Simple Interest Questions Answers

  • 22. A lent Rs. 5000 to B for 2 years and Rs 3000 to C for 4 years on simple interest at the same rate of interest and received Rs 2200 in all from both of them as interest. The rate of interest per annum is

    1. 9%
    2. 10%
    3. 11%
    4. 12%
    Answer And Explanation

    Answer: Option B

    Explanation:

    Let R% be the rate of simple interest then,

    from question we can conclude that

    \begin{aligned}
    (\frac{5000*R*2}{100}) + (\frac{3000*R*4}{100}) = 2200 \\

    <=> 100R + 120R = 2200 \\
    <=> R = 10\%


    \end{aligned}

  • 23. We have total amount Rs. 2379, now divide this amount in three parts so that their sum become equal after 2, 3 and 4 years respectively. If rate of interest is 5% per annum then first part will be ?

    1. 818
    2. 828
    3. 838
    4. 848
    Answer And Explanation

    Answer: Option B

    Explanation:

    Lets assume that three parts are x, y and z.
    Simple Interest, R = 5%

    From question we can conclude that, x + interest (on x) for 2 years = y + interest (on y) for 3 years = z + interest (on y) for 4 years

    \begin{aligned}
    \left( x + \frac{x*5*2}{100} \right) = \left( y + \frac{y*5*3}{100} \right) = \left( z + \frac{z*5*4}{100} \right)\\
    \left( x + \frac{x}{10} \right) = \left( y + \frac{3y}{20} \right) = \left( z + \frac{z}{5} \right) \\
    => \frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\

    \text{lets assume k = }\frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\
    \text{then }x = \frac{10k}{11} \\
    y = \frac{20k}{23}\\
    z = \frac{5k}{6}\\
    \text{we know x+y+z = 2379}\\
    => \frac{10k}{11} + \frac{20k}{23} + \frac{5k}{6} = 2379\\
    \text{10k*23*6+20k*11*6+5k*11*23=2379*11*23*6}\\
    \text{1380k+1320k+1265k=2379*11*23*6}\\
    \text{3965k=2379*11*23*6}\\
    k = \frac{2379*11*23*6}{3965}\\
    \text{by putting value of k we can get x} \\
    x = \frac{10k}{11} \\
    =>x = \frac{10}{11}*\frac{2379*11*23*6}{3965}\\
    =>x = \frac{10*2379*23*6}{3965}\\
    = \frac{2*2379*23*6}{793}\\
    = 2 * 3 * 23 * 6 = 828
    \end{aligned}