Question

A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. 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?

Answer 1

Answer:

(37.8 ) times speed  of Original Speed

Explanation:

Let say Distance = X Yard   = (2 π * 100/2) = 100 π Yard

let say Tortoise Speed = T Yard/sec

& Hare Speed = H Yard/Sec

Time taken to complete circle by Tortoise = X/T  sec

Time taken to complete 1/5 of its distance = X/5T

Hare & Tortoise meet when Hare covered 1/8 of Distance = X/8

Time Taken by Hare = X/8H

Distance Covered by Tortoise in such Time =  X - X/8  - X/5

= 27X/40

Time taken by Tortoise for 27X/40  =   27X/40T sec

27X/40T = X/8H

=> H = 5T/27

Now Distnace to be covered by Tortoise = X/8

Time taken by Tortoise = X/8T

Distance to be Covered by Hare = 7X/8

With in time frame of X/8T

Speed required =  (7X/8)/(X/8T) = 7T

7T/(5T/27)  = 189/5 (37.8 ) times speed  of Original Speed

Increase in speed = 7T - (5T/27)  =  184T/27 =

% increased in speed =  (184T/27)/(5T/27) * 100 = 3680 %

Answer 2

Explanation:

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