Unit Test (Solutions): Exponents and Powers | Mathematics (Maths) Class 7 (Old NCERT) PDF Download

Time: 1 hour M.M. 20 

Attempt all questions. 

• Question numbers 1 to 5 carry 1 mark each. 
• Question numbers 6 to 8 carry 2 marks each. 
• Question numbers  9 to 11 carry 3 marks each. 

Q1: Evaluate: (4)³  (1 Mark)
(a) 64
(b) 16
(c) 8
(d) 4
Ans: (a)
Sol:
(4)³ = 4 × 4 × 4
= 64

Q2: Simplify: (3⁻²) × (3⁻³)    (1 Mark)
(a) 1/3
(b) 1/9
(c) 1/27
(d) 1/81
Ans: (c)
Sol:
Using the property of exponents aᵐ × aⁿ = aᵐ⁺ⁿ:
(3⁻²) × (3⁻³) = 3⁻² + (⁻³) = 3⁻⁵
= 1 / 3⁵
= 1 / (3 × 3 × 3 × 3 × 3)
= 1 / 243

Q3: Evaluate: (13)⁰    (1 Mark)
(a) 0
(b) -1
(c) 1
(d) 3
Ans: (c)
Sol: Any non-zero number raised to the power of 0 is 1. So, (13)⁰ = 1.

Q4: Simplify: (5⁻²) × (5⁻³)    (1 Mark)
(a) 1/25
(b) 1/125
(c) 1/5
(d) 1/50
Ans: (b)
Sol: Using the property of exponents aᵐ × aⁿ = aᵐ⁺ⁿ:
(5⁻²) × (5⁻³) = 5⁻² + (⁻³) = 5⁻⁵
= 1 / 5⁵

Q5: a^m × aⁿ = _______?    (1 Mark)
(a) a^m × aⁿ
(b) a^{m - n}
(c) a^m+n
(d) a^mn
Ans: (c) a^m+n
Sol: Using the property of exponents,
a^m × aⁿ = a^m+n. 

Q6: Express the following in exponential form 3 × 3 × b × b × b.  (2 marks)
Ans: Given: 3 × 3 × b × b × b
We need to write the given expression as an exponential form. A number can be written in its exponential form if we raise the power of the number by the exponent. Therefore, the exponential form of 3 × 3 × b × b × b is:
= 3 × 3 × b × b × b
= 3² × b³
= 9b³

Q7: Simplify:  10² × 10⁴ × 10^-3 × 10⁶ × 10^-2 × 10^3.   (2 marks)
Ans:
Given:
10² × 10⁴ × 10^-3 × 10⁶ × 10^-2 × 10³
We know that x^m × xⁿ = x^m+n, so simplifying using this as below:
= 10² × 10⁴ × 10^-3 × 10⁶ × 10^-2 × 10³
= 10^{(2+4+(-3)+6+(-2)+3)}
= 10⁽¹⁰⁾
= 10¹⁰

Q8: Express the following information in the standard form: (2 marks)
The distance from Earth to the Moon is 384400000 m.
Ans: The distance from Earth to the Moon is 384400000 m.
In standard form,
The distance from Earth to the Moon is 3.844 × 10⁸ m.

Q9: Simplify the following expression:  (3 marks)
(5⁶a³b⁵) / (5²a²b³)
Ans:
Given:
We need to find the value of the given expression using laws.
We know that
aᵐ / aⁿ = aᵐ⁻ⁿ
aᵐ × aⁿ = aᵐ⁺ⁿ
Therefore, the value of (5⁶a³b⁵) / (5²a²b³) will be:
= (5⁶a³b⁵) / (5²a²b³)
= 5⁶⁻² × a³⁻² × b⁵⁻³
= 5⁴ × a¹ × b²
= 625a × b²

Q10: Evaluate: (4⁻³ × 6² × p⁻⁴) / (4⁶ × 6⁻³ × p²)
Ans:
Given:
(4⁻³ × 6² × p⁻⁴) / (4⁶ × 6⁻³ × p²)
We can simplify using the exponent rules:
= 4⁻³⁻⁶ × 6²⁻⁻³ × p⁻⁴⁻²
= 4⁻⁹ × 6⁵ × p⁻⁶
Now, simplify:
= 4⁻⁹ × 6⁵ × p⁻⁶
= (1 / 4⁹) × 6⁵ × (1 / p⁶)
= 6⁵ / (4⁹ × p⁶)

Q11. Using the laws, find
(a) ((4³ × 5²) ÷ 4⁵)
(b)  ((2⁴ × 3³) ÷ 2⁶)
Ans: (a) ((4³ × 5²) ÷ 4⁵)
We need to find the value of a given expression using laws.
We know that
aᵐ ÷ aⁿ = aᵐ⁻ⁿ
aᵐ × aⁿ = aᵐ⁺ⁿ
Therefore, the value of ((4³ × 5²) ÷ 4⁵) will be
= (4³ × 5²) ÷ 4⁵
= (4³ ÷ 4⁵) × 5²
= 4³⁻⁵ × 5²
= 4⁻² × 5²
= (1 / 4²) × 5²
= (1 / 16) × 25
= 25 / 16

(b) ((2⁴ × 3³) ÷ 2⁶)
We need to find the value of a given expression using laws.
We know that
aᵐ ÷ aⁿ = aᵐ⁻ⁿ
aᵐ × aⁿ = aᵐ⁺ⁿ
Therefore, the value of ((2⁴ × 3³) ÷ 2⁶) will be
= (2⁴ × 3³) ÷ 2⁶
= 2⁴ ÷ 2⁶ × 3³
= 2⁴⁻⁶ × 3³
= 2⁻² × 3³
= (1 / 2²) × 3³
= (1 / 4) × 27
= 27 / 4