Mona invests ₹P in scheme A (offering `12%` pa simple interest) for 2 years and ₹2P in scheme B (offering `15%` pa simple interest) for 3 years. If the difierence between the interests earned from both the schemes is ₹1320, what is the value of P ?

Options

A. 6000
B. 5000
C. 2000
D. 4000

Answer

To solve the problem step by step, we will calculate the simple interest earned from both schemes and set up an equation based on the difference given.

### Step 1: Calculate Simple Interest for Scheme A
- **Principal (P)**: ₹P
- **Rate of Interest**: 12% per annum
- **Time**: 2 years

The formula for Simple Interest (SI) is:
\[ SI = \frac{P \times R \times T}{100} \]
For Scheme A:
\[ SI_A = \frac{P \times 12 \times 2}{100} = \frac{24P}{100} = \frac{6P}{25} \]

### Step 2: Calculate Simple Interest for Scheme B
- **Principal (2P)**: ₹2P
- **Rate of Interest**: 15% per annum
- **Time**: 3 years

For Scheme B:
\[ SI_B = \frac{2P \times 15 \times 3}{100} = \frac{90P}{100} = \frac{9P}{10} \]

### Step 3: Set Up the Equation Based on the Difference in Interests
According to the problem, the difference between the interests earned from both schemes is ₹1320:
\[ SI_B - SI_A = 1320 \]
Substituting the values we calculated:
\[ \frac{9P}{10} - \frac{6P}{25} = 1320 \]

### Step 4: Find a Common Denominator
The least common multiple (LCM) of 10 and 25 is 50. We will convert both fractions to have a denominator of 50:
\[ \frac{9P}{10} = \frac{9P \times 5}{10 \times 5} = \frac{45P}{50} \]
\[ \frac{6P}{25} = \frac{6P \times 2}{25 \times 2} = \frac{12P}{50} \]

### Step 5: Substitute Back into the Equation
Now we substitute these back into our equation:
\[ \frac{45P}{50} - \frac{12P}{50} = 1320 \]
This simplifies to:
\[ \frac{33P}{50} = 1320 \]

### Step 6: Cross Multiply to Solve for P
Cross multiplying gives us:
\[ 33P = 1320 \times 50 \]
Calculating the right side:
\[ 33P = 66000 \]

### Step 7: Solve for P
Now, divide both sides by 33:
\[ P = \frac{66000}{33} = 2000 \]

### Final Answer
The value of P is ₹2000.

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