If ₹1,875 becomes ₹2,625 in 4 years, what will ₹24,000 become at the end of 9 years at the same rate of interest, under simple interest?

Correct — C. First recover the rate from the pair you are given.

Interest = 2625 − 1875 = ₹750 over 4 years
R = (750 × 100) ⁄ (1875 × 4) = 75000 ⁄ 7500 = 10% per annum

Now apply 10% to ₹24,000 for 9 years:
SI = (24000 × 10 × 9) ⁄ 100 = ₹21,600
Amount = 24,000 + 21,600 = ₹45,600 → option (c).

Why the others are wrong

• (a)
  ₹43,200 — ₹43,200 is the amount after 8 years, not 9 — interest of ₹19,200 on ₹24,000 at 10%. The term in the question is nine years.
• (b)
  ₹21,600 — ₹21,600 is the interest by itself. The question asks what ₹24,000 becomes, so the principal has to be added back to it.
• (d)
  ₹34,800 — ₹34,800 implies interest of ₹10,800, which is 45% of ₹24,000 and so a rate of 5% a year. The first sum fixes the rate at 10%, since ₹750 on ₹1,875 is 40% over four years.

Simple interest is linear: SI = P × R × T ⁄ 100 and Amount = P + SI. The growth factor 1 + RT⁄100 does not depend on the principal, so a rate learnt from one sum transfers straight to another.

That is what makes this a two-stage question. Stage one reads the rate out of ₹1,875 becoming ₹2,625 in four years. Stage two applies that rate to ₹24,000 for nine years.

The shortcut version skips the rate formula: the factor is 1.4 over four years, so 0.1 a year, so 1.9 over nine years, and 24,000 × 1.9 = 45,600.

Key facts

• SI = P × R × T ⁄ 100, and the amount is the principal plus that interest.
• Under simple interest the growth factor 1 + RT⁄100 is identical for every principal.
• ₹1,875 growing to ₹2,625 is a factor of 1.4, so RT = 40 and the rate is 10% over four years.
• At 10% for nine years the factor is 1.9, so ₹24,000 becomes ₹45,600.

Common traps

• Answering the interest ₹21,600 instead of the amount.
• Carrying the four-year term over to the second sum.
• Treating the growth as compound, which would take ₹24,000 past ₹56,000 in nine years.

No directly related past PYQ was found.