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# The Z train leaves station A moving at a constant speed, and

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Manager
Joined: 10 Sep 2012
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The Z train leaves station A moving at a constant speed, and [#permalink]

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30 Oct 2012, 20:45
1
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4
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Difficulty:

35% (medium)

Question Stats:

75% (01:57) correct 25% (02:30) wrong based on 271 sessions

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The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.
[Reveal] Spoiler: OA

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Director
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Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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30 Oct 2012, 20:57
4
KUDOS
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

1.8m
6m
7m
9m
12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

The reason it is failing for you is that you chose incorrect numbers. If the question says it took 7 hrs to reach from A to B and 5 hrs to reach from B to C at a constant speed. It shows that distance AB and BC should be in ratio of 7/5.

If you take such numbers you can solve problem.
AB = 7, BC=5
Therefore
AB-BC = 2

But from question, AB-BC =m
=> m=2

Now total distance = AB+BC= 12
Substitute 12 to get answer in terms of m
Total distance =12 =6m

Ans B

So you get right answer not by plugging in numbers but by 'plugging in right numbers'

Hope it helps and if does, kudos is right there <<
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Kudos [?]: 676 [4], given: 23

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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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29 Mar 2013, 02:33
4
KUDOS
Expert's post
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

Use a variable, plug in numbers or simply look at the big picture:

Draw the diagram as you read the question.

A.....................7 hrs......................B.................5 hrs............C

A to B the distance is m km extra and the train takes 2 hrs extra to cover m kms. Then, what is the speed of the train? It is (m/2) kms/hr
So the distance between A and C = Speed* Time = (m/2)*12 = 6m
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Kudos [?]: 18153 [4], given: 236

Intern
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Joined: 07 Jan 2013
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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29 Mar 2013, 00:43
3
KUDOS
The above methods are much faster, but I did it the long way.

From the question distance from A to B is "m" miles longer.

Distance from B to C = x
Distance from A to B = x + m

since speed is constant you have two equations equivalent to each other:

Speed = $$\frac{distance}{time}$$

So: $$\frac{x+m}{7}$$ = $$\frac{x}{5}$$

7x = 5x + 5m
x = 2.5m

Add all the distance together = x + x + m = 2.5m + 2.5m + m = 6m, Answer B

Kudos [?]: 43 [3], given: 18

Manager
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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23 Feb 2013, 07:03
2
KUDOS
Time taken from A to B = 7 hours

Time taken from B to C = 5 hours

Distance from A to B is $$m$$ miles more than distance between B to C. Since the average speed is constant, time taken to travel $$m$$ miles is 7-5 = 2 hours.

Total time taken for the entire trip (A to C) = 7 + 5 = 12hours

in 12 hours you can travel $$\frac{12}{2}*m$$ miles = 6m miles.

Kudos [?]: 292 [2], given: 31

Current Student
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Concentration: Finance
Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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26 Nov 2013, 09:12
1
KUDOS
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

Let's say 'r' is the constant rate
Then we have Distance from A to B = 7r
Also distance from B to C = 5r

The difference between the distanced 7r - 5r = m
2r = m

Total distance is 12r (7r+5r)
So in terms of m---> 12 (1/2) = 6m

Hope it helps
Cheers
J

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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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07 Dec 2015, 02:20
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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07 Dec 2015, 13:02
d=distance from A to B
d-m=distance from B to C
2d-m=distance from A to C
d/7=(d-m)/5
d=7m/2
2d-m=(14m/2)-m=6m

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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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19 Apr 2017, 19:54
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: The Z train leaves station A moving at a constant speed, and   [#permalink] 19 Apr 2017, 19:54
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