Question

The simple interest accrued on an amount of Rs.25,000 at the end of three years is Rs.7,500. What would be the compound interest accrued on the same amount at the same rate in the same period?

• A. 7750
• B. 8275
• C. 8500
• D. 8250

Answer 1

The Correct Option is B

The correct answer is B:8275
Given: 
Principal amount (P)=Rs.25,000 
Time period (T)=3 years 
Simple interest=Rs.7,500 
First,let's find the rate of interest (R) using the simple interest formula: 
Simple Interest (SI)=\(\frac{(P \times R \times T)}{100} \)
7500=\(\frac{(25000 \times R \times 3)}{100}\) 
R=\(\frac{(7500 \times 100)}{(25000 \times 3) }\)
R=\(10\%\) 
Now that we have the rate of interest (R),we can calculate the compound interest using the compound interest formula: 
Compound Interest (CI)=A- P 
A=\(P \times (1+(\frac{R}{100}))^T \)
Substitute the values: 
A=\(25000 \times (1+(\frac{10}{100}))^3 \)
A=\(25000 \times 1.331\) 
A≈33275 
Now,calculate the compound interest: 
CI=33275- 25000 
CI=8275 
So,the compound interest accrued on the same amount at the same rate over the same period would be approximately Rs.8,275. 

Answer 2

Given,
Principal \((P) = Rs. 25,000\)
Rate of Interest \((r) = 10%\)
Time \((t) = 3\) years
Formula for compound interest:
\(CI=P[(1+\frac {r}{100}​)^n−1]\)

\(CI=25000[(1+\frac {10}{100​})^3−1]\)
\(CI=25000[(1+0.1)^3−1]\)
\(CI=25000[(1.1)^3−1]\)
\(CI=25000[(1.1×1.1×1.1)−1]\)
\(CI=25000×(1.331−1)\)
\(CI=25000×0.331\)
\(CI=8275\)

Therefore, the correct option is \((B):8275\)