In how many years will ₹50,000 become ₹66,550 at 10% compound interest per annum?
- (a)4
- (b)5
- (c)6
- (d)3
Answer
Why
Correct — D. Compare the growth factor with powers of 1.1.
Amount ÷ principal: 66,550 ÷ 50,000 = 1.331
Powers of 1.1: 1.1² = 1.21, 1.1³ = 1.331
So n = 3 years → option (d)
Year by year: ₹50,000 → ₹55,000 → ₹60,500 → ₹66,550
Why the others are wrong
- (a)4 — 4 years would give 50,000 × 1.1⁴ = 50,000 × 1.4641 = ₹73,205, well past ₹66,550.
- (b)5 — 5 years gives 50,000 × 1.61051 = ₹80,525.50. The sum reaches ₹66,550 two years before that.
- (c)6 — 6 years gives 50,000 × 1.771561 ≈ ₹88,578, far above ₹66,550.
Concept
Amount = P × (1 + r⁄100)ⁿ. At 10% a year the factor is 1.1 for each year, so after n years ₹50,000 becomes 50,000 × 1.1ⁿ.
Divide the amount by the principal to isolate the factor: 1.1ⁿ = 1.331. Knowing 1.1² = 1.21 and 1.1³ = 1.331 by sight settles n without logarithms.
Simple interest at 10% adds ₹5,000 a year and reaches only ₹65,000 in 3 years. The extra ₹1,550 is interest on earlier interest: ₹500 in year 2 and ₹1,050 in year 3.
Key facts
- 1.1² = 1.21, 1.1³ = 1.331, 1.1⁴ = 1.4641.
- At 10% compound interest, each year's amount is the previous year's × 1.1.
- On ₹50,000 at 10% for 3 years, compound interest exceeds simple interest by ₹1,550.
Study next
Common traps
- Using simple interest: ₹16,550 at ₹5,000 a year gives 3.31 years, not a whole number.
- Miscounting powers: 1.331 is 1.1³, while 1.1⁴ is 1.4641.
Compound growth read as a power also decides 12 Sep 2025, 09:00, Quant Q.10: a sum that doubles in 5 years becomes 8 times itself, 2³, in 15 years.
14 Sep 2025, 09:00, Quant Q.10 runs it backwards: ₹3,136 ÷ 1.12² = ₹2,500 invested, and 140% of that is ₹3,500.
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