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Let be the time Sam takes to drive from A to B, and
Let be the time Richard takes to drive from B to A.
We are given
= , or = (1)
Hence, the ratio of their rates per minute is reciprocal ratio
= , or = = . (2)
The total distance equation is
+ = d (3)
Replace here = based on (2). You will get
+ = d, or
+ = d
= d
which implies that the time = 72 minutes.
Thus, Richard completes his trip in 72 minutes: = 72 minutes.
Then the time by Sam is = see (1) = = = 90 minutes.
Thus Sam reaches B 90-72 = 18 minutes after Richard reaches A. ANSWER
Solved.
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Surely, this problem is two levels higher than a regular Math/Physics problem in any high school.
It is, actually, an Olympiad level problem in Physics.
Could you tell me please where is it from, from which source ?
Do not forget to post your "THANKS" to me for my teaching.
Driving between City A and City B, the ratio of the time Sam takes to the time Richard takes is 5 : 4. If Sam leaves City A and Richard leaves City B at the same time toward to each other, they meet in 40 minutes, and they continue to drive. After Richard arrives at City A, how many minutes later will Sam arrives at City B?
Let multiplicative factor be x, and distance from A to B, or B to A, D
Then times Sam and Richard take to complete journey are 5x and 4x, respectively
Therefore, Sam’s speed = , and Richard’s is:
With them meeting in 40 minutes, we get:
------- Factoring out GCF, D, in numerators
4 + 5 = 30x ------ Mulltiplying by LCD, 30x
9 = 30x
Multiplicative factor, or
Therefore, the difference in times for them to complete their respectively journeys is