The C.I. and S.I. on a sum of money for 2 years is Rs. 210 and Rs. 200 respectively. If the rate of the interest is same, then sum will be

Options

A. a. Rs 2200
B. b. Rs. 500
C. c. Rs . 1000
D. d. Rs. 2000

Answer

To solve the problem, we need to find the principal sum (P) given the compound interest (C.I.) and simple interest (S.I.) for 2 years. The given values are:

- C.I. = Rs. 210
- S.I. = Rs. 200

### Step-by-Step Solution:

**Step 1: Understand the relationship between C.I. and S.I.**
The difference between the compound interest and simple interest for 2 years can be expressed as:
\[ \text{C.I.} - \text{S.I.} = \text{P} \cdot \frac{R^2}{100^2} \]
Where P is the principal, and R is the rate of interest.

**Step 2: Calculate the difference between C.I. and S.I.**
Using the values provided:
\[ 210 - 200 = 10 \]
So, we have:
\[ \text{P} \cdot \frac{R^2}{100^2} = 10 \]

**Step 3: Calculate S.I. using the formula**
The formula for simple interest is:
\[ \text{S.I.} = \frac{P \cdot R \cdot T}{100} \]
Given that S.I. = Rs. 200 and T = 2 years, we can write:
\[ 200 = \frac{P \cdot R \cdot 2}{100} \]
Rearranging gives:
\[ P \cdot R = 100 \]

**Step 4: Substitute P.R into the equation**
From the previous step, we know:
\[ P \cdot R = 100 \]
Now, we can express R in terms of P:
\[ R = \frac{100}{P} \]

**Step 5: Substitute R into the difference equation**
Substituting R into the equation from Step 2:
\[ P \cdot \frac{\left(\frac{100}{P}\right)^2}{100^2} = 10 \]
This simplifies to:
\[ P \cdot \frac{10000}{P^2 \cdot 10000} = 10 \]
Which simplifies further to:
\[ \frac{1}{P} = 10 \]
Thus:
\[ P = \frac{1}{10} \]

**Step 6: Solve for P**
From the equation \( P = 10 \), we find:
\[ P = 1000 \]

### Final Answer:
The sum (P) is Rs. 1000.