MCQ Questions for Class 8 Maths Exponents And Powers Quiz 2

The value of

${ \left[ { \left( { 3 }^{ 2 } \right) }^{ 2 } \right] }^{ -1 }$ is-

0%
$81$
0%
$-81$
0%
$-0.0123$
0%
$0.0123$

Explanation

$\cfrac { 1 }{ { \left( { 3 }^{ 2 } \right) }^{ 2 } }$

$\because (a^m)^n=a^{mn}$

$=\cfrac { 1 }{ { 3 }^{ 4 } }$

$=\cfrac { 1 }{ 81 }$

$=0.0123$

The thickness of paper is

$0.004\ m$ and that of another paper is

$0.008\ m$ . Compare their sizes.

0%
Paper-1 is double thicker than paper-2.
0%
Thickness of paper-1 is half the thickness of paper-2.
0%
Thickness of paper-1 is one-fourth of the thickness of paper-2.
0%
Paper-1 and paper-2 are both equally thick.

Explanation

$\textbf{Step1: Given}$

The thickness of paper-1

$=0.004\ m$

Thickness of paper-2

$=0.008\ m$

$\textbf{Step2: To find}$

We have to compare the thickness of papers

$\dfrac{Thickness\ of\ paper-1}{Thickness\ of\ paper-2}=\dfrac{0.004}{0.008}=\dfrac{1}{2}$

$Thickness\ of\ paper-1$

$=\dfrac{1}{2}$ .

$(Thickness\ of\ paper-2)$

$\textbf{Step3: Conclusion}$

Hence, the thickness of paper-1 is half the thickness of paper-2 is the right answer.

$For \ a \ natural \ number \ n , \ 2n(n-1)!\leqslant n^n, if$ :

0%
$n<2$
0%
$n>2$
0%
$n\ge2$
0%
Never

The value of

$(-2)^{-5}$

0%
$-32$
0%
$\dfrac{-1}{32}$
0%
$32$
0%
$\dfrac{1}{32}$

Explanation

${\textbf{Step - 1: Use negative exponent rule}}$

${\text{If the number written in the power of a number is negative like }}{{\text{x}}^{ - n}}$ ${\text{ then we write it as }}\dfrac{1}{{{x^n}}}.$

$\therefore{( - 2)^{ - 5}} = \dfrac{1}{{{{( - 2)}^{^5}}}}$

${\textbf{Step - 2: Find the value}}$

$\dfrac{1}{{{{( - 2)}^{^5}}}} =\dfrac{1}{{{{- 2}^{^5}}}}= - \dfrac{1}{{32}}$

${\textbf{Hence, the correct option is B.}}$

The value of

$(-3)^{-4}$ is

0%
$12$
0%
$81$
0%
$\dfrac{-1}{12}$
0%
$\dfrac{1}{81}$

Explanation

Given

$(-3)^{-4}$

We know according to the law of exponents

$a^{-m}={ \left( \dfrac { 1 }{ { a }^{ m } } \right) }$

We can write given equation as

$(-3)^{-4} ={ \left( \dfrac { 1 }{ { -3 }^{ 4 } } \right) }$

But we also know that

$3^4=81$

substituting in above we get

$(-3)^{-4}=\dfrac{1}{81}$

[

$\therefore$ Since the multiplication of negative sign is even so it becomes positive]

The value of

$\left(\dfrac{2}{5}\right)^{-3}$ is

0%
$\dfrac{-8}{125}$
0%
$\dfrac{25}{4}$
0%
$\dfrac{125}{8}$
0%
$\dfrac{-2}{5}$

Explanation

Given

$\left(\dfrac{2}{5}\right)^{-3}$

We know according to the law of exponents

${ \left( \dfrac { a }{ b } \right) }^{ -m } = { \left( \dfrac { b }{ a } \right) }^{ m }$

We can write given equation as

$\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3}$

And also we know that

$2^3$ is

$8$ and

$5^3$ is

$125$

By substituting in above we get

$\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3} = \dfrac{125}{8}$

Mark tick against the correct answer in each of the following:

$(1/2)^{-2} +(1/3)^{-2} + (1/4)^{-2}=$ ?

