The compound interest on a certain sum for 3 years at 20% per annum is ₹864. Find the simple interest on the same sum at the same rate and period.
- (a)₹612.54
- (b)₹618.79
- (c)₹712.09
- (d)₹512.45
Answer
Why
Correct — C.
CI for 3 years = P × (1.2³ − 1)
1.2³ = 1.728, so CI = 0.728P
0.728P = 864
P = 864 ⁄ 0.728 = ₹1186.81
SI = P × 20% × 3 = 0.6 × 1186.81
= ₹712.09 → option (c).
Why the others are wrong
- (a)₹612.54 — Back-solve: an SI of ₹612.54 means P = 612.54 ⁄ 0.6 = ₹1020.90, whose 3-year CI is 0.728 × 1020.90 = ₹743.22, not ₹864.
- (b)₹618.79 — An SI of ₹618.79 means P = 618.79 ⁄ 0.6 = ₹1031.32, and that sum earns 0.728 × 1031.32 = ₹750.80 of CI, not ₹864.
- (d)₹512.45 — An SI of ₹512.45 means P = 512.45 ⁄ 0.6 = ₹854.08, which earns only ₹621.77 of CI. The ₹864 of CI needs P = ₹1186.81.
Concept
For the same sum, rate and time, CI exceeds SI by the interest earned on interest. At 20% for 3 years, CI is 1.2³ − 1 = 0.728 of the sum and SI is 3 × 0.2 = 0.6 of it.
So SI : CI = 0.6 : 0.728 = 75 : 91, and SI = 864 × 75⁄91 = ₹712.09 without finding P at all.
The sum works out to ₹108000⁄91 ≈ ₹1186.81, not a round figure, which is why every option carries paise. Keep the fraction, or at least two decimals, until the last step.
Key facts
- CI = P[(1 + r)ⁿ − 1] and SI = P × r × n.
- At 20% for 3 years, CI = 0.728P and SI = 0.6P.
- For 3 years, CI − SI = P × r² × (3 + r), here 1186.81 × 0.04 × 3.2 = ₹151.91.
- Year by year on ₹1186.81 the CI is ₹237.36, ₹284.84 and ₹341.80, which add to ₹864.00.
Study next
Common traps
- Treating ₹864 as the amount rather than the interest. 864 ⁄ 1.728 = ₹500 then passes for the principal, and its SI of ₹300 is not offered.
- Rounding P to ₹1187 before the last step. 0.6 × 1187 = ₹712.20, which misses the printed ₹712.09.
The same CI-to-SI conversion over 3 years is keyed at 13 Sep 2025, 12:30, Quant Q.11 (16⅔%, CI ₹1270 → SI ₹1080). There the rate is 1⁄6 and the sum comes out whole, ₹2160.
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