Latest TCS Aptitude Question SOLUTION: a hare and a tortoise a race along a circle of 100 yards dia.the tortoise goes in 1 direction n the hare in te other.the hare starts after tortoise has covered 1/5 its distance .th

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a hare and a tortoise a race along a circle of 100 yards dia.the tortoise goes in 1 direction n the hare in te other.the hare starts after tortoise has covered 1/5 its distance .the hare n tortoise meet when te hare has covrd 1/4 of te distanc.by wat factor should hare increase its speed so as to tie te race?
options
8
37
45
6.6
need a easy way to solve frnds:)

Asked In TCS shreya (12 years ago)

Unsolved Read Solution (7)

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hariharan  (12 years ago)

see the option which one is decimal that is ans for this type of ques in TCS

Is this solution Helpfull? Yes (6) | No (1)  

manas ranjan panda  (12 years ago)

formula:(h-1)(h-1-h/t)
h=4;t=5;
then ans 6.6

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jai Narayan Mishra  (12 years ago)

the answer is 6.6
Just forget 100 yards given in problem...
Multiply 5 by 4(numbers given in denominator of fraction) we get 20 yards...
Now Hare starts when tortoise had covered (1/5) of 40 yards i.e 4 yard...
tortoise meet the hare when hare has covered (1/4) of 40 yards i.e 5 yards..
Now suppose speed of Tortoise is "T" and that of Hare is "H"..therefore
(H/T)= (5/11)..[in the same time hare travels distance of 5 yards while tortoise travels 11 yards]..
now to tie the race hare must have increase it's speed...
to tie the race the ratio of H & T must be (15/5)i.e 3...
suppose by factor 'x' hare has to increase it's speed..
therefore,
(5/11)*x = 3..
x = 3*(11/5)
x= 6.6

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chandra  (12 years ago)

ans is 6.6 always see for . result

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Dhandayuthapani  (12 years ago)

distance of hare distance of tortoise
0 (rest) 20yards(1/5)
25yds(1/4) 55yds(100-45)
now hare crossd now tortoise
25 yds 75yds
hare want 2 go 75 yds and tortoise just 25yds so hare want 2 speed up
75t/25h=25/55
h=6.6t ans 6.6

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SHRI KANT  (12 years ago)

vandana says,
6.6 will be answer here.. :)

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chintan Makwana  (12 years ago)

in this problem at first find the ratio of the distance covered by the turtle with compare to the hare
means-(100(total)-25(cover by hare)-20(all ready covered by turtle))/25
=(55/25)=2.2
and than fine distance left to cover by the hare.
then (100-25(all ready coverd))/25=75/25=3
and multiply these two 2.2*3=6.6

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