Question

The difference between Simple interest and compound interest at 20% per annum at the end of 3 years is Rs.1,024. What is the sum?
A. Rs.6,000
B. Rs.5,000
C. Rs.8,000
D. Rs.9,000

Answer 1

Hint: Here we substitute the values of rate of interest and time period in formulas of simple interest and compound interest separately. Subtract the equation of amount from simple interest from equation of amount from compound interest. Equate the value obtained after subtraction to the given difference.
* Simple Interest: \[A = P(1 + rt)\]
Where A is the amount obtained after applying simple interest, P is the principal amount, R is the rate of interest and t is the time period.
* Compound interest: \[A = P{(1 + r)^t}\]
Where A is the amount obtained after applying compound interest, P is the principal amount, R is the rate of interest and t is the time period.

Complete step-by-step answer:
We are given that rate of interest is 20%
\[ \Rightarrow r = 20\% \]
Using the concept of percentage we can write rate of interest in fraction form as
\[ \Rightarrow r = \dfrac{{20}}{{100}}\]
Also, we are given time period is 3 years
\[ \Rightarrow t = 3\]
Let us assume the principal amount as P
Also, amount obtained after simple interest be A
Then we can write from the formula of simple interest:
 \[A = P(1 + rt)\]
Substitute the value of \[r = \dfrac{{20}}{{100}}\]and\[t = 3\]
\[ \Rightarrow A = P\left( {1 + \dfrac{{20}}{{100}} \times 3} \right)\]
Multiply the whole number to numerator in RHS of the equation.
\[ \Rightarrow A = P\left( {1 + \dfrac{{60}}{{100}}} \right)\]
Take LCM in the bracket
\[ \Rightarrow A = P\left( {\dfrac{{100 + 60}}{{100}}} \right)\]
\[ \Rightarrow A = P\left( {\dfrac{{160}}{{100}}} \right)\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow A = P \times \dfrac{8}{5}\] … (1)
The amount obtained after compound interest be B
Then we can write from the formula of compound interest
\[B = P{(1 + r)^t}\]
Substitute the value of \[r = \dfrac{{20}}{{100}}\]and\[t = 3\]
\[ \Rightarrow B = P{\left( {1 + \dfrac{{20}}{{100}}} \right)^3}\]
Take LCM in the bracket
\[ \Rightarrow B = P{\left( {\dfrac{{100 + 20}}{{100}}} \right)^3}\]
\[ \Rightarrow B = P{\left( {\dfrac{{120}}{{100}}} \right)^3}\]
Cancel the same factors from numerator and denominator in the bracket
\[ \Rightarrow B = P \times {\left( {\dfrac{6}{5}} \right)^3}\]
Solve the bracket
\[ \Rightarrow B = P \times \left( {\dfrac{{6 \times 6 \times 6}}{{5 \times 5 \times 5}}} \right)\]
\[ \Rightarrow B = P \times \dfrac{{216}}{{125}}\] … (2)
Now subtract equation (1) from equation (2)
\[ \Rightarrow B - A = \left( {P \times \dfrac{{216}}{{125}}} \right) - \left( {P \times \dfrac{8}{5}} \right)\]
Take P common from both the brackets
\[ \Rightarrow B - A = P\left( {\dfrac{{216}}{{125}} - \dfrac{8}{5}} \right)\]
Take LCM in RHS of the equation
\[ \Rightarrow B - A = P\left( {\dfrac{{216 - 8 \times 25}}{{125}}} \right)\]
\[ \Rightarrow B - A = P\left( {\dfrac{{216 - 200}}{{125}}} \right)\]
Calculate the difference in numerator inside the bracket
\[ \Rightarrow B - A = P\left( {\dfrac{{16}}{{125}}} \right)\] … (3)
We are given the value of difference as Rs.1,024. Equate the difference obtained in equation (3) to this value.
\[ \Rightarrow 1024 = P\left( {\dfrac{{16}}{{125}}} \right)\]
Cross multiply the values in RHS to LHS of the equation
\[ \Rightarrow 1024 \times \dfrac{{125}}{{16}} = P\]
Now we write \[1024 = 16 \times 64\]
\[ \Rightarrow \dfrac{{16 \times 64 \times 125}}{{16}} = P\]
Cancel same terms from numerator and denominator
\[ \Rightarrow 64 \times 125 = P\]
\[ \Rightarrow P = 8000\]
So, the Principal sum is Rs.8,000

Therefore, option C is correct.

Note: Students might get confused with the statement of the question where we have to find the sum and they might try to find the sum of both the amounts obtained from Simple interest and compound interest. Keep in mind the Principal amount is called principal sum. Also, if writing the formula in terms of r divided by hundred then don’t take the value of r in decimals, just take the numerical value as the percentage part is already converted into fraction in the formula.