Question

A sum of money was borrowed at 6% per annum simple interest. At the end of first year Rs.68000 was paid off and the rate of interest on the balance was reduced to $ 5\% $ per annum. If the interest for the second year was $ \dfrac{{11}}{{20}} $ of the interest for the first year what was the original sum borrowed.
a) Rs. $ 10000 $
b) Rs. $ 12000 $
c) Rs. $ 17000 $
d) Rs. $ 15000 $

Answer 1

Hint: We are asked to find simple interest for two years with different interest rate for the first one year and the next one year. We are not given the principal amount, so will use the other given data to find the principal amount, by using the simple interest formula.
Formula used:
Simple interest for a given principal amount ( $ P $ ), time ( $ T $ ) and rate of interest ( $ T $ ) is given by the formula:
 $ SI = \dfrac{{P \times T \times R}}{{100}} $

Complete step by step solution:
We are not given the initial sum of money that was borrowed.
Let us determine this by using algebra.
Let, the given sum of money be Rs. $ x $ .
So the principal amount is $ P = x $
We are also given that the rate of interest for the first annum is $ 6\% $
So, the rate of interest is $ R = 6\% $ and time is $ T = 1 $ year.
So the simple interest is given by:
 $ {I_1} = \dfrac{{x \times 1 \times 6}}{{100}} = \dfrac{{3x}}{{50}} $
Now the amount of money after one year becomes:
 $ {P_1} = x + \dfrac{{3x}}{{50}} = \dfrac{{53x}}{{50}} $
According to the question, Rs. $ 6800 $ was paid after one year. So, the new principal amount after a year is:
 $ P = {P_1} - 6800 = \dfrac{{53x}}{{50}} - 6800 $
 $ \Rightarrow P = \dfrac{{53x - 340000}}{{50}} $
Now, we have a rate of interest as $ 5\% $ for the next year.
So applying the simple interest formula here we get:
 $ {I_2} = \dfrac{{(53x - 340000) \times 5 \times 1}}{{50 \times 100}} $
We are also given in the question that $ {I_1} \times \dfrac{{11}}{{20}} = {I_2} $ , so
 $ \dfrac{{3x}}{{50}} \times \dfrac{{11}}{{20}} = \dfrac{{(53x - 340000) \times 5 \times 1}}{{50 \times 100}} $
 $ \Rightarrow \dfrac{{33x}}{{1000}} = \dfrac{{(53x - 340000)}}{{1000}} $
 $ \Rightarrow 33x = 53x - 340000 $
 $ \Rightarrow 20x = 340000 $
 $ \Rightarrow x = 17000 $
So, the correct answer is “Option C”.

Note: The question addresses a bunch of facts within it. Identifying the required data from the asked question and understanding all facts elaborately gives a clear picture of what is asked in the question. So, before proceeding to the solution read the question carefully.