Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The Z train leaves station A moving at a constant speed, and [#permalink]
30 Oct 2012, 20:45

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

77% (02:58) correct
23% (02:58) wrong based on 96 sessions

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]
30 Oct 2012, 20:57

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]
29 Mar 2013, 02:33

2

This post received 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.

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]
26 Nov 2013, 09:12

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

Answer is B Hope it helps Cheers J

gmatclubot

Re: The Z train leaves station A moving at a constant speed, and
[#permalink]
26 Nov 2013, 09:12

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...