Latest TCS Aptitude Question SOLUTION: a hare & a tortoise have a race along a circle of 100 yards diameter.the tortoise goes in 1 direction & hare in another.after the tortoise has covered 1/5th of its distance & that

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a hare & a tortoise have a race along a circle of 100 yards diameter.the tortoise goes in 1 direction & hare in another.after the tortoise has covered 1/5th of its distance & that is too leisurely.the hare & tortoise meet when the hare has covered only 1/8 th of the distance.by what factor should the hare increase its speed so as it to tie the race?
5
8
37.80
40

please explain

Solution

let total distance to cover=x
nw first tortoise goes x/5
after dat when they meet together,
hare goes x/8
and so tortoise goes x-(x/5+x/8)=27x/40
so, ratio of tortoise and hare=27x/40 : x/8=27:5
nw to tie up the incresed fctor of spped of hare be v
then,

(x/8)/27=(x-x/8)/5v
or, 1/27=7/5v
or, v=(27*7)/5=37.80

option (c) is correct
this answer is given by BISWAJIT DAS (6 MONTH AGO) and is correct answer

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Asked In TCS bristi (13 years ago)

Solved Read Solution (5)

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Read Solution (5) :

animesh paul  (13 years ago)

assume tortoise as T
hare as H
T upon covering 1/5th,H starts running!
after H covered 1/8 that is T covered 7/8 they meet.
let the time when H starts to when he reaches at 1/8th of the total as 't',that is equal to T's covering of (7/8-1/5)=27/40
T covers 27/40 at 't' time.
then T covers remaining 1/8th in (40/27)t/8 t=5/27t

now in this time H has to reach the finishing point.
initially H goes 1/8th in t
then 7/8 th in 7t.

thus speed should be increased=7t/(5/27)t =37.8 :)

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Prateek Soham Mahanti  (13 years ago)

37.80

Is this solution Helpfull? Yes (2) | No (6)  

saurabh kolhe  (13 years ago)

The trick for this question is that the answer with the decimal is wright because the software for the online written paper of TCS has some flaw which gives a decimal in this question. But it’s not with other questions.
so the answer is 37.80

Is this solution Helpfull? Yes (1) | No (1)  

Ankit  (12 years ago)

As here when hare start the race tortoise has cover 1/5 th of distance.
at the point when they meet hare cover 1/8 and tortoise cover 7/8 of the distance.
the distance travel by tortoise when hare cover its 1/8 distance = 4/5-1/8
= 27/40.
let speed of tortoise is t and hare is h before the point they meet then.
27/40t=1/8h
t= 27/5h
after the point they meet distance travel by tortoise is 1/8 and by hare is 7/8 to finish the race
let new speed of hare= H( after the point when they meet)
tortoise=t
7/8H=1/8t
H=7t
H=7*27/5h
H=189h/5
H=37.8h

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RAVI  (13 years ago)

d=100
2*pie*r=314.15
h covers=.5*314.15=62.83
they both meet at=.8*314.15=39.26
so
h need to cover 212.06(314.15-(62.83+39.26)) at t
1 at t/212.o6
39.26 at 39.26t/212.06----------A

again similarly
for T
39.26 covers at t
1 covers at t/39.26
314-39.26=274.89 covers at .. 274.89t/39.26.........B

divide equation B/A= 37.80

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