Quote:
Two cyclists start moving simultaneously from opposite ends of a straight track towards each other back and forth. Cyclists' speeds are constant, but one is faster than the other. First time, the cyclists meet each other at a distance of x meters from the nearest end of the track. Second time, on the way back, they meet y meters from the other end of the track. What is the length of the track?
(1) x = 720 meters
(2) y = 400 meters
As per the given problem,
c1---x------|----(d-x)-------c2
While returning c2------(d-y)-----|---y---c1
let speed of c2>speed of c1. If the they meet after "t"
Speed of \(c2=\frac{(d-x)}{t}\) and Speed of c1=\(\frac{x}{t}\)
After first meeting they will meet when the total distance covered by them will be 2d after time 2t.
So c1 will cover a distance of d-x+y. Speed=\(\frac{(d-x+y)}{2t}\)
So, \(\frac{(d-x+y)}{2t}=\frac{x}{t}\)
\(=>d-x+y=2x\)
\(=>d=3x-y\)
Sufficient I:We don't know the value of y
Sufficient II:We don't know the value x
Combining I and II:d=3x-y
We will know the value of "d".
Ans: CPosted from my mobile device