Aman invested a certain sum in scheme A at `15%` pa for 2 years and earned Rs 1950 as simple interest. He increased his sum by X and invested in another scheme B at `10%` pa CI for 2 years and received Rs 1680 as compound interest. Find the value of 'X'.

Options

A. 1750
B. 1500
C. 1250
D. none of these

Answer

To solve the problem step by step, we will first determine the principal amount invested by Aman in scheme A and then use that to find the value of X for scheme B.

### Step 1: Calculate the Principal Amount for Scheme A
We know that the formula for Simple Interest (SI) is given by:
\[ SI = \frac{P \times R \times T}{100} \]
Where:
- \(SI\) = Simple Interest
- \(P\) = Principal amount
- \(R\) = Rate of interest (in %)
- \(T\) = Time (in years)

For scheme A:
- \(SI = 1950\)
- \(R = 15\%\)
- \(T = 2\)

Substituting the values into the formula:
\[ 1950 = \frac{P \times 15 \times 2}{100} \]
\[ 1950 = \frac{30P}{100} \]
\[ 1950 = 0.3P \]
Now, solve for \(P\):
\[ P = \frac{1950}{0.3} = 6500 \]

### Step 2: Determine the Principal for Scheme B
In scheme B, Aman increases his principal amount by \(X\). Therefore, the principal for scheme B is:
\[ P_B = 6500 + X \]

### Step 3: Calculate the Compound Interest for Scheme B
The formula for Compound Interest (CI) for 2 years is:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]
Where:
- \(CI\) = Compound Interest
- \(P\) = Principal amount
- \(R\) = Rate of interest (in %)
- \(T\) = Time (in years)

For scheme B:
- \(CI = 1680\)
- \(R = 10\%\)
- \(T = 2\)

Substituting the values into the formula:
\[ 1680 = (6500 + X) \left(1 + \frac{10}{100}\right)^2 - (6500 + X) \]
\[ 1680 = (6500 + X) \left(1.1^2\right) - (6500 + X) \]
Calculating \(1.1^2\):
\[ 1.1^2 = 1.21 \]
Now substituting back:
\[ 1680 = (6500 + X)(1.21) - (6500 + X) \]
\[ 1680 = (6500 + X)(1.21 - 1) \]
\[ 1680 = (6500 + X)(0.21) \]
Now, solving for \(6500 + X\):
\[ 6500 + X = \frac{1680}{0.21} = 8000 \]

### Step 4: Solve for \(X\)
Now, we can find \(X\):
\[ X = 8000 - 6500 = 1500 \]

### Final Answer
The value of \(X\) is \(1500\).

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