0%
$(61/144)$
0%
$29$
0%
$(144/61)$
0%
none of these

Explanation

We know that,

$\Rightarrow (1/2)^{-2}=(2/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$
also,

$\Rightarrow (1/3)^{-2}=(3/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$
and

$\Rightarrow (1/4)^{-2}=(4/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$
Now add,

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

$=4+9+16$

$=29$

$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }$

0%
$\dfrac{61}{144}$
0%
$\dfrac{144}{61}$
0%
$29$
0%
$\dfrac{1}{29}$

Explanation

Given

$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }$

We know according to the law of exponents

$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$

$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }=\left( \dfrac { 2 }{ 1 } \right) ^{ 2 }+\quad \left( \dfrac { 3 }{ 1 } \right) ^{ 2 }+\left( \dfrac { 4 }{ 1 } \right) ^{ -2 } =2^2+3^2+4^2=4+9+16=29$

The value of

$\left(\dfrac 25 \right)^{-1}$ is

0%
$\dfrac {2}{5}$
0%
$\dfrac {5}{2}$
0%
$-\dfrac {5}{2}$
0%
$-\dfrac {2}{5}$

Explanation

We will use, law of exponent

$a^{-m}=\dfrac {1}{a^m}$ where

$a=$ non-zero integer

$\left(\dfrac {2}{5}\right)^{-1}=\dfrac {1}{\left(\dfrac {2}{5}\right)^1} =\dfrac {5}{2}$

${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=?$

0%
$\dfrac{5}{3}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{-3}{5}$
0%
none of these

Explanation

Given

${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }$

We know according to the law of exponents

$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$

By applying the above law we get

${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=\dfrac{1}{{ \dfrac { -5 }{ 3 } }^{ -1 }}$

On rearranging the above we get

${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=\dfrac{-3}{5}$

$(-3/7)^{-1} = ?$

0%
$7/3$
0%
$-7/3$
0%
$3/7$
0%
None of these

Explanation

$(-3/7)^{-1}$

We know,

$x^{-1}=\dfrac{1}{x}$

$\Rightarrow$

$(-3/7)^{-1}$

$=\dfrac{1}{\dfrac{-3}{7}}$

$=1\times\dfrac{7}{-3}$

$=\dfrac{-7}{3}$

$\therefore$

$(-3/7)^{-1}$

$=\dfrac{-7}{3}$

${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=?$

0%
$\dfrac{-1}{6}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{8}$
0%
$-\dfrac{1}{8}$

Explanation

Given

${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }$

We know according to the law of exponents

$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$

By applying the above law we get

${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=\dfrac{-2}{1}^3$

But we know that

$(-2)^3=-8$

By substituting this we get

${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=-\dfrac{1}{8}$

$\left\{ \left( \dfrac { 1 }{ 3 } \right) ^{ -3 }-\quad \left( \dfrac { 1 }{ 2 } \right) ^{ -3 } \right\} \div \quad \left( \dfrac { 1 }{ 4 } \right) ^{ -3 }=?$

0%
$\dfrac{19}{64}$
0%
$\dfrac{27}{16}$
0%
$\dfrac{64}{19}$
0%
$\dfrac{16}{25}$

Explanation

Given

$\left\{ \left( \dfrac { 1 }{ 3 } \right) ^{ -3 }-\quad \left( \dfrac { 1 }{ 2 } \right) ^{ -3 } \right\} \div \quad \left( \dfrac { 1 }{ 4 } \right) ^{ -3 }$

We know according to the law of exponents

$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$

$\left\{ \left( \dfrac { 1 }{ 3 } \right) ^{ -3 }-\quad \left( \dfrac { 1 }{ 2 } \right) ^{ -3 } \right\} \div \quad \left( \dfrac { 1 }{ 4 } \right) ^{ -3 }=\left\{ \left( \dfrac { 3 }{ 1 } \right) ^{ 3 }-\quad \left( \dfrac { 2 }{ 1 } \right) ^{ 3 } \right\} \div \quad \left( \dfrac { 4 }{ 1 } \right) ^{ 3 }=(3^3-2^3) \div 4^3$

But we know that

$3^3$ is

$27,2^3$ is

$8$ and

$4^3$ is

$64$ , on substituting

$\Rightarrow (3^3-2^3) \div 4^3=(27-8) \div \ 64$

$\Rightarrow 19\ \div 64=\dfrac{19}{64}$

Mark tick against the correct answer in each of the following:

$(6^{-1}-8^{-1})^{-1}=$ ?

