A runs 1 3 2 ​ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Options

A. a. 200
B. b. 300m
C. c. 270 m
D. d. 160 m

Answer

To solve the problem, we need to determine how far the winning post must be so that A and B reach it at the same time, given that A runs \( \frac{3}{2} \) times as fast as B and gives B a start of 80 meters.

### Step-by-Step Solution:

1. **Define the Speeds of A and B**:
Let the speed of B be \( 2x \). Then, the speed of A, which is \( \frac{3}{2} \) times the speed of B, will be:
\[ \text{Speed of A} = \frac{3}{2} \times 2x = 3x \]

2. **Set Up the Distance Equation**:
Let the distance to the winning post be \( D \) meters. Since B has a start of 80 meters, the distance B has to run is:
\[ D - 80 \text{ meters} \]

3. **Use the Time Formula**:
The time taken by A to reach the winning post is:
\[ \text{Time for A} = \frac{D}{3x} \]
The time taken by B to reach the winning post is:
\[ \text{Time for B} = \frac{D - 80}{2x} \]

4. **Set the Times Equal**:
Since A and B reach the winning post at the same time, we set the two time equations equal:
\[ \frac{D}{3x} = \frac{D - 80}{2x} \]

5. **Cross Multiply to Solve for D**:
Cross multiplying gives us:
\[ 2xD = 3x(D - 80) \]
Simplifying this, we get:
\[ 2D = 3D - 240 \]
Rearranging gives:
\[ 3D - 2D = 240 \implies D = 240 \text{ meters} \]

### Final Answer:
The distance to the winning post must be **240 meters**.