Latest Infosys Aptitude Question SOLUTION: Jack and Jill went up and down a hill. They started from the bottom and Jack met jill again 20 miles from the top while returning. Jack completed the race 1 min a head of jill. if

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(#M40040111) INFOSYS question Keep an EYE

Jack and Jill went up and down a hill. They started from the bottom and Jack
met jill again 20 miles from the top while returning. Jack completed the race
1 min a head of jill. if the hill is 440 mts high and their speed while down
journey is 1.5 times the up journey. how long it took for the Jack to
complete the race ?

Asked In Infosys Namita (9 years ago)

Solved Aniruddha Mishra Read Solution (2)

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Read Solution (2) :

Aniruddha Mishra  (9 years ago) Best Solution

(20 miles is greater than 440 meter. Assuming that height of the hill is 440 miles )

Let speed of Jack when going up = x miles/minute
and speed of Jill when going up =y miles/minute

Then speed of Jack when going down = 1.5x miles/minute
and speed of Jill when going up =1.5y miles/minute

Case 1 : Time taken for Jack to travel 440 miles up and 20 miles down
= Time taken for Jill to travel 440-20=420 miles up

440/x + 20/1.5x = 420/y
68/1.5x = 420/y
68y = 63x
y = 63x/68 ---(1)

Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down
= Time taken for Jill to travel 440 miles up and 440 miles down - 1

440/x + 440/1.5x = 440/y + 440/1.5y - 1
1100/1.5x = 1100/1.5y - 1
1100/1.5x = 1100/(1.5 * 63x/68) - 1
1100/1.5x = 68*1100/(1.5 * 63x) - 1
1100/1.5x - 68*1100/(1.5 * 63x) = -1
1100/1.5x (1 - 68/63) = -1
1100/1.5x (-5/63) = -1
1100/1.5x = 63/5
x = 1100*5/(1.5*63) = 11000/189

Time taken by Jack to complete the race = 440/x + 440/1.5x
= 1100/1.5x = 1100/(1.5*11000/189) = 189/15 = 12.6 minutes

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Payal Jain  (9 years ago)

Let speed and time of Jack be s1 & t1 & Jill having s2 & t2.

Time taken for Jack to travel 440 miles up and 20 miles down
= Time taken for Jill to travel 440-20=420 miles up

440/s1 + 20/1.5s1 = 420/s2
68/1.5s1= 420/s2
68s2= 63s1
s2 = 63s1/68 ---(1)

Now from the second given condition- Jack reaches 1 min before Jill. So
t2-t1=1
880/s2 - 880/s1 = 1
1/s2 - 1/s2 = 1/880
from equation (1) put value of s2
Solving and putting value of s1 in t1=d/s1 (d=880)
t1 = 12.6 min

Is this solution Helpfull? Yes (5) | No (1)  

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