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# A man borrowed Rs.40000 at 8% simple interest per year. At the end of second year he paid back a certain amount and at the end of fifth year he paid back Rs. 35960 and cleared the debt. What is the amount that he paid back after the second year?A.Rs.16200B.Rs.17400C.Rs.18600D.Rs.19200

Last updated date: 12th Aug 2024
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Hint: We need to find the Simple Interest and the amount for the first two years. And since a certain amount is being paid at the end of second year, the new principal for next three years will be the difference of the amount for the first two years and the amount paid at the end of the second year. On this principle we need to find the amount for the last three years which will be equal to the amount paid by him at the end of the fifth year and accordingly we can find the amount paid at the end of the second year.
Formula:
For a given Principal (P), Rate of interest (R) & Time Period (T), Simple Interest (S.I) is given by:
$S.I = \dfrac{{P \times R \times T}}{{100}}$ and $Amount = S.I + P$

Given, A man borrowed Rs.40000 at rate R = 8% per annum at simple interest for 5 yrs. The man paid a certain amount at the end of second year and again he paid Rs. 35960 at the end of fifth year and cleared the debt.
Hence, Principal (P)= 40000
Rate of interest per annum(R)= 8%
No of years= 5
But since he paid a certain value at the end of second year we need to find the amount for the first two years.
So, $S.I = \dfrac{{P \times R \times T}}{{100}}$ $= \dfrac{{40000 \times 8 \times 2}}{{100}}$ $= 6400$
S.I = 6400
Now, Amount at the end of second year is:
$\Rightarrow Amount = S.I + P$ $\Rightarrow$ $A = 6400 + 40000$
$\Rightarrow A = 46400$
Since, the man paid some amount at the end of second year, therefore his principal would change and interest for the next three years is to be calculated on this principal and since he paid Rs.35960 at the end of fifth year and cleared his debt. So, the amount for the last three years will be Rs.35960.
Now, let the Principal (P) for the next three years be Rs.x, R = 8% and T = 3 years.
So,
$\Rightarrow S.I = \dfrac{{P \times R \times T}}{{100}}$ $\Rightarrow$ $S.I = \dfrac{{x \times 8 \times 3}}{{100}}$
$\Rightarrow S.I = \dfrac{{6x}}{{25}}$
Now, Amount at the end of fifth year is:
$\Rightarrow Amount({A_1}) = S.I + P$ $\Rightarrow$ ${A_1} = x + \dfrac{{6x}}{{25}}$
${A_1} = \dfrac{{31x}}{{25}}$
Hence, ${A_1} = \dfrac{{31x}}{{25}} = 35960$
$x = 35960 \times \dfrac{{25}}{{31}}$
$x = 29000$
x = 29000
Hence, the amount paid by the man at the end of second year $({A_2})$ = Amount for the first two years (A) – Principal at which interest is being calculated for the last three years (x).
Hence, ${A_2} = A - x$ $\Rightarrow$ ${A_2} = 46400 - 29000$
${A_2} = 17400$
Thus, amount he paid back after the end of second year is $Rs. 17400$
So, the correct answer is “Option B”.

Note: Kindly memorize the formulas and try to understand the concept, it will help you to solve easily. Kindly don’t confuse between Simple Interest and Compound Interest because the concept and the procedure for solving is completely different in both the cases.