Free Exponents and Powers 02 Practice Test - 8th Grade
Question 1
Find the value of 7−3.
343
1343
121
149
SOLUTION
Solution : B
We know that a−m=1am for any non-zero integer a and integer m.
⟹7−3=173
=17×7×7=1343
Question 2
Find the multiplicative inverse of 10−7.
10−2
102
107
106
SOLUTION
Solution : C
Multiplicative inverse of a non-zero x is 1x; so that when these two are multiplied, the product value is 1. Therefore, the multiplicative inverse of 10−7should be a number which when multiplied with 10−7 gives 1.
Let the multiplicative inverse of 10−7 be x.
Then, 10−7×x=1.
⇒x=110−7=107
(∵a−n=1an⟹1a−n=11an=an)
Thus, the multiplicative inverse of 10−7 is 107.
Question 3
Simplify and write in exponential form: 43×4−2×162
43
44
45
46
SOLUTION
Solution : C
Given expression is
43×4−2×162=43×4−2×((4)2)2=43×4−2×44 (∵(am)n=amn)=43−2+4 (∵am×an=am+n)=45
Question 4
Simplify (15)−3−(13)−4(14)−3.
1213
1312
1116
1611
SOLUTION
Solution : C
∵a−n=1an for a≠0
⟹(15)−3=1(15)3=(51)3=53=125
Similarly, (13)−4=1(13)4=(31)4=34=81 and
(14)−3=1(14)3=(41)3=43=64
⟹(15)−3−(13)−4(14)−3=125−8164=4464=1116
Question 5
The standard exponential form of 1,540,000,000 is 1.54×109.
True
False
SOLUTION
Solution : A
1,540,000,000=1.54×1000000000=1.54×109
Question 6
Identify the base in the given example: 53
Base:
SOLUTION
Solution :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 7
If 3676.48 can be written in expanded form as:
3×10a+b×102+7×10c+6×100+4×10d+e×10−2 .
Find the value of a+b−c−d−e.
SOLUTION
Solution :3676.48 can be written in expanded form as:
3×103+6×102+7×101+6×100+4×10−1+8×10−2.
a=3,b=6,c=1,d=−1,e=8a+b−c−d−e=3+6−1+1−8=1
Question 8
If (2)m+1×25=4−2 the value of m is
SOLUTION
Solution :Given, (2)m+1×25=4−2
⇒ (2)m+1×25=((2)2)−2
⇒ (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 9
Express in usual form: 5.06x10−4
SOLUTION
Solution :5.06×10−4=5.0610000=0.000506
Question 10
The distance between Earth and Mars is 2.2253x108 kms. The distance between Mars and Saturn is 1.19666×109 kms. A spaceship starts from Earth and travels to Saturn via Mars. Find the total distance covered by the spaceship in metres.
1.41919x1012 m
SOLUTION
Solution : A and C
The total distance travelled by the spaceship between Earth and Saturn = Distance between (Earth & Mars + Mars & Saturn)
2.2253x108 kms + 1.19666 x 109 kms
= 2.2253x108 kms + 11.9666 x 108 kms
=14.1919 x108 kms
= 1.41919x109×103 m
= 1.41919x 1012m=1419190000000 m