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Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is

Solution

✅ Correct Option: 2

Anil: Rs 22000 for 6 years at 4% compounded half-yearly

Sunil: Unknown amount for 5 years at 4% compounded half-yearly, then reinvest everything for 1 year at 10% simple interest. Both end up with equal amounts after 6 years


Rate adjustment: 4% per year → 2% per half-year period

Time adjustment: 6 years → 12 half-year periods

Using compound interest formula: A=P(1+r100)nA = P(1 + \frac{r}{100})^n

Where:

P = Rs 22000 (principal)

r = 2% (rate per half-year)

n = 12 (number of half-year periods)

Anil's final amount = 22000×(1+2100)1222000 \times (1 + \frac{2}{100})^{12}

= 22000×(1.02)1222000 \times (1.02)^{12}


Finding Sunil's final amount:

Let Sunil's initial investment = P

Phase 1 (Years 1-5): Compound Interest

Rate = 2% per half-year

Time = 10 half-year periods

Amount after 5 years = P×(1.02)10P \times (1.02)^{10}

Phase 2 (Year 6): Simple Interest

Principal for year 6 = P×(1.02)10P \times (1.02)^{10}

Rate = 10% per year

Time = 1 year

Using simple interest formula: A=P(1+rt100)A = P(1 + \frac{rt}{100})

Final amount = P×(1.02)10×(1+10×1100)P \times (1.02)^{10} \times (1 + \frac{10 \times 1}{100})

= P×(1.02)10×(1.1)P \times (1.02)^{10} \times (1.1)


Since both final amounts are equal:

22000×(1.02)12=P×(1.02)10×(1.1)22000 \times (1.02)^{12} = P \times (1.02)^{10} \times (1.1)

22000×(1.02)12(1.02)10=P×(1.02)10×(1.1)(1.02)10\frac{22000 \times (1.02)^{12}}{(1.02)^{10}} = \frac{P \times (1.02)^{10} \times (1.1)}{(1.02)^{10}}

22000×(1.02)2=P×(1.1)22000 \times (1.02)^2 = P \times (1.1)


(1.02)2=1.02×1.02=1.0404(1.02)^2 = 1.02 \times 1.02 = 1.0404

22000×1.0404=P×1.122000 \times 1.0404 = P \times 1.1

22888.8=P×1.122888.8 = P \times 1.1

Therefore: P=22888.81.1=20808P = \frac{22888.8}{1.1} = 20808

Answer: Sunil's initial investment = Rs 20808

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