MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 6

Find the missing term in the following problem.

{(\frac{3 x}{4} - \frac{4 y}{3})}^{2} = \frac{9 x^{2}}{16} + . . . . . . . . . + \frac{16 y^{2}}{9}

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

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

Explanation

The L.H.S. of the algebraic expression is of the form

{(a - b)}^{2}

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

where,

a = \frac{3 x}{4} & b = \frac{4 y}{3}

$a=\dfrac { 3x }{ 4 } \quad \& \quad b=\dfrac { 4y }{ 3 }$ .

${ \left( a-b \right) }^{ 2 }={ a }^{ 2 }-2ab+{ b }^{ 2 }\\ \implies \left(\dfrac{3x}{4}-\dfrac{4y}{3}\right)={ \left( \dfrac { 3x }{ 4 } \right) }^{ 2 }-2\times \dfrac { 3x }{ 4 } \times \dfrac { 4y }{ 3 } +{ \left( \dfrac { 4y }{ 3 } \right) }^{ 2 }\\ =\dfrac { 9{ x }^{ 2 } }{ 16 } -2xy+\dfrac { 16{ y }^{ 2 } }{ 9 }$ .

So the middle term

$-2xy$ is missing.

Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively

$(p, q); (10m, 5n); (20x^2, 5y^2); (4x, 3x^2); (3mn, 4np)$

0%
$100x^2y^2 ;12x^3 ; 12mn^2p$
0%
$100x^2y^2 ;12x^2 ; 12mn^2p$
0%
$100x^2y ;12x^3 ; 12mn^2p$
0%
$100x^2y^2 ;12x^3 ; 12mnp$

Explanation

We know that the area of a rectangle = I

$\times$ b, where I = length and b = breadth.
Therefore, the areas of rectangles with pair of monomials

$(p, q), (10m, 5n); (20x^2, 5y^2); (4x, 3x^2)$ and

$(3mn, 4np)$ as their lengths and breadths are given by

$p\times q = pq$
1.

$10m \times n =(10 \times5) \times (m \times n) = 50mn$
2.

$20x^2 \times 5y^2= (20 \times 5) \times (x^2 \times y^2) = 100x^2y^2$
3.

$4x \times 3x^2 = (4 \times 3) \times (x \times x^2)$
=

$12x^3$
and,
4.

$3mn \times 4np = (3\times4) \times (m \times n \times n \times p)$
=

$12mn^2p$

Multiply the binomials

$(2x + 5)$ and

$(4x - 3)$

0%
$8x^2+ 14x -15$
0%
$7x^2-14x -15$
0%
$x^2+ 14x -18$
0%
None of these

Explanation

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

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

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

Obtain the product of:

$rn, -mn, mnp$

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

Explanation

product of rn,mn,mnp

$rn \times -mn \times mnp = (1\times -1 \times 1) \times (r \times m \times m \times n \times n \times n \times p)$
=

$-1 \times r \times m^2 \times n^3 \times p$
=

$-rm^2n^3p$

Using identities, evaluate

$8.9^2$ .

0%
$74.53$
0%
$73.42$
0%
$79.21$
0%
$73.21$

Explanation

Given,

$8.9^2$

$= (9 -0.1)^2$ .

We know,

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

Then,

$8.9^2$

$= (9 -0.1)^2$

$= (9)^2 -2(9) (0.1) + (0.1)^2$

$= 81 -1.8 + 0.01 = 79.20+0.01=79.21$ .

Therefore, option

$C$ is correct.

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

$a, 2b, 3c$

0%
$6abc$
0%
$3ab$
0%
$2abc$
0%
$9ac$

Explanation

volume of rectangular box= length

$\times$ breadth

$\times$ height
volume =

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

$(1 \times 2 \times 3) \times (a \times b \times c)$

$= 6abc$

Obtain the product of

$xy, yz, zx$ .

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

Explanation

product of xy, yz, zx is

$xy \times yz \times zx = x \times x \times y \times y \times z \times z$
=

$x^{1 +1} \times y^{1+1} \times z^{1+1}$
=

$x^2y^2z^2$

Carry out the multiplication of the expressions in the following pair:

$a^2 -9, 4a$

0%
$a^3 -12a$
0%
$4a^3 -22a$
0%
$a^3 -16$
0%
$4a^3 -36a$

Explanation

multiplication of

$a^2 -9, 4a$
=

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

$4a^3 -36a$

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

$5a, 3a^2, 7a^4$

0%
$102a^7$
0%
$105a^7$
0%
$192a^7$
0%
$606a^7$

Simplify:

$(7m -8n)^2 + (7m + 8n)^2$ .

