# Free Exponents and Powers Subjective Test 02 Practice Test - 7th grade

### Question 1

Express 472 as a power of its prime factors. [1 MARK]

#### SOLUTION

Solution :

472=(2)×(2)×(2)×(59)

⇒472=23×59

### Question 2

A man has many chocolates and he takes 37 chocolates and distributes it to 34 students. How many chocolates do each student get if each student should get an equal number of chocolates? [2 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Answer: 1 Mark

Given that,

A man has many chocolates and he takes 37 chocolates.

He has to distribute them equally between 34 students.

The number of chocolates a student gets is:

=3734

=37−4 [∵aman=am−n]

=33

=27

So, each student will get 27 chocolates.

### Question 3

Simplify: [2 MARKS]

25×3×42×5

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

Given equation is:

25×3×42×5

=25×3×(22)2×5

=25×3×24×5

∵(am)n=amn

=25+4×3×5

∵am×an=am+n

=29×3×5

=29×3×5

=7680

### Question 4

What is the difference between 43 and 32? [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

43=(4)×(4)×(4)

⇒43=64

32=(3)×(3)

⇒32=9

The difference between given numbers is:

=64−9=55

The difference between 43 and 32 is 55.

### Question 5

The distance between two cities is 3245678 km. Convert it into meters and express in standard form. [2 MARKS]

#### SOLUTION

Solution :Conversion: 1 Mark

Standard form: 1 Mark

Given distance that the distance between two cities 3245678 km.

Now, 1 km = 1000 m

3245678 km = (3245678 × 1000) m = 3245678000 m.

Standard form is,

3245678000m=3.245678×109

### Question 6

In a stck there are 5 books each of thickness 20mm and 5 paper sheets of thickness 0.016mm. What is the total thickness of the stack? (1mm = 0.001m) [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Application: 1 Mark

Answer: 1 Mark

Thickness of each book =20mm

Hence, thickness of 5 books =5×20=100mm

Thickness of each paper sheet =0.016mm

Hence, thickness of 5 paper sheets =5×0.016=0.080mm

Total thickness of the stack

= Thickness of 5 books + Thickness of 5 paper sheets

=(100+0.080)mm

=100.08mm

=1.0008×102mm

### Question 7

Simplify : [3 MARKS]

38×45×32×54×7539×44×75

#### SOLUTION

Solution :Concept: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Given,

38×45×32×54×7539×44×75

=38+2×45×54×7539×44×75

[ ∵am×an=am+n]

=310−9×45−4×54×75−5

[ ∵aman=am−n]

=31×41×54×70

=3×4×54 [ ∵a0=1]

=7500

### Question 8

Write the following numbers in the expanded form:

(i) 279404

(ii) 3006194

(iii) 2806196

[3 Marks]

#### SOLUTION

Solution :Each option: 1 Mark

i) 279404 = 2,00,000 + 70,000 + 9,000 + 400 + 00 +4

=2×100000+7×10000+9×1000+4×100+0×10+4×1

= 2×105+7×104+9×103+4×102+0×101+4×100

ii) 3006194 = 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4

=3×1000000+0×100000+0×10000+6×1000+1×100+9×10+4×1

=3×106+0×105+0×104+6×103+1×102+9×101+4×100

iii) 2806196 = 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6

= 2×1000000+8×100000+0×10000+6×1000+1×100+9×10+6×1

=2×106+8×105+0×104+6×103+1×102+9×101+6×100

### Question 9

Simplify and express in exponential form:

[(52)3×54]÷57 [3 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Answer: 1 Mark

=[(52)3×54]÷57

=[56×54]÷57 [∵(am)n=am×n]

=[56+4]÷57 [∵am×an=am+n]

=510÷57

=510−7 [∵am÷an=am−n]

=53

### Question 10

The area of a certain number of triangles is equal to the sum of the exponents of the prime factors of the number 1628, and each prime factor represents a triangle. Find the sum of areas of the triangles and find the number of the triangles. [4 MARKS]

#### SOLUTION

Solution :Prime factorization: 1 Marks

Number of Triangles: 1 Mark

Sum of the area: 1 Mark

Steps: 1 Mark

Given that,

The area of a certain number of triangles is equal to the exponents of the prime factors of the number 1628 and each prime factor represents a triangle.

