leap plzzzzzzz solve
Thu, 2011-08-25 08:24
a hare makes 4 leap in the same time in which a dog makes 3 but 2 leaps of the dog cover as much distance as 3 leaps of hare; how many leaps will the dog have to make before catching the hare,supposing the hare to have a start of 60 leaps.
ans is 360
in unit time hare makes 4 hops and dog makes 3 hops
distance covered by hare in single hop = h
-do- dog -do- = d
2d=3h given
hence, hare speed per unit time = 4h/1 = 4h
hence, dog speed per unit time = 3d/1 = 3d
dog catches hare in time t
60h + 4h*t = 3d*t
60h + 4h*t = 3*(3/2)*h*t
60 + 4t = 9/2 * t
t = 120
hops done by dog in unit time = 3
hops done by dog in 120 time unit = 3*120 = 360