Exponents And Powers(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

EXPONENTS AND POWERS MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. Express in usual form: 5.06x

10^{- 4}

Discuss Question

5.06 \times 10^{- 4} = \frac{5.06}{10000} = 0.000506

Question 22. Identify the base in the given example:

5^{3}

Base: ___

Discuss Question

:
The number written at the bottom is the base whereas the number at the superscript is known as the exponent. In the given example, 5 is the base and 3 is the exponent.

Question 23. The standard exponential form of 1,540,000,000 is

1.54 \times 10^{9}

True
False
1116
1611

Discuss Question

Answer: Option A. -> True
:
A

1, 540, 000, 000 = 1.54 \times 1000000000 = 1.54 \times 10^{9}

Question 24. Simplify

\frac{(\frac{1}{5})^{- 3} - (\frac{1}{3})^{- 4}}{(\frac{1}{4})^{- 3}} .

1213
1312
1116
1611

Discuss Question

Answer: Option C. -> 1116
:
C

∵ a^{- n} = \frac{1}{a^{n}} f o r a \neq 0

⟹ (\frac{1}{5})^{- 3} = \frac{1}{(\frac{1}{5})^{3}} = (\frac{5}{1})^{3} = 5^{3} = 125

Similarly,

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

and

(\frac{1}{4})^{- 3} = \frac{1}{(\frac{1}{4})^{3}} = (\frac{4}{1})^{3} = 4^{3} = 64

⟹ \frac{(\frac{1}{5})^{- 3} - (\frac{1}{3})^{- 4}}{(\frac{1}{4})^{- 3}} = \frac{125 - 81}{64} = \frac{44}{64} = \frac{11}{16}

Question 25. If

(2)^{m + 1} \times 2^{5} = 4^{- 2}

the value of

m

is
___

Discuss Question

:
Given,

(2)^{m + 1} \times 2^{5} = 4^{- 2}

\Rightarrow

(2)^{m + 1} \times 2^{5} = {((2)^{2})}^{- 2}

\Rightarrow

(2)^{m + 1 + 5} = (2)^{- 4}

Since the bases on both the sides are equal, therefore the exponents must also be equal. This means,

m + 1 + 5 = - 4

, or

m = - 10

Question 26. The value of (

4^{- 7}

4^{- 10}

) ×

4^{- 5}

16^{- 1}

True
False
25716
27118

Discuss Question

Answer: Option A. -> True
:
A
The value of (

4^{- 7}

4^{- 10}

) ×

4^{- 5}

16^{- 1}

(4^{- 7}

4^{- 10}) \times 4^{- 5}

(\frac{4^{- 7}}{4^{- 10}}) \times 4^{- 5}

4^{- 7 + 10} \times 4^{- 5} (∵ \frac{a^{m}}{a^{n}} = a^{m - n})

= 4^{3} \times 4^{- 5}

\Rightarrow 4^{3 - 5} (∵ a^{m} \times a^{n} = a^{m + n})

= 4^{- 2} = \frac{1}{4^{2}}

= \frac{1}{16}

= 16^{- 1}

Question 27.

4^{- 2} + 4^{- 1} + 4^{0} + 4^{2}

= ___

27716
27718
25716
27118

Discuss Question

Answer: Option A. -> 27716
:
A
Note that

a^{0} = 1

for any non-zero integer

a

.
Also, for integers

a (\neq 0)

and

m,

a^{- m} = \frac{1}{a^{m}} .

⟹ 4^{- 2} = \frac{1}{4^{2}} = \frac{1}{16}

and

4^{- 1} = \frac{1}{4^{1}}

∴ 4^{- 2} + 4^{- 1} + 4^{0} + 4^{2} = \frac{1}{16} + \frac{1}{4} + 1 + 16

= \frac{1 + 4 + 16 + 256}{16}

= \frac{277}{16}

Question 28. Between

2^{10}

and

10^{2}

10^{2}

is greater.

True
False
123
27

Discuss Question

Answer: Option B. -> False
:
B

2^{10}

(2^{5})^{2}

(2 \times 2 \times 2 \times 2 \times 2)^{2}

= 32×32 = 1024

10^{2}

= 10×10 = 100
Since1024 > 100so,

2^{10}

is greater than

10^{2}

Question 29.

If $m$ and $n$ are integers, $a$ is a non-zero integer and $a^{m}$ × $a^{n}$ = $a^{x}$ , then the value of $x$ is _____.

$m n$
$m + n$
$m - n$
$\frac{m}{n}$

Discuss Question

Answer: Option B. ->

m + n

:
B
We have

a^{m} \times a^{n} = a^{m + n},

where

a

m

and

n

are integers,

a \neq 0.

Given

a^{m} \times a^{n} = a^{x}

⟹ a^{m + n} = a^{x}

⟹ x = m + n

Question 30.

In standard form 21600000 is written as___________.

$2.16 \times 10^{7}$
$216 \times 10^{7}$
$2.16 \times 10^{5}$
$216 \times 1000000$

Discuss Question

Answer: Option A. ->

2.16 \times 10^{7}

:
A

21600000 = 2.16 x 10000000
Therefore, in standard form 21600000 is written as :
$2.16 \times 10^{7}$

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