The price of pertrol is increased by `14.(2)/(7)%`. If the expenditure should not be increased for a students, she will have to reduce his consumption of petrol by the percentage (a) 30% (b) 40% (C) 25/2 % (d) 60%

Options

A. `30%`
B. `40%`
C. `(25)/(2)%`
D. `60%`

Answer

To solve the problem, we need to determine how much the student needs to reduce their petrol consumption in order to keep their expenditure constant after the price increase. Let's break this down step by step.

### Step-by-Step Solution

1. **Understanding the Percentage Increase**:
The price of petrol is increased by \( 14 \frac{2}{7}\% \). This can be converted to an improper fraction:
\[ 14 \frac{2}{7} = \frac{14 \times 7 + 2}{7} = \frac{98 + 2}{7} = \frac{100}{7} \approx 14.2857\% \]

2. **Let the Original Price be \( P \)**:
Assume the original price of petrol per liter is \( P \). After the increase, the new price \( P' \) will be:
\[ P' = P + \left(\frac{100}{7} \times P\right) = P \left(1 + \frac{100}{700}\right) = P \left(1 + \frac{1}{7}\right) = P \left(\frac{8}{7}\right) \]

3. **Expenditure Calculation**:
Let’s denote the original quantity of petrol consumed as \( Q \). The original expenditure \( E \) is given by:
\[ E = P \times Q \]
After the price increase, the new expenditure \( E' \) will be:
\[ E' = P' \times Q' = \left(\frac{8}{7}P\right) \times Q' \]

4. **Keeping Expenditure Constant**:
To keep the expenditure constant, we set \( E = E' \):
\[ P \times Q = \left(\frac{8}{7}P\right) \times Q' \]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ Q = \frac{8}{7} Q' \]

5. **Finding the New Quantity**:
Rearranging gives:
\[ Q' = \frac{7}{8} Q \]
This means the student can only consume \( \frac{7}{8} \) of the original quantity \( Q \).

6. **Calculating the Reduction in Consumption**:
The reduction in consumption is:
\[ \text{Reduction} = Q - Q' = Q - \frac{7}{8} Q = \frac{1}{8} Q \]

7. **Finding the Percentage Reduction**:
The percentage reduction in consumption is:
\[ \text{Percentage Reduction} = \left(\frac{\text{Reduction}}{Q}\right) \times 100 = \left(\frac{\frac{1}{8} Q}{Q}\right) \times 100 = \frac{1}{8} \times 100 = 12.5\% \]

8. **Final Answer**:
The percentage reduction in petrol consumption is:
\[ 12.5\% = \frac{25}{2}\% \]

### Conclusion
The student will have to reduce their consumption of petrol by \( \frac{25}{2}\% \) to keep their expenditure constant.