Question

Shankar takes a loan of $ Rs.10,000 $ at a compound interest rate of $ 10\% $ per annum (p.a) .
A. Find the compound interest after one year.
B. Find the compound interest for $ 2 $ years.
C. Find the sum of money required to clean the debt at the end of $ 2 $ years.
D. Find the difference between the compound interest and the simple interest at the same rate for $ 2 $ years.

Answer 1

Hint: As we know that the above question is related to compound interest. We know that the compound interest is an addition of interest to the principal sum of a loan or a deposit or we can say interest on the interest. The general formula of compound interest is Amount i.e. $ A = P{\left( {1 + \dfrac{R}{{100}}} \right)^n} $ , where $ P $ is the principal, $ n $ is the time period and $ R $ is the rate of interest.

Complete step by step solution:
As per the given question we have Principal i.e. $ P = 10,000 $ , rate of interest $ (R) = 10\% $ .
(a)In the first question we have $ P = 10,000 $ , rate of interest $ (R) = 10\% $ and $ n = 1 $ .
Compound interest after one year will be $ C.I = A - P $ .
 By putting the values in the formula we have: $ C.I = 10000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1} $ . Now we will solve it, $ 10000\left( {\dfrac{{100 + 10}}{{100}}} \right) = 10000 \times \dfrac{{110}}{{100}} $ . It gives us the value $ C.I = 11,000 $ .
So C.I after one year is $ 11000 - 10000 = 1000 $ .

(b)In the second question we have, $ P = 10,000 $ , rate of interest $ (R) = 10\% $ and $ n = 2 $ . By putting the values in the formula we have: $ C.I = 10000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2} $ . Now we will solve it, $ 10000{\left( {\dfrac{{100 + 10}}{{100}}} \right)^2} = 10000 \times \dfrac{{110}}{{100}} \times \dfrac{{110}}{{100}} $ . It gives us the value $ C.I = 12,100 $ .
So the C.I after two years is $ 12100 - 10000 = 2100 $

(c)In the third question, we have to find the sum of money to clear the debt of $ 2 $ years .
From the above we have $ p = 10000 $ and $ C.I = 2100 $ (of two years).
So the sum of money required to clear the debt at the end of two years is $ 10000 + 2100 = 12100 $ Rs

(d) Here we have C.I of two years $ = 2100,p = 10000,r = 10 $ and $ n = 2 $ .
So the Simple Interest i.e. $ S.I = \dfrac{{10000 \times 10 \times 2}}{{100}} = 2000Rs $
Therefore the difference between the compound interest and simple interest at the same rate for two years is $ C.I - S.I = 2100 - 2000 $ .
The required difference is $ Rs100 $ .

Note: Before solving this kind of question we should have the proper knowledge of compound interest and simple interest and their formulas. We should know that the major difference between the compound and simple interest is that simple interest is based on the principal of deposit or loan whereas the compound interest is based on the principal and interest that accumulates in every period of time.