(x-y) = xy - xz -yz + xy -zx + zy$

$2xy -2xz$

$2x(y-z)$

$\displaystyle \left( \frac { 1 }{ 5 } x-\frac { 1 }{ 6 } y \right) \left( 5x+6y \right)$ is equal to

0%
$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } -{ y }^{ 2 }$
0%
$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } -{ y }^{ 2 }$
0%
$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } +{ y }^{ 2 }$
0%
$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } +{ y }^{ 2 }$

Explanation

Required multiplication

$=\displaystyle \left( \frac { 1 }{ 5 } x-\frac { 1 }{ 6 } y \right) \left( 5x+6y \right)$

$\displaystyle =\frac { 1 }{ 5 } x\left( 5x+6y \right) -\frac { 1 }{ 6 } y\left( 5x+6y \right)$

$\displaystyle =\left( \frac { 1 }{ 5 } x \right) \left( 5x \right) +\left( \frac { 1 }{ 5 } x \right) \left( 6y \right) +\left( -\frac { 1 }{ 6 } y \right) \left( 5x \right) +\left( -\frac { 1 }{ 6 } y \right) \left( 6y \right)$

$\displaystyle ={ x }^{ 2 }+\frac { 6 }{ 5 } xy-\frac { 5 }{ 6 } xy-{ y }^{ 2 }$

$\displaystyle ={ x }^{ 2 }+\frac { 36xy-25xy }{ 30 } -{ y }^{ 2 }$

$={ x }^{ 2 }+\dfrac { 11xy }{ 30 } -{ y }^{ 2 }$

Find the product:

$\displaystyle \left( -3axy \right) \times \left( -15{ a }^{ 2 }{ xy }^{ 2 }z \right)$

0%
$\displaystyle -45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$
0%
$\displaystyle -45{ a }^{ 2 }{ x }^{ 2 }{ y }^{ 2 }$
0%
$\displaystyle 45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$
0%
$\displaystyle -45{ a }{ x }^{ 2 }{ y }^{ 2 }z$

Explanation

Required product

$=\displaystyle \left( -3axy \right) \times \left( -15{ a }^{ 2 }{ xy }^{ 2 }z \right)$
=

$\displaystyle \left( -3\times -15 \right) \times \left( a\times { a }^{ 2 } \right) \times \left( x\times x \right) \times \left( y\times { y }^{ 2 } \right) \times z$

$\displaystyle =45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$ .

$\displaystyle \left ( z^{2}+13 \right )\left ( z^{2}-5 \right )$ is equal to

0%
$\displaystyle 2z^{4}+18z^{2}-8$
0%
$\displaystyle z^{4}+8z^{2}-65$
0%
$\displaystyle z^{4}-8z^{2}-65$
0%
$\displaystyle z^{4}+8z^{2}+65$

Explanation

$( z^{2}+13) ( z^{2}-5)$

$( z^{2})^{2}+ ( 13+ ( -5 ) )z^{2}+ ( 13\times -5 )$

$\displaystyle z^{4}+8z^{2}-65$

Simplify:

$\displaystyle \left ( p-q \right )^{2}+4pq$ .

0%
$\displaystyle p^{2}-q^{2}$
0%
$\displaystyle \left ( p+q \right )^{2}$
0%
$\displaystyle \left ( 2p-q \right )^{2}$
0%
$\displaystyle \left ( 2p-2q \right )^{2}$

Explanation

Given,

$(p-q)^2+4pq$ .

We know,

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

and

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

Therefore,

$(p-q)^2+4pq$

$=p^2-2pq+q^2$ + 4

$pq$ [Since,

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

$=p^2 + 2pq +q^2$

$=(p+q)^2$ . [Since,

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

Therefore option

$B$ is correct.

