Latest HCL Aptitude Question SOLUTION: A man starts walking at 3 pm, He walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill, 6 km/hr downhill and then 4 km/hr on level ground to reach home at

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A man starts walking at 3 pm, He walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill, 6 km/hr downhill and then 4 km/hr on level ground to reach home at 9 pm. What is the distance covered on one way?

Asked In HCL sushmita (11 years ago)

Solved Garima Read Solution (6)

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Read Solution (6) :

Garima  (11 years ago)

if x km is level ground and y km is hill road, then
x/4 + y/3 +y/6+x/4= 6
6(x+y)=72
x+y=12 km

so one side distance is 12 kms.

Is this solution Helpfull? Yes (84) | No (24)  

sneha yedle  (11 years ago)

as the man contineously with diff speed
find avg of speed by
no. of speed/(1/s1+1/s2+1/s3+1/s4)=avg speed
4/(1/4+1/3+1/6+1/4)=4km/hr
time=9-3=6
6*4=24kms

Is this solution Helpfull? Yes (27) | No (7)  

Aninda Ghosh  (11 years ago)

dear garima,you have taken x twice in the equation and also y....so keeping that frame in mind your total distance in one side should be 2(x+y)= 24 kms

Is this solution Helpfull? Yes (17) | No (19)  

abhishek  (9 years ago)

incomplete ques.as distance on plane is not given to be same

Is this solution Helpfull? Yes (3) | No   

narendra  (9 years ago)

here level ground and hill road

let
level ground=x
hill road =y
starting time=3 p.m
ending time=9 p.m
therefore time=9-3
=6
let eq
x/4+y/3+y/6+x/4=6
(6x+6y)/12=6
6x+6y=70
6(x+y)=70
x+y=70/6
+y=12kms

Is this solution Helpfull? Yes (2) | No   

Souradip Sen  (5 years ago)

The question is " A man starts walking ***from home*** at 3 pm . ha walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill , 6 km/hr downhill and then 4 km/hr on level ground to reach ***back*** home at 9 pm. What is the distance covered on one way?"

Now the ans is 12 as GARIMA explained...
the question was not properly explained

Is this solution Helpfull? Yes (1) | No   

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