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# John and Steve are speedwalkers in a race. John is 15 meters behind

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Math Expert
Joined: 02 Sep 2009
Posts: 55272
John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 04:46
12
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Difficulty:

25% (medium)

Question Stats:

78% (02:04) correct 22% (02:53) wrong based on 185 sessions

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John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 05:07
3
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

Let t be the time that John spent for his final push.

Thus, per the question,

4.2t = 3.7t+15+2 ---> 0.5t = 17 ---> t = 34 seconds.

D is the correct answer.
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Joined: 10 Jun 2015
Posts: 118
Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 05:23
1
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

let Steve's distance during his final push be 'd'
d/3.7 =(d+17)/4.2
solving we get d=125.3m
time taken by steve is 125.3/3.7 = 34 seconds.
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 06:10
2
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

J is behind S by 15 m
Relative speed=4.2-3.7=0.5m/s(meaning in 1 s, J covers 0.5 m more than S)
Therefore, to cover a distance of 15m, J will take 15/0.5=30s
Also, to cover additional 2 m, J will take 4 s more
Total time taken=34 S
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 12:36
3
Total distance = (15+2) metres
relative speed =0.5 m/s

final push time =17/0.5 =34 seconds
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 14:57
2
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

Total dist covered by John = 15+2 = 17[m]
relative speed= 4.2 - 3.7[m/s] = 0.5 [m/s]

speed = dist/time = 17/0.5 = 34 sec

D it is
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 16:48
1
John gains .5 m/s on Steve.
John needs to gain 17 meters.
17/.5 = 34 seconds
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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21 Aug 2015, 23:53
1
see that relative speed approach is faster

I used complicated algebra:

(15+x)/4.2=(x-2)/3.7 => x=127.8 is the distance from Steve to finish

John's time=(127.8+15)/4.2=34 sec.

D
Math Expert
Joined: 02 Sep 2009
Posts: 55272
Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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23 Aug 2015, 11:42
1
1
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

Economist GMAT Tutor Official Solution:

There seems to be a lot going on here. John and Steve are both in motion, and John must overtake Steve and move ahead to the finish line. However, this problem can be simplified into a single equation.

Focus first on the rate at which the gap is being closed. The only rate you need to know to solve this problem is the rate at which John is overtaking Steve: 0.5 m/s (4.2 m/s - 3.7 m/s).

Next, treat the distance as a single unit. If we use Steve as our fixed point, we can say that John moves from -15 m (behind Steve) to 2 m (in front of Steve), or a total distance of 17 m.

Finally, calculate the time it takes to move 17 m at a speed of 0.5 m/s (time = distance/rate) to get a total of 34 seconds.

This race sounds like a nail biter, but with a little strategy, GMAT rate problems can be a walk in the park.
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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25 Mar 2019, 05:22
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

relative speed of John - 4.2-3.7 ; .5
distance = 15+2 = 17
time 17/.5 ; 34 sec
IMO D
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Re: John and Steve are speedwalkers in a race. John is 15 meters behind  [#permalink]

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27 Mar 2019, 18:59
Bunuel wrote:
John and Steve are speed walkers in a race. John is 15 meters behind Steve when he begins his final push. John blazes to the finish at a pace of 4.2 m/s, while Steve maintains a blistering 3.7 m/s speed. If John finishes the race 2 meters ahead of Steve, how long was John’s final push?

A. 13 seconds
B. 17 seconds
C. 26 seconds
D. 34 seconds
E. 51 seconds

Kudos for a correct solution.

This is a catch up and pass problem, so we can use the formula:

time = (change in distance)/(change in rate)

time = 17/(4.2 - 3.7) = 17/0.5 = 34 seconds

Alternate Solution:

The distance between John and Steve is decreasing (and after John catches Steve, increasing) by 4.2 - 3.7 = 0.5 meters every second. Since John starts 15 meters behind and finishes 2 meters ahead, he needs to travel 15 + 2 = 17 meters more than Steve. The time to cover this distance is 17/0.5 = 34 seconds.

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Re: John and Steve are speedwalkers in a race. John is 15 meters behind   [#permalink] 27 Mar 2019, 18:59
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# John and Steve are speedwalkers in a race. John is 15 meters behind

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