0%
$49m^2+164n^2$
0%
$98m^2+128n^2$
0%
$49m^2+64n^2$
0%
$128m^2+64n^2$

Explanation

Given,

$(7m -8n)^2 + (7m + 8n)^2$ .

We know,

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

and

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

Then,

$(7m -8n)^2 + (7m + 8n)^2$

$= ((7m)^2-2(7m)(8n) + (8n)^2)+((7m)^2 + 2(7m)(8n) + (8n)^2)$

$= (49m^2-112mn + 64n^2)+(49m^2 + 112mn + 64n^2)$

$= 49m^2-112mn + 64n^2+49m^2 + 112m + 64n^2$

$= 98m^2 + 128n^2$ .

Therefore, option

$B$ is correct.

Using identities, evaluate

$5.2^2$ .

0%
$26.04$
0%
$2704$
0%
$27.04$
0%
$2604$

Explanation

Given,

$(5.2)^2$

We know,

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

Then,

$(5.2)^2$

$=(5+0.2)^2$

$=5^2+(0.2)^2+2(5)(0.2)$

$=25+0.04+2(1)$

$=25+0.04+2$

$=27.04$ .

Therefore, option

$C$ is correct.

The product of (x -y) and (y -x) is equal to

0%
$x^2+y^2 -2xy$
0%
$2xy + x^2+y^2$
0%
$2xy-x^2-y^2$
0%
$x^2-2xy+y^2$

Explanation

$(x -y) (y -x) = x(y -x) -y(y -x)$

$=xy-x^2-y^2+yx$

$=2xy-x^2-y^2$

The product of (2x + 3) and (3x + 2) is equal to

0%
$6x^2+6$
0%
$6x^2+5x+6$
0%
$6(x^2+x+1)$
0%
$6x^2+13x+6$

Explanation

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

$= 6x^2 + 4x + 9x +6$

$=6x^2 + 13x +6$

Using

$(x + a) (x + b) = x^2+(a+b)x+ab$ , find

$5.1 \times 5.2$ .

0%
$17.61$
0%
$22.41$
0%
$13.32$
0%
$26.52$

Explanation

Given,

$5.1 \times 5.2 = (5 + 0.1)(5 + 0.2)$ .

We know,

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

Then,

$5.1 \times 5.2$

$= (5 + 0.1)(5 + 0.2)$

$= (5)^2 + (0.1 + 0.2) (5)+ (0.1)(0.2)$

$= 25 + (0.3) (5)+ 0.02$

$=25 + 1.5 + 0.02$

$= 26.52$ .

Therefore, option

$D$ is correct.

Using identities, evaluate

$12.1^2 -7.9^2$ .

0%
$84$
0%
$72$
0%
$69$
0%
$87$

Explanation

Given,

$(12.1)^2-(7.9)^2$ .

We know,

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

Then,

$(12.1)^2-(7.9)^2$

$=(12.1+7.9)(12.1-7.9)$

$=20\times 4.2$

$=84$ .

Therefore, option

$A$ is correct.

The value of

$(x + 3y)^2 + (x - 3y)^2$ is:

0%
$2x^2 + 18 y^2$
0%
$2x^2 - 18 y^2$
0%
$2x^2 + 18y^2 - 2xy$
0%
None of these

Explanation

We know,

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

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

Then,

$(x + 3y)^2 + (x - 3y)^2=[(x)^2+(3y)^2+2\times x\times3y]+[(x)^2+(3y)^2-2\times x\times3y]$

$=x^2+9y^2+6xy+x^2+9y^2-6xy$

$=2x^2+18y^2$ .

Hence, Option

$A$ is correct.

Using

$(x + a) (x + b) = x^2+(a+b)x+ab$ , find

$103 \times 104$ .

0%
$14982$
0%
$11992$
0%
$10712$
0%
$11482$

Explanation

Given,

$103 \times 104$

$=(100+3)(100+4)$ .

Using,

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

Then,

$103 \times 104$

$=(100+3)(100+4)$

$=(100)^2+(3+4)(100) +(3)(4)$

$=(100)^2+(7)(100) +12$

$= 10000 +700 + 12$

$= 10712$ .

Therefore, option

$C$ is correct.

