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

Simplify:

$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$

0%
${m}^{5}{p}^{8}{r}^{3}$
0%
${m}^{9}{p}^{8}{r}^{7}$
0%
${m}^{9}{p}^{7}{r}^{4}$
0%
${m}^{9}{p}^{8}{r}^{6}$

Explanation

$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$

$\Rightarrow$

$\cfrac { { m }^{ 12 }{ p }^{ 9 }{ r }^{ 1 5} }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{1 } }$

$\Rightarrow$

$m ^9 p ^8 r ^{ 6 }$ (option D)

$\left( 3-\sqrt { 7 } \right) \left( 3+\sqrt { 7 } \right) =$ ?

0%
$4$
0%
$2$
0%
$6$
0%
$8$

Explanation

Given,

$({3}-{\sqrt { 7 } })$

$({3}+{\sqrt { 7 } })$ .

We know,

$(x-y)(x+y)$

$={x}^{2}-{y}^{2}$ .

Thus,

$({3}-{\sqrt { 7 } })$

$({3}+{\sqrt { 7 } })$

$={3}^{2}-{(\sqrt { 7 } )}^{2}$

$=9-7=2$ .

Therefore, option

$B$ is correct.

Evaluate:

$-3z^2(-4x^2+10z^4-4xz^5)$

0%
$12x^2z^2-30z^6+12z^10$
0%
$12x^2z^2-30z^8+12z^10$
0%
$12x^4-30z^6+12z^7$
0%
$12x^2z^2-30z^6+12xz^7$

Explanation

$-3z^2(-4x^2+10z^4-4xz^5)$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.

$=$

$-3z^2.(-4x^2)+-3z^2.10z^4-(-3z^2).4xz^5$

$=$

$12x^2z^2-30z^6+12xz^7$

Using binomial identities, expand the following expression

$(t+3)^2$

0%
$t^2-6t-9$
0%
$t^2+6t-9$
0%
$t^2+6t+9$
0%
$t^2-6t+9$

Explanation

We need to expand

$(t+3)^2$
We know that binomial identities for

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

$a = t, b = 3$

$=$

$t^2+2\times 3t+3^2$

$=$

$t^2+6t+9$

Find the value of

$(2x+5)(2x-5)$ .

0%
$4x-25$
0%
$4x^2-25$
0%
$2x^2-25$
0%
$4x^2-5$

Explanation

We know that binomial identities for

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

$\therefore (2x+5)(2x-5) = (2x)^2-5^2$

$=$

$4x^2-25$

Multiply:

$6x^5(-3x^4-2x^3+x^2)$

0%
$-18x^9-12x^8+6x^7$
0%
$-18x^5-12x^8+6x^7$
0%
$-18x^9-12x^5+6x^7$
0%
$-18x^9-12x^8+7x^7$

Explanation

$6x^5(-3x^4-2x^3+x^2)$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.

$=$

$6x^5.-3x^4-6x^5.2x^3+6x^5.x^2$

$=$

$-18x^9-12x^8+6x^7$

Simplify the following expression:

$(s-r)^2-(s+r)^2$

0%
$2sr$
0%
$2s^2+2r^2$
0%
$-4sr$
0%
$2s^2-2r^2$

Explanation

We know that binomial identities for

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

$(x-y)^2=x^2-2xy+y^2$
So,

$(s-r)^2-(s+r)^2$

$=$

$s^2-2sr+r^2-s^2-2sr-r^2$

$=$

$-4sr$

The value of

$-3x(5k^2+y)$ is

0%
$-9k^2x-3xy$
0%
$-15k^2x-3xy$
0%
$-15k^3x-3xy$
0%
$-5k^2x-3xy$

Explanation

We need to find value of

$-3x(5k^2+y)$
Multiply

$-3x$ to the inside bracket, we get
=

$-15k^2x-3xy$

Multiply:

$4x^2z(3x^2y^2-4xz)$

0%
$2x^4y^2z-16x^3z^2$
0%
$12x^4y^2z-16x^3z^2$
0%
$12x^4yz-16x^3z^2$
0%
$12x^4y^2z-16x^2z^2$

Explanation

$4x^2z(3x^2y^2-4xz)$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
=

$4x^2z.3x^2y^2-4x^2z.4xz$
=

$12x^4y^2z-16x^3z^2$

Simplify:

$7a^3(-4a^4+6a^5)$

0%
$14a^7$
0%
$14a^8$
0%
$-28a^7+42a^8$
0%
$-28a^7+42a^{15}$

Explanation

$7a^3(-4a^4+6a^5)$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.

$=$

$7a^3.-4a^4+7a^3.6a^5$

$=$

$-28a^7+42a^8$

Reduce the following expression using binomial identities:

$(3x+5)(3x+10)$ .

