An investment grew by 8% in the first year and by 12.5% in the second year, each year’s growth being on the value at the start of that year. If the investment was worth Rs 19440 after two years, what was the sum invested?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Rs 16000
Each year the value is multiplied by its own growth factor, so the final value is the sum times both factors.
Factors: \(1+\frac{8}{100}=1.08\) and \(1+\frac{12.5}{100}=1.125\)
Sum \(\times1.08\times1.125=19440\)
Sum \(=19440\div1.215=16000\)
The sum invested was Rs 16000.
Why the other options are incorrect
Option (1) Rs 16500 — Incorrect. Rs 16500 would grow to \(16500\times1.08\times1.125=20047.5\), not Rs 19440.
Option (2) Rs 17280 — Incorrect. Rs 17280 would grow to \(17280\times1.08\times1.125=20995.2\), not Rs 19440.
Option (4) Rs 15200 — Incorrect. Rs 15200 would grow to \(15200\times1.08\times1.125=18468\), not Rs 19440.