MCQ Questions for Class 8 Maths Exponents And Powers Quiz 3

State true or false.

$10^{-2}=\dfrac{1}{100}$

0%
True
0%
False

Explanation

True.
Using

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

$10^{-2}=\dfrac{1}{10^{2}}=\dfrac{1}{10\times 10}=\dfrac{1}{100}$

$3^{-2}$ can be written as

0%
$3^2$
0%
$\dfrac {1}{3^2}$
0%
$\dfrac {1}{3^{-2}}$
0%
$-\dfrac {2}{3}$

Explanation

We will use, law of exponent

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

$a=$ non zero integer
Therefore,

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

Mark tick against the correct answer of the following:

$\left(\dfrac23\right)^{-5}=$ ?

0%
$\dfrac{32}{243}$
0%
$\dfrac{243}{32}$
0%
$-\dfrac{32}{243}$
0%
$-\dfrac{243}{32}$

Explanation

We know that,

$\left(\dfrac ab\right)^{-n}=\left(\dfrac ba\right)^{n}$

$\therefore\left(\dfrac23\right)^{-5}\\=\left(\dfrac32\right)^5$

$=\dfrac{3^5 }{2^5}$

$\left[\because \left(\dfrac ab\right)^{n}=\left(\dfrac{a^n}{b^n}\right)\right]$

$=\dfrac{243}{32}$

Hence the correct answer is

$Opt:[B]$

The value of

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

0%
$16$
0%
$8$
0%
$\dfrac {1}{16}$
0%
$\dfrac {1}{8}$

Explanation

We will use, law of exponent

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

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

The value of

$\left(\dfrac 25 \right)^{-2}$ is

0%
$\dfrac 45$
0%
$\dfrac {4}{25}$
0%
$\dfrac {25}{4}$
0%
$\dfrac {5}{2}$

Explanation

We will use, law of exponent

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

$a=$ non-zero integer

$\left(\dfrac {2}{5}\right)^{-2}=\dfrac {1}{\left(\dfrac {2}{5}\right)^2} \\=\dfrac {1}{\dfrac {4}{25}}\\=\dfrac {25}{4}$

The value of

$(-2)^{2\times 3-1}$ is

0%
$32$
0%
$64$
0%
$-32$
0%
$-64$

Explanation

To find the value of

$(-2)^{2\times 3-1}$

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

$=(-2)^5$

$=(-2)\times (-2)\times (-2)\times (-2)\times (-2)$

$=(4)\times (4)\times (-2)$

$=(16)\times (-2)$

$=-32$

Hence, the result

$=-32$

The value of

$(7^{-1}-8^{-1})^{-1} -(3^{-1}-4^{-1})$ is

0%
$44$
0%
$56$
0%
$68$
0%
$12$

Explanation

Using law of exponents,

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

$\Rightarrow \ (7^{-1}-8^{-1})^{-1}-(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{7}-\dfrac {1}{8}\right)^{-1}-\left(\dfrac {1}{3}-\dfrac {1}{4}\right)^{-1}$

$\Rightarrow \ \left(\dfrac {1\times 8-1\times 7}{56}\right)^{-1}- \left(\dfrac {1\times 4-1\times 3}{12}\right)^{-1}$

$=\left(\dfrac {8-7}{56}\right)^{-1} -\left(\dfrac {4-3}{12}\right)^{-1}$

$=\left(\dfrac {1}{56}\right)^{-1} -\left(\dfrac {1}{12}\right)^{-1}$

Again we will use

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

$\Rightarrow (7^{-1} -8^{-1})^{-1} -(3^{-1} -4^{-1})^{-1}=\left(\dfrac {1}{56}\right)^{-1}-\left(\dfrac {1}{12}\right)^{-1}$

$=56-12=44$ .

On multiplying ________ by

$2^{-5}$ we get

$2^5$

0%
$1$
0%
$2$
0%
$-1$
0%
$0$

Explanation

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

$\left(-\dfrac 57 \right)^{-5}$ is equal to

0%
$\left(\dfrac 57 \right)^{-5}$
0%
$\left(\dfrac 57 \right)^{5}$
0%
$\left(\dfrac 75 \right)^{5}$
0%
$\left(-\dfrac 75 \right)^{5}$

Explanation

Using law of exponent

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

[Where,

$a$ is non-zero integer]

$\left(\dfrac {-5}{7}\right)^{-5}=\dfrac {1}{\left(\dfrac {-5}{7}\right)^{5}}=\left(-\dfrac {7}{5}\right)^{5}$

$x$ be any integer different from zero and

$m$ be any integers then

$x^{-m}$ is equal to

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

Explanation

Using law of exponents,

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

[Where,

$a$ is non-zero integer]

Similarly,

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

$\left(\dfrac {-7}{5} \right)^{-1}$ is equal

0%
$\dfrac 57$
0%
$-\dfrac 57$
0%
$\dfrac 75$
0%
$\dfrac {-7}{5}$

Explanation

Using law of exponent

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

[Where,

$a$ is non-zero integer]

$\left(\dfrac {-7}{5}\right)^{-1}=\dfrac {1}{\left(\dfrac {-7}{5}\right)}=\left(-\dfrac {5}{7}\right)$

The expression for

$3^5$ with a negative exponent is _________.

0%
$\dfrac{1}{3^{-5}}$
0%
$-\dfrac{1}{3^{-5}}$
0%
${3^{-5}}$
0%
$\dfrac{1}{3^{5}}$

Explanation

Using

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

$3^5 = \dfrac{1}{3^{-5}}$

The value of

$[2^{-1} \times\ 3^{-1}]^{2}$ is ______.

0%
$6^2$
0%
$\dfrac{1}{6^2}$
0%
$-6^2$
0%
$-\dfrac{1}{6^2}$

Explanation

$[2^{-1} \times\ 3^{-1}]^{2} =\dfrac{1}{(2 \times 3)^{2}}=\dfrac{1}{6^{2}}$

State whether the following statement is true (T) or false (F):
The value of

$(-2)^{-3}$ is

$\dfrac{-1}{8}$

0%
True
0%
False

Explanation

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

Choose the correct answer from the given four options:
The value of

$(6^{-1}-8^{-1})^{-1}$ is

0%
$\dfrac{-1}{2}$
0%
$-2$
0%
$\dfrac{1}{24}$
0%
$24$

Explanation

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

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

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

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

$=24$

Choose the correct answer from the given four options:

$\bigg(-\dfrac{3}{2}\bigg)^{-1}$ is equal

0%
$\dfrac{2}{3}$
0%
$-\dfrac{2}{3}$
0%
$\dfrac{3}{2}$
0%
$\dfrac{4}{9}$

Explanation

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

If the coefficients of

$x^{7} \& x^{8}$ , in the expansion of

$\left[2+\frac{x}{3}\right]^{n}$ are equal, then the value of n is:

0%
15
0%
45
0%
55
0%
56

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Exponents And Powers Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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