Simplify:

$(x + 5)(x + 4)$

0%
$x^2+27x+20$
0%
$x^2+9x+20$
0%
$x^2+18x+20$
0%
$x^2+x+20$

Explanation

$(x + 5)(x + 4)$ , using the formula

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

$=x^2+x(5+4)+5\times 4$

$=x^2+9x+20$

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

0%
12
0%
0
0%
6x
0%
4x+12

Explanation

$\textbf{Step 1: Multiply the factors by distribution law}$

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

$\Rightarrow x(x+3)+2(x+3)-(x(x-2)+3(x-2))$

$\Rightarrow x^2+3x+2x+6-(x^2-2x+3x-6)$

$\Rightarrow x^2+3x+2x+6-x^2+2x-3x+6$

$\text{[multiplying with $$(-)$$ will change the sign]}$

$\Rightarrow x^2+5x+6-x^2-x+6$

$\Rightarrow 4x+12$

$\textbf{Hence, Option D is correct}$

Find the product of

$\left( 3x-5 \right)$ and

$\left( x+5 \right)$ .

0%
${ x }^{ 2 }-25x+10$
0%
${ x }^{ 2 }+10x-22$
0%
$3{ x }^{ 2 }+10x-25$
0%
$3{ x }^{ 2 }-22x-25$

Explanation

Product of

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

$=3x^2+15x-5x-25$

$=3x^2+10x-25$
Option C is correct.

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

0%
$2ab$
0%
$4ab$
0%
$6ab$
0%
$-4ab$

Explanation

Given,

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

We know,

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

and

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

Then,

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

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

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

$= -4ab$ .

Therefore, option

$D$ is correct.

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

0%
$3x^2+y^2+5xy$
0%
$3x^2+2y^2+5xy$
0%
$3x^2+2y^2+6xy$
0%
$3x^2+2y+5xy$

Explanation

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

$=3x\times x+2y\times y+3x\times y+2y\times x$

$= 3x^2+2y^2+5xy$

The product of a monomial and a trinomial is a

0%
monomial
0%
binomial
0%
trinomial
0%
none of these

Explanation

The product of a monomial and trinomial results trinomial.

For example: Suppose a monomial

$2x$ and a trinomial

$2x+3y+3$ .

Their product will be

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

The result is a trinomial.

Hence, option C is correct.

Find the value of

$1.03^2$ using identity.

0%
$1.0409$
0%
$1.0609$
0%
$1.0009$
0%
$1.0309$

Explanation

Given,

$1.03^2 = (1+0.03)^2$ .

We know,

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

Then,

$1.03^2 = (1+0.03)^2$

$=1^2+(0.03)^2+2\times 1\times 0.03$

$=1+0.0009+0.06$

$=1.0609$ .

Therefore, option

$B$ is correct.

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

0%
$12x^{2} + 6x$
0%
$6 (2x^{2} + x)$
0%
$x (12x + 6)$
0%
All of the above

Explanation

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

$= 12x^{2} + 6x$

Hence option A is correct.

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

$=3x \times 2 (2x+1)$

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

Hence option B is correct.

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

$= 12x^{2} + 6x$

$=x(12x+6)$

Hence option C is correct.

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

0%
$16x^{2} + 8x + 2$
0%
$16x^{2} + 8x + 1$
0%
$16x^{2} - 1$
0%
$16x^{2} + 1$

Explanation

Given that:

$(4x+1)\times(4x-1)$

We can see that:

$(4x+1)\times(4x-1)=$

$4x\times(4x-1)+1\times (4x-1)$

$\Rightarrow (4x\times 4x)-(4x\times 1)+(1\times 4x)-(1\times 1)$

$\Rightarrow 16x^2-4x+4x-1$

$\Rightarrow 16x^2-1$

$-2x\times 10y\times -3z=$

0%
$-60xyz$
0%
$60xz$
0%
$-60xz$
0%
$60xyz$

Explanation

$-2x\times 10y\times -3z= -2\times 10\times -3\times x\times y\times z$

$=60xyz$

$-8yz\times -2xy=$

0%
$16xyz$
0%
$-16xy^2z$
0%
$16xy^2z$
0%
$-16x^2y^2z^2$

Explanation

$-8yz\times -2xy = (-8\times -2)\times yz\times xy$

$= 16\times y^2zx=16xy^2z$

$2xy\times 7yz\times 9xz$

0%
$126xyz$
0%
$126x^2y^2z^2$
0%
$126xyz^2$
0%
$126x^2yz^2$

Explanation

$2xy\times 7yz\times 9xz= (2\times 7\times 9)\times xy\times yx\times xz$

$=126x^2y^2z^2$

Find the product of

$x^2yz\times xy^2z^3$ .

0%
$x^3y^3z^4$
0%
$x^3y^3z^3$
0%
$x^3y^4z^3$
0%
$x^3y^3z$

Explanation

$x^2yz\times xy^2z^3 = x^2\times y\times z\times x\times y^2\times z^3$

$x^2\times x\times y\times y^2\times z\times z^3=x^3y^3z^4$

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 6 - MCQExams.com

0:0:4

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 6 - MCQExams.com

0:0:4

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.