amanvermagmat
Two soldiers started racing from point P to point Q, a distance of 10 miles. Soldier 1 ran the first half of the distance at a constant speed of X miles per hour and the second half at a constant speed of Y miles per hour. Soldier 2 ran the entire race at a constant speed of Z miles per hour. X and Y are distinct multiples of 10. If neither of them stopped before reaching point Q, which soldier reached point Q first?
(1) Y is 50% more than X.
(1) Z is the average of X and Y.
Time taken by S1 = 10/[5/X + 5/Y] = 2XY/(X+Y)
Time taken by S2 = 10/Z
1) Nothing is given about Z
Insufficient
2) Time taken by S2 = 10/(X+Y)
And time taken by S1 is 2XY/(X+Y)
now we look at above equations, we are already told that X & Y are multiple of 10. Look at time taken by S2, i can write it as 1/(x+y) by taking 10 common and cancelling.
time taken by S1 can be represented as 2xY/(x+y) in this case i can take 10 common either from X or Y in numerator.
Clearly time taken by S1 is more as it has a larger numerator. Hence, B is sufficient.