Maths-
General
Easy

Question

Ranbir borrows  Rs20,000 at 12% per annum compound interest. If he repays  Rs8400 at the end of the first year and  Rs9680 at the end of the second year. Find the amount of loan outstanding at the beginning of the third year.

Hint:

Find the interest every year, add it with the principal amount to get total amount and from that subtract the repaid amount every year.

The correct answer is: Rs 6000


    Complete step by step solution:
    Money borrowed by Ranbir at 12% compound interest, that is  = Rs 20000
    For first year,
    We know that compound interest = Total amount - principal amount that is, CI = A - P …(i)
    Here we have A equals P open parentheses 1 plus R over 100 close parentheses to the power of T…(ii)
    Here, we have T = 1 years, P = 20000 , R = 12% and A = ?
    On substituting the known values in (ii), we get A equals 20000 open parentheses 1 plus 12 over 100 close parentheses to the power of blank
    On substituting the known values in (i), we get C I equals 20000 open parentheses 1 plus 12 over 100 close parentheses to the power of blank minus space 20000
    not stretchy rightwards double arrow C I equals 20000 cross times 28 over 25 minus 20000
    not stretchy rightwards double arrow C I equals 2400 Rs
    Thus total amount after 1 year = 20000+2400=22400 Rupees.
    Money repaid = Rs 8400
    ∴ Balance = 22400 - 8400 = 14000 Rupees.
    For second year,
    Here, we have T = 1 years, P = 14000 , R = 12% and A = ?
    On substituting the known values in (ii), we get A equals 14000 open parentheses 1 plus 12 over 100 close parentheses
    On substituting the known values in (i), we get C I equals 14000 open parentheses 1 plus 12 over 100 close parentheses minus 14000
    not stretchy rightwards double arrow C I equals 1680 Rs
    Thus total amount after 1 year = 14000+1680=15680 Rupees.
    Money paid at the end of second year by Ranbir = Rs 9680
    ∴ Loan at the beginning of third year = 15680 - 9680 = Rs 6000

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