8th Grade > Mathematics
EXPONENTS AND POWERS MCQs
Total Questions : 58
| Page 2 of 6 pages
Answer: Option A. -> (35)2
:
A
We know that for anon-zero integer a and any integer m,
a−m=1am.
∴6−2=162
⟹16−2=62=36
Similarly, 13−3=33=27 and 15−2=52=25.
⟹(16−2−13−3)÷15−2=(36−27)÷25
=925
=3252
=(35)2(∵ambm=(ab)m
:
A
We know that for anon-zero integer a and any integer m,
a−m=1am.
∴6−2=162
⟹16−2=62=36
Similarly, 13−3=33=27 and 15−2=52=25.
⟹(16−2−13−3)÷15−2=(36−27)÷25
=925
=3252
=(35)2(∵ambm=(ab)m
Answer: Option A. -> 3
:
A
((2)2)2× (33 ÷42) ÷ 32
=42 ×2716× 19(∵(am)n=amn)
= 16 × 2716× 19
=279
= 3
:
A
((2)2)2× (33 ÷42) ÷ 32
=42 ×2716× 19(∵(am)n=amn)
= 16 × 2716× 19
=279
= 3
Answer: Option A. -> 8.78×1012
:
A
8,780,000,000,000=878×1010
=8.78×102×1010
=8.78×1012
=87.8×10−1×1012
=87.8×1011
(Using am×an=am+n)
:
A
8,780,000,000,000=878×1010
=8.78×102×1010
=8.78×1012
=87.8×10−1×1012
=87.8×1011
(Using am×an=am+n)
Answer: Option C. -> 10−6 m
:
C
1 micron = 10−6m.
:
C
1 micron = 10−6m.
Answer: Option A. -> 128
:
A
27 is 2 multiplied by itself, 7 times.
So, 27=2×2×2×2×2×2×2=128
:
A
27 is 2 multiplied by itself, 7 times.
So, 27=2×2×2×2×2×2×2=128
Answer: Option D. -> 100000
:
D
10−5
= 1105
= 1100000
When a number is multiplied with its multiplicative inverse, the result is 1. To make the given number, which is 1100000, equal to 1, we need to multiply it with 100000, which is the reciprocal of 1100000.
Therefore, 100000, or 105 is the multiplicative inverse of 10−5.
:
D
10−5
= 1105
= 1100000
When a number is multiplied with its multiplicative inverse, the result is 1. To make the given number, which is 1100000, equal to 1, we need to multiply it with 100000, which is the reciprocal of 1100000.
Therefore, 100000, or 105 is the multiplicative inverse of 10−5.
Answer: Option B. -> False
:
B
Note that a0=1 for any a≠0
Thus, the value of1000∘is 1.
:
B
Note that a0=1 for any a≠0
Thus, the value of1000∘is 1.
Answer: Option B. -> m+n
:
B
We have am×an=am+n, where a, m and n are integers, a≠0.
Given am×an=ax
⟹am+n=ax
⟹x=m+n
:
B
We have am×an=am+n, where a, m and n are integers, a≠0.
Given am×an=ax
⟹am+n=ax
⟹x=m+n
Answer: Option D. -> 512
:
D
x5 × x4, for x = 2
Substituting 2 in place of x
= 25× 24
= 29(∵am×an=am+n) = 512
:
D
x5 × x4, for x = 2
Substituting 2 in place of x
= 25× 24
= 29(∵am×an=am+n) = 512
:
4×103+3×1+1×10−1+6×100=(4×1000)+(3×1)+110+(6×1)=4000+3+0.1+6=4009.1
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