Latest TCS Aptitude Question SOLUTION: A father and his son go out for a 'walk-and-run' every morning around a track formed by an equilateral triangle. The father's walking speed is 2 mph and his running speed is 5 mph.

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A father and his son go out for a 'walk-and-run' every morning around a track formed by an equilateral triangle. The father's walking speed is 2 mph and his running speed is 5 mph. The son's walking and running speeds are twice that of his father. Both start together from one apex of the triangle, the son going clockwise and the father anti-clockwise. Initially the father runs and the son walks for a certain period of time. Thereafter, as soon as the father starts walking, the son starts running. Both complete the course in 45 minutes. For how long does the father run? Where do the two cross each other?

Solution

Let father runs t hours and walks (0.75-t) hours, then according to the given condition
5t+2(0.75-t)=x (for father)
10(0.75-t)+4t=x (for son)
on comparing these two we will get t=2/3 hrs=40 min
on taking t=2/3 hrs, x=3.5 miles
so father will run for 2/3 hrs = 40 min

now let the two will cross each other after time x then according to the given condition 5x+4x=3.5 so x = 7/18 hrs = 70/3 min in this much time father will run
5*7/18 = 35/18 miles and son will walk (7/2)-(35/18) = 28/18 = 14/9 miles


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Asked In TCS Yashar (13 years ago)

Solved Read Solution (1)

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Read Solution (1) :

frnzrkut  (13 years ago)

Let the perimeter of the equilateral triangle = x
Let xf1,xf2,xs1 and xs2 be the distance covered by father and son in Act 1 and Act 2 resp. (The definition of Act is given below)
So, x= xf1+xf2 = xs1+xs2
First Act: Son walks and Father runs for time = t1
Second Act: Son runs and Father walks for time = t2
t = t1 + t2
So, 45 = xf1/5 + xf2/2 = xs1/4 + xs2/10
But xf1/5 = xs1/4 and xf2/2 = xs2/10

Solving, we get,
xf1 = 200
xf2=10
xs1=160
xs2=50

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