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varun
What annual payment will discharge a debt of $\text{Rs.}6450$ due in $4$ years at $5\%$ simple interest?(without using standard formula)
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Javed Khan
commented 2016-06-05 19:06:07
@Leela, it is not like that. Since we are making payments at the end of each year, corresponding interest need to be reduced from the total interest to be paid.

Can you try to solve the question based on your understanding and post the answer here?
0 0
Leela
commented 2016-06-05 17:54:11
I thought we have to pay interest amount to the buyer for opting EMI option. But he is earning interest on our installment by some other means. Is that right sir?
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Javed Khan
There are two kind of questions here which may be slightly confusing. It's important to understand the difference between these. Let me use this opportunity to explain both of them here.

a. What annual payment will discharge a debt of $\text{Rs.6450}$ due in $4$ years at $5\%$ simple interest?

In this case, $6450$ is considered as the amount which needs to paid if the person keeps the borrowed money for the entire period of $4$ years. This is the question asked here.

Let $x$ be the annual installment.

Total effective repayment
$=x$ after $3$ years at $5\%$ simple interest
$+x$ after $2$ year at $5\%$ simple interest
$+x$ after $1$ year at $5\%$ simple interest
$+x$

$= \left[x+\dfrac{x × 5 × 3}{100}\right]+\left[x+\dfrac{x × 5 × 2}{100}\right]$ $+\left[x+\dfrac{x × 5 × 1}{100}\right]+\left[x\right]$
$=x+\dfrac{15x}{100}+x+\dfrac{10x}{100}+x+\dfrac{5x}{100}+x\\ =4x+\dfrac{3x}{10}$

$4x+\dfrac{3x}{10}=6450\\ 40x+3x=64500\\ 43x=64500\\ x = \dfrac{64500}{43} = \text{Rs. }1500$

[ Formula 2 explains this]

Now consider this question.
b. A sum of $\text{Rs.6450}$ is borrowed at $5\%$ simple interest and is paid back in $4$ equal annual installments. What is amount of each installment?

In this case, the person needs to repay
$6450+\dfrac{6450 × 5 × 4}{100}$
$=6450+1290 =\text{Rs.}7740$ in $4$ years

Total effective repayment
$=x$ after $3$ years at $5\%$ simple interest
$+x$ after $2$ year at $5\%$ simple interest
$+x$ after $1$ year at $5\%$ simple interest
$+x$

$=\left[x+\dfrac{x × 5 × 3}{100}\right]+\left[x+\dfrac{x × 5 × 2}{100}\right]$ $+\left[x+\dfrac{x × 5 × 1}{100}\right]+\left[x\right]$
$=x+\dfrac{15x}{100}+x+\dfrac{10x}{100}+x+\dfrac{5x}{100}+x\\ =4x+\dfrac{3x}{10}$

$4x+\dfrac{3x}{10}=7740\\ 40x+3x=77400\\ 43x=77400\\ x=\dfrac{77400}{43}=\text{Rs. }1800$

I have answered a similar question here.
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