In how much time (in years, rounded off to 2 decimal places) will ₹10,800 amount to ₹24,800 if invested at simple interest at the rate of 15.5% per annum?

Correct — D. The interest is the amount minus the principal:
SI = 24,800 − 10,800 = ₹14,000

Rearrange SI = P × R × T ⁄ 100:
T = (SI × 100) ⁄ (P × R)
= (14,000 × 100) ⁄ (10,800 × 15.5)

Denominator: 10,800 × 15.5 = 1,67,400
T = 14,00,000 ⁄ 1,67,400 = 8.3632…

Rounded to two decimal places, T = 8.36 years → option (d).

Why the others are wrong

• (a)
  7.72 — The interest is ₹1,674 a year (15.5% of ₹10,800). In 7.72 years that is ₹12,923, leaving the amount at ₹23,723, short of ₹24,800.
• (b)
  9.54 — 9.54 years earns ₹15,970 of interest, pushing the amount to about ₹26,770 — well past the ₹24,800 given.
• (c)
  6.89 — 6.89 years earns only ₹11,534, an amount of ₹22,334. Multiplying any option by ₹1,674 and adding the principal settles it in seconds.

Simple interest is linear: the same rupee figure is added every year, calculated on the original principal alone.

Here that figure is 15.5% of ₹10,800, which is ₹1,674 a year. The question is then just how many of those it takes to build ₹14,000 of interest.

The formula SI = PRT⁄100 rearranges to T = 100 × SI ⁄ (PR). Start by separating interest from amount: the ₹24,800 in the stem is P + SI, not SI.

Key facts

• SI = P × R × T ⁄ 100, and amount A = P + SI.
• 15.5% of ₹10,800 is ₹1,674, the interest added in each year under simple interest.
• T = (SI × 100) ⁄ (P × R) = 14,00,000 ⁄ 1,67,400 = 8.3632 years.
• Under simple interest the yearly addition never changes, unlike compound interest where it grows.

Common traps

• Feeding the amount ₹24,800 into the formula in place of the interest ₹14,000
• Rounding 8.3632 to 8.4 when two decimal places are asked for
• Treating the growth as compound, which would need less time than 8.36 years

No directly related past PYQ was found.