timing word problem

eric beans

Junior Member
"Two bus companies started service between Lake Geneva and Chicago. The buses take the same highway route and travel at constant speed. Speedy Coach buses leave every 10 minutes and take 90 minutes to reach Chicago. Leisure Tour buses leave every 5 minutes and take 60 minutes to reach Chicago. Leisure Tour's driver Travis, leaves from Lake Geneva 5 minutes after the previous Speedy Coach bus has departed.

How many Speedy Coach buses will driver Travis have passed as he arrives in Chicago?"

I'm stuck on how to solve this. What is a fast and intuitive way to solve this? What type of word problem would this be called? I always get stuck on problems like this that have moving things, and timing/synchronizing. I don't know what to do.

jalexandre

New member
Are we assuming that we're counting how many Speedy buses Travis passed going in the opposite direction of him? Because Travis will not be passing any speedy buses going the same direction as him. They would be passing Travis.

If yes, then are we assuming that the Speedy buses leave Chicago at the exact same 10-minute intervals as they leave Lake Geneva? For example, exactly 10 minutes after a bus arrives in Chicago, another one leaves Chicago, which would mean two buses are leaving Chicago and Geneva at the same time?

I'm also a little confused about the phrasing "as he arrives in Chicago" because that implies to me that we're looking at who is he passing upon his arrival, not on his whole journey.

I understand what this question is trying to ask, but I feel like the reader is supposed to make several assumptions that aren't explicitly stated. Is this the exact phrasing of the question, or is there more?

lev888

Senior Member
Let D be the distance. What's the difference in speeds? How many slow buses are on the road when Travis departs? Reduce both speeds by the difference. Now the slow buses are stationary. How many of them will Travis pass in an hour?

lev888

Senior Member
Are we assuming that we're counting how many Speedy buses Travis passed going in the opposite direction of him? Because Travis will not be passing any speedy buses going the same direction as him. They would be passing Travis.

If yes, then are we assuming that the Speedy buses leave Chicago at the exact same 10-minute intervals as they leave Lake Geneva? For example, exactly 10 minutes after a bus arrives in Chicago, another one leaves Chicago, which would mean two buses are leaving Chicago and Geneva at the same time?

I'm also a little confused about the phrasing "as he arrives in Chicago" because that implies to me that we're looking at who is he passing upon his arrival, not on his whole journey.

I understand what this question is trying to ask, but I feel like the reader is supposed to make several assumptions that aren't explicitly stated. Is this the exact phrasing of the question, or is there more?
They travel in the same direction. The names are pointlessly confusing. Speedy Coach is actually slower than Leisure Tour.

JeffM

Elite Member
I believe students are way too quick to look for formulas and equations. The hardest part of a problem usually involves understanding what we want to know and how to use what we do know to get there. Usually, we know a lot of extraneous material, and part of this step is identifying what we do know that is relevant. Moreover, to think, humans need names for things.

We want to know how many Speedy buses Travis passes. He cannot pass any Speedy Bus that arrives in Chicago before he arrives in Chicago. Nor can he pass any Speedy bus that leaves after him because he drives faster.

The only creative steps needed to solve this problem are (a) to determine the time that the last bus to arrive in Chicago before Travis's arrival left L. Geneva, and (b) relative to that time when did Travis leave L. Geneva.

We are talking relative time so, to avoid negative numbers, we can NAME as ZERO the time of departure from L. Geneva of the bus that was the last to arrive in Chicago before Travis.

Therefore that bus arrived at minute 90. Now we could mess around demonstrating it, but it is fairly obvious then that Travis arrived at minute 95.

Therefore, he left at minute (95 - 60) or 35.

Our problem reduces to: how many Speedy buses left after 0 and before 35.

You could work all day with formulas and equations to solve a fundamentally simple problem. But it becomes that simple only after thinking about what we want to know, what we know that is relevant, and naming things.

EDIT: As lev showed, there are frequently a number of ways to solve a problem.

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lev888

Senior Member
Let D be the distance. What's the difference in speeds? How many slow buses are on the road when Travis departs? Reduce both speeds by the difference. Now the slow buses are stationary. How many of them will Travis pass in an hour?
Correction: reduce both speeds by the slow speed.

eric beans

Junior Member
I believe students are way too quick to look for formulas and equations. The hardest part of a problem usually involves understanding what we want to know and how to use what we do know to get there. Usually, we know a lot of extraneous material, and part of this step is identifying what we do know that is relevant. Moreover, to think, humans need names for things.

We want to know how many Speedy buses Travis passes. He cannot pass any Speedy Bus that arrives in Chicago before he arrives in Chicago. Nor can he pass any Speedy bus that leaves after him because he drives faster.

The only creative steps needed to solve this problem are (a) to determine the time that the last bus to arrive in Chicago before Travis's arrival left L. Geneva, and (b) relative to that time when did Travis leave L. Geneva.

We are talking relative time so, to avoid negative numbers, we can NAME as ZERO the time of departure from L. Geneva of the bus that was the last to arrive in Chicago before Travis.

Therefore that bus arrived at minute 90. Now we could mess around demonstrating it, but it is fairly obvious then that Travis arrived at minute 95.

Therefore, he left at minute (95 - 60) or 35.

Our problem reduces to: how many Speedy buses left after 0 and before 35.

You could work all day with formulas and equations to solve a fundamentally simple problem. But it becomes that simple only after thinking about what we want to know, what we know that is relevant, and naming things.

