Question

.Mr ram borrows
rupees 20000 for 2 years compounded annually the rate of interest for the two successive your are 9% and 10% respectively if we replace rupees 1200 at the end of first year 2016 at the end of second year find the amount outstanding at the beginning of the third year ​

Answer 1

Mr Ram borrows Rs. 20000 for 2 years compounded annually. The rate of interest for the two successive years are 9% and 10% respectively. If he repays Rs. 1200 at the end of first year and Rs. 2016 at the end of second year, find the amount outstanding at the beginning of third year.

P = 20000 , $r_{1}$ = 9% , $r_{2}$ = 10% , n = 1 (as it is compounded annually)

Amount at the end of first year, $A_{1}$ = $P (1 + \frac{r_{1} }{100} )^{n}$ = $20000 (1 + \frac{9}{100} )^1 = Rs. 21800$

Amount repaid = Rs. 1200

Amount at the end of first year = 21800 - 1200 = Rs. 20600

Amount at the end of second year, $A_{2}$ = $20600 ( 1 + \frac{10}{100} )^1 = Rs. 22660$

But amount paid = Rs. 2016

Hence, amount outstanding at the beginning of third year

= Rs. (22660 - 2016)

= Rs. 20644

Answer 2

Given : Mr.Ram borrows Rs20000 compounded annually. rate of interest for two successive years are 9% and 10% respectively.

To find :  outstanding Amount at the beginning of the third year ​

Solution:

P = 20000

R = 9 %

Interest for 1st year = 20000 * 9 * 1 /100 = Rs 1800

Amount Paid = Rs 1200

Amount remain  after 1 year = 20000 + 1800 - 1200

= 20600 Rs

Interest for 2nd Year = 20600 * 10 * 1 /100

= Rs 2060

Paid at 2nd year  = Rs 2016

Amount Balance after 2 year = 20600 + 2060 - 2016

= Rs 20644

outstanding Amount at the beginning of the third year ​ = Rs 20644

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