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Joined: 10 Sep 2012
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The Z train leaves station A moving at a constant speed, and
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30 Oct 2012, 21:45
Question Stats:
78% (02:41) correct 22% (02:54) wrong based on 280 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=183=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 AC. Since the speed is constant, divide the distances of AB and BC as 7 km and 5 km, respectively, making m equal 75=2 additional kilometers in the first leg. The question asks for the value of the distance AC, 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.
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Re: The Z train leaves station A moving at a constant speed, and
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30 Oct 2012, 21:57
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=183=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 AC. Since the speed is constant, divide the distances of AB and BC as 7 km and 5 km, respectively, making m equal 75=2 additional kilometers in the first leg. The question asks for the value of the distance AC, 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 ABBC = 2 But from question, ABBC =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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Re: The Z train leaves station A moving at a constant speed, and
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23 Feb 2013, 08:03
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 75 = 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.



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Re: The Z train leaves station A moving at a constant speed, and
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29 Mar 2013, 01:43
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



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Re: The Z train leaves station A moving at a constant speed, and
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29 Mar 2013, 03:33
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=183=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 AC. Since the speed is constant, divide the distances of AB and BC as 7 km and 5 km, respectively, making m equal 75=2 additional kilometers in the first leg. The question asks for the value of the distance AC, 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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Re: The Z train leaves station A moving at a constant speed, and
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26 Nov 2013, 10:12
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=183=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 AC. Since the speed is constant, divide the distances of AB and BC as 7 km and 5 km, respectively, making m equal 75=2 additional kilometers in the first leg. The question asks for the value of the distance AC, 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



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Re: The Z train leaves station A moving at a constant speed, and
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07 Dec 2015, 14:02
d=distance from A to B dm=distance from B to C 2dm=distance from A to C d/7=(dm)/5 d=7m/2 2dm=(14m/2)m=6m



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Re: The Z train leaves station A moving at a constant speed, and
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19 Apr 2017, 20:54
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Re: The Z train leaves station A moving at a constant speed, and &nbs
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