7th Grade > Mathematics
EXPONENTS AND POWERS MCQs
Total Questions : 102
| Page 5 of 11 pages
:
Concept: 1 Mark
Answer:1 Mark
Given that,
A man has many chocolates and he takes 37chocolates.
He has to distribute them equally between34students.
The number of chocolates a student gets is:
=3734
=37−4 [∵aman=am−n]
=33
=27
So, each student will get 27 chocolates.
:
Formula: 1 Mark
Value of n: 1 Mark
Numerical value: 1 Mark
Steps: 1 Mark
Given that,
(am)2n=a2m
a2mn=a2m [∵(am)n=amn]
Since bases are same, their exponents also should be equal.
⇒2mn=2m
⇒n=2m2m
⇒n=1
As per question,
a=2 and m=4
On substituting the values, we get:
a2m=22×4
=28
=2×2×2×2×2×2×2×2
=256
Hence, the required value is 256.
Answer: Option C. -> 4
:
C
Third power of 14=(14)3
Fourth power of 4 is equal to 44.
(14)3×44
= 14×14×14×4×4×4×4
= 4×4×4×44×4×4
= 4443
= 4
:
C
Third power of 14=(14)3
Fourth power of 4 is equal to 44.
(14)3×44
= 14×14×14×4×4×4×4
= 4×4×4×44×4×4
= 4443
= 4
Answer: Option B. -> 5.126×106
:
B
In standard form, any number is expressed in powers of 10. The decimal number must be in between 1 and 10.
5.126×1000000=5.126×106
:
B
In standard form, any number is expressed in powers of 10. The decimal number must be in between 1 and 10.
5.126×1000000=5.126×106
:
Notice that here the two terms43 and 23have different bases, but the same exponents.
43×23=(4×4×4)×(2×2×2)
=(4×2)×(4×2)×(4×2)=83=512
Answer: Option C. -> 64
:
C
On substitutinga = 2 and b = 3, we get
(23)2=23×2[∵(xm)n=x(m×n)]
=26
=64
:
C
On substitutinga = 2 and b = 3, we get
(23)2=23×2[∵(xm)n=x(m×n)]
=26
=64
Answer: Option A. -> 28×35
:
A
We can write:
256=(2×2×2×2×2×2×2×2)=28
243=(3×3×3×3×3)=35
∴256×243=28×35
:
A
We can write:
256=(2×2×2×2×2×2×2×2)=28
243=(3×3×3×3×3)=35
∴256×243=28×35
Answer: Option D. -> 3020.403
:
D
3×1000+2×10+410+31000
=3000+20+0.4+0.003
=3020.403
:
D
3×1000+2×10+410+31000
=3000+20+0.4+0.003
=3020.403
Answer: Option B. ->
73
:
B
:
B
343 can be written as
343=7×7×7
so, 343=73