If ₹5000 amounts to ₹6050 over 2 years with compound interest, what is the annual interest rate?

Correct — A. Compound amount = principal × (1 + r⁄100)ⁿ.

Growth factor: 6050 ÷ 5000 = 1.21
Over 2 years: (1 + r⁄100)² = 1.21
Square root: 1 + r⁄100 = √1.21 = 1.1

Rate: r = 10% → option (a)

Check: ₹5000 grows to ₹5500 after year 1 and ₹6050 after year 2.

Why the others are wrong

• (b)
  9% — 9% falls short: 5000 × 1.09² = 5000 × 1.1881 = ₹5940.50. The amount must reach ₹6050, so the rate is higher.
• (c)
  8% — 8% falls further short: 5000 × 1.08² = 5000 × 1.1664 = ₹5832, which is ₹218 below the ₹6050 the question gives.
• (d)
  11% — 11% overshoots: 5000 × 1.11² = 5000 × 1.2321 = ₹6160.50, which is ₹110.50 more than ₹6050.

Under compound interest each year's interest joins the principal, so the amount grows by the same factor every year: A = P × (1 + r⁄100)ⁿ.

To find r, divide the amount by the principal for the total growth factor, then take the n-th root. Here 1.21 is a perfect square, 1.1², so the yearly factor is 1.1 and the rate 10%.

Key facts

• Compound amount: A = P × (1 + r⁄100)ⁿ.
• At 10% a year compounded annually, a sum grows by a factor of 1.21 in 2 years.
• Over 2 years, compound interest exceeds simple interest by P × (r⁄100)², which is ₹50 here.

Common traps

• Spreading the ₹1050 interest evenly to get 10.5%, the simple-interest rate. Compounding reaches ₹6050 at a lower rate, 10%.
• Reading 1.21 as 21% a year. That is the growth over both years together.

No directly related past PYQ was found.