A certain sum is invested for certain time. It amount to 3500 at 10% pen annum. But when invested at 8% per annum. It amounts to 3000 the find time?

Options

A. 16 year
B. 20 year
C. 30 year
D. 25 year

Answer

To solve the problem step by step, we will follow the process of calculating the principal amount and the time period based on the given interest rates and amounts.

### Step 1: Define the Variables
Let the principal amount be \( P \) and the time period be \( T \) (in years).

### Step 2: Set Up the Equations
From the problem, we have two scenarios:

1. When the amount is \( 3500 \) at \( 10\% \) per annum:
\[ P + \frac{P \times 10 \times T}{100} = 3500 \]
This can be simplified to:
\[ P \left(1 + \frac{10T}{100}\right) = 3500 \]
Let's call this Equation (1).

2. When the amount is \( 3000 \) at \( 8\% \) per annum:
\[ P + \frac{P \times 8 \times T}{100} = 3000 \]
This can be simplified to:
\[ P \left(1 + \frac{8T}{100}\right) = 3000 \]
Let's call this Equation (2).

### Step 3: Express \( P \) from Both Equations
From Equation (1):
\[ P = \frac{3500}{1 + \frac{10T}{100}} = \frac{3500}{1 + 0.1T} \]

From Equation (2):
\[ P = \frac{3000}{1 + \frac{8T}{100}} = \frac{3000}{1 + 0.08T} \]

### Step 4: Set the Two Expressions for \( P \) Equal
Since both expressions represent the same principal amount \( P \), we can set them equal to each other:
\[ \frac{3500}{1 + 0.1T} = \frac{3000}{1 + 0.08T} \]

### Step 5: Cross-Multiply to Solve for \( T \)
Cross-multiplying gives us:
\[ 3500(1 + 0.08T) = 3000(1 + 0.1T) \]
Expanding both sides:
\[ 3500 + 280T = 3000 + 300T \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 3500 - 3000 = 300T - 280T \]
\[ 500 = 20T \]

### Step 7: Solve for \( T \)
Dividing both sides by 20:
\[ T = \frac{500}{20} = 25 \]

### Final Answer
The time period \( T \) is \( 25 \) years.

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