# Free Exponents and Powers Subjective Test 01 Practice Test - 7th grade

### Question 1

Express 625 as a power of 5. [1 MARK]

#### SOLUTION

Solution :

625=(5)×(5)×(5)×(5)

⇒625=54

### Question 2

Which of the numbers 726 and 938, when expressed as factors has greater power of 2?

[2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

The prime factorisation of 726 and 938 are given below.

726=(2)×(3)×(11)×(11)

⇒726=21×31×112

938=(2)×(7)×(67)

⇒938=21×71×67

Both have the same powers of 2 which is 1.

### Question 3

Find the difference between 45 and 54? [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

45=(4)×(4)×(4)×(4)×(4)

⇒45=1024

54=(5)×(5)×(5)×(5)

⇒54=625

The difference between 45 and 54 is

45−54=1024−625

⇒45−54=399

The difference between 45 and 54 is 399.

### Question 4

Write (8)×(8)×(8)×(8)×(8) in the exponential form, taking 2 as base. [2 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Answer: 1 Mark

Here we know that for any non zero number if the base is same and the number is multiplied, then the exponents are added.

(8)×(8)×(8)×(8)×(8)

=23×23×23×23×23

=23+3+3+3+3

∵ab×ac = ab+c

=215

### Question 5

Is 30=(1000)0 true or false? Justify your answer. [2 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Answer: 1 Mark

Consider, 30=(1000)0

L.H.S. is 30=1 [ ∵a0 = 1]

and R.H.S. (1000)0=1 [ ∵a0 = 1]

⇒ L.H.S. = R.H.S.

⇒30=(1000)0.

∴ It is true.

We should remember that any non-zero number raised to the power 0 is always equal to 1.

### Question 6

Simplify: 84×143×5×3445×73×62 [3 MARKS]

#### SOLUTION

Solution :Application: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Consider,

84×143×5×3445×73×62

84=(23)4=212

143=(2×7)3=23×73

45=(22)5=210

62=(2×3)2=22×32

On substituting the values we get:

212×23×73×5×34210×73×22×32

=212+3−10−2×73−3×5×34−2 [∵ aman=am−n]

=23×5×32 [∵a0=1]

=8×5×9

=360

### Question 7

When a number is multiplied with 25 it gives 78125. Find the number and express it as a power of 5. What is the product if the number is multiplied by (−5)×(5)? [3 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Application: 1 Mark

Answer: 1 Mark

Let the number be x.

Given that,

(x)×(25)=78125

x=3125

⇒ The required number is 3125

Now,

3125=5×5×5×5×5

⇒3125=55

Now, if the number was multiplied by (−5)×(5) then the product=(−5)×(5)×3125 = −78125

### Question 8

A shopkeeper A has Rs 34265 in his counter and shopkeeper B has 432500 paise in his counter. Express the difference in rupees in standard form. [ 3 MARKS]

#### SOLUTION

Solution :Conversion: 1 Mark

Difference: 1 Mark

Standard form: 1 Mark

As per the question

The amount shopkeeper A has = Rs 34265.

The amount shop keeper B has = 432500 paise ⇒ Rs 4325.

Difference = Rs 34265 - Rs 4325 = Rs 29940.

The standard form of Rs 29940.

29940=2.994×104

The difference in rupees is 2.994×104

### Question 9

Express each of the following numbers using the exponential notation:

(i) 512

(ii) 343

(iii) 729

#### SOLUTION

Solution :Each option: 1 Mark

i)512=2×2×2×2×2×2×2×2×2=29

ii)343=7×7×7=73

iii)729=3×3×3×3×3×3=36

### Question 10

A ball is released from a height of x meters. Given that it bounces back to half of its original height. Find the product of heights after eight bounces. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Application: 1 Mark

