If the compound interest on a certain sum at 16⅔% per annum for 3 years is ₹1270, find the simple interest on the same sum at the same rate and for the same period.

Correct — A. Write the rate as a fraction: 16⅔% = 1⁄6.

3-year growth factor = (7⁄6)³ = 343⁄216
CI factor = (343 − 216) ⁄ 216 = 127⁄216
127⁄216 × P = 1,270, so P = 10 × 216 = ₹2,160

SI = P × 3 × 1⁄6 = 2,160 ÷ 2
= ₹1,080 → option (a)

Why the others are wrong

• (b)
  ₹1000 — Here SI is half the principal, so ₹1,000 needs P = ₹2,000. CI on ₹2,000 is 2,000 × 127⁄216 ≈ ₹1,176, not ₹1,270.
• (c)
  ₹1500 — Over 3 years at the same rate, CI exceeds SI because interest also earns interest. ₹1,500 is above the ₹1,270 CI, so it cannot be the SI.
• (d)
  ₹2000 — It sits above the ₹1,270 CI, which SI over 3 years cannot do. It would also need P = ₹4,000, whose CI is about ₹2,352.

Rates like 16⅔%, 12½% and 33⅓% are fractions in disguise: 1⁄6, 1⁄8 and 1⁄3. Written that way, 3-year growth is a clean cube: (1 + 1⁄6)³ = 7³ ⁄ 6³ = 343⁄216.

CI is that cube minus one, 127⁄216 of the principal. SI for 3 years is 3 × 1⁄6 = 1⁄2 of the principal, or 108⁄216.

So CI : SI = 127 : 108, and SI = 1,270 × 108⁄127 = ₹1,080 in a single line.

Key facts

• 16⅔% = 1⁄6, 12½% = 1⁄8, 33⅓% = 1⁄3 and 8⅓% = 1⁄12.
• 7³ = 343 and 6³ = 216.
• For the same principal and rate over 3 years, CI : SI = [(1 + r)³ − 1] : 3r, with r as a fraction.
• SI = P × R × T ⁄ 100.

Common traps

• Forgetting that CI beats SI over more than one year: that check alone removes ₹1,500 and ₹2,000.
• Cubing 1.1667 instead of 7⁄6: it reaches ₹2,160, but the decimals cost time the fraction does not.

No directly related past PYQ was found.