A sum becomes ₹6,600 in 2 years and ₹7,920 in 3 years at compound interest. What is the original principal?

Correct — C. Consecutive compound amounts differ by one year's interest on the earlier amount.

Interest in year 3 = 7,920 − 6,600 = ₹1,320
Rate = 1,320 ÷ 6,600 = 0.2 = 20%
Growth over 2 years = 1.2² = 1.44

Principal = 6,600 ÷ 1.44 = ₹4,583.33 → option (c).

Check forward: 4,583.33 → 5,500 → 6,600 → 7,920, each step × 1.2.

Why the others are wrong

• (a)
  ₹4,000.33 — Grown two years at 20%, ₹4,000.33 × 1.44 ≈ ₹5,760 — well short of the ₹6,600 the stem gives for the end of year 2.
• (b)
  ₹5,583.33 — ₹5,583.33 × 1.44 ≈ ₹8,040 after two years, not ₹6,600. Undo both years of growth: one year back from ₹6,600 is only ₹5,500, the amount at the end of year 1.
• (d)
  ₹6,583.33 — ₹6,583.33 is barely below the year-2 amount of ₹6,600, which would mean almost no interest in two years — yet year 3 alone earns ₹1,320. At 20% it grows to about ₹9,480.

When a compound-interest question gives amounts for two consecutive years, the rate falls out without the principal.

The later amount is the earlier one times (1 + r), so r = (A₃ − A₂) ÷ A₂. The ₹1,320 difference is interest on ₹6,600, not on the principal.

With r known, divide the year-2 amount by (1 + r)² to undo two years of growth and recover P.

Key facts

• Compound amount after n years = P(1 + r)ⁿ.
• The rate equals the difference of consecutive amounts divided by the earlier amount: 1,320 ÷ 6,600 = 20%.
• At 20% a year, two years of growth multiply a sum by 1.44.

Common traps

• Dividing the ₹1,320 difference by the principal instead of by ₹6,600 — year-3 interest is earned on the year-2 amount.
• Discounting ₹6,600 by one year only, which gives ₹5,500, the amount after year 1.
• Subtracting 2 × ₹1,320 from ₹6,600 as if the interest were simple, which gives ₹3,960.

No directly related past PYQ was found.