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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels

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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   06 Apr 2015, 07:46
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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   06 Apr 2015, 10:31
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Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


Let the total distance between Birmingham and Newcastle be D, and let x be the rate of Train B.

A travels at 100km/hr so total travel time of A is D/100 hours. B travels at x km/hr so total travel time of B is D/x hours.

\(\frac{D}{100} + \frac{D}{x} = 2\)

During the first hour, A travels 100 km. B travels x kilometers/hour * (1/6 hours) = x/6 kilometers. The total distance traveled by either A or B is then
\(100 + \frac{x}{6} = D.\)

This is a very nasty quadratic to solve which is much harder than typical GMAT quadratics.

The solutions to (D,x) are (133.3, 200) and (150,300).

(1) Train B arrived at Newcastle before train A arrived at Birmingham.

If (D,x) = (133.3, 200) then Train B traveled 133.3 km at 200km/hr, so B traveled for 2/3 hours, or 40 minutes and arrived at 4:30. A traveled 133.33 km at 100km/hr, so A traveled for 1.33 hours, or 80 minutes, and arrived at 4:20.

If (D,x) = (150,300) then Train B traveled 150km at 300 km/hr, so B traveled for 30 minutes and arrived at 4:20. A traveled for 1.5 hours and arrived at 4:30.

If B has to arrive before A, then only (150,300) fits. Sufficient.

(2) The distance between Newcastle and Birmingham is greater than 140 km.

This says that D>140. Only (D,x) = (150,300) fits. Sufficient.

Therefore, either option is sufficient. Answer: D
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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   Updated on: 07 Apr 2015, 05:20
Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


I am not sure about my approach, but can't find mistake in it. 

We know that first train took 60 minutes before meet train B and train B took 10 minutes before this event. So we have sum 70 minutes. And we know that their common sum of time equal to 120 minutes. So the rest of distance took in common 50 minutes

1) From this statement we know that B time for last part < A time for last part. 
So B < A and A + B = 50

Let's try some variants: 
B = 10 minutes for 100 km, so first part of distance equal to last = 100 km. Train A need 60 minutes to arrive in Birmingham. Wrong variant
B = 15 minutes for 100 km, so first part of distance equal to 2/3 = near 66 km. Train A need near 40 minutes to arrive in Birmingham. Wrong variant but closer.
B = 20 minutes for 100 km, so first part of distance equal to 1/3 = 50 km. Train A need near 30 minutes to arrive in Birmingham. Correct variant.
And this is sufficient.

2) Two variants are possible:
Train A took 20 minutes for last part and drive 100/3 km
So train B took 30 minutes and drive 100 km

Second variant
Train A took 30 minutes for last part and drive 50 km
So train B took 20 minutes and drive 50 km

But in first variant total distance equal to 133,33 km and this is less than 140
And in second variant distance equal to 150 km
Sufficient

Answer is D

Originally posted by Harley1980 on 06 Apr 2015, 12:18.
Last edited by Harley1980 on 07 Apr 2015, 05:20, edited 1 time in total.
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Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   07 Apr 2015, 04:35
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\(V_2\) is our variable.
\(\frac{1}{6}+\frac{100}{V_2} + 1 + \frac{V2}{(6*100)} = 2\)
\(\frac{100}{V_2} + \frac{V_2}{600} = \frac{5}{6}\)
\(\frac{120}{V_2} + \frac{V_2}{500} = 1\)
\(V_2^2 - 500*V_2 + 120*500 = 0\)
\(V_2^2 - (200+300)*V_2 + 200*300 = 0\)
Vieta's formula easily gives us 2 values of \(V_2\) which are 200 and 300.
Now lets use our given options

#1 - Train B arrived at Newcastle before train A arrived at Birmingham.
This one we can interpret as this: time it took travel B to travel 100 km was less than it took train A to travel \(\frac{V_2}{6}\) km
So, for \(V_2 = 200\) we get \(\frac{100}{200} < \frac{200}{(6*100)}\) => \(\frac{1}{2} < \frac{1}{3}\) which is incorrect
For \(V_2 = 300\) we get \(\frac{100}{300} < \frac{300}{(6*100)}\) => \(\frac{1}{3}< \frac{1}{2}\) which is CORRECT

This leaves us with 1 option for \(V_2\) and thus with only 1 option for everything else including arrival time, Sufficient

#2 The distance between Newcastle and Birmingham is greater than 140 km.
To use this option we need to check if \(\frac{V2}{6} > 40\). Only if \(V_2 = 300\) the inequality holds, thus, yet again, we are left with just one option for \(V_2\) and thus we can answer our question about arrival time with no problems. Sufficient

D that is then.
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   13 Apr 2015, 07:36
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Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


Check similar question HERE.
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   26 Jun 2019, 11:49
eaze wrote:
Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


Let the total distance between Birmingham and Newcastle be D, and let x be the rate of Train B.

A travels at 100km/hr so total travel time of A is D/100 hours. B travels at x km/hr so total travel time of B is D/x hours.

