Question

Arun borrowed a sum of money from Jayant at the rate of $ 8\% $ simple interest for the first four years, $ 10\% $ p.a. for the next $ 6 $ years and $ 12\% $ p.a. beyond $ 10 $ years. If he pays a total of $ Rs.12160 $ as interest only at the end of $ 15 $ years, how much money did he borrow?
a) $ Rs.12000 $
b) $ Rs.10000 $
c) $ Rs.8000 $
d) \[Rs.9000\]

Answer 1

Hint: We shall analyze the given information so that we are able to solve the problem. We are asked to calculate the money borrowed by Arun. We shall use the given rate of interest and the number of years to obtain the simple interest. And then we need to compare the obtained simple interests with the total interest.
Formula to be used:
The formula to calculate the simple interest is as follows.
Simple interest, $ S.I = \dfrac{{P \times N \times R}}{{100}} $
Where, $ P $ is the principal amount, $ N $ is the number of years (time), and $ R $ is the rate of interest.

Complete step by step answer:
Let us assume the money borrowed by Arun as $ Rs.x $ .
Let $ S.{I_1} $ , $ S.{I_2} $ and $ S.{I_3} $ be the given simple interests.
We shall calculate these simple interests separately.
Let $ {P_1} $ , $ {P_2} $ and $ {P_3} $ denote the money borrowed by Arun.
Then, $ {P_1} = {P_2} = {P_3} = Rs.x $
It is given that $ {R_1} = 8\% $ and $ {n_1} = 4years $ .
Using the formula $ S.I = \dfrac{{P \times N \times R}}{{100}} $ , we have
 $ S.{I_1} = \dfrac{{{P_1} \times {N_1} \times {R_1}}}{{100}} $
           $ = \dfrac{{x \times 4 \times 8}}{{100}} $
           $ = \dfrac{{32x}}{{100}} $
Thus $ S.{I_1} = \dfrac{{32x}}{{100}} $
Also, it is given that $ {R_2} = 10\% $ and $ {n_2} = 6years $
Using the formula $ S.I = \dfrac{{P \times N \times R}}{{100}} $ , we have
 $ S.{I_2} = \dfrac{{{P_2} \times {N_2} \times {R_2}}}{{100}} $
           $ = \dfrac{{x \times 6 \times 10}}{{100}} $
           $ = \dfrac{{60x}}{{100}} $
Thus $ S.{I_2} = \dfrac{{60x}}{{100}} $
Also, it is given that $ {R_3} = 12\% $ .
Now Arun borrowed $ 12\% $ p.a. beyond $ 10 $ years. Also, it is given that Arun pays the interest for $ 15 $ years.
Hence, we need to subtract it from the total years and the obtained is the required $ {n_3} $ .
That is $ {n_3} = total{\text{ }}years - 15 $
 $ \Rightarrow {n_3} = \left( {4 + 6 + 10} \right) - 15 $
 $ \Rightarrow {n_3} = 5years $
Using the formula $ S.I = \dfrac{{P \times N \times R}}{{100}} $ , we have
 $ S.{I_3} = \dfrac{{{P_3} \times {N_3} \times {R_3}}}{{100}} $
           $ = \dfrac{{x \times 12 \times 5}}{{100}} $
           $ = \dfrac{{60x}}{{100}} $
Thus $ S.{I_3} = \dfrac{{60x}}{{100}} $
We are given that the total interest is $ Rs.12160 $ .
We need to add the obtained simple interests with the total interest.
Thus, total interest $ = S.{I_1} + S.{I_2} + S.{I_3} $
We shall substitute the calculated simple interests.
Hence, total interest $ = \dfrac{{32x}}{{100}} + \dfrac{{60x}}{{100}} + \dfrac{{60x}}{{100}} $
 $ \Rightarrow Rs.12160 = \dfrac{{32x}}{{100}} + \dfrac{{60x}}{{100}} + \dfrac{{60x}}{{100}} $
 $ \Rightarrow Rs.12160 = \dfrac{{152x}}{{100}} $
 $ \Rightarrow 1216000 = 152x $
 $ \Rightarrow x = \dfrac{{1216000}}{{152}} $
 $ \Rightarrow x = \dfrac{{608000}}{{76}} $
 \[ \Rightarrow x = \dfrac{{152000}}{{19}}\]
 \[ \Rightarrow x = 8000\]
Hence, Arun borrowed $ Rs.8000 $

So, the correct answer is “Option c”.

Note: We observe that Arun borrowed $ 12\% $ p.a. beyond $ 10 $ years. Also, it is given that Arun pays the interest for $ 15 $ years.
Hence, we need to subtract it from the total years and the obtained is the required $ {n_3} $ .
That is $ {n_3} = total{\text{ }}years - 15 $
If we assume $ {n_3} = 10 $ without using the condition, we will obtain the wrong answer.