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

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30 Oct 2012, 21:45

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This post received KUDOS

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Difficulty:

25% (medium)

Question Stats:

79% (02:47) correct
21% (02:43) wrong based on 209 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.

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.

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

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30 Oct 2012, 21:57

4

This post received 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.

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 << _________________

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

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23 Feb 2013, 08:03

2

This post received 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.

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.

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:

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 _________________

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

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26 Nov 2013, 10:12

1

This post received 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.

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

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

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07 Dec 2015, 03:20

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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).

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