Free Exponents and Powers 03 Practice Test - 8th Grade
Question 1
Simplify (25÷28)5 × 2−5.
1210
1220
12(−15)
12(−5)
SOLUTION
Solution : B
(25÷28)5×2−5=(2528)5×2−5
=(25−8)5×2−5 (∵aman=am−n)
=(2−3)5×2−5
=2−15×2−5 (∵(am)n=amn)
=2−15−5 (∵am×an=am+n)
=2−20
=1220 (∵a−m=1am)
Question 2
4−2+4−1+40+42 = ___
27716
27718
SOLUTION
Solution : A
Note that a0=1 for any non-zero integer a.
Also, for integers a (≠0) and m,
a−m=1am.
⟹4−2=142=116 and 4−1=141
∴4−2+4−1+40+42=116+14+1+16
=1+4+16+25616
=27716
Question 3
The value of (4−7 ÷ 4−10 ) × 4−5 is 16−1.
True
False
SOLUTION
Solution : A
The value of (4−7 ÷ 4−10 ) × 4−5 is 16−1.
(4−7÷ 4−10)×4−5
=(4−74−10) ×4−5
=4−7+10×4−5 (∵aman= am−n)
=43×4−5
⇒43−5 (∵am×an= am+n)
=4−2=142
=116
=16−1
Question 4
The exponent of c after simplifying (3ab)2(−5a2bc4)2 is
SOLUTION
Solution :(3ab)2(−5a2bc4)2
= (9a2b2) × (25a4b2c8)
= 225a6b4c8
Exponent of c = 8
Question 5
Simplify :(16−2−13−3)÷15−2
(35)2
(925)2
(53)2
(259)2
SOLUTION
Solution : A
We know that for a non-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
Question 6
8,780,000,000,000= _______.
8.78×1012
8.78×10−12
87.8×1012
87.8×10−11
SOLUTION
Solution : 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)
Question 7
1 micron equals to _________.
110000
106 m
10−6 m
10−5 m
SOLUTION
Solution : C
1 micron = 10−6 m.
Question 8
The value of 27 is _____.
128
64
256
512
SOLUTION
Solution : A
27 is 2 multiplied by itself, 7 times.
So, 27=2×2×2×2×2×2×2=128
Question 9
((2)2)2× (33 ÷42) ÷ 32 is equal to
3
1
SOLUTION
Solution : A
((2)2)2× (33 ÷42) ÷ 32
= 42 ×2716× 19 (∵(am)n=amn)
= 16 × 2716× 19
= 279
= 3
Question 10
Between 210 and 102 , 102 is greater.
True
False
SOLUTION
Solution : B
210= (25)2 = (2×2×2×2×2)2 = 32×32 = 1024
102 = 10×10 = 100
Since 1024 > 100 so, 210 is greater than 102.