At 6 PM, Renee and Eva begin cycling up a 10 kilometre slope

1) IF rene and eva move up at different speeds and their speeds are constant.so,if eva was ahead of rene for the first 14 kms then how can rene take over eva???
this statement is not sufficient.

(2)this one is also not sufficient.i think u must know either of their speed to answer that.
i solved it mathematically.this is what i came up with.
let speed of rene = x km/min
speed of eva= y

time taken by eva=10/y
timetaken by rene=10/y+40=10/x
hence 4xy=x-y

now when rene has already reached the top,distance covered by eva=speed*time=y*10/x=10y/x

and now rene takes 10/2x time to reach at the bottom.
so eva has 10/2x to climb to top and then descent whatec\ver distance she can.

while climbing up,distace left frr eva=10-10y/x
speed=y
therefore time taken by eva to climb up = (10-10y/x)/y =(10x-10y)/xy

so now time remaining for eva while she comes down the slope= (10/2x) -(10x-10y)/xy=(15y-10x)/xy

speed =2y

hence distance travelled by eva in this much time =(15y-10x)/xy *2y=(30y-20x)/x

right now rene is at the bottom,so the distance between the two =

10-(30y-20x)/x=(30x-30y)/x=30(x-y)/x

we can substitute x-y=4xy as derived earlier on,
we get,

distance between the two=30(x-y)/x=30*4xy/x=120y
since y isstill unknown,we can not determine the distance between the two.


i hope this is not the solution or this type doesn't come in actual gmat!

A

Statement 1:

(10/R) + (4/2R) = 6/E

6R = 12E.

When Renee has covered 10 miles at R and 10 miles at 2R, E would have covered 7.5 miles.
So, 7.5 miles apart. SUFF

Statement 2: Insufficient

Is their a consenus on the correct answer? If so, please explain.

tx anand,

i just missed the thought that rene could meet eva on the way back.

Kevin, i will choose A.

The gap will be 7.5...........