0%
$-\dfrac{1}{24}$
0%
$-24$
0%
$\dfrac{1}{24}$
0%
$24$

Explanation

We know that,

$=(6)^{-1}=\left(\dfrac{1}{6}\right)^{1}\quad [\because \left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n]$

$=(8)^{-1}=\left(\dfrac{1}{8}\right)^{1}\quad$
Now

$(6^{-1}-8^{-1})^{-1}=\left(\dfrac{1}{6}-\dfrac{1}{8}\right)^{-1}$

$=\left(\dfrac{4-3}{24}\right)^{-1}$

$=\left(\dfrac{1}{24}\right)^{-1}$

$=24$

Mark tick against the correct answer in each of the following:

$(-1/2)^{-6}=$ ?

0%
$-64$
0%
$64$
0%
$(1/64)$
0%
$(-1/64)$

Explanation

We know that,

$=(-1/2)^{-6}=(-2/1)^6\quad [\because (a/b)^{-n}=(b/a)^n]$

$=(-2)^6$

$=64$

Mark tick against the correct answer in each of the following:

$\left\{(3/4)^{-1}+(1/4)^{-1}\right\}^{-1}=$ ?

0%
$(3/8)$
0%
$(3/16)$
0%
$(8/3)$
0%
$(-8/3)$

Explanation

We know that,

$=(3/4)^{-1}=(4/3)^1\quad [\because (a/b)^{-n}=(b/a)^n]$

$=(1/4)^{-1}=(4/1)^1 \quad [\because (a/b)^{-n}=(b/a)^n]$
Now subtract,

$=\left\{(4/3)+(4/1)\right\}^{-1}$

$=\left\{(4+12)/3\right\}^{-1}$

$=\left\{16/3\right\}^{-1}$

$=\left\{3/16\right\}$

State whether the statements are true (T) or false (F).
The reciprocal of

$x^{-1}$ is

$\dfrac{1}{x}.$

0%
True
0%
False

Explanation

The given statement is false.

The reciprocal of

$x^{-1}$ is

$x$ .

For a fixed base(10) if the exponent decreases by

$1$ , the number becomes

0%
One-tenth of the previous number.
0%
Ten times of the previous number
0%
Hundred of the previous number
0%
Hundred times of the previous number

Explanation

If the exponent is decreased by

$1$ , then for the fixed base, the number becomes one-tenth of the previous number.
For

$10^5$ , exponent decreases by

$1$

$\Rightarrow \ 10^{5-1}=10^4$

$\Rightarrow \ \dfrac {10^4}{10^5}=\dfrac {1}{10}$

Mark tick against the correct answer in each of the following:

$\left\{6^{-1} +(3/2)^{-1}\right\}^{-1}$

0%
$(2/3)$
0%
$(5/6)$
0%
$(6/5)$
0%
None of these

Explanation

We know that,

$=(6)^{-1}=(1/6)\quad [\because (a/b)^{-n}=(b/a)^n]$

$=(3/2)^{-1}=(2/3)\quad [\because (a/b)^{-n}=(b/a)^n]$

Now add,

$=\left\{(1/6)+(2/3)\right\}^{-1}$

$=\left\{(1+4)/6\right\}^{-1}$

$=\left\{5/6\right\}^{-1}$

$=\left\{6/5\right\}$

Mark tick against the correct answer in each of the following:

$[\left\{(-1/2) ^{2}\right\}^{-2}]^{-1}=$ ?

0%
$(1/16)$
0%
$16$
0%
$(-1/16)$
0%
$-16$

Explanation

$[\left\{(-1/2) ^{2}\right\}^{-2}]^{-1}=[\left\{(-1^2/2) ^{2}\right\}^{-2}]^{-1}=$

$=[\left\{1/4\right\}^{-2}]^{-1}$

$=[\left\{4\right\}^{2}]^{-1}$

$=[16^{-1}$

$=[1/16]$

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 2 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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