Latest TCS Aptitude Question SOLUTION: a hare and tortoise hav a race along a 100 yards diameter .tortoise goes in one n hare in other.hare starts aftr tortoise has covered 1/4 of the distnce.they meet when hare has cov

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Time Distance and Speed

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40026410) TCS Numerical Ability question Keep an EYE

a hare and tortoise hav a race along a 100 yards diameter .tortoise goes in one n hare in other.hare starts aftr tortoise has covered 1/4 of the distnce.they meet when hare has covered only 1/5 of the distance.by wat factor should hare increase his speed so as to tie the race.?

Asked In TCS asxa asxaxax (10 years ago)

Unsolved Read Solution (10)

Is this Puzzle helpful?   (1)   (1) Submit Your Solution Time Distance and Speed

Read Solution (10) :

Ankit Upadhyay  (10 years ago)

11 times.
Assume the total distance to be 200.
distance covered by tortoise before hare start the race - 50.
so, now remaining distance to be covered by tortoise is 150.
1/5 of the distance ( 40 ) , which is covered by hare, when they meet, In the mean time distance covered by tortoise is 110.
Ratio of speed of tortoise and hare- 11:4
Remaining distance to be covered by tortoise to end the race - 40.
Time taken by Tortoise to cover remaining distance - 40/11.
Speed to maintained by hare to tie the race (x)-> x*40/11=160;
x=4*11;
x=44;
so hare has to increase his speed by the factor of 11 to tie the race.

Is this solution Helpfull? Yes (24) | No (6)  

amit kumar burnwal  (10 years ago)

as given,remaining distance cover for tortoise = 1/5 of total distance.
remaining distance cover for hare = 1-1/5 = 4/5 of total distance.
so according to question,
by wat factor should hare increase his speed so as to tie the race,
let factor = x
so, x*1/5 = 4/5
hence, x=4
i.e by a factor of 4.

Is this solution Helpfull? Yes (22) | No (6)  

Deepika Jain  (10 years ago)

Assume the circumference of the circle is 200 meters. Hare and tortoise started at the same point but moves in the opposite direction. It is given that by that time tortoise covered 40 m (1/5th of the distance), Hare started and both met after hare has covered 25. This implies, in the time hare has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.
So Hare : tortoise speeds = 25 : 135 = 5 : 27
Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
As time =Distance/Speed=25/27=175/5×K
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8

Is this solution Helpfull? Yes (15) | No (21)  

Rajat Arora  (10 years ago)

4 times
is it correct??

Is this solution Helpfull? Yes (10) | No (5)  

palak  (10 years ago)

Assume the circumference of the circle is 200 meters. Hare and tortoise started at the same point but moves in the opposite direction. It is given that by that time tortoise covered 40 m (1/5th of the distance), Hare started and both met after hare has covered 25. This implies, in the time hare has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.
So Hare : tortoise speeds = 25 : 135 = 5 : 27
Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
As time =Distance/Speed=25/27=175/5×K
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8

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

Nishant Tyagi  (10 years ago)

10 times..

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

ajit kumar mahato  (10 years ago)

i think 4 times

Is this solution Helpfull? Yes (1) | No   

aditya jain  (9 years ago)

when hare complete its 1/5th=20 yards both are meet that means hare complete its 4/5th =80 yards
so hare is 20 yard complete
tortoise is 80 yard complete
so if hare have to finish as same tortoise so it cover remaining 80 distant in same time as tortoise
so
80/x=20/y
x=4y.
x= speed of hare
y=speed of tortoise

Is this solution Helpfull? Yes (1) | No   

Sourabh Sharma  (10 years ago)

solving gives that tortoise is already 3 times faster than hare.
moreover, many friends have solved for the factor of 4.

Hence, hare has to increase the speed by 4 times to beat distance and 3 times to beat the speed of tortoise.
Therefore, answer could be...4X4=12 times the current speed.

Is this solution Helpfull? Yes | No (6)  

anshu sangal  (10 years ago)

take 100m circum
hare cover 20m and tortoise cover 55m so we can take tortoise speed 55m/sec and hare speed 20m/sec
now to tie the match hare have to cover 50m in time of in which tortoise cover 25m
so time taken by tortoise=25/55
x*(25/55)=50
x=110 becoz hare previously speed was 20m/sec
and know hare have to increase up to 110m/sec
110/20=5.5

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.