speed distance time question

[rishianand7]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

[vinay1983]

Is the answer 5.40 p.m?B

What is the source?

[ganeshrkamath]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

Speed of train A = 60 mph
Train B leaves after 40 minutes.
Distance between the two trains at this time = speed * time = 60*(40/60) = 40 miles

Time taken for the two trains to meet after train B starts
= distance/(relative speed)
= 40/(75-60)
= 40/15 hours
= 40/15*60 minutes = 160 minutes
Train A left 40 minutes before B started.
So, train B meets train A (160+40) = 200 minutes after train A leaves.

3.30pm + 200 minutes = 6.50pm
Choose D

Cheers

[Brent@GMATPrepNow]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

Here's one approach.

Train A had a 40-minute head start (i.e., 2/3 of an hour head start).
Distance = (time)(speed)
So, head start distance = (2/3)(60) = 40-mile head start.

At 4:10, train B leaves the station, traveling at 75 mph
Train A is traveling at 60 mph

IMPORTANT: The 15mph difference in their speeds (75 - 60 = 150) means that, for EVERY HOUR, the distance between the trains shrinks by 15 MILES. In other words, we might say that the "speed of shrinkage" = 15 mph.

So, how long will it take to shrinks the initial 40-mile distance between them?
Time = distance/speed
So, time = 40/15 = 8/3 = 2 2/3 hours = 2 hours 40 minutes

The time when train B overtakes A = 4:10 + (2 hours 40 minutes) [spoiler]= 6:50 = D[/spoiler]

Cheers,
Brent

[Brent@GMATPrepNow]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

Another approach is to begin with a "word equation".
When train B overtakes train A, we know that they have both traveled the same (equal) distance.
So, we can write: Train A's travel distance = Train B's travel distance

Distance = (speed)(time)
We know the speeds of the trains, but what about the times?

IMPORTANT: Let's base travel times from the time train B left the station
So, when train B overtakes train A we can say:
Train B's travel time = t hours
Train A's travel time = t + 2/3 hours (since train A had already traveled for 40 minutes before train B left the station)

Okay, we're ready to transform the word equation into an equation:

Train A's travel distance = Train B's travel distance
(60)(t + 2/3 hours) = (75)(t)
60t + 40 = 75t
40 = 15t
t = 40/15 = 8/3 = 2 2/3 hours.

Since we based the travel times from the time train B left the station (4:10), the time when train B overtakes A = 4:10 + (2 2/3 hours)[spoiler] = 6:50 = D [/spoiler]

Cheers,
Brent

[Brent@GMATPrepNow]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

IMPORTANT: In my approach (using a "word equation"), I based both travel times from the time train B left the station. I want to point out that I don't necessarily have to do this in order to solve the question. Instead I could have based the travel times from the time train A left the station. My answer will still be the same.

I'll show you.

When train B overtakes train A, we know that they have both traveled the same (equal) distance.
So, we can write: Train A's travel distance = Train B's travel distance

Train A's travel time = t hours
Train B's travel time = t - 2/3 hours (since train B did not travel for the first 40 minutes after train A left the station)

Okay, we're ready to transform the word equation into an equation:
Train A's travel distance = Train B's travel distance
(60)(t) = (75)(t - 2/3)
60t = 75t - 50
50 = 15t
t = 50/15 = 10/3 = 3 1/3 hours.

Since we based the travel times from the time train A left the station (3:30), the time when train B overtakes A = 3:30 + (3 1/3 hours)[spoiler] = 6:50 = D [/spoiler]

Cheers,
Brent

[GMATGuruNY]

Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

A's rate = 60mph, B's rate = 75mph.

In the 40 minutes from 3:30pm to 4:10pm, the distance traveled by A = r*t = 60 * 2/3 = 40 miles.

Every hour thereafter, A travels 60 more miles, while B travels 75 miles.
At 5:10pm, A = 40+60 = 100 miles, B = 75 miles.
At 6:10pm, A = 100+60 = 160 miles, B = 75+75 = 150 miles.
At 7:10pm, A = 160+60 = 220 miles, B = 150+75 = 225 miles.
Since B is 5 miles ahead at 7:10pm, B must have overtaken A between 6:10pm and 7:10pm.

The correct answer is D.

[abhasjha]

Fantatstic Mitch!I love the way you utilise answer options . Your Approach is refreshing...

[DamianDavila]

Here is my approach:

B leaves later than A so it first needs to catch up to A, and because it is going faster than A, B will eventually overtake A.

At 4:10 pm, A has covered 2/3 of what it would cover in 1 hour (40/60 minutes), so 60 x 2/3 = 40 miles.

Let's do a table of the trains after every hour:
[Time lapsed] [A]
[0 hours] [40] [0]
[1 hour] [100] [75]
[2 hours] [160] [150]
[3 hours] [220] [225]

As you can see, B is already 5 miles ahead of A after 3 hours, so that means that B overtakes A a couple minutes before 3 hours have passed. The closest answer choice is [spoiler](D) 2 hours 40 minutes[/spoiler]. Remember that the start time of B is 4:10pm.

[Mathsbuddy]

Here's a graphical interpretation.
Please see the diagram.