Question

Sanjay borrowed a certain sum from Anil at a certain rate of simple interest for 2 years. He lent this sum to Ram at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 4200 as compound interest but paid Rs. 4000 only as simple interest. Find the rate of interest.

• A. 15\(\%\)
• B. 20\(\%\)
• C. 35\(\%\)
• D. 10\(\%\)

Answer 1

The Correct Option is D

The correct option is (D): 10\(\%\)
Let's solve the problem step by step.

Step 1: Let the principal amount be \( P \) and the rate of interest be \( r \%\).

- **Simple Interest (SI):** Sanjay pays simple interest of Rs. 4000 after 2 years. The formula for simple interest is:
\[ SI = \frac{P \times r \times t}{100}\]
 Where \( t = 2 \) years. So,
 \[4000 = \frac{P \times r \times 2}{100} \implies 4000 = \frac{2Pr}{100} \implies 4000 = \frac{Pr}{50}\]
 Therefore, 
 \[Pr = 4000 \times 50 = 200000\]

Step 2: **Compound Interest (CI):** Sanjay receives compound interest of Rs. 4200 after 2 years. The formula for compound interest is:
 \[CI = P \left(1 + \frac{r}{100}\right)^t - P\]
 Where \( t = 2 \) years. Substituting \( CI = 4200 \), we get:
 \[4200 = P \left(1 + \frac{r}{100}\right)^2 - P\]
 Simplifying this:
 \[4200 = P \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]\]

Step 3: **Solving for \( P \) and \( r \):** We now have two equations:
 - \( Pr = 200000 \)
 - \( 4200 = P \left[\left(1 + \frac{r}{100}\right)^2 - 1\right] \)

Substituting \( P = \frac{200000}{r} \) into the second equation:
 \[4200 = \frac{200000}{r} \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]\]

By solving this equation numerically for \( r \), we find that:

\[r = 10\%\]

### Conclusion: The rate of interest is **10%**.

Thus, the correct option is:

\[\boxed{10\%}\]