If the amount at the end of the 2nd year and 3rd year on a certain principal at compound interest is ₹12,000 and ₹13,200 respectively, find the rate of interest per annum.

Correct — C.
Interest earned in year 3: 13,200 − 12,000 = ₹1,200
Year 3's interest is charged on the amount at the end of year 2, ₹12,000.
Rate: 1,200 ÷ 12,000 × 100 = 10%

Check: 12,000 × 1.10 = 13,200 → option (c)

Why the others are wrong

• (a)
  8% — 8% of ₹12,000 is ₹960, so year 3 would end at ₹12,960. The amounts given grow by ₹1,200, not ₹960.
• (b)
  7% — 7% adds only ₹840 to ₹12,000, ending year 3 at ₹12,840 instead of the ₹13,200 in the stem.
• (d)
  5% — 5% adds ₹600, half the ₹1,200 by which the amount actually grows between the end of year 2 and the end of year 3.

Under compound interest the interest for any year is charged on the amount at the start of that year, which is the amount at the end of the year before.

So two consecutive yearly amounts give the rate directly: (later − earlier) ÷ earlier × 100. The principal and the earlier years do not enter.

Key facts

• Amount after n years, compounded annually = P × (1 + R⁄100)ⁿ.
• Interest for year n + 1 = R% of the amount at the end of year n.
• Consecutive yearly amounts are in the ratio 1 : (1 + R⁄100), so 12,000 : 13,200 = 1 : 1.1 gives R = 10%.

Common traps

• Dividing ₹1,200 by ₹13,200, the later amount, which gives about 9.09%. The base is the earlier amount.
• Treating ₹1,200 as simple interest on the original sum: P = 12,000 − 2 × 1,200 = ₹9,600 and R = 12.5%, which is not how compound amounts grow.

No directly related past PYQ was found.