A man borrows Rs. 8,000 at 10 percent Compound interest. If he repays Rs. 1,500 at the end of first year and Rs. 3,200 at the end of second year, find the amount of loan outstanding at the beginning of the third year.

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Solution

Amount to be paid at the end of two years will be $= A m o u n t a f t e r 2 y e a r s = 8000 {(1 + \frac{10}{100})}^{2}$ $= 8000 {(1.1)}^{2}$ $= 8000 (1.21)$ $= 9680$ Amount already paid in 2 years is $(3200 + 1500) = 4700$ Therefore amount of loan outstanding at the beginning of third year will be $(9680 - 4700) = 4980$

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A man borrows Rs. 8,000 at 10 percent Compound interest. If he repays Rs. 1,500 at the end of first year and Rs. 3,200 at the end of second year, find the amount of loan outstanding at the beginning of the third year.

Amount to be paid at the end of two years will be =Amountafter2years=8000(1+10100)2 =8000(1.1)2 =8000(1.21) =9680 Amount already paid in 2 years is(3200+1500)=4700 Therefore amount of loan outstanding at the beginning of third year will be(9680−4700)=4980