Question

a man borrows rupees 15000 at 12% per annum compounded annually. if he paid rupees 4400 at the end of each year find the amount outstanding against him at the beginning of the third year?​

Answer 1

Answer:

Amount to be paid at the end of two years will be =Amountafter2years=8000(1+

100

10

)

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 follow me ✌️ keep samile ✌

Answer 2

$\bf\underline{Solutions:-}$

• $\: \: \: \: \: \leadsto \: \: \sf{Principal \: = \: {1500}}$

• $\: \: \: \: \: \leadsto \: \: \sf{rate \: \: = \: \: {{12} \: \%}}$

• $\: \: \: \: \: \leadsto \: \: \sf{Amount \: \: = \: \: {4400}}$

$\bf\underline{To \: \: find:-}$

• $\: \: \: \: \: \sf{The \: \: amount \: \: outstanding \: \: against \: \: him \: \: at \: \: the \: \: beginning \: \: of \: \: the \: \: third \: \: year.}$

$\bf\underline{Solutions:-}$

• $\: \: \: \: \: \sf{simple \: \: interest \: \: = \: \: \dfrac{principal \: \times \: rate \: \times \: time}{100}}$

$\: \: \: \: \: \leadsto \: \: \dfrac{15000 \: \times \: 12 \: \times \: 1}{100}$

$\: \: \: \: \: \leadsto \: \: \dfrac{15000 \: \times \: 12 \: \times}{100}$

$\: \: \: \: \: \leadsto \: \: {150 \: \times \: 12}$

$\: \: \: \: \: \leadsto \: \: {1800}$

• $\: \: \: \: \: \sf{Amount \: \: at \: \: the \: \: end \: \: of \: \: first \: \: year \: \: \leadsto \: \: p \: + \: {1}}$

$\: \: \: \: \: \leadsto \: \: {15000}\: + \: {1800}$

$\: \: \: \: \: \leadsto \: \: {16800}$

$\: \: \: \: \: \sf{Amount \: \: left \: \: after \: \: repaying \: \: \leadsto \: \: {16800} \: - \: {4400}}$

$\: \: \: \: \: \leadsto \: \: {12400}$

• $\: \: \: \: \: \sf{principal \: \: for \: \: the \: \: second \: \: year:-}$

$\: \: \: \: \: \leadsto \: \: \dfrac{{12400} \: \times \: {12} \: \times \: {1}}{100}$

$\: \: \: \: \: \leadsto \: \: \dfrac{12400 \: \times \: 12}{100}$

$\: \: \: \: \: \leadsto \: \: {1800} \: \times \: {12}$

$\: \: \: \: \: \leadsto \: \: {1488}$

$\: \: \: \: \: \sf{Amount \: \: with \: \: interest \: \: \leadsto {12400} \: - \: {1488}}$

$\: \: \: \: \: \leadsto \: \: {13888}$

$\: \: \: \: \: \sf{Amount \: \: after \: \: repaying \: \: \leadsto {13888} \: - \: {4400}}$

$\: \: \: \: \: \leadsto \: \: {9488}$

$\: \: \: \: \: \sf{Hence, \: the \: \: third \: \: year \: \: is \: \: {9488}}$