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Option C is Correct.

Let's suppose total distance is L.
When 1st time they meet.
Faster travelled F*T=L-720----(I)
Shower one =S*T=720----(II)
Deviding both we will get.
F/S=L-720/720.---(III).

Similarly in 2nd case.
F/S=(2L-400)/L+400.--+(iv)
As eqn(iii)=(iv)

So, upon solving {(L-720)/720}={(2L-400)/L+400}.

We will get L=1760.

Hence both statements require to ans.

C

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Ans: C

Let the two cyclists be A and B, where A is the faster one
Let the track length be t
So the first round distance covered by A:B = t-x:x
Similarly, the second round distance covered by A:B = 2t-y:t+y

Since the speed is constant, (t-x)/x = (2t-y)/(t+y)
To find t we need value of both x and y

Statement 1 alone is not sufficient
Statement 2 alone is not sufficient

Both statements together will give the value of t

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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?

Let Va and Vb be speeds of A and B
Let B be faster than A
let t1 be first meeting time and
t2 be second meeting time
Let L be total track length

(1) x = 720 meters

\(Va*t=720\)
\(Vb*t=L-720\)
\(\frac{Va}{Vb}=\frac{720}{(L-720)}\)
we cant derive anything else so INSUFFICIENT

(2) y = 400 meters..........they can meet in two scenario

1. B reached other end and came back to meet A who is still reaching the end

we get one more equation similarly in terms of Va/Vb=L-400/2L-400

1. B reached other end and came back to meet A who is coming back after reaching the other end

we get one more equation similarly in terms of Va/Vb=L+400/2L-400

still INSUFFICIENT as we cant solve for L

combing (1) and (2) we cant solve for L
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TOTAL DISTANCE - Z



(1) x = 720 meters - INSUFFICIENT



Z-720/S1 = 720/S2

S2/S1 = 720/Z-720



(2) y = 400 meters



320+Z/S1 = Z-320/S2

S2/S1 = Z-320/320+Z



INSUFFICIENT



(1) + (2) -



Z-320/320+Z = 720/Z-720



Z CAN BE FOUND OUT.



SUFFICIENT



ANSWER - C

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Hey All, it seems to me that all the comments above explaining why (C) is the answer don't seem to be accounting for the case where the slow cyclist is so slow that they don't make it to the other end before the fast one catches up to them. Can you explain what in the question allows us to eliminate this case from consideration?
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