In a 1 km race A, B and C are the three participants. A can give B a start of 50 m and C a start of 69 m. Find the start, which B can allow C.

Answer

To solve the problem, we need to determine the distance that participant B can give to participant C in a 1 km race based on the information provided about participant A's performance relative to B and C.

### Step-by-Step Solution:

1. **Understanding the Information Given**:
- A can give B a start of 50 meters. This means when A finishes the 1 km race, B has run only 950 meters.
- A can give C a start of 69 meters. This means when A finishes the 1 km race, C has run only 931 meters.

2. **Finding the Speeds of A, B, and C**:
- Let the speeds of A, B, and C be represented as \( v_A, v_B, \) and \( v_C \) respectively.
- When A finishes the race (1000 meters), B has covered 950 meters:
\[ \frac{v_B}{v_A} = \frac{950}{1000} = \frac{19}{20} \]
- When A finishes the race (1000 meters), C has covered 931 meters:
\[ \frac{v_C}{v_A} = \frac{931}{1000} \]

3. **Expressing B's Speed in Terms of A's Speed**:
- From the first equation, we can express B's speed in terms of A's speed:
\[ v_B = \frac{19}{20} v_A \]

4. **Expressing C's Speed in Terms of A's Speed**:
- From the second equation, we can express C's speed in terms of A's speed:
\[ v_C = \frac{931}{1000} v_A \]

5. **Finding the Ratio of Speeds Between B and C**:
- To find the start that B can give to C, we need the ratio of their speeds:
\[ \frac{v_B}{v_C} = \frac{\frac{19}{20} v_A}{\frac{931}{1000} v_A} = \frac{19}{20} \cdot \frac{1000}{931} = \frac{19000}{18620} \]
- Simplifying this gives:
\[ \frac{v_B}{v_C} = \frac{1900}{1862} \approx \frac{950}{931} \]

6. **Calculating the Start B Can Give C**:
- If B runs 1000 meters, we can find out how far C would run:
\[ \text{Distance covered by C} = 1000 \cdot \frac{931}{950} = \frac{931000}{950} \approx 980.0 \text{ meters} \]
- Therefore, the distance that C is behind B when B finishes the race is:
\[ \text{Start B can give C} = 1000 - 980 = 20 \text{ meters} \]

### Final Answer:
B can allow C a start of **20 meters**.