The product of

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

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

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

Explanation

$\displaystyle \left ( 4a^{2} \right )\left ( -6b^{2} \right )\left ( 3a^{2}b^{2} \right )$
=

$\displaystyle 4\times \left ( -6 \right )\times 3\times a^{2+2}.b^{2+2}$
=

$\displaystyle -72a^{4}b^{4}$

When

$\displaystyle a^{2}b\left ( a^{3}-a+1 \right )-ab\left ( a^{4}-2a^{2}+2a \right )-b\left ( a^{3}-a^{2}-1 \right )$ is simplified, the answer is

0%
$\displaystyle -a^{2}b$
0%
$ab$
0%
$b$
0%
$0$

Explanation

Given expression is,

$a^2b(a^3-a+1)-ab(a^4-2a^2+2a)-b(a^3-a^2-1)$

$=(a^5b-a^3b+a^2b)-(a^5b-2a^3b+2a^2b)-(a^3b-a^2b-b)$

$=a^5b-a^3b+a^2b-a^5b+2a^3b-2a^2b-a^3b+a^2b+b$

$=(a^5b-a^5b)+(2a^3b-a^3b-a^3b)+(a^2b-2a^2b+a^2b)+b$

$=0+0+0+b$

$=b$
Option C is correct.

The value of

$\displaystyle \left ( x+4 \right )(x+3)-\left ( x-4 \right )\left ( x-3 \right )$ is equal to

0%
$\displaystyle 2x^{2}-14x+24$
0%
$\displaystyle 2x^{2}+14x-24$
0%
$14x$
0%
$24$

Explanation

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

$\displaystyle ( x^{2}+3x+4x+12)-( x^{2}-3x-4x+12 )$
=

$\displaystyle ( x^{2}+7x+12 )- ( x^{2}-7x+12 )$
=

$\displaystyle x^{2}+7x+12-x^{2}+7x-12$
=

$7x + 7x = 14x$

$\displaystyle \left ( \frac{2}{5}ab+c \right )\left ( \frac{2}{5}ab-c \right )$ is equal to

0%
$\displaystyle \frac{4}{25}a^{2}b^{2}-\frac{4}{5}abc+c^{2}$
0%
$\displaystyle \frac{4}{25}a^{2}b^{2}+\frac{4}{5}abc+c^{2}$
0%
$\displaystyle \frac{4}{25}a^{2}b^{2}-c^{2}$
0%
$\displaystyle \frac{4}{25}a^{2}b^{2}+c^{2}$

Explanation

$\displaystyle \left ( \frac{2}{5}ab+c \right )\left ( \frac{2}{5}ab-c \right )$

$\dfrac 25 ab \left( \dfrac 25ab -c\right) + c \left(\dfrac 25 ab - c\right)$

$\dfrac{4}{25}a^2b^2 - \dfrac 25 abc + \dfrac 25 abc - c^2$

$\dfrac{4}{25}a^2b^2-c^{2}$

$\displaystyle \left ( -a-b \right )\left ( b-a \right )$ is equal to

0%
$\displaystyle a^{2}+b^{2}$
0%
$\displaystyle b^{2}-a^{2}$
0%
$\displaystyle a^{2}-b^{2}$
0%
$\displaystyle -\left ( b^{2}+a^{2} \right )$

Explanation

Given,

$(-a-b)(b-a)$
Taking the minus sign outside and rearranging

$=-(b+a)(b-a)$
Using the identity,

$x^2 - y^2 = (x+y)(x-y)$

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

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

$= a^2 - b^2$

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$ is equal to:

0%
$\displaystyle x^{2}-64$
0%
$\displaystyle x^{4}-64$
0%
$\displaystyle x^{4}-256$
0%
$\displaystyle x^{2}-256$

Explanation

Given,

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$ .

We know,

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

Then,

$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$
=

$\displaystyle \left ( x^{2}-4^{2} \right )\left ( x^{2}+16 \right )$
=

$\displaystyle \left ( x^{2}-16 \right )\left ( x^{2}+16 \right )=\left ( x^{2} \right )^{2}-\left ( 16 \right )^{2}$
=

$\displaystyle x^{4}-256$ .

Therefore, option

$C$ is correct.

