MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 8 Maths Exponents And Powers Quiz 1 - MCQExams.com
CBSE
Class 8 Maths
Exponents And Powers
Quiz 1
The value of $$\dfrac{1}{4^{-2}}$$ is equal to $$16$$
Report Question
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.
$$a^{m}=\dfrac{1}{a^{-m}}$$
Report Question
0%
True
0%
False
Explanation
True.
Using law of exponents, $$x^{-m}=\dfrac{1}{x^{m}}$$
$$LHS=RHS$$
Simplify:
$$\left(\dfrac{27}{125}\right)^{-1}=$$
Report Question
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 _____.
Report Question
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,
$$\left(\dfrac {2^5}{3^5}\right)^{-3/5}$$
$$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$$
$$\left \{ \left (\dfrac {3}{4}\right )^{-1} - \left (\dfrac {1}{4}\right )^{-1}\right \}^{-1} = ?$$
Report Question
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 : $$(-4)^{-2}$$---
Report Question
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
Report Question
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?
Report Question
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.
Report Question
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.
Report Question
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:
Report Question
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: $$1 \div 7 \div 7$$
Report Question
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:
Report Question
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
Report Question
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?
Report Question
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 $$25$$.
Report Question
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:
$$0.000008$$ _______ $$0.000016$$
Report Question
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
Report Question
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:
$$0.000000038$$ _______ $$\displaystyle 3\cdot 8\times 10^{-8}$$
Report Question
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.
$$0.00000998$$ ______ $$0.0000116$$
Report Question
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 $$\displaystyle 9\cdot 89\times 10^{-6}< 1\cdot 16\times 10^{-5}$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 8 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
