Latest TCS Aptitude Question SOLUTION:  Hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely3. A hare and a tortoise have a race alongÂ

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(#M40115755) TCS Numerical Ability question time and dist Keep an EYE

 Hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely3. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?

Asked In TCS harish (9 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

sampita ghosh  (9 years ago)

40-5=35
35-8=27
35*27/25=37.8

Is this solution Helpfull? Yes (8) | No (1)  

Divya  (8 years ago)

After tortoise covers 1/5th distance hare starts the race
When hare covers 1/8th of distance tortoise meets hare
So distance covered by tortoise = 1-(1/5 + 1/8)
= 27/40
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
27/40t = 1/8h
h = (40/(8*27) )* t = 5/27 * t
Now for the next part hare has to cover 7/8 th distance when tortoise covers 1/8 th distance
so we get
1/8t = 7/8h
h = 7t
so the factor by which h's speed increases = 7t/ (5/27 * t)
which we get as 37.8

Is this solution Helpfull? Yes (4) | No   

chandan goswami  (9 years ago)

let ,speed of hare be Vh and tortoise be Vt
initially,
Vh:Vt=25:(100pi-20pi-12.5pi)
but to Tie the race ratio should be,
Vh:Vt=125:25=5:1
so 25x/(100pi-20pi-12.5pi)=5/1;
x=42.39 ans

Is this solution Helpfull? Yes (1) | No (2)  

Roshan singh  (9 years ago)

first distance = 1/x
second distance = 1/y

let z = xy-x
required factor will be = z*(z-y)/x^2

in this question..
x=5 , y=8
z=5*8-5=35
z-y = 35-8 = 27
required factor = 35*27/25 = 37.8 ans

Is this solution Helpfull? Yes (1) | No   

Akshara  (9 years ago)

Total circumfrance = pie*diameter
tortoise has covered = 25pie
remaining distance = 75 pie
1/8th of total distance = 12.5 pie
let the velocity of tortoise = u and that of hare =v
on time comparison we get
62.5 pie/u = 12.5 pie/v
5v=u so v=u/5

now for remaining journey time taken by tortoise = 12.5 pie/u
so velocity of hare to make a tie = 7*12.5 pie u / 12.5 pie
=7u
so 7u/u/5 = 35 times answer

Is this solution Helpfull? Yes | No (3)  

Reena  (8 years ago)

@diva plz explain
So distance covered by tortoise = 1-(1/5 + 1/8)????

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