$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )$ =_____

0%
$\displaystyle m^{2}x^{2}+2mxny-n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}-2mxny-n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$
0%
$\displaystyle m^{2}x^{2}+2mxny+n^{2}y^{2}$

Explanation

$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )=\left ( mx-ny \right )^{2}$
=

$\displaystyle \left ( mx \right )^{2}-2mx\times ny+\left ( ny \right )^{2}$
=

$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$

The product of

$\displaystyle \left ( \frac{1}{5}x^{2}-\frac{1}{6}y^{2} \right )$ and

$\displaystyle \left ( 5x^{2}+6y^{2} \right )$ is

0%
$1$
0%
$\displaystyle x^{4}+\frac{11}{60}x^{2}y^{2}+y^{4}$
0%
$\displaystyle x^{4}+\frac{11}{30}x^{2}y^{2}-y^{4}$
0%
$\displaystyle x^{4}-\frac{11}{30}x^{2}y^{2}-y^{4}$

Explanation

$\displaystyle \left ( \frac{1}{5}x^{2}-\frac{1}{6}y^{2} \right )\left ( 5x^{2}+6y^{2} \right )$

$\displaystyle \frac{1}{5}x^{2}\left ( 5x^{2}+6y^{2} \right )-\frac{1}{6}y^{2}\left ( 5x^{2}+6y^{2} \right )$

$\displaystyle x^{4}+\frac{6}{5}x^{2}y^{2}-\frac{5}{6}x^{2}y^{2}-y^{4}$

$\displaystyle x^{4}+\frac{36x^{2}y^{2}-25x^{2}y^{2}}{30}-y^{4}$

$\displaystyle x^{4}+\frac{11x^{2}y^{2}}{30}-y^{4}$

The product of

$(3x^2\, -\, 5x\, +\, 6)$ and

$-8x^3$ when

$x=0$ is

0%
$\displaystyle\frac{1}{2}$
0%
$2$
0%
$1$
0%
$0$

Explanation

$(3x^2\,-\,5x\,+\,6)\, \times\, -8x^3$

$-24x^5\,+\, 40x^4\,-\, 48x^3$

Put

$x=0,$ we get value as

$0$

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$ =______

0%
$\displaystyle \left ( 2a+2b \right )$
0%
$\displaystyle \left ( 2a-2b \right )$
0%
$\displaystyle 4ab$
0%
$\displaystyle -4ab$

Explanation

Given,

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$ .

We know,

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

Then,

$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$
=

$\displaystyle \left \{ (a+b)+\left ( b-a \right ) \right \}\left \{ (a+b)-\left ( b-a \right ) \right \}$

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

$= 2b \times 2a = 4ab$ .

Therefore, option

$C$ is correct.

The value of

$(501)^2\, -\, (500)^2$ is :

0%
$1$
0%
$101$
0%
$1,001$
0%
None of these

Explanation

Given,

$(501)^{2}- (500)^{2}$ .

We know,

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

Then,

$(501)^{2}- (500)^{2}$

$=(501+500)(501-500)$

$=(1001)(1)$

$= 1001$ .

Therefore, option

$C$ is correct.

Find the product of the following pairs of monomials:

$4, 7p$

0%
$13p^2$
0%
$22p^2$
0%
$28p$
0%
$8p$

Explanation

$4\times 7 p=(4\times 7)p=28p$

Find the product of the following pairs of monomials:

$-4p, 7p$

0%
$-28p^2$
0%
$14p^2$
0%
$-12p^2$
0%
$16p^2$

Explanation

After multiplying the above two monomials we get

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

$= - 28p^2$

$=-28p^2$

$a^{2} + 4a + 4$ =

0%
$(a + 2)^{2}$
0%
$(a + 1)^{2}$
0%
$(a - 2)^{2}$
0%
$(a - 1)^{2}$

Explanation

$a^{2} + 4a + 4 = a^{2} + 2(a) (2)+ (2)^{2}$

$=(a + 2)^{2}$

Find the missing term in the following problem:

$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, =\, \displaystyle \frac{9x^2}{16}\, +\, ..........\, +\, \displaystyle \frac{16y^2}{9}$ .

0%
$2xy$
0%
$- 2xy$
0%
$12xy$
0%
$- 12xy$

Explanation

Given, $\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\,$ .

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

Then,

$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\,$

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

$=\, \displaystyle \frac{9x^2}{16}\, -\, 2xy\, +\, \displaystyle \frac{16y^2}{9}$

$=\, \displaystyle \frac{9x^2}{16}\, +\, (-\, 2xy)\, +\, \displaystyle \frac{16y^2}{9}$ .

Hence, the missing term is $-2xy$ .

Therefore, option $B$ is correct.

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 5 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 5 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.