Ramesh invested two equal amounts in two different schemes. In the first scheme.amount is invested at 8% p.a. on SI for t years and SI received is Rs. 2000 while in second scheme. amount is invested at 10%p.a. for 2 years at CI and the compound interest received is Rs.1050.The value of t is

Options

A. 4 yr
B. 8 yr
C. 6 yr
D. 5 yr

Explanation

To solve the problem step by step, we will break down the information provided and apply the formulas for Simple Interest (SI) and Compound Interest (CI).

### Step 1: Understand the Problem
Ramesh invested equal amounts in two different schemes. We need to find the time period \( t \) for the first scheme where the interest is calculated using Simple Interest (SI).

### Step 2: Set Up the Equations
1. **First Scheme (Simple Interest)**:
- Rate = 8% per annum
- Simple Interest (SI) = Rs. 2000
- Time = \( t \) years
- Let the principal amount be \( P \).

The formula for Simple Interest is:
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the known values:
\[ 2000 = \frac{P \times 8 \times t}{100} \]
Rearranging gives:
\[ P \times t = \frac{2000 \times 100}{8} = 25000 \quad \text{(Equation 1)} \]

2. **Second Scheme (Compound Interest)**:
- Rate = 10% per annum
- Time = 2 years
- Compound Interest (CI) = Rs. 1050
- Let the principal amount be \( P \) (same as in the first scheme).

The formula for Compound Interest is:
\[ CI = P \left(1 + \frac{R}{100}\right)^T - P \]
Substituting the known values:
\[ 1050 = P \left(1 + \frac{10}{100}\right)^2 - P \]
Simplifying:
\[ 1050 = P \left(1.1^2 - 1\right) \]
\[ 1.1^2 = 1.21 \quad \Rightarrow \quad 1050 = P(1.21 - 1) = P(0.21) \]
Rearranging gives:
\[ P = \frac{1050}{0.21} = 5000 \quad \text{(Equation 2)} \]

### Step 3: Substitute and Solve for \( t \)
Now, substitute the value of \( P \) from Equation 2 into Equation 1:
\[ 5000 \times t = 25000 \]
Solving for \( t \):
\[ t = \frac{25000}{5000} = 5 \]

### Final Answer
The value of \( t \) is **5 years**.

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