\(\frac{D}{100} + \frac{D}{x} = 2\)

During the first hour, A travels 100 km. B travels x kilometers/hour * (1/6 hours) = x/6 kilometers. The total distance traveled by either A or B is then
\(100 + \frac{x}{6} = D.\)

This is a very nasty quadratic to solve which is much harder than typical GMAT quadratics.

The solutions to (D,x) are (133.3, 200) and (150,300).

(1) Train B arrived at Newcastle before train A arrived at Birmingham.

If (D,x) = (133.3, 200) then Train B traveled 133.3 km at 200km/hr, so B traveled for 2/3 hours, or 40 minutes and arrived at 4:30. A traveled 133.33 km at 100km/hr, so A traveled for 1.33 hours, or 80 minutes, and arrived at 4:20.

If (D,x) = (150,300) then Train B traveled 150km at 300 km/hr, so B traveled for 30 minutes and arrived at 4:20. A traveled for 1.5 hours and arrived at 4:30.

If B has to arrive before A, then only (150,300) fits. Sufficient.

(2) The distance between Newcastle and Birmingham is greater than 140 km.

This says that D>140. Only (D,x) = (150,300) fits. Sufficient.

Therefore, either option is sufficient. Answer: D



Hi,

Could you please show the working out of how you got D= 150 and x = 300 & D = 200 and x= 200
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   28 Jul 2019, 08:04
eaze wrote:
Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


Let the total distance between Birmingham and Newcastle be D, and let x be the rate of Train B.

A travels at 100km/hr so total travel time of A is D/100 hours. B travels at x km/hr so total travel time of B is D/x hours.

\(\frac{D}{100} + \frac{D}{x} = 2\)

During the first hour, A travels 100 km. B travels x kilometers/hour * (1/6 hours) = x/6 kilometers. The total distance traveled by either A or B is then
\(100 + \frac{x}{6} = D.\)

This is a very nasty quadratic to solve which is much harder than typical GMAT quadratics.

The solutions to (D,x) are (133.3, 200) and (150,300).

(1) Train B arrived at Newcastle before train A arrived at Birmingham.

If (D,x) = (133.3, 200) then Train B traveled 133.3 km at 200km/hr, so B traveled for 2/3 hours, or 40 minutes and arrived at 4:30. A traveled 133.33 km at 100km/hr, so A traveled for 1.33 hours, or 80 minutes, and arrived at 4:20.

If (D,x) = (150,300) then Train B traveled 150km at 300 km/hr, so B traveled for 30 minutes and arrived at 4:20. A traveled for 1.5 hours and arrived at 4:30.

If B has to arrive before A, then only (150,300) fits. Sufficient.

(2) The distance between Newcastle and Birmingham is greater than 140 km.

This says that D>140. Only (D,x) = (150,300) fits. Sufficient.

Therefore, either option is sufficient. Answer: D


Hi, when I try to make a quadratic equation, I get \((100+x/6)/100 + (100 + x/6) / x = 2\)
After I bring the equation to final state, I get \(x^2-500x+60000=0\)
And then I get in discriminant negative number.
Please can you show me where is the flaw in my solution?

Thank you
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   19 Aug 2019, 03:34
RusskiyLev wrote:
eaze wrote:
Bunuel wrote:
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

(1) Train B arrived at Newcastle before train A arrived at Birmingham.
(2) The distance between Newcastle and Birmingham is greater than 140 km.


Kudos for a correct solution.


Let the total distance between Birmingham and Newcastle be D, and let x be the rate of Train B.

A travels at 100km/hr so total travel time of A is D/100 hours. B travels at x km/hr so total travel time of B is D/x hours.

\(\frac{D}{100} + \frac{D}{x} = 2\)

During the first hour, A travels 100 km. B travels x kilometers/hour * (1/6 hours) = x/6 kilometers. The total distance traveled by either A or B is then
\(100 + \frac{x}{6} = D.\)

This is a very nasty quadratic to solve which is much harder than typical GMAT quadratics.

The solutions to (D,x) are (133.3, 200) and (150,300).

(1) Train B arrived at Newcastle before train A arrived at Birmingham.

If (D,x) = (133.3, 200) then Train B traveled 133.3 km at 200km/hr, so B traveled for 2/3 hours, or 40 minutes and arrived at 4:30. A traveled 133.33 km at 100km/hr, so A traveled for 1.33 hours, or 80 minutes, and arrived at 4:20.

If (D,x) = (150,300) then Train B traveled 150km at 300 km/hr, so B traveled for 30 minutes and arrived at 4:20. A traveled for 1.5 hours and arrived at 4:30.

If B has to arrive before A, then only (150,300) fits. Sufficient.

(2) The distance between Newcastle and Birmingham is greater than 140 km.

This says that D>140. Only (D,x) = (150,300) fits. Sufficient.

Therefore, either option is sufficient. Answer: D


Hi, when I try to make a quadratic equation, I get \((100+x/6)/100 + (100 + x/6) / x = 2\)
After I bring the equation to final state, I get \(x^2-500x+60000=0\)
And then I get in discriminant negative number.
Please can you show me where is the flaw in my solution?
Thank you


I have the exact same issue..

Can anyone help?
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink] New post   18 Aug 2022, 05:38
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Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels [#permalink]
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