Two men left simultaneously two places A and B.One of them left A for ...
Problem Statement:
Two men left simultaneously two places A and B. One of them left A for B while the other left B for A. Both travel each at his own uniform velocity. The first person on reaching B returns to A and then again travels back to B and so on. What will be the distance covered by the first person when they meet for the third time given the ratio of the speed of the first person to that of the second person is 3:2, the distance between A and B is 500 m?
Solution:
Let's assume the speed of the first person to be 3x and the second person's speed to be 2x.
Let the distance covered by the first person until they meet for the third time be D.
Let's calculate the time taken by both persons to meet at the first meeting point.
- Calculation of time taken by both persons to meet at the first meeting point:
Let's assume the distance covered by the first person from A to B be x.
Then the distance covered by the second person from B to A would be (500 - x) [As the distance between A and B is 500m]
We know that, time = distance / speed
- Time taken by the first person: x / (3x)
- Time taken by the second person: (500 - x) / (2x)
Since both the persons started simultaneously, the time taken by both persons to meet at the first meeting point would be the same. Therefore, we can equate the above two equations.
x / (3x) = (500 - x) / (2x)
Solving the above equation, we get x = 300 m
Therefore, the time taken by both persons to meet at the first meeting point = x / (3x) = 1/3 hours
- Calculation of distance covered by the first person until they meet for the third time:
We know that the first person traveled from A to B and then back to A before the first meeting point.
Therefore, the distance covered by the first person until the first meeting point = 2x
After the first meeting point, the first person has to travel the distance between A and B twice to meet the second person for the third time.
Therefore, the distance covered by the first person from the first meeting point until the third meeting point = 2(500 m) = 1000 m
The distance covered by the first person until they meet for the third time = distance covered by the first person until the first meeting point + distance covered by the first person from the first meeting point until the third meeting point
= 2x + 1000
= 2(3x) + 1000
= 6x + 1000
= 6(300 m) + 1000
= 1800 m
Therefore, the distance covered by the first person until they meet for the third time is 1800m.
Two men left simultaneously two places A and B.One of them left A for ...
Please explain how it come 1500