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    Subject

    Quantitative Aptitude

    Topic

    Interest

    The simple interest accrued on a certain principal is rupees 6,00,000 in 5 years at a rate of 12% per annum. What would be the compound interest accrued on that principal at a rate of 3% per annum in 2 years?

    ক)
    60900
    খ)
    25830
    গ)
    26000
    ঘ)
    37230
    ঙ)
    32450

    Explanation

    Simple interest accrued on a certain principal is rupees 6,00,000 in 5 years at a rate of 12% per annum. We know that, S.I. = (P × R × T)/100    where, P = principal, R = rate % per annum, T = time in years Here, S.I. = 6,00,000, R = 12%, T = 5 years ⇒ 6,00,000 = (P × 12 × 5)/100 ⇒  P  =  6,00,000×1005×12  =  10,00,000\Rightarrow {\rm{\;P\;}} = {\rm{\;}}\frac{{6,00,000 \times 100}}{{5 \times 12}}\; = {\rm{\;}}10,00,000 Now let’s find the compound interest on principal amount P = 10,00,000, R = 3%, T = 2 years Compound interest =  P(1+r100)n  −  P= {\rm{\;}}P{\left( {1 + \frac{r}{{100}}} \right)^{n\;}}-\;P Compound interest =P[(1+r100)n−1]=1000000[(1+3100)2−1]=10,00,000[(1+3100)2−1]=  1000000[(103100)2−1]=  10,00,000[1060910000−1]=  1000000[60910000]=60900\begin{array}{l} = {\rm{P}}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] = 1000000\left[ {{{\left( {1 + \frac{3}{{100}}} \right)}^2} - 1} \right]\\ = 10,00,000\left[ {{{\left( {1 + \frac{3}{{100}}} \right)}^2} - 1} \right] = \;1000000\left[ {{{\left( {\frac{{103}}{{100}}} \right)}^2} - 1} \right]\\ = {\rm{\;}}10,00,000\left[ {\frac{{10609}}{{10000}} - 1} \right] = \;1000000\left[ {\frac{{609}}{{10000}}} \right] = 60900 \end{array}

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