0%
$9x^2+45x+50$
0%
$9x^2-45x+50$
0%
$9x^2+45x-50$
0%
$9x^2+45-50$

Explanation

We know that binomial identities for

$(x+y)(x+z)=x^2+(y+z)x+yz$
So,

$x = 3x, y = 5, z = 10$

$=$

$(3x)^2+(5+10)3x+5\times 10$

$=$

$9x^2+45x+50$

What is the product of

$(x-2x)^2$ ?

0%
$2x^2$
0%
$x-2x^2$
0%
$x^2$
0%
$8x^2$

Explanation

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

$=$

$x.x-x.2x-2x.x+2x.2x$

$=$

$x^2-2x^2-2x^2+4x^2$

$=x^2$

$x\Box y=(x+y)^2-(x-y)^2$ . Then

$\sqrt 4\Box \sqrt 5=$ .

0%
$0$
0%
$8\sqrt5$
0%
$10\sqrt5$
0%
$15$
0%
$20$

The product of

$4{ a }^{ 2 }$ ,

$-6{ b }^{ 2 }$ and

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

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

Explanation

$4a^{2}\times (-6b^{2})\times 3a^{2}b^{2}$

$=4\times a\times a\times -6\times b\times b\times 3\times a\times a\times b\times b$

$=4\times -6\times 3\times a\times a\times a\times a\times b\times b\times b\times b$

$=-72 a^{4}b^{4}$

The value of the product

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

$a=1$ ,

$b=-2$ is

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

Explanation

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

$=36a^{2}b^{2}+27b^{3}+16a^{3}+12ab$

Put the value of

$a=1$ and

$b=-2$ , we get

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

$=36\times (1)^{2}\times (-2)^{2}+27\times (-2)^{3}+16\times (1)^{3}+12\times (1)\times (-2)\\$

$=36\times 4-27\times 8+16\times 1-12\times 2\\$

$=144-216+16-24=-80$

The product of the monomials

$\displaystyle \left( 2{ x }^{ 4 }y \right)$ and

$\left( 3{ x }^{ 5 }{ y }^{ 8 } \right)$ is :

0%
$\displaystyle 5{ x }^{ 9 }{ y }^{ 9 }$
0%
$\displaystyle 6{ x }^{ 9 }{ y }^{ 8 }$
0%
$\displaystyle 6{ x }^{ 9 }{ y }^{ 9 }$
0%
$\displaystyle 5{ x }^{ 20 }{ y }^{ 8 }$
0%
$\displaystyle 6{ x }^{ 20 }{ y }^{ 8 }$

Explanation

The product of monomials

$2x^4y$ and

$3x^5y^8$ is

$2x^4y\times 3x^5y^8= 2(3).x^4(x^5).y^1(y^8)$

$\Rightarrow 6x^{4+5}y^{1+8}$

$\Rightarrow 6x^9y^9$

Number of terms in the expansion

$\left( a+b \right) \left( c+d \right)$ is ________.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

given,

$(a+b)(c+d)$

$=ac+bc+ad+bd$

In above expression the number of terms are Four (4)

Multiply the binomials

$\displaystyle \left( 4z+3 \right)$ and

$\left( z-2 \right)$ :

0%
$\displaystyle 4{ z }^{ 2 }-5$
0%
$\displaystyle 4{ z }^{ 2 }-6$
0%
$\displaystyle 4{ z }^{ 2 }-3z-5$
0%
$\displaystyle 4{ z }^{ 2 }-5z-6$
0%
$\displaystyle 4{ z }^{ 2 }+5z-6$

Explanation

On multiplying,

$(4z+3)(z-2)$

$\Rightarrow 4z(z-2)+3(z-2)$

$\Rightarrow 4z^2-8z+3z-6$

$\Rightarrow 4z^2-5z-6$

$-4p\left( 3q-5p \right) =$

0%
$-12pq-20{ p }^{ 2 }$
0%
$-12pq+20{ p }^{ 2 }$
0%
$-7pq+20{ p }^{ 2 }$
0%
$-7pq-20{ p }^{ 2 }$

Explanation

Multiplying

$-4p$ with

$(3q-5p)$

$-4p(3q) = -12pq$

$-4p(-5p)$

$= -12pq+20p^2$

What is the product

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

0%
$2x-6$
0%
$3x-2$
0%
$x^2-x-6$
0%
$x^2-6x$

Explanation

$(x+a)(x+b)$

$=$

$x^2+(a+b)x+ab$
In this formula, taking

$x=x$

$a= 2$

$b= (-3)$

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

$x^2+[(2+ (-3)]x+2(-3)$

$\Rightarrow$

$x^2-1x-6$

$\Rightarrow$

$\boxed{x^2-x-6}$

The value of

$(7.2)^2$ is (use an identity to expand):

0%
$49.9$
0%
$14.4$
0%
$51.84$
0%
$49.04$

Explanation

We will use,

$(a+b)^2$ =

$a^2+2ab+b^2$ to find the required result.

Substitute

$7$ for

$a$ and

$0.2$ for

$B$ in the above identity,

$(7+0.2)^2=(7)^2+2(7)(0.2)+(0.2)^2$

$=49+2.8+0.04$

$=51.84$

Hence, option

$C$ is correct.

