Question

A man borrows Rs 6000 for 2 years at the rate of 12% per annum compounded annually he repays rupees 3000 at the end of each year how much does he still over after the second repayment?

Answer 1

Dear Student,

● Answer -

Remaining payment = 1526 Rs

◆ Explanation -

Total amount man should have paid till the end of 2 years -

Returns = P (1 + r)^t

Returns = 6000 (1 + 12/100)^2

Returns = 6000 × (1.12)^2

Returns = 6000 × 1.2544

Returns = 7526 Rs

After 2 payments, he have paid 3000×2 = 6000 Rs.

Remaining payment = 7526 - 6000

Remaining payment = 1526 Rs

Thanks dear. Hope this helps you...

Answer 2

Answer:

$\red { Remaining \: payment } \green {= Rs \:1166.40}$

Step-by-step explanation:

$Principal (P) = Rs \:6000$

$Rate \:of \: interest (R) = 12\%\:p.a$

$Amount \:after \: \: end \:of \:the \: first \:year$

$number \:of \:times \: Interest \:paid = n$

$Amount \: at \: the \: end \: of \: first \:year \\= P\left( 1 + \frac{R}{100}\right)^{n}$

$= 6000\left( 1+ \frac{12}{100}\right)^{1}$

$= 6000 \times \frac{112}{100}$

$= 60\times 112$

$= Rs \: 6720$

$After \: paying \: Rs \:3000\: end \:of \: the \\first \:year , next \: year \: principal\\ = 6720-3000\\=Rs \:3720$

$Amount \: at \: the \: end \: of \: second\:year \\= 3720 \left( 1+\frac{12}{100}\right)^{1}$

$= 3720 \times \frac{112}{100}\\=Rs\:4166.40$

$After \: paying \: Rs \:3000\: end \:of \: the \\second \:year , Remaining\: payment \\= Rs\:4166.40 - Rs\:3000 = Rs \: 1166.40$

Therefore.,

$\red { Remaining \: payment } \green {= Rs \:1166.40}$

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