MCQ Questions for Class 8 Maths Exponents And Powers Quiz 1

The value of

$\dfrac{1}{4^{-2}}$ is equal to

$16$

0%
True
0%
False

Explanation

Using law of exponents,

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

$\Rightarrow \dfrac{1}{4^{-2}}=4^{2}=4\times 4=16$

State true or false.

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

0%
True
0%
False

Explanation

True.
Using law of exponents,

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

$LHS=RHS$

Simplify:

$\left(\dfrac{27}{125}\right)^{-1}=$

0%
$\dfrac{3}{5}$
0%
$\left(\dfrac{3}{5}\right)^{-1}$
0%
$\left(\dfrac{3}{5}\right)^{-2}$
0%
$\left(\dfrac{5}{3}\right)^{3}$

Explanation

${\left(\dfrac{125}{27}\right)}^{1}$ using

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

$=\dfrac{{5}^{3}}{{3}^{3}}$

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

The value of

$\left (\dfrac {32}{243}\right )^{-3/5}$ is _____.

0%
$\dfrac {27}{8}$
0%
$\dfrac {8}{27}$
0%
$\dfrac {16}{27}$
0%
$\dfrac {27}{16}$

Explanation

We need to find value of

$\left (\dfrac {32}{243}\right )^{-3/5}$

It can be written as,

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

$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$

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

0%
$\dfrac {3}{8}$
0%
$\dfrac {-3}{8}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {-8}{3}$

Explanation

We need to find value of

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

It can be written as

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

$=$

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

$=$

$-\dfrac {3}{8}$

Evaluate :

$(-4)^{-2}$ ---

0%
$\displaystyle \frac{1}{-16}$
0%
$\displaystyle \frac{1}{16}$
0%
$-16$
0%
$16$

Explanation

$(-4)^{-2}=\dfrac{1}{(-4)^2}$

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

$=\dfrac{1}{-4*-4}$

$=\dfrac1{16}$

$\therefore(-4)^{-2}=\dfrac1{16}$

$3^{-3}$ can also be expressed as

0%
$\dfrac{1}{27}$
0%
$\dfrac{-1}{27}$
0%
$27$
0%
$-27$
0%
None of these

Explanation

$a^{-b}$ is expressed as:

$\dfrac1{a^b}$

In the same manner,

$3^{-3}$ can be expressed as:

$\dfrac1{3^3} = \dfrac1{27}$

Which of the following has an exponent with negative index?

0%
$-3^{4}$
0%
$4^{3}$
0%
$\dfrac {1}{3^{4}}$
0%
$-3$

Explanation

$\dfrac {1}{3^{4}} = 3^{-4}$

$\therefore \dfrac {1}{3^{4}}$ has a negative index.

So, option

$C$ is correct.

If the exponent of a negative integer is even then the result is a ............ integer.

0%
Positive
0%
Negative
0%
0
0%
None

Explanation

If the exponent of a negative integer is even, then the result is a

$positive$ integer.

$\Rightarrow$

$(-2)^{4}=(-2)\times (-2)\times (-2)\times(-2)=16$

$\Rightarrow$ Here,

$-2$ is negative base and

$4$ is even power.

$\Rightarrow$ Then result

$16$ is a positive integer.

If the exponent of a negative integer is odd, then the result is a .......... integer.

0%
positive
0%
negative
0%
zero
0%
None of these

Explanation

If the exponent of a negative integer is odd, then the result is a negative integer.

$\Rightarrow$

$(-2)^{3}=(-2)\times (-2)\times (-2)=-8$

$\Rightarrow$ Here,

$-2$ is negative base and

$3$ is odd power.

$\Rightarrow$ Then result

$-8$ is a negative integer.

$-2^{-3}$ can also be expressed as:

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

Explanation

$-2^{-3} = \dfrac {1}{-2 \times -2 \times -2} = -\dfrac 18$

So, option B is correct.

Simplify:

$1 \div 7 \div 7$

0%
$7^{-2}$
0%
$7^{-1}$
0%
$0.7$
0%
$0.777$

Explanation

$1 \div 7 \div 7 = \dfrac 17 \div 7 = \dfrac 17 \times \dfrac 17$

$=\dfrac {1}{7^2} = 7^{-2}$

So, option

$A$ is correct.

$\dfrac{1}{5}$ can also be expressed as:

0%
$5^{-1}$
0%
$\dfrac{1}{25}$
0%
$\dfrac{-1}{25}$
0%
None of these

Explanation

$\dfrac 15 = \dfrac {1}{5^1}=5^{-1}$

$5^{-2}$ can also be expressed as

0%
$\dfrac{1}{25}$
0%
$\dfrac{1}{5^{-2}}$
0%
$\dfrac{1}{5}$
0%
None of these

Explanation

$5^{-2}=5^{-1*2}$

$=(5^2)^{-1}$

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

$=\dfrac{1}{5^2}$

$\because a^{-n}=\dfrac{1}{a^n}$

$=\dfrac{1}{25}$

Which of the following has a negative index?

0%
$\dfrac { 1 }{ { x }^{ -3 } }$
0%
${ x }^{ 3 }$
0%
${ -x }^{ 3 }$
0%
$\dfrac { 1 }{ { x }^{ 3 } }$

Explanation

$\dfrac{1}{x^3}=x^{-3}$ . Index of

$x$ is

$(-3)$ , which is negative.

Hence, option

$D$ is correct.

The value of

$5^{-2}$ is equal to

$25$ .

0%
True
0%
False

Explanation

$5^{-2}=\dfrac{1}{5^{2}}$

$\left[\text{By using }a^{-n}=\dfrac1{a^n}\right]$

$=\dfrac{1}{25}$

Hence, the given statement is false.

Fill in the blanks:

$0.000008$ _______

$0.000016$

0%
is half of
0%
is double of
0%
is one-fourth of
0%
is one-third of

Difference of the greatest

$7$ digit number and the smallest

$5$ digit number is

0%
$9,98,999$
0%
$99,89,999$
0%
$99,899$
0%
$9,98,099$

Explanation

$\textbf{Step-1: Find greatest 7 digit number and smallest 5 digit number}$

$\text{We know greatest n digit number is= 9999..n times }$

$\text{And smallest n digit number is= 10000..(n-1) times }$

$\text{ So 7- digit greatest number}$

$=99,99,999$

$\text{And 5-digit smallest number}$

$=10,000$

$\textbf{Step-2: Find Difference}$

$\text{Difference}$

$=99,99,999-10,000=99,89,999$

$\textbf{Hence Option B is correct.}$

Compare the folllowing:

$0.000000038$ _______

$\displaystyle 3\cdot 8\times 10^{-8}$

0%
$<$
0%
$>$
0%
$=$
0%
None of these

Explanation

$\displaystyle 0\cdot 000000038= 3\cdot 8\times 10^{-8}$

Fill in blank with an appropriate comparison symbol.

$0.00000998$ ______

$0.0000116$

0%
$<$
0%
$>$
0%
$=$
0%
None of these

Explanation

$\displaystyle 0\cdot 00000998= 9\cdot 89\times 10^{-6}$

$\displaystyle 0\cdot 0000116= 1\cdot 16\times 10^{-5}$
Thus

$\displaystyle 9\cdot 89\times 10^{-6}< 1\cdot 16\times 10^{-5}$

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 1 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 1 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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