EDIT: As lev showed, there are frequently a number of ways to solve a problem.

the answer says it's 3. is there any easier way? i'm still somewhat confused. can you draw this out? i'm a visual learner and it helps me more when i can see the problem on paper. i tried to draw this out with lines on top of lines and still got stuck.

lev888

Senior Member
the answer says it's 3. is there any easier way? i'm still somewhat confused. can you draw this out? i'm a visual learner and it helps me more when i can see the problem on paper. i tried to draw this out with lines on top of lines and still got stuck.
Draw a line between Lake Geneva and Chicago. On the line mark the location of each slow bus as of the moment Travis left. The line represents 90 min. So the first slow bus will be 5 min away from the start. Post the image.

eric beans

Junior Member
Draw a line between Lake Geneva and Chicago. On the line mark the location of each slow bus as of the moment Travis left. The line represents 90 min. So the first slow bus will be 5 min away from the start. Post the image.

something like this??? i'm not sure if i did it right.

lev888

Senior Member
View attachment 25992

something like this??? i'm not sure if i did it right.
Looks good. Now under each of those buses write how many minutes it would take them to get to Chicago.

eric beans

Junior Member
Looks good. Now under each of those buses write how many minutes it would take them to get to Chicago.

like this?

eric beans

Junior Member
Yes.
So, given that Travis will be in Chicago in 60 min, which ones of those buses he will pass?

wouldn't it be all the ones between when he started and when 60 minutes is up?

jonah2.0

Full Member
Beer soaked speculation follows.
"Two bus companies started service between Lake Geneva and Chicago. The buses take the same highway route and travel at constant speed. Speedy Coach buses leave every 10 minutes and take 90 minutes to reach Chicago. Leisure Tour buses leave every 5 minutes and take 60 minutes to reach Chicago. Leisure Tour's driver Travis, leaves from Lake Geneva 5 minutes after the previous Speedy Coach bus has departed.

How many Speedy Coach buses will driver Travis have passed as he arrives in Chicago?"

I'm stuck on how to solve this. What is a fast and intuitive way to solve this? What type of word problem would this be called? I always get stuck on problems like this that have moving things, and timing/synchronizing. I don't know what to do.
Let $$\displaystyle d>0$$ be the distance from Lake Geneva to Chicago.
At 5 minutes, the Speedy Coach bus that left 5 min. earlier that Travis must first pass has a headstart of (5/60)*d/90. Since Leisure Tour buses take 60 minutes to reach Chicago, it follows that its speed or rate is d/60. The first Speedy Coach bus that Travis passes is at the point when (5 min. headstart) + d/90*x = d/60*x where x is the time when Travis is side by side with the bus with the 5 min. headstart. This is the time when both buses have covered the same distance.

The next Speedy Coach bus that Travis must pass has a head start of (5 + 10 min.). They will be side by side at the point when (5 + 10 min headstart) + d/90*x = d/60*x. Again, same distance.

When you reach a value of x that's greater than 1, you'd know that you've gone past Chicago.

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lev888

Senior Member
View attachment 25997

wouldn't it be all the ones between when he started and when 60 minutes is up?
Travis will be in Chicago in 60 min.
A particular bus will in Chicago in x min (the second number under the bus).
How do we determine whether Travis will pass this bus?

JeffM

Elite Member
the answer says it's 3. is there any easier way? i'm still somewhat confused. can you draw this out? i'm a visual learner and it helps me more when i can see the problem on paper. i tried to draw this out with lines on top of lines and still got stuck.
The buses leave every 10 minutes

0
10
20
30
40

How many are higher than 0 but lower than 35?

eric beans

Junior Member
Travis will be in Chicago in 60 min.
A particular bus will in Chicago in x min (the second number under the bus).
How do we determine whether Travis will pass this bus?

all these are under 60 minutes so wouldn't he pass them before the 60 minutes? wouldn't the answer be 6 buses not 3?

lev888

Senior Member
View attachment 25998

all these are under 60 minutes so wouldn't he pass them before the 60 minutes? wouldn't the answer be 6 buses not 3?
Travis will be in Chicago in 60 min. The last bus on the right will be in Chicago in 5min. How do you think Travis will pass it??
So, how should the bus time compare to 60 in order for Travis to pass the bus?

eric beans

Junior Member
Travis will be in Chicago in 60 min. The last bus on the right will be in Chicago in 5min. How do you think Travis will pass it??
So, how should the bus time compare to 60 in order for Travis to pass the bus?
oh. because the 3 left ones will take longer than 60 minutes, travis will pass them because those 3 will still be on the road after 60 minutes.

SUPER SUPER difficult.

is there another way to look at this type of problem? ain't no way in a million years i would have gotten that on my own. i was trying to imagine all those buses moving and i was trying to use my erasers and pencils trying to act it out but it was impossible.

how did you know to set up the diagram like that with time intervals?

when i tried to draw it out, this is what i had:

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lev888

Senior Member
oh. because the 3 left ones will take longer than 60 minutes, travis will pass them because those 3 will still be on the road after 60 minutes.

SUPER SUPER difficult.

is there another way to look at this type of problem? ain't no way in a million years i would have gotten that on my own. i was trying to imagine all those buses moving and i was trying to use my erasers and pencils trying to act it out but it was impossible.
D - distance.
Fast bus speed: D/1h = D mph
Slow: D/1.5 mph
Difference: D - D/1.5h = D/3 mph
So, the fast bus moves with the speed of D/3 mph relative to the slow buses, which are "stationary".
At this speed after 1 hour the fast bus will be 1/3 of the total distance away from the start. Which of the 9 buses will it pass?
The ones that are to the left of 1/3 mark. 1/3 is 30/90. Look at your diagram, that would be the 3 buses on the left: 5/90, 15/90, 25/90.
I think this solution is more complicated than JeffM's.