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Anil invests Rs. 2200022000 for 66 years in a certain scheme with 4%4 \% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 55 years, and then reinvests the entire amount received at the end of 55 years for one year at 10%10 \% simple interest. If the amounts received by both at the end of 66 years are same, then the initial investment made by Sunil, in rupees, is

Entered answer:

Solution

✅ Correct Answer: 20808

We have Anil's Investment:

Amount: Rs. 22000

Duration: 6 years

Interest: 4% per annum, compounded half-yearly

We have Sunil's Investment:

Amount: Unknown (we call it 'x')

First 5 years: 4% per annum, compounded half-yearly

Year 6: Reinvests everything at 10% simple interest

Both receive the same final amount after 6 years.


When interest is compounded half-yearly, we adjust both the rate and time:

Rate becomes: 4% ÷ 2 = 2% per half-year

Time becomes: 6 years × 2 = 12 half-years

Formula for Compound Interest: A=P(1+r100)nA = P(1 + \tfrac{r}{100})^n

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

= 22000(1.02)1222000(1.02)^{12}


Phase 1 (First 5 years): Compound Interest

Rate per half-year = 2%

Number of half-years = 5 × 2 = 10

Amount after 5 years = x(1.02)10x(1.02)^{10}

Phase 2 (6th year): Simple Interest

Principal for 6th year = x(1.02)10x(1.02)^{10} (the entire amount from Phase 1)

Rate = 10% per annum

Time = 1 year

Simple Interest Formula: SI=P×R×T100SI = \tfrac{P \times R \times T}{100}

Simple Interest earned = x(1.02)10×10×1100=0.1×x(1.02)10\tfrac{x(1.02)^{10} \times 10 \times 1}{100} = 0.1 \times x(1.02)^{10}

Final Amount for Sunil:

Amount after 6 years = x(1.02)10+0.1×x(1.02)10x(1.02)^{10} + 0.1 \times x(1.02)^{10}

= x(1.02)10(1+0.1)x(1.02)^{10}(1 + 0.1)

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


Since both receive the same amount:

22000(1.02)12=x(1.02)10×1.122000(1.02)^{12} = x(1.02)^{10} \times 1.1


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

Using the Law of Exponents: (1.02)12÷(1.02)10=(1.02)12−10=(1.02)2(1.02)^{12} ÷ (1.02)^{10} = (1.02)^{12-10} = (1.02)^2

x=22000(1.02)21.1x = \dfrac{22000(1.02)^2}{1.1}

We calculate (1.02)2(1.02)^2:

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

x=22000×1.04041.1x = \dfrac{22000 \times 1.0404}{1.1}

x=22888.81.1x = \dfrac{22888.8}{1.1}

x=20808x = 20808


Sunil's initial investment = Rs. 20808

Half-yearly compounding means dividing the annual rate by 2 and multiplying the time by 2. When solving such problems, we always set up equations based on the condition that final amounts are equal. The Law of exponents helps simplify calculations: am÷an=am−na^m ÷ a^n = a^{m-n}.

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