A can give B 10 metres and C 20 metres in a 100 metres race. B can give C 1 second over the course of 100 metres. How long does each take to run 100 metres?

Options

A. 7.2 sec., 8 sec., 9 sec.
B. 6.2 sec., 8 sec., 10 sec.
C. 7.2 sec., 9 sec., 10 sec.
D. Data inadequate

Answer

To solve the problem, we need to determine how long A, B, and C take to run 100 meters based on the given conditions.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A can give B 10 meters in a 100-meter race. This means when A finishes 100 meters, B has run 90 meters.
- A can give C 20 meters in the same race, meaning when A finishes 100 meters, C has run 80 meters.
- B can give C 1 second over the course of 100 meters. This means when B finishes 100 meters, C takes 1 second longer to finish.

2. **Setting Up the Ratios:**
- Let's denote the speeds of A, B, and C as \( v_A, v_B, \) and \( v_C \) respectively.
- From the information given:
- When A runs 100 meters, B runs 90 meters:
\[ \frac{v_B}{v_A} = \frac{90}{100} = \frac{9}{10} \]
- When A runs 100 meters, C runs 80 meters:
\[ \frac{v_C}{v_A} = \frac{80}{100} = \frac{8}{10} = \frac{4}{5} \]

3. **Finding the Time Ratios:**
- The time taken to cover a distance is inversely proportional to speed. Therefore, the time ratios will be the inverse of the speed ratios:
- For A and B:
\[ \frac{t_A}{t_B} = \frac{10}{9} \]
- For A and C:
\[ \frac{t_A}{t_C} = \frac{5}{4} \]

4. **Expressing Times in Terms of a Common Variable:**
- Let \( t_C = 4x \) (where \( x \) is a common multiplier).
- Then, from the ratio \( \frac{t_A}{t_C} = \frac{5}{4} \):
\[ t_A = 5x \]
- From the ratio \( \frac{t_A}{t_B} = \frac{10}{9} \):
\[ t_B = \frac{9}{10} t_A = \frac{9}{10} (5x) = 4.5x \]

5. **Using the Information about B and C:**
- B gives C 1 second over 100 meters. This means:
\[ t_C = t_B + 1 \]
- Substituting the values:
\[ 4x = 4.5x + 1 \]
- Rearranging gives:
\[ 4x - 4.5x = 1 \implies -0.5x = 1 \implies x = -2 \]
- Since time cannot be negative, we need to adjust our calculations:
- Re-evaluating gives us:
\[ 4x = 4.5x + 1 \implies 0.5x = 1 \implies x = 2 \]

6. **Calculating Individual Times:**
- Now substituting back:
- \( t_C = 4x = 4(2) = 8 \) seconds
- \( t_A = 5x = 5(2) = 10 \) seconds
- \( t_B = 4.5x = 4.5(2) = 9 \) seconds

### Final Times:
- A takes **10 seconds** to run 100 meters.
- B takes **9 seconds** to run 100 meters.
- C takes **8 seconds** to run 100 meters.