# Free Exponents and Powers 03 Practice Test - 7th grade

### Question 1

Express 343 as a power of 7.

72

73

74

70

#### SOLUTION

Solution :B

343 can be written as

343=7×7×7

so, 343=73

### Question 2

Express 32000 as a product of powers of prime factors.

28×52

28×53

27×54

28×54

#### SOLUTION

Solution :B

32000 can be written as

32000=(2×2×2×2×2)×1000=25×103

(as 32=2×2×2×2×2)= 25×(2×2×2×5×5×5)

= 25×23×53

= 28×53

### Question 3

If a = 2 and b = 3. Find the value of ab×ba.

56

72

64

8

#### SOLUTION

Solution :B

Putting a=2 and b=3

We get:

23=2×2×2=8 and 32=9^{ }

Hence, 8×9=72

### Question 4

Find the value of 33×32.

35

343

31

243

#### SOLUTION

Solution :A and D

Here the two terms have 33 and 32 have different exponents, but the same base.Hence by adding the powers we get,

33×32=33+2 = 35=243

### Question 5

Express the given number in standard form:

21700000

2.1×107

2.17×107

1.17×107

1.1701×107

#### SOLUTION

Solution :B

21700000 can be written as

2.17×10000000=2.17×107

### Question 6

The value of 33×23 is:

216

215

201

219

#### SOLUTION

Solution :A

Expanding the terms:

(3×2)3=63=216

### Question 7

(102)2 =

#### SOLUTION

Solution :From the law of exponent, we have,

(am)n=(a)mn

Therefore,

(102)2=104=10000

### Question 8

In 104, 10 is the base and 4 is the

#### SOLUTION

Solution :Here, 4 is called the exponent.

### Question 9

(25)2=−(25)2

True

False

#### SOLUTION

Solution :B

2252 = 425

−2252 = −425

Hence, the given statement is false.

### Question 10

am×bm=(ab)m

True

False

#### SOLUTION

Solution :A

From the law of exponents, we know that

am×bm=(ab)m

Thus, the given statement is true.