Latest TCS Aptitude Question SOLUTION: A hare and tortoise a race along a circle of 100 yardds diameter . The tortoise goes in one direction and the hare in the other . The hare starts after tortoise has covered 1/5 its

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A hare and tortoise a race along a circle of 100 yardds diameter . The tortoise goes in one direction and the hare in the other . The hare starts after tortoise has covered 1/5 its distance and that leisurely . The hare and tortoise meet when the hare has covered only ¼ of the distance . by what factor should the hare increase its speed so as to tie the race?
a)8 b)37 c)45 d)6.6

Asked In TCS jeetendra kumar (12 years ago)

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Dr.Dipin Singh  (12 years ago)

6.6

After tortoise covers 1/5th distance, Hare starts the race.
When hare covers 1/4th of distance tortoise meets hare
So distance covered by tortoise = 1-(1/5 + 1/4)
= 11/20
Time taken by both is same
time = dist/speed
So (dist/ speed) of tortoise = (dist / speed) of hare
let speed of tortoise be t and of hare be h
11/20 t = 1/4 h
h = (20/(4*11) )* t = (5/11) * t
Now for the next part hare has to cover 3/4 th distance when tortoise covers 1/4 th distance.
Let new speed of Hare = H
so we get
1/4t = 3/4H
H= 3t
so the factor by which h's speed increases = 3t/ (5/11 * t)
which we get as 33/5 = 6.6

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Sindhu Siva  (12 years ago)

1/5,1/4
5*4=20
20-5=15
15-4=11
15*11/5^2=
6.6

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sandeep de  (12 years ago)

After tortoise covers 1/5rd distance hare starts the race
When hare covers 1/4th of distance tortoise meets hare
So distance covered by tortoise = 1-(1/5 + 1/4)
= 11/20
Time taken by both is same
time = dist/speed
So (dist/ speed) of tortoise = (dist / speed) of hare
let speed of tortoise be t and of hare be h
11/20t = 1/4h
h = (20/(11x4) )x t = 5/11 * t
Now for the next part hare has to cover 3/4 the distance when tortoise covers 1/4 the distance
so we get
1/4t = 3/4h
h = 3t
so the factor by which h's speed increases = 3t/ (5t/11)
which we get as 33/5…so ans is 6.6(d)

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