The question is in details.?
Two riders on horseback simultaneously leave Village A. They proceed with different but constant speeds to Village B and then return without stopping. On her way back to Village A, one of the riders overtakes the other, meeting her at a point located m miles from Village B. Upon returning to Village A, she leaves...
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answers:
jibz say: Let D be the distance in miles between the two villages. On their first encounter the faster rider has traveled a distance of

D+m

miles while the slower rider has traveled a distance of

D–m

miles. On their second encounter the faster rider has traveled a distance of (2 + 1/3)D i.e.

(7/3)D

miles while the slower rider has traveled a distance of (1 + 2/3)D i.e.

(5/3)D

miles. So the faster rider is (7/3) / (5/3) i.e. 7/5 times more ground than the slower rider. Hence

D+m = (7/5)(D–m).

Solving for D we get D = 6m.
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Ian H say: Use a numerical example to get the idea and convert that to algebra.
Imagine u has speed 3 and v has speed 5 with distance d = 10 miles
After 2 hours v is at B, u is at 6 from A which is 4 from B
After d/v hours v is at B, u is at ud/v from A or (v – u)d/v from B
That gap gets divided in ratio 3 to 5 or u to v
u is at 6 + (3/8)*4 = 7.5 while v is at 10 - (5/8)*4 = 7.5
u is at (u/v)d + [u/(v + u)]*[(v – u)d/v] = 2ud/(v + u) = m
v is at d – [v/(v + u)]* (v – u)d/v = 2ud/(v + u) = m

After 4 hours v is at A while u has gone 10 + 2 miles so he is 8 from A
After 2d/v hours v is at A while u is 2(v – u)d/v from A
That gap of 8 gets divided in ratio 3 to 5 or u to v
v is at (5/8)*8 = 5 from A
u is at 8 - (3/8)*8 = 5 from A
In the created example she meets other at 5/10 or one HALF of the distance between Village A and Village B, whereas your given aim is to meet at a THIRD.
Complete the algebra equivalent and equate it to 1/3 to work out answer.
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Andrew say: It's your homework assignment, kid, not mine. I graduated already and my days of doing homework are over. Enjoy.
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qcmkg say: nfyhksos
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Sean say: .
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