A man borrows Rs.6500 at 10 % per annum compound interest payable… - Vidyadip

Question

A man borrows $Rs.6500$ at $10\%$ per annum compound interest payable half$-$yearly. He repays $Rs.2000$ at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest rupee.

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Answer

For $1^{\text {st }}$ half year: $P=R s, 6500, R=10 \%$ and $T=\frac{1}{2}$ year
Interest $=\text { Rs. } \frac{6500 \times 10 \times 1}{100 \times 2}$
$=\text { Rs. } 325$
Amount
$=\text { Rs. } 6500+\text { Rs. } 325$
$=\text { Rs. } 6825$
Money paid at the end of $1^{st}$ half year $= Rs. 2000$
Balance money for $2^{\text {nd }}$ half year
$=\text { Rs. } 6825-\text { Rs. } 2000$
$=\text { Rs. } 4825$
For $2^{\text {nd }}$ half year : $P=$ Rs. $4825 ; R=10 \%$ and $T=\frac{1}{2}$ year
$\text { Interest }=\text { Rs. } \frac{4825 \times 10 \times 1}{100 \times 2}$
$=\text { RS. } 241.25$
Amount
$=\text { Rs. } 4825+\text { Rs. } 241.25$
$=\text { Rs. } 5066.25$
Money paid at the end of $2^{\text {nd }}$ half year $=Rs. 2000$
Balance money for $3^{\text {rd }}$ half year
$=\text { Rs. } 5066.25-\text { Rs. } 2000$
$=\text { Rs. } 3066.25$
For $3^{\text {rd }}$ half year: $P=R s .3066 .25 ; R=10 \%$ and $T=\frac{1}{2}$ year
$\text { Interest }=\text { Rs. } \frac{3066.25 \times 10 \times 1}{100 \times 2}$
$=\text { Rs. } 153.3125$
Amount
$=\text { Rs. } 3066.25+\text { Rs. } 153.3125$
$=\text { Rs. } 3219.5625$
Money paid at the end of $3^{\text {rd }}$ half year $= Rs. 2000$
Amount outstanding at the end of $3^{\text {rd }}$ payment
$=\text { Rs. } 3219.5625-\text { Rs. } 2000$
$=\text { Rs. } 1219.5625$
$=\text { Rs. } 1220($nearest rupee$)$

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