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Question

Mathematics Question on SI & CI

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

10%\%

Explanation

Solution

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

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

- Simple Interest (SI): Sanjay pays simple interest of Rs. 4000 after 2 years. The formula for simple interest is:
SI=P×r×t100SI = \frac{P \times r \times t}{100}
Where t=2t = 2 years. So,
4000=P×r×2100    4000=2Pr100    4000=Pr504000 = \frac{P \times r \times 2}{100} \implies 4000 = \frac{2Pr}{100} \implies 4000 = \frac{Pr}{50}
Therefore,
Pr=4000×50=200000Pr = 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(1+r100)tPCI = P \left(1 + \frac{r}{100}\right)^t - P
Where t=2t = 2 years. Substituting CI=4200CI = 4200, we get:
4200=P(1+r100)2P4200 = P \left(1 + \frac{r}{100}\right)^2 - P
Simplifying this:
4200=P[(1+r100)21]4200 = P \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]

Step 3: Solving for PP and rr: We now have two equations:
- Pr=200000Pr = 200000
- 4200=P[(1+r100)21]4200 = P \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]

Substituting P=200000rP = \frac{200000}{r} into the second equation:
4200=200000r[(1+r100)21]4200 = \frac{200000}{r} \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]

By solving this equation numerically for rr, we find that:

r=10%r = 10\%

Conclusion: The rate of interest is 10%.

Thus, the correct option is:

10%\boxed{10\%}