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Yard = 60 metres = Distance

Let speed (empty hand) be = 12 mps

Time taken = \(\frac{60 }{12} = 5 sec\)

Return speed (heavy ball) = \(\frac{12}{2} = 6 mps\)

Time taken = \(\frac{60 }{ 6} = 10 sec.\)

Answer E
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DST164

for this kind of problems i like to use a technique i saw a couple of times on the forum.

in this case the distance is the same, so we equate times:
distance speed time
without heavy ball 60 v60/v
with heavy ball 60 v/2 60/v/2 = 120/v

now we know that when he is faster he takes 5 less seconds, meaning that the higher time (when he is slower), minus the lower time (when he is faster) = 5
in this case:

120/v-60/v = 5

solving for v --> v = 12

we know time = 120/v with the ball --> time = 120/12 = 10


i find it so much easier and less confusing. in general i like to convert every mutation of this problem in such a way that is solvable with this strategy. i made a little collection of similar problem to practice it!

https://gmatclub.com/forum/in-a-100-met ... 80581.html
https://gmatclub.com/forum/harvey-runs- ... 99049.html
https://gmatclub.com/forum/a-trip-of-90 ... 54768.html
https://gmatclub.com/forum/today-dave-d ... 95328.html
https://gmatclub.com/forum/if-a-man-wal ... 14111.html
https://gmatclub.com/forum/alex-and-ben ... 33915.html
https://gmatclub.com/forum/a-and-b-star ... 19329.html
https://gmatclub.com/forum/tom-runs-emp ... 56140.html
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