A certain amount becomes 2200 at rate of 5% in certain time period and same amount becomes 2320 at rate of 8% then find the required amount ?

Options

A. Rs 2000
B. Rs 1900
C. Rs 2100
D. Rs 1800

Answer

To solve the problem step by step, we need to find the principal amount (P) that leads to the two different amounts at different interest rates.

### Step 1: Understand the problem
We have two scenarios:
1. The amount becomes 2200 at a rate of 5% per annum.
2. The same amount becomes 2320 at a rate of 8% per annum.

We need to find the principal amount (P).

### Step 2: Set up the equations
Using the formula for Simple Interest:
\[ A = P + SI \]
Where:
- \( A \) is the total amount after interest.
- \( P \) is the principal amount.
- \( SI \) is the simple interest.

The simple interest can also be calculated as:
\[ SI = \frac{P \times r \times t}{100} \]
Where:
- \( r \) is the rate of interest.
- \( t \) is the time period.

From the first scenario:
\[ 2200 = P + \frac{P \times 5 \times t}{100} \]
This can be rearranged to:
\[ 2200 = P(1 + \frac{5t}{100}) \]
(Equation 1)

From the second scenario:
\[ 2320 = P + \frac{P \times 8 \times t}{100} \]
This can be rearranged to:
\[ 2320 = P(1 + \frac{8t}{100}) \]
(Equation 2)

### Step 3: Solve the equations
Now we have two equations:
1. \( 2200 = P(1 + \frac{5t}{100}) \)
2. \( 2320 = P(1 + \frac{8t}{100}) \)

We can express \( P \) from both equations:
From Equation 1:
\[ P = \frac{2200}{1 + \frac{5t}{100}} \]
From Equation 2:
\[ P = \frac{2320}{1 + \frac{8t}{100}} \]

### Step 4: Set the two expressions for P equal to each other
\[ \frac{2200}{1 + \frac{5t}{100}} = \frac{2320}{1 + \frac{8t}{100}} \]

### Step 5: Cross-multiply to eliminate the fractions
\[ 2200(1 + \frac{8t}{100}) = 2320(1 + \frac{5t}{100}) \]

### Step 6: Expand both sides
\[ 2200 + \frac{17600t}{100} = 2320 + \frac{11600t}{100} \]

### Step 7: Simplify the equation
\[ 2200 + 176t = 2320 + 116t \]

### Step 8: Move all terms involving t to one side
\[ 176t - 116t = 2320 - 2200 \]
\[ 60t = 120 \]

### Step 9: Solve for t
\[ t = \frac{120}{60} = 2 \text{ years} \]

### Step 10: Substitute t back into one of the equations to find P
Using Equation 1:
\[ P = \frac{2200}{1 + \frac{5 \times 2}{100}} \]
\[ P = \frac{2200}{1 + 0.1} \]
\[ P = \frac{2200}{1.1} \]
\[ P = 2000 \]

### Final Answer
The required amount (principal) is **2000 rupees**.

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