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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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Bunuel
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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eaze
Bunuel
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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eaze
Bunuel
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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RusskiyLev
eaze
Bunuel
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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