Free Exponents and Powers 03 Practice Test - 8th Grade

Simplify (25÷28)5 × 25.

A.

1210

B.

1220

C.

12(15)

D.

12(5)

SOLUTION

Solution : B

(25÷28)5×25=(2528)5×25
=(258)5×25   (aman=amn)
=(23)5×25
=215×25   ((am)n=amn)
=2155   (am×an=am+n)
=220
=1220  (am=1am)

42+41+40+42 = ___

A.

27716

B.

27718

C. 25716
D. 27118

SOLUTION

Solution : A

Note that a0=1 for any non-zero integer a.
Also, for integers a (0) and m,
am=1am.
42=142=116 and 41=141
42+41+40+42=116+14+1+16
=1+4+16+25616
=27716

The value of (47 ÷ 410 ) × 45 is 161.

A.

True

B.

False

SOLUTION

Solution : A

The value of (47 ÷ 410 ) × 45 is 161.
(47÷ 410)×45

=(47410) ×45
=47+10×45    (aman= amn)
=43×45
435    (am×an= am+n)
=42=142
=116
=161

The exponent of c after simplifying (3ab)2(5a2bc4)2 is

___

SOLUTION

Solution :

(3ab)2(5a2bc4)2
= (9a2b2) × (25a4b2c8)
= 225a6b4c8
Exponent of c = 8

Simplify :(162133)÷152

A.

(35)2

B.

(925)2

C.

(53)2

D.

(259)2

SOLUTION

Solution : A

We know that for a non-zero integer a and any integer m,
am=1am.
62=162
162=62=36
Similarly, 133=33=27 and 152=52=25.
(162133)÷152=(3627)÷25
=925
=3252
=(35)2    (ambm=(ab)m

8,780,000,000,000= _______.

A.

8.78×1012

B.

8.78×1012

C.

87.8×1012

D.

87.8×1011

SOLUTION

Solution : A

8,780,000,000,000=878×1010
=8.78×102×1010
=8.78×1012
=87.8×101×1012
=87.8×1011
(Using am×an=am+n)

1 micron  equals  to _________.

A.

110000

B.

106  m

C.

106  m

D.

105 m

SOLUTION

Solution : C

1 micron = 106 m.

The value of  27 is _____.

A.

128

B.

64

C.

256

D.

512

SOLUTION

Solution : A

27 is 2 multiplied by itself, 7 times.

So, 27=2×2×2×2×2×2×2=128

((2)2)2× (33 ÷42) ÷ 32 is equal to

A.

3

B.

1

C. 123
D. 27

SOLUTION

Solution : A

((2)2)2× (33 ÷42) ÷ 32
42 ×2716× 19   ((am)n=amn)
= 16 × 2716× 19
279
= 3

Between 210 and 102 , 102  is greater.

A.

True

B.

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.