GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 07 Dec 2019, 09:04

### GMAT Club Daily Prep

#### 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

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

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59588
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

06 Apr 2015, 07:46
2
32
00:00

Difficulty:

95% (hard)

Question Stats:

25% (02:49) correct 75% (03:03) wrong based on 263 sessions

### HideShow timer Statistics

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.

_________________
Current Student
Joined: 23 May 2013
Posts: 183
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

06 Apr 2015, 10:31
3
2
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
##### General Discussion
Retired Moderator
Joined: 06 Jul 2014
Posts: 1215
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

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

_________________

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.
Manager
Joined: 17 Mar 2015
Posts: 113
Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

07 Apr 2015, 04:35
A picture
$$V_2$$ 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.
Math Expert
Joined: 02 Sep 2009
Posts: 59588
Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

13 Apr 2015, 07:36
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.
_________________
Manager
Joined: 07 May 2018
Posts: 61
Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

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
Manager
Joined: 26 Mar 2019
Posts: 107
Concentration: Finance, Strategy
Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

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
Manager
Joined: 02 Nov 2018
Posts: 54
Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels  [#permalink]

### Show Tags

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?
Re: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels   [#permalink] 19 Aug 2019, 03:34
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne