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This problem involves two runners starting from points \( A \) and \( B \), running towards each other. Their speeds are given in the ratio \( 7:3 \). Additionally, it is mentioned that the distance between their points of intersection, as they meet twice, is 45 meters.

Let's denote:
- The speed of the first runner as \( 7x \),
- The speed of the second runner as \( 3x \),
- The total distance between points \( A \) and \( B \) as \( d \).

### Key idea:
When two runners run towards each other, the sum of their speeds determines how fast they approach each other. The first time they meet, they have together covered the entire distance \( d \). After continuing, they meet again at a distance of \( 45 \) meters after their first intersection.

### Approach:
1. The ratio of their speeds is \( 7:3 \), so they cover distances in the same ratio. Therefore, after the first intersection:
   - The first runner has covered \( \frac{7}{7+3} = \frac{7}{10} \) of the total distance \( d \),
   - The second runner has covered \( \frac{3}{10} \) of the total distance \( d \).

2. After their first meeting, they continue running until they meet again, but this second meeting happens after they have collectively run another 45 meters (the given distance between the first and second meeting points).

The total distance run during one full cycle (two meetings) is twice the distance \( d \). Since the sum of their speeds is \( 7x + 3x = 10x \), and they meet again after 45 meters, we can set up the following proportion:

\[
\frac{45}{d} = \frac{3}{10}
\]

This proportion arises because the relative distances between meetings follow the same ratio as their speeds. Solving this equation for \( d \):

\[
d = \frac{45 \times 10}{3} = 150 \, \text{meters}
\]

However, the answer marked as correct in the image is \( 100 \, \text{m} \), so there might be an alternative interpretation or calculation method being used.

### Clarification:
Do you need further details, or would you like to go over this solution again? 

Here are 5 related questions for further practice:
1. If the speeds were in the ratio \( 5:4 \), and the distance between the second and third intersections was 30 meters, what would be the total distance between \( A \) and \( B \)?
2. How long would it take for the runners to meet again if their speeds were increased proportionally to \( 9x \) and \( 4x \)?
3. What if both runners started at the same point and ran in opposite directions—what would change in the intersection pattern?
4. How would you calculate the time taken to reach the first meeting point given their individual speeds?
5. What other conditions would affect the time intervals between meetings?

**Tip:** When solving ratio-based problems, always check if the final steps use all the provided conditions, such as distance, time, or speed, to ensure no critical information is overlooked.

Two runners are running to and from points A and B respectively. Their speeds are in the ratio 7:3. If the distance between the first and second points of intersection is 45 meters, what is the distance between point A and B?

Learn how to solve a problem where two runners are running from points A and B with speeds in the ratio 7:3. Find the total distance between points A and B using the given data and relative motion formulas. This problem involves key concepts such as ratios, speed-time-distance relationships, and relative speed. Suitable for Grades 9-12.

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