A runs 100 metres in 11 seconds and B runs 100 metres in 12 seconds. The head start which must be given to B for race to be completed in 11 seconds.

Options

A. 8 m
B. ` 8 (1)/(4)` m
C. ` 8 (1)/(3)` m
D. ` 8 (1)/(2)m`

Answer

To solve the problem of how much head start should be given to B so that he finishes the race in 11 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the speeds of A and B:**
- A runs 100 meters in 11 seconds.
- B runs 100 meters in 12 seconds.

To find their speeds, we use the formula:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
- Speed of A:
\[ \text{Speed of A} = \frac{100 \text{ m}}{11 \text{ s}} \approx 9.09 \text{ m/s} \]
- Speed of B:
\[ \text{Speed of B} = \frac{100 \text{ m}}{12 \text{ s}} \approx 8.33 \text{ m/s} \]

2. **Calculate the distance B can cover in 11 seconds:**
- Using B's speed, we find out how far he can run in 11 seconds:
\[ \text{Distance covered by B in 11 seconds} = \text{Speed of B} \times \text{Time} = 8.33 \text{ m/s} \times 11 \text{ s} = \frac{275}{3} \text{ m} \approx 91.67 \text{ m} \]

3. **Determine the head start for B:**
- Since A runs the full 100 meters in 11 seconds, we need to find out how much less distance B covers in the same time:
\[ \text{Head start} = \text{Distance A runs} - \text{Distance B runs in 11 seconds} \]
\[ \text{Head start} = 100 \text{ m} - \frac{275}{3} \text{ m} = 100 - 91.67 = \frac{25}{3} \text{ m} \approx 8.33 \text{ m} \]

4. **Conclusion:**
- The head start that must be given to B for the race to be completed in 11 seconds is:
\[ \text{Head start} = \frac{25}{3} \text{ meters} \approx 8.33 \text{ meters} \]