Suneeta borrowed certain sum from Reena for 2 yr at simple interest. Suneeta lent this sum to Venu at the same rate for 2 yr at compound interest. At the end of 2 yr, she received ? 110 as compound interest but paid ? 100 as simple interest. Find the sum and the rate of interest.

Options

A. ?250, rate 10% pa
B. ? 250, rate 20% pa
C. ? 250, rate 25% pa
D. ? 250 , rate 30% pa

Answer

To solve the problem step by step, we need to find the principal amount (sum borrowed) and the rate of interest.

### Step 1: Understand the Given Information
- Suneeta borrowed a certain sum (let's denote it as \( P \)) from Reena at simple interest for 2 years.
- She lent the same sum to Venu at compound interest for 2 years at the same rate.
- The simple interest paid to Reena after 2 years is \( 100 \).
- The compound interest received from Venu after 2 years is \( 110 \).

### Step 2: Calculate the Rate of Interest Using Simple Interest Formula
The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]
Where:
- \( SI \) = Simple Interest
- \( P \) = Principal amount
- \( R \) = Rate of interest
- \( T \) = Time in years

From the problem, we know:
\[ SI = 100, \quad T = 2 \]
Substituting these values into the formula:
\[ 100 = \frac{P \times R \times 2}{100} \]
Rearranging gives:
\[ 10000 = P \times R \quad \text{(Equation 1)} \]

### Step 3: Calculate the Compound Interest
The formula for compound interest is:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]
Where:
- \( CI \) = Compound Interest

From the problem, we know:
\[ CI = 110, \quad T = 2 \]
Substituting these values into the formula:
\[ 110 = P \left(1 + \frac{R}{100}\right)^2 - P \]
This simplifies to:
\[ 110 = P \left(\left(1 + \frac{R}{100}\right)^2 - 1\right) \]
Expanding the square:
\[ \left(1 + \frac{R}{100}\right)^2 = 1 + 2\frac{R}{100} + \left(\frac{R}{100}\right)^2 \]
Thus:
\[ 110 = P \left(2\frac{R}{100} + \left(\frac{R}{100}\right)^2\right) \]
Rearranging gives:
\[ 110 = P \left(\frac{2R + \frac{R^2}{100}}{100}\right) \]
Multiplying through by 100:
\[ 11000 = P(2R + \frac{R^2}{100}) \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we have:
\[ P = \frac{10000}{R} \]
Substituting this into Equation 2:
\[ 11000 = \left(\frac{10000}{R}\right) \left(2R + \frac{R^2}{100}\right) \]
Simplifying:
\[ 11000R = 10000(2R + \frac{R^2}{100}) \]
Expanding:
\[ 11000R = 20000R + 100R^2 \]
Rearranging gives:
\[ 100R^2 + 9000R = 0 \]
Factoring out \( R \):
\[ R(100R + 9000) = 0 \]
Thus, \( R = 0 \) or \( R = -90 \) (not possible), so we need to solve:
\[ 100R + 9000 = 0 \Rightarrow R = 90 \]

### Step 5: Find the Principal Amount \( P \)
Using \( R = 20 \) in Equation 1:
\[ P \times 20 = 10000 \Rightarrow P = \frac{10000}{20} = 500 \]

### Conclusion
The principal amount \( P \) is \( 500 \) and the rate of interest \( R \) is \( 20\% \).