Systems of equations word problems
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Example 4: Solving a word problem with substitution
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Mixture problems 1
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Mixture problems 2
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Mixture problems 3
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Systems and rate problems
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Systems and rate problems 2
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Systems and rate problems 3
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Officer on Horseback
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Two Passing Bicycles Word Problem
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Passed Bike Word Problem
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Passing Trains
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Overtaking Word Problem
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Problem Solving Word Problems 2
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Systems of equations word problems
Passed Bike Word Problem Fun algebra word problem involving a train passing a bike.
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- Here's another interesting word problem that was
- sent by [? Cortagio ?].
- A woman cycles to work alongside a railroad track
- at 6 kilometers per hour.
- That's an interesting piece of data.
- 6 kilometers per hour.
- Every day she arrives at a crossing the same
- time that a train does.
- OK.
- One day she was 50 minutes late, and was overtaken by
- the train 6 kilometers from the crossing.
- In how many minutes will the train reach the crossing.
- Let's see if we can draw a diagram here that can somewhat
- describe what's going on in this problem.
- So what's on a normal day?
- On a normal day, she leaves from home, she leaves from
- home, and she goes and she travels alongside
- a railroad track.
- Not too different than my own commute on my own bicycle.
- Actually, I actually do travel on a bike path that's right
- along a railroad track.
- Anyway, let's say she leaves here at time is equal to 0.
- Time is equal to 0.
- And right at this point right here, maybe that's where the
- railroad crossing is, right there, the train passes her up.
- And she's traveling at 6 kilometers per hour.
- 6 kilometers per hour.
- 6 kilometers per hour.
- And what else does it tell us?
- It doesn't tell us how long it takes her, or
- how far she travels.
- So let's just say she arrives here at t is equal to a hours.
- t is equal to a.
- We'll do time in hours, since we're in kilometers per hour.
- So this path takes her a hours to travel.
- So it takes her a hours.
- If we care about the time, a hours.
- Now on this next day, she's 50 minutes late.
- So on this day, she has to travel the same distance in
- time, well in both distance and time.
- But this is this day.
- She leaves not a t equals 0.
- She leaves 50 minutes late.
- Since we're dealing with hours, let's write
- 50 minutes in hours.
- So she leaves at-- if this was time equal 0, that's when she
- normally leaves --she's now leaving at that time is equal
- to 50 minutes, or time is equal to 5/6 hours.
- Right?
- 50 minutes is just 50/60 hours or 5/6 hours of time is
- equal to 5 over 6 hours.
- This time is equal to 0 hours.
- And let's see, it says that she was overtaken by the train 6
- kilometers from the crossing.
- So she was overtaken by the train some place over
- here, and this distance.
- Right this is where the crossing is, that's
- that crossing.
- This is 6 kilometers.
- 6 kilometers.
- In how many minutes will the train reach the crossing?
- So if we can figure out some way of expressing this time, if
- we can figure out with this time is, we know that the train
- reaches the crossing a t is equal to a.
- So if we know this time, we can just find the difference
- between these two times and we would have solved the problem.
- So let's see if we can figure out this time.
- So we know that it normally takes her a hours to travel
- this whole distance.
- It doesn't matter when she leaves it takes her a hours.
- So in this case it should also take her a hours.
- And how long does she have to go when the train passes up on
- this day that she was late?
- How long will it take her to go this distance right there?
- Well she's traveling at 6 kilometers per hour, she
- has to go 6 kilometers.
- So in time, this is going to take her 1 hour.
- To get-- you know, the train will have long passed her --but
- to get back to the crossing, she's going to have to
- travel for another hour.
- So if her whole route takes her a hours, this whole length
- in time is a hours.
- Then this distance right here is going to have
- to take her, what?
- She has an hour left, her whole trip is going to be a hours, so
- this is going to be a minus 1 hours.
- Let me scroll down a little bit.
- This is going to take her a minus 1 hours.
- So if on the day she was late it took her a minus 1 hours to
- get passed by the train, what time are we talking
- about right here?
- What time is this?
- This is t is equal to-- well we started at 5/6
- --t is equal to 5/6.
- That's where we started.
- Plus the amount of hours she traveled plus a minus 1.
- So there we, I think, we now have all the information we
- need to solve this problem.
- On the day she is late, the train passes or at time is
- equal to 5/6 plus a minus 1 hours, where a is the amount of
- time it normally takes her to travel her entire route
- to the crossing.
- We know that the train passes this point at t is equal to a.
- So if we want to know how many minutes does it take the train
- to go from this point to this point, we just have to
- subtract the two times.
- So the minutes it takes, well the hours because
- we're doing it hours.
- And remember I said 50 minutes was 5/6 of an hour, so the
- minutes it takes is a, that a, minus this.
- Minus 5/6 plus a minus 1.
- Right?
- That's just the difference in time for the train, in hours.
- So this is a minus 5/6 minus a plus 1.
- Luckily the a's cancel out.
- a minus a, and you're left with 1 minus 5/6 hours, which is
- just equal to 1/6 hours.
- There are 60 minutes in an hour, so 1/6 of
- that is 10 minutes.
- So it takes the train 10 minutes to get to the
- crossing from the point that it passed her.
- Another neat problem, and I thank [? Cortagio ?].
- for that.
- And then you can actually use this information if you want
- figure out how fast the train is going, or you could even
- think of other types of derivative problems
- off of this one.
- Anyway, another good problem.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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