Time: 1 hour
M.M. 30
Attempt all questions.
Q1. Which of the following is equal to 2³ × 2⁴? (1 Mark)
a) 2⁷
b) 2¹²
c) 2⁸¹
d) 2¹
Answer: a) 2⁷
Same base → add exponents → 2³ × 2⁴ = 2³⁺⁴ = 2⁷
Q2. The value of (3⁴) ÷ (3²) is: (1 Mark)
a) 3⁸
b) 3²
c) 3⁶
d) 3¹
Answer: b) 3²
Same base → subtract exponents → 3⁴ ÷ 3² = 3⁴⁻² = 3²
Q3. Which of the following is NOT correct? (1 Mark)
a) am × an = am+n
b) (am)n = amn
c) am ÷ an = am+n
d) a0 = 1
Answer: c) am ÷ an = am+n
The rule is am ÷ an = am-n, not m+n
Q4. The value of 2⁻³ is: (1 Mark)
a) -8
b) 8
c) 1/8
d) 1/-8
Answer: c) 1/8
Negative exponent rule → a⁻ⁿ = 1/aⁿ
Q5. If 5ˣ = 125, then the value of x is: (1 Mark)
a) 2
b) 3
c) 4
d) 5
Answer: b) 3
125 = 5³, so on comparing powers x = 3.
Q6: Express 4-3 as a power with base 2. (2 Marks)
Solution:
4-3 can be written as:
4-3 = (22)-3
Now, by using exponential law i.e. (am)n = amn
4-3 = 2-6 (which is in base 2 form).
Q7: Express 0.00000000837 in standard form. (2 Marks)
Solution:
0.00000000837
= 0.00000000837 x 109 / 109
= 8.37 ×10-9
Q8: Simplify the expression: (4² × 4⁻³). (2 Marks)
Solution: Using the product rule aᵐ × aⁿ = aᵐ⁺ⁿ
Now, by applying the rule in given expression 4² × 4⁻³ we get,
= 4²⁺⁽⁻³⁾
= 4⁻¹Now the outcome is in the form of negative exponent rule a⁻ⁿ = 1/aⁿ.
∴ 4⁻¹ = 1/4¹Therefore, the simplified expression is 1/4
Q9: Solve (3 Marks)
Solution:
Given:On solving:
Q10: Find value of (3 Marks)
Solution: (i)
(ii)
Q11: Express the following numbers in standard form. (3 Marks)
(i) 0.0000000015
(ii) 0.00000001425
(iii) 102000000000000000
Solution:
(i) 0.0000000015 = 1.5 ×10-9
(ii) 0.00000001425 = 1.425×10-8
(iii) 102000000000000000 = 1.02×1017
Q12: Simplify the following (5 Marks)
Solution:
Q13: The number of books in a library increases by 5 times every 2 years. A library starts with 100 books. How many books will there be in the library after:
(a) 6 years
(b) 10 years
Solution: Initial number of books = 100
The number of books increases by a factor of 5 every 2 years.
(a) After 6 years (3 periods of 2 years):
Number of books after 6 years = 100 × 5³ = 100 × 125 = 12500 books
- After 6 years: 12500 books
(b) After 10 years (5 periods of 2 years):
Number of books after 10 years = 100 × 5⁵ = 100 × 3125 = 312500 books
After 10 years: 312500 books
21 videos|177 docs|11 tests
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