Free Exponents and Powers 01 Practice Test - 7th grade
Question 1
3×1000+2×10+410+31000 is the expanded form of the number ____.
3200.430
3002.403
32.430
3020.403
SOLUTION
Solution : D
3×1000+2×10+410+31000
=3000+20+0.4+0.003
=3020.403
Question 2
In power notation, 8116 can be expressed as ___.
(3424)
(34)3
(32)4
(32)3
SOLUTION
Solution : A and C
We can write:
81=3×3×3×3
16=2×2×2×2
So, 8116 = (3424)
And we know that, (ambm)=(ab)m
∴8116 = (3424) =(32)4
Question 3
Third power of 14× fourth power of 4 equals to ______.
42
43
4
40
SOLUTION
Solution : 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
Question 4
If a = 2 and b = 3, find the value of (ab)a.
16
32
64
128
SOLUTION
Solution : C
On substituting a = 2 and b = 3, we get
(23)2=23×2 [∵(xm)n=x(m×n)]
=26
=64
Question 5
The value of 43×23 is
SOLUTION
Solution :Notice that here the two terms 43 and 23 have different bases, but the same exponents.
43×23=(4×4×4)×(2×2×2)
=(4×2)×(4×2)×(4×2)=83=512
Question 6
Express 5126000 in standard form.
0.5126×107
5.126×106
51.26×105
5126×103
SOLUTION
Solution : 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
Question 7
Express 256×243 in exponential form.
28×35
27×34
26×33
22×32
SOLUTION
Solution : A
We can write:
256=(2×2×2×2×2×2×2×2)=28243=(3×3×3×3×3)=35
∴256×243=28×35
Question 8
Find the value of (−2 5)7÷(−2 5)6.
(−2 5)2
25
−2 5
45
SOLUTION
Solution : C
abac=ab−c [a is a non-zero integer]
Using this law, we get,
(−2 5)7÷(−2 5)6(−2 5)7−6=(−2 5)1=−2 5
Question 9
23×33
SOLUTION
Solution :23×33=(2×3)3=63=216
Question 10
The value of (30−20)×52 is:
1
25
0
9
SOLUTION
Solution : C
∵ For any number a ≠ 0, a0 = 1.
∴30=20=1
⇒(30−20)×52=(1−1)×25 =0