A, B and C are three contestants in a 500 m race. If A can give B a start of 20 m and A can give C a start of 32 m, how many metres start can B give to C in the same race?

Options

A. 12 m
B. 14 m
C. 12.5 m
D. 13.5

Answer

To solve the problem step by step, we need to analyze the information given about the three contestants A, B, and C in a 500 m race.

### Step 1: Understand the given information
- A can give B a start of 20 m. This means that when A runs 500 m, B runs only 480 m.
- A can give C a start of 32 m. This means that when A runs 500 m, C runs only 468 m.

### Step 2: Determine the speeds of A, B, and C
Let's denote the speeds of A, B, and C as \( v_A \), \( v_B \), and \( v_C \) respectively.

From the information:
- When A runs 500 m, B runs 480 m:
\[ \frac{v_B}{v_A} = \frac{480}{500} = \frac{48}{50} = \frac{24}{25} \]

- When A runs 500 m, C runs 468 m:
\[ \frac{v_C}{v_A} = \frac{468}{500} = \frac{234}{250} = \frac{117}{125} \]

### Step 3: Express the speeds of B and C in terms of A's speed
From the above ratios:
- \( v_B = \frac{24}{25} v_A \)
- \( v_C = \frac{117}{125} v_A \)

### Step 4: Find the ratio of speeds of B and C
To find how much start B can give C, we need to find the ratio of their speeds:
\[ \frac{v_B}{v_C} = \frac{\frac{24}{25} v_A}{\frac{117}{125} v_A} = \frac{24}{25} \times \frac{125}{117} = \frac{24 \times 125}{25 \times 117} = \frac{600}{117} = \frac{200}{39} \]

### Step 5: Calculate the distance B can give C in a 500 m race
If B runs 500 m, we can find out how far C would run using the ratio:
\[ \text{Distance C runs} = 500 \times \frac{39}{200} = \frac{19500}{200} = 97.5 \text{ m} \]

### Step 6: Calculate the start B can give C
The start that B can give C is the difference between the total distance and the distance C runs:
\[ \text{Start B can give C} = 500 - 97.5 = 402.5 \text{ m} \]

### Final Answer
Thus, B can give C a start of **12.5 m** in a 500 m race.

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