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An escalator moves downward from street level to a subway platform at [#permalink]
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26 Mar 2017, 04:05
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An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.) (A) Between 50 and 100 steps (B) Between 100 and 200 steps (C) Between 200 and 500 steps (D) Between 500 and 1000 steps (E) Over 1000 steps Source: Manhattan GMAT Challenge question
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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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26 Mar 2017, 05:10
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iMyself wrote: An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.) (A) Between 50 and 100 steps (B) Between 100 and 200 steps (C) Between 200 and 500 steps (D) Between 500 and 1000 steps (E) Over 1000 steps Source: Manhattan GMAT Challenge question SOLUTION: Let the level distance between street to Subway = D Let the speed at which Wesley moves when escalator is still= u Let the speed at which escalator moves up or down = v When escalator is stand still he needs 70 steps Therefore D/u = 70 => u = D/70 ............. (I) Also when both escalator and Wesley moves in same direction he needs 36 steps Therefore D/(u +v) = 36 => u + v = D/36 ........... (Ii) Solving both (i) & (ii) we get v= 17D/(35x36) ..........(iii) Hence u  v = D/(35x36) Therefore Number of steps required when escalator and Wesley are moving in opposite directions = D/(uv) = D/ (D/(35x36)) = 35x36 = 1,260 (which is greater than 1000 steps). Hence E



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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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27 Mar 2017, 23:33
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Shubhranil88 wrote: iMyself wrote: An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.) (A) Between 50 and 100 steps (B) Between 100 and 200 steps (C) Between 200 and 500 steps (D) Between 500 and 1000 steps (E) Over 1000 steps Source: Manhattan GMAT Challenge question SOLUTION: Let the level distance between street to Subway = D Let the speed at which Wesley moves when escalator is still= u Let the speed at which escalator moves up or down = v When escalator is stand still he needs 70 steps Therefore D/u = 70 => u = D/70 ............. (I) Also when both escalator and Wesley moves in same direction he needs 36 steps Therefore D/(u +v) = 36 => u + v = D/36 ........... (Ii) Solving both (i) & (ii) we get v= 17D/(35x36) ..........(iii) Hence u  v = D/(35x36) Therefore Number of steps required when escalator and Wesley are moving in opposite directions = D/(uv) = D/ (D/(35x36)) = 35x36 = 1,260 (which is greater than 1000 steps). Hence E Can you elaborate more on step iii?



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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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04 Apr 2017, 06:56
Is there a easy way to solve this question.?
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An escalator moves downward from street level to a subway platform at [#permalink]
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17 Aug 2017, 10:15
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sarugiri wrote: Is there a easy way to solve this question.? When the person is not aided by Escalator movement..he takes 70 steps...so total distance in terms of steps is 70 When his movement is aided by escalator ..he has to take only 36 steps, i.e., rest of the 34 steps are taken care by escalator movement. Or we can say that for every 36 steps taken by person..escalator is helping him with 34 steps.. Now..when he tries to move up in a downward escalator..his same movement will now be opposed by escalator, So...for each of his 36 steps, escalator will oppose him with 34 steps..so resultant will be 2 steps only.. As the total distance in terms of steps is 70...and for each of his 36 steps..person is able to cover only 2 steps due to downward motion of escalator..so total no. of steps required by person will be (36*70/2)=1260 steps...Hence option E Give kudos plz if you find above solution useful..
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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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26 Aug 2017, 03:40
devarshi9283 wrote: sarugiri wrote: Is there a easy way to solve this question.? When the person is not aided by Escalator movement..he takes 70 steps...so total distance in terms of steps is 70 When his movement is aided by escalator ..he has to take only 36 steps, i.e., rest of the 34 steps are taken care by escalator movement. Or we can say that for every 36 steps taken by person..escalator is helping him with 34 steps.. Now..when he tries to move up in a downward escalator..his same movement will now be opposed by escalator, So...for each of his 36 steps, escalator will oppose him with 34 steps..so resultant will be 2 steps only.. As the total distance in terms of steps is 70...and for each of his 36 steps..person is able to cover only 2 steps due to downward motion of escalator..so total no. of steps required by person will be (36*70/2)=1260 steps...Hence option E Give kudos plz if you find above solution useful.. Amzing solution. How did you come up with this?



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An escalator moves downward from street level to a subway platform at [#permalink]
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26 Aug 2017, 05:55
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iMyself wrote: An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.) (A) Between 50 and 100 steps (B) Between 100 and 200 steps (C) Between 200 and 500 steps (D) Between 500 and 1000 steps (E) Over 1000 steps Source: Manhattan GMAT Challenge question This is how I usually go with questions that involve rate. Whether it is Time, Speed, and distance or time and work. I don't know my total distance. So I assume it to be the LCM(70,36)=1260. Let us just give a unit for distance, 1260 miles. Wesley covers 1260 miles in 70 steps. His speed is \(\frac{1260}{70}\)=18miles/step. Wesley and the escalator together cover 1260 miles in 36 steps. So the combined speed is \(\frac{1260}{36}\) = 35miles/step This means the speed of escalator is 3518 = 17 miles/step. Now when he is going up, the escalator's motion is against his motion. So the net effect of rates will be Wesley's speed  Escalator speed = 1817=1mile/step. Distance=1260 miles. Speed = 1 mile/step. So number of steps = \(\frac{1260}{1}\) = 1260.



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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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16 Sep 2017, 06:47
Intuitively speaking, it seems that the speed of the escalator and the speed of Wesley are equal (while descending he covers 2 steps when the escalator is on). So, if the speed of the escalator and Wesley are unchanged when he ascends, every upward step Wesley takes will be cancelled by the downward movement of the escalator. In that case (E) looks like the best answer.



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Re: An escalator moves downward from street level to a subway platform at [#permalink]
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19 Sep 2017, 04:07
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Answer Assume that the escalator is 70 steps long, since it takes Wesley 70 steps to descend when the escalator isn’t moving. When the escalator is turned on, Wesley takes only 36 steps to descend, which means the escalator is doing the work of 70 – 36 = 34 steps in the time that it takes Wesley to descend. For every 36 steps Wesley takes, then, the escalator “takes” 34 steps. If Wesley reverses his trip and walks upward, against the escalator, then Wesley “gains” 2 steps on the escalator in the period of time it would normally take to descend to the platform. That is, for every 36 steps that Wesley takes, he actually moves 2 steps up the escalator. Since he has to “gain” a total of 70 steps in order to make it to the top of the escalator, he must gain a total of 2 steps 35 times. In total, then, he takes (36)(35) steps. That number is greater than 1,000. (The exact number is 1,260 steps, but don’t do math that you don’t have to do!) The correct answer is (E).
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