Prime factors of 1628 are:

1628=22×3×7×19.

Since there are 5 prime factors,

⇒ The number of given triangles are = 5

The area of the triangles is the sum of powers of the prime factors.

⇒The sum of areas of the triangle = 2 + 1 + 1 + 1

= 5 square units

The number of triangles is 5 and the sum of areas of the triangle is 5 square units.

### Question 11

Find the value of 'n' if (am)2n=a2m. Find the numerical value of a2m if a=2 and m=4. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Value of n: 1 Mark

Numerical value: 1 Mark

Steps: 1 Mark

Given that,

(am)2n=a2m

a2mn=a2m [∵(am)n=amn]

Since bases are same, their exponents also should be equal.

⇒2mn=2m

⇒n=2m2m

⇒n=1

As per question,

a=2 and m=4

On substituting the values, we get:

a2m=22×4

=28

=2×2×2×2×2×2×2×2

=256

Hence, the required value is 256.

### Question 12

The value of a man's property is greater than the multiplication of cube and square of properties of A and B respectively.

What is the minimum value of his property, if the value of properties A and B are Rs 20 and Rs 400 respectively?

The actual value of his property was Rs 2,00,00,00,000 but due to recession, the value of the property decreased by 30%.

Find the present value of the property. [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Minimum Value: 1 Mark

Present Value: 1 Mark

Given that,

A's property = Rs 20

⇒ cube of A's property =203

B's property = Rs 400

⇒ Square of B's property =4002

The minimum value of the man's property should be

=Rs 203×4002

=Rs 128×107

=Rs 1.28×109

Given that

The actual value of the property is Rs2,00,00,00,000.

The value of the property depreciated by 30%.

The present value of the property=2000000000−2000000000×30100

=2000000000−600000000

=1400000000

Hence, the present value of the property is Rs 1400000000

### Question 13

Find the value of 'a' satisfying the equation.

a) (42)a=(7a)2.

b) (2a)5=24×43

[4 MARKS]

#### SOLUTION

Solution :Each option: 2 Marks

a) Given that,

(42)a=(7a)2

42a=72a [amn=amn]

Given, 42a=72a

Since 4≠7 the only case where the above equation is true is when both the exponents are zero.

⇒ a=0

⇒42a=72a=1 [∵40=1, 70=1]

b) The given equation is

(2a)5=24×43

⇒2a×5=24×(22)3

∵(am)n=am×n

⇒2a×5=24×(22×3)

⇒2a×5=24×(26)

⇒2a×5=24+6=210

∵am×an=am+n

Since their bases are same and they are equal, threfore their powers must be same,

So, 5×a=10

⇒a=105

⇒a=2

So, the value of a = 2.

### Question 14

Simplify:

(i)0.005×102−5×10−1

(ii)(25)2×7383×7

[4 MARKS]

#### SOLUTION

Solution :Each question: 2 Marks

i) It's easy to subtract powers when we convert them into normal form.

0.005×102=0.5 (move the decimal point 2 places to the right)

5×10−1=0.5 (move the decimal point 1 place to the left)

Now,

0.005×102−5×10−1

=0.5−0.5

=0

ii)

(25)2×7383×7

= (25)2×7383×7

= (25×2)×73(23)3×7

= 210×7329×7

= 210−9×73−1=2×72

= 2×49=98

### Question 15

Find the value of m and n, if 6−5×5−2×63×5m×6n=1. [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Application: 1 Mark

Answer: 1 Mark

6−5×5−2×63×5m×6n=1

6−5×63×6n×5−2×5m=1

6−5+3+n×5−2+m=1

6−2+n×5−2+m=1

The bases of both the numbers are not equal.

So, the value will be equal to 1, when the exponents of these bases will be equal to zero.

Any non - zero number raised to the power zero is 1.

So, the exponents of both 6 and 5 are zero.

Therefore,

−2+m=0 ⇒m=2

−2+n=0 ⇒n=2

∴m=n