The expansion of

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

0%
$2x^2+3y^2+6xy$
0%
$4x^2+9y^2-12xy$
0%
$2x^2+3y^2-6xy$
0%
$4x^2+9y^2+12xy$

Explanation

Given,

$(2x-3y)^2$ .

We know,

$(a-b)^2$

$=$

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

Then,

$(2x-3y)^2$

$=$

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

$=$

$4x^2-12xy+9y^2$ .

Therefore, option

$B$ is correct.

Which of the following relation is correct.

0%
$3(x - 9) = 3x - 27$
0%
$3(x - 9) = 3x - 24$
0%
$3(x - 8) = 3x - 27$
0%
None

Explanation

Let us multiply the given monomial

$3$ by a binomial

$x-9$ as shown below:

$3(x-9)=(3\times x)-(3\times 9)=3x-27$

Hence,

$3(x-9)=3x-27$ .

The product of the following monomial:

$-3a,4a,b,-6c,d$ is

$72{a}^{2}bcd$

0%
True
0%
False

Explanation

We multiply the given monomials

$-3a,4a,b,-6c$ and

$d$ as shown below:

$(-3a)\times 4a\times b\times (-6c)\times d\\ =(-3\times a)\times 4a\times b\times (-6\times c)\times d\\ =(-3\times 4\times (-6))\times (a\times a)\times b\times c\times d\\ =72\times a^{ 2 }\times bcd\\ =72a^{ 2 }bcd$

Hence, the product of

$-3a,4a,b,-6c$ and

$d$ is

$72a^2bcd$ .

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

0%
$4ab$
0%
$2ab$
0%
$a^{2} + 2ab + b^{2}$
0%
$2(a^{2} + b^{2})$

Explanation

Given,

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

We know,

$(a+b)^{2}$

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

and

$(a-b)^{2}$

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

Then,

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

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

$=a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}\\=4ab$ .

Therefore, the required answer is

$4ab$ .

Hence, option

$A$ is correct.

$A=xy, B=yz$ and

$C=zx$ then find ABC=

${x}^{2}{y}^{2}{z}^{2}$

0%
True
0%
False

Explanation

We multiply the given binomials

$A=xy,B=yz$ and

$C=zx$ as shown below:

$(xy)\times (yz)\times (zx)\\ =(x\times y)\times (y\times z)\times (z\times x)\\ =(x\times x)\times (y\times y)\times (z\times z)\\ =x^{ 2 }\times y^{ 2 }\times z^{ 2 }\\ =x^{ 2 }y^{ 2 }z^{ 2 }$

Hence, the product of

$A=xy,B=yz$ and

$C=zx$ is

$x^2y^2z^2$ .

On multiplying the binomials:

$2a-9$ and

$3a+4$ , we get

$6{a}^{2}-19a-36$

0%
True
0%
False

Explanation

We multiply the given binomials

$(2a-9)$ and

$(3a+4)$ as shown below:

$(2a-9)(3a+4)\\ =2a(3a+4)-9(3a+4)\\ =6a^{ 2 }+8a-27a-36\\ =6a^{ 2 }+(8a-27a)-36\quad \quad \quad \quad \quad \quad \left( \text{Combining like terms} \right) \\ =6a^{ 2 }+a(8-27)-36\\ =6a^{ 2 }-19a-36$

Hence,

$(2a-9)(3a+4)=6a^{ 2 }-19a-36$ .

So, the given statement is true.

State whether the statement is True or False.
Product of

$(kl+lm)$ and

$(k-l)$ is equal to

${k}^{2}l-k{l}^{2}+{l}^{2}m-klm$

0%
True
0%
False

Explanation

Product of

$(kl+lm)$ and

$(k−l)$ is ,

$(kl+lm)\times (k−l)=kl(k-l)+lm(k-l)\\=k^2l-kl^2+klm-l^2m$

Clearly

$k^2l-kl^2+klm-l^2m\neq k^2l-kl^2-klm+l^2m$

Hence, The given statement is false.

Find the product of the following pairs:

$-3l,-2m$

0%
$6lm$
0%
$-6lm$
0%
$4lm$
0%
None of these

Explanation

We multiply the given monomials

$-3l$ and

$-2m$ as shown below:

$(-3l)\times (-2m)\\ =(-3\times -2)\times (l\times m)\\ =6\times lm\\ =6lm$

Hence, the product of

$-3l$ and

$-2m$ is

$6lm$ .

Find the product of the following pairs:

$6,7k$

0%
$42k$
0%
$21k$
0%
$23k$
0%
None of these

Explanation

We multiply the given monomials

$6$ and

$7k$ as shown below:

$\Rightarrow 6\times 7k\\ =(6\times 7)\times k\\ =42\times k\\ =42k$

Hence, the product of

$6$ and

$7k$ is

$42k$ .

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

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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