Answer: 2 Marks

Given that

The ball is released from a height of x m

Height reached after first bounce =x2

Height reached after second bounce =x4=x22

Height reached after third bounce =x8=x23

Height reached after fourth bounce =x16=x24

Height reached after fifth bounce =x32=x25

Height reached after sixth bounce =x64=x26

Height reached after seventh bounce =x128=x27

Height reached after eight bounce =x256=x28

∴ Product of first 9 heights =x×x2×x22×x23×x24×x25×x26×x27×x28

=x(1+1+1+1+1+1+1+1+1)2(1+2+3+4+5+6+7+8)

(∵ab×ac = ab+c)

=x9236

The product of heights after 8 bounces =x9236

### Question 11

A man X has a land whose area is 34 meter square and a man Y has 43 meter square of land. If the land costs Rs 50 per meter square, who has higher property value? [4 MARKS]

#### SOLUTION

Solution :Step: 2 Marks

Application: 1 Mark

Answer: 1 Mark

The area of land owned by X is

=34 m2=81 m2

The value of given property is

=(81)×(50)=Rs.4050

The area of land owned by Y is

43 m2=64 m2

The value of given property is

=(64)×(50)=Rs.3200

⇒ Value of property owned by X is greater than property owned by Y.

### Question 12

Simplify: [4 MARKS]

i)(58)−7×(85)−5

ii)(25)2×7383×7

#### SOLUTION

Solution :Each: [2 Marks]

i)(58)−7×(85)−5

=5−78−7×8−55−5

=5−75−5×8−58−7

=5(−7)−(−5)×8(−5)−(−7)

(∵ aman=am−n)

=5−2×82=8252=6425

ii)(25)2×7383×7

(25)2×7383×7

= (25×2)×73(23)3×7

(∵(am)n=am×n)

= 210×7329×7

=210−9×73−1=2×72

=2×49=98

### Question 13

If abc = 1, then find the value of:

11+a+b−1+11+b+c−1+11+c+a−1 [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given that,

abc=1

So, c=1÷ab=(ab)−1

Similarly,

a=(bc)−1

b=(ac)−1

And,

ab=c−1

ac=b−1

bc=a−1

11+a+b−1+11+b+c−1+11+c+a−1

=11+a+b−1+b−1b−1+bb−1+b−1c−1+aa+ac+aa−1

[Multiplying 2nd term by b−1b−1 and 3rd term by aa]

=11+a+b−1+b−1b−1+1+a+aa+b−1+1

∵ aman=am−n

=1+b−1+a1+a+b−1

=1

### Question 14

If x=ya, y=zb, z=xc then find abc. If a = 12, b = 1288, then find the value of c. [4 MARKS]

#### SOLUTION

Solution :Steps: 1 Marks

Application: 1 Mark

Value of abc: 1 Mark

Value of c: 1 Mark

It is given that

x=ya, y=zb, z=xc

x=ya

=(zb)a (∵y=zb)

=zab

=(xc)ab (∵xc=z)

x=xabc ∵((am)n=amn)

⇒x1=xabc

As bases are equal, exponents are equal.

⇒abc=1

Now, it is also given that,

a = 12, b = 1288

a×b×c = 1

c = 1a×b

On substituting the values we get:

c = 112×1288

c = 28812

c = 24

Hence, the value of c is 24.

### Question 15

In a race Karan covers 1m in 1st second, in the next second he covers twice the distance he covered in 1st second, in 3rd second he covers twice the distance which he covered in 2nd second and he continues to run like this till six seconds. After that he slows down, as in 7th second he covers only half the distance which he covered in 6th second and in 8thsecond he covers half the distance which he covered in 7thsecond and he continuous in that manner. If Karan completes the race in 9 seconds, find the length of given track. [4 MARKS]

#### SOLUTION

Solution :Application: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Here are the distances covered by Karan in each second

1st second = 1m

2nd second =1×2=2m

Hence, the length of the track is 2 m.

3rd second =2×2= 4m

4th second =4×2= 8m

5th second = 8×2=16m

6th second = 16×2=32m

7th second =32÷2= 16m

8th second =16÷2= 8m

9th second =8÷2= 4m

Length of the given track = (1+2+4+8+16+32+16+8+4) m

⇒Length of the given track = 91 m

Hence, the length of the track is 91 m.