A runs `1 2/3` times as fast as b. if A gives B a start of 4 m , how far must the wining post be , so that A and B might reach it at the same time ?

Options

A. 75 m
B. 200 m
C. 10 m
D. 125 m

Answer

To solve the problem step by step, 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 \(1 \frac{2}{3}\) times as fast as B and gives B a 4-meter head start.

### Step 1: Establish the speed ratio
A runs \(1 \frac{2}{3}\) times as fast as B. We can convert this mixed number into an improper fraction:
\[ 1 \frac{2}{3} = \frac{5}{3} \]
Thus, the speed ratio of A to B is:
\[ \frac{A}{B} = \frac{5}{3} \]

### Step 2: Set up the relationship between distance and speed
Since A and B will finish the race at the same time, the ratio of the distances they run will be the same as the ratio of their speeds. Let the distance A runs be \(d_A\) and the distance B runs be \(d_B\). We know that:
\[ \frac{d_A}{d_B} = \frac{A}{B} = \frac{5}{3} \]

### Step 3: Express B's distance in terms of A's distance
Since B has a 4-meter head start, we can express B's distance as:
\[ d_B = d_A - 4 \]
Now, substituting this into the distance ratio:
\[ \frac{d_A}{d_A - 4} = \frac{5}{3} \]

### Step 4: Cross-multiply to solve for \(d_A\)
Cross-multiplying gives us:
\[ 3d_A = 5(d_A - 4) \]
Expanding the right side:
\[ 3d_A = 5d_A - 20 \]
Now, rearranging the equation:
\[ 20 = 5d_A - 3d_A \]
\[ 20 = 2d_A \]
Dividing both sides by 2:
\[ d_A = 10 \]

### Step 5: Calculate the total distance of the race
Since \(d_A\) is the distance A runs, we need to find the total distance of the race. Since B runs \(d_B = d_A - 4\):
\[ d_B = 10 - 4 = 6 \]
Thus, the total distance of the race is:
\[ d_A + d_B = 10 + 6 = 16 \text{ meters} \]

### Step 6: Verify the solution
To ensure that A and B reach the winning post at the same time, we can check the time taken by both:
- Time taken by A:
\[ \text{Time}_A = \frac{d_A}{A} = \frac{10}{\frac{5}{3}} = 10 \times \frac{3}{5} = 6 \text{ seconds} \]
- Time taken by B:
\[ \text{Time}_B = \frac{d_B}{B} = \frac{6}{1} = 6 \text{ seconds} \]
Both A and B take the same time to reach the winning post.

### Final Answer
Thus, the total distance of the race must be **10 meters**.

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