If the simple interest on ₹7,200 in 3 years at the rate of 16% per annum equals the simple interest on ₹9,600 at the rate of x% per annum in 4 years. The value of x is equal to:

Correct — A. Work out the first interest, set the second equal to it, then solve.

SI = P × R × T ⁄ 100
First: 7200 × 16 × 3 ⁄ 100 = ₹3,456
Second: 9600 × x × 4 ⁄ 100 = 384x

384x = 3456
x = 3456 ⁄ 384 = 9 → option (a)

Faster, since the 100 cancels on both sides: 7200 × 16 × 3 = 9600 × 4 × x, so x = 345600 ⁄ 38400 = 9.

Why the others are wrong

• (b)
  10 — 384 × 10 = ₹3,840, which overshoots the ₹3,456 the first loan earns. A round 10% only looks right because the numbers are large.
• (c)
  11 — 384 × 11 = ₹4,224, well above ₹3,456. The second sum is bigger and runs longer, so its rate has to be well below 16%, not close to it.
• (d)
  8 — 384 × 8 = ₹3,072, which is ₹384 short. 3456 ÷ 384 = 9 exactly, so there is no rounding here that could justify 8.

Simple interest never earns on itself. Each year the same principal produces the same interest, so SI = P × R × T ⁄ 100 with no compounding anywhere in it.

When two simple interests are set equal, the 100 cancels from both sides and the whole question reduces to P₁R₁T₁ = P₂R₂T₂. Rate, principal and time trade off against each other directly: double the principal and the rate needed halves.

Here the second principal is 4⁄3 of the first and runs 4⁄3 as long, so its rate must be (3⁄4) × (3⁄4) × 16 = 9%.

Key facts

• Simple interest = P × R × T / 100, and the principal never changes.
• Two equal simple interests satisfy P₁R₁T₁ = P₂R₂T₂ once the 100 cancels.
• Here 7200 × 16 × 3 = 345,600 and 9600 × 4 = 38,400, so x = 9.
• Amount = principal + simple interest, which is a different quantity from the interest.

Common traps

• Equating the two amounts instead of the two interests
• Swapping the two time periods, using 4 years on ₹7,200 and 3 on ₹9,600
• Cancelling 7200 against 9600 to 3 : 4 and then forgetting to carry the 16

No directly related past PYQ was found.