A town's population was 100,000 and is now 121,000. If the annual increase is 10%, in how many years did this increase occur? MCQ with Answer and Explanation

Question

A town's population was 100,000 and is now 121,000. If the annual increase is 10%, in how many years did this increase occur?

A. 2

B. 1

C. 4

D. 3

Answer: Option A

Solution (By JKSSB Mock Tests)

121000 = 100000 * (1.10)^t. 1.21 = (1.10)^t. Since 1.10^2 = 1.21, t = 2 years.

This question belongs to: Maths Percentage

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Question #1

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A man gave 35% of money to wife, 40% of remainder to son, rest to daughter. Daughter got Rs. 3900. Total money:

A. 8000

B. 9000

C. 12000

D. 10000

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Correct Answer: Option D

Explanation:
Wife 35%, left 65%, son 40% of 65% = 26%, left 39% = 3900 → total = 10000.

This question belongs to: Maths Percentage

Question #2

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Price decreased by 10% then increased by 10%. Net change is:

A. 1% increase

B. 2% decrease

C. 1% decrease

D. No change

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Correct Answer: Option C

Explanation:
0.9 × 1.1 = 0.99, so 1% decrease.

This question belongs to: Maths Percentage

Question #3

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In a town, 70% of people speak English, 65% speak Hindi, and 27% speak neither. What percentage of people speak both languages?

A. 60%

B. 70%

C. 62%

D. 52%

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Correct Answer: Option C

Explanation:
Speak at least one = 100% - 27% = 73%. n(E U H) = n(E) + n(H) - n(E ∩ H). 73 = 70 + 65 - n(E ∩ H). n(E ∩ H) = 135 - 73 = 62%.

This question belongs to: Maths Percentage

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