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by sanju09 » Fri May 14, 2010 2:29 am
It takes a child 90 seconds to climb the 60-meter length of an escalator which is not working. When in operation, the escalator lifts a passenger from bottom to top in 60 seconds. How many seconds does it take the child to cover the 60 meter, if she walks on the moving escalator?
(A) 18
(B) 24
(C) 28
(D) 36
(E) 72
The mind is everything. What you think you become. -Lord Buddha



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by jusgaurav » Fri May 14, 2010 3:00 am
is it Option D?
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by sanju09 » Fri May 14, 2010 3:07 am
jugaurav wrote:is it Option D?
please explain, always
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by kevincanspain » Fri May 14, 2010 3:07 am
sanju09 wrote:It takes a child 90 seconds to climb the 60-meter length of an escalator which is not working. When in operation, the escalator lifts a passenger from bottom to top in 60 seconds. How many seconds does it take the child to cover the 60 meter, if she walks on the moving escalator?
(A) 18
(B) 24
(C) 28
(D) 36
(E) 72
You could think of this problem in terms of combined work. The child takes 3/2 minutes to complete the task of climbing, whereas the escalator takes 1 minutes. Thus in 1 minute, the fraction of the task completed is 2/3 + 1 = 5/3. Therefore, the child will take 3/5 of 1 minute (i.e. 36 seconds)
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by sanju09 » Fri May 14, 2010 3:10 am
kevincanspain wrote:
sanju09 wrote:It takes a child 90 seconds to climb the 60-meter length of an escalator which is not working. When in operation, the escalator lifts a passenger from bottom to top in 60 seconds. How many seconds does it take the child to cover the 60 meter, if she walks on the moving escalator?
(A) 18
(B) 24
(C) 28
(D) 36
(E) 72
You could think of this problem in terms of combined work. The child takes 3/2 minutes to complete the task of climbing, whereas the escalator takes 1 minutes. Thus in 1 minute, the fraction of the task completed is 2/3 + 1 = 5/3. Therefore, the child will take 3/5 of 1 minute (i.e. 36 seconds)
mind blowing
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by jusgaurav » Fri May 14, 2010 3:19 am
Applying the formula -> speed = distance/ time can help one sail thru; something which Kevin has illustrated. One twist that can be applied to this question is as follows....
What if the child starts at the top of the elevator and decides to come down the elevator, which means (s)he is moving against the direction in which elevator is moving. In this case, how much time would the child take to come down?
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by gmatjedi » Fri May 14, 2010 3:24 am
d=rt
t= d/r
combine rates

t=60 m/[60m/90s+ 60m/60s]
t=36 seconds
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by sanju09 » Fri May 14, 2010 3:46 am
jugaurav wrote:Applying the formula -> speed = distance/ time can help one sail thru; something which Kevin has illustrated. One twist that can be applied to this question is as follows....
What if the child starts at the top of the elevator and decides to come down the elevator, which means (s)he is moving against the direction in which elevator is moving. In this case, how much time would the child take to come down?
As the child is faster than the escalator, so she'll surely get to the bottom end sometime, in the changed case. On this occasion, the effective speed would be the difference in the two speeds. But, the climbing rate of child cannot be expected to be same as her descend speed on the stationary escalator, and I could go one step further to doubt the same thing in the descend of the escalator as well. So, supply the missing info, and give us a chance.
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by jusgaurav » Fri May 14, 2010 4:15 am
sanju09 wrote:
jugaurav wrote:Applying the formula -> speed = distance/ time can help one sail thru; something which Kevin has illustrated. One twist that can be applied to this question is as follows....
What if the child starts at the top of the elevator and decides to come down the elevator, which means (s)he is moving against the direction in which elevator is moving. In this case, how much time would the child take to come down?
As the child is faster than the escalator, so she'll surely get to the bottom end sometime, in the changed case. On this occasion, the effective speed would be the difference in the two speeds. But, the climbing rate of child cannot be expected to be same as her descend speed on the stationary escalator, and I could go one step further to doubt the same thing in the descend of the escalator as well. So, supply the missing info, and give us a chance.
i agree with you. the idea was to just to illustrate the calculation of effective speed (which you've correctly pointed out as the difference in the two speeds).
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by akdayal » Fri May 14, 2010 8:46 am
use relative velocity concept.
Vc,e = 60/90 = 2/3
Ve = 60/60 = 1
Vc = Vc,e + Ve (these all are in same direction in question)
Vc = 1 + 2/3 = 5/3
t = d/Vc = 60/(5/3) = 36 sec
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by harshavardhanc » Fri May 14, 2010 11:37 am
sanju09 wrote:It takes a child 90 seconds to climb the 60-meter length of an escalator which is not working. When in operation, the escalator lifts a passenger from bottom to top in 60 seconds. How many seconds does it take the child to cover the 60 meter, if she walks on the moving escalator?
(A) 18
(B) 24
(C) 28
(D) 36
(E) 72
one thing to observe is that the distance is constant Hence, speed will be inversely proportional to time. OR speed ratios in any two cases will be inverse of time ratios.

In the first case , (speed of child)/(speed of escalator) = 60/90 = 2/3 (inverse ratio of their times)

now in the second case, when both are moving, you can think that the speeds get added or (2 units are added to the 3 : by above ratio).
So, (combined speed)/(speed of escalator) = (2+3)/3 = 5/3 .

And as said before, the ratio of times will be inverse or will be 3/5 which will be equal to
(time when both are moving)/(time when escalator moves alone)

OR

3/5 = x/60 => x = 36
Regards,
Harsha
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