A person borrowed some money on simple interest. After 4 years, he returned 9⁄5 of the money to lender. What was the rate of interest?

Correct — D. The 9⁄5 is the amount handed back, so the interest is what is left after the principal is taken out.

Let the principal be P.
Amount = 9P⁄5, so SI = 9P⁄5 − P = 4P⁄5

Now use SI = P × R × T ⁄ 100 with T = 4:
4P⁄5 = P × R × 4 ⁄ 100
4⁄5 = 4R⁄100 → R = 20 → option (d)

Check: 20% for 4 years is 80% of P, and P + 0.8P = 1.8P = 9P⁄5.

Why the others are wrong

• (a)
  19% p.a. — At 19% the four-year interest is 76% of P, so the borrower would return 1.76P. The stem needs 1.8P, because 9⁄5 = 1.8.
• (b)
  18% p.a. — At 18% the interest over 4 years is 72% of P and the amount returned is 1.72P — short of 9P⁄5.
• (c)
  17% p.a. — At 17% the amount comes to 1.68P. Only R = 20 makes 4R⁄100 equal the required 4⁄5.

Simple interest keeps the principal fixed: SI = PRT⁄100, and the amount is A = P + SI.

Most simple-interest questions that look hard are really about which of those two the stem hands you. Returning 9⁄5 of the money is an amount, so the interest is 9⁄5 − 1 = 4⁄5 of the principal.

Had the stem said the interest paid was 9⁄5 of the money, the interest itself would be 9P⁄5 and the rate would work out at 45%.

The principal cancels either way, which is why no rupee figure is given and none is needed.

Key facts

• SI = PRT⁄100, and the amount A = P + SI — simple interest never earns interest on itself.
• If a sum grows to k times itself in T years at simple interest, then R = 100(k − 1)⁄T.
• Here k = 9⁄5 and T = 4, so R = 100 × (4⁄5) ⁄ 4 = 20.

Common traps

• Reading 9⁄5 of the money as the interest rather than the amount, which yields 45%.
• Dropping the 4 years and treating the 4⁄5 interest as a one-year rate of 80%.
• Hunting for a principal to plug in when none is given and none is required.

No directly related past PYQ was found.