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

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

#### SOLUTION

Solution :

472=(2)×(2)×(2)×(59)

472=23×59

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

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

=374   [aman=amn]

=33

=27

So, each student will get 27 chocolates.

Simplify: [2 MARKS]

25×3×42×5

#### SOLUTION

Solution : Steps: 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

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

#### SOLUTION

Solution :

Steps: 1 Mark

43=(4)×(4)×(4)

43=64

32=(3)×(3)

32=9

The difference between given numbers is:

=649=55

The difference between 43 and 32 is 55.

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

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

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

Simplify : [3 MARKS]

38×45×32×54×7539×44×75

#### SOLUTION

Solution :

Concept: 1 Mark
Steps: 1 Mark

Given,

38×45×32×54×7539×44×75

=38+2×45×54×7539×44×75
[ am×an=am+n]

=3109×454×54×755
[ aman=amn]

=31×41×54×70

=3×4×54   [ a0=1]

=7500

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

Simplify and express in exponential form:

[(52)3×54]÷57  [3 MARKS]

#### SOLUTION

Solution : Steps: 2 Marks

=[(52)3×54]÷57

=[56×54]÷57          [(am)n=am×n]

=[56+4]÷57                 [am×an=am+n]

=510÷57

=5107                [am÷an=amn]

=53

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.

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.

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=20000000002000000000×30100
=2000000000600000000
=1400000000
Hence, the present value of the property is Rs 1400000000

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 47 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.

Simplify:

(i)0.005×1025×101

(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)

5×101=0.5         (move the decimal point 1 place to the left)

Now,

0.005×1025×101

=0.50.5

=0

ii)
(25)2×7383×7

= (25)2×7383×7

= (25×2)×73(23)3×7

= 210×7329×7

= 2109×731=2×72

= 2×49=98

Find the value of m and n,  if 65×52×63×5m×6n=1. [4 MARKS]

#### SOLUTION

Solution :

Steps: 2 Marks
Application: 1 Mark

65×52×63×5m×6n=1

65×63×6n×52×5m=1

65+3+n×52+m=1

62+n×52+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