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
Two Passing Bicycles Word Problem Algebra word problem involving two bicycles passing each other.
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- This is the last of the word problems that Kortaggio had
- sent, and they're all interesting because they're
- all involving distances and rates and times.
- They all involve you having to do a little bit of implicit
- algebra without having to explicitly solve for values of
- variables, and that's why they're fun.
- So this one says Ann and Betty, two cyclists,
- start simultaneously.
- Ann at point A and Betty at point B.
- They travel to each other at constant speeds.
- So let's say that Ann is traveling with velocity A
- and Betty is traveling with velocity B.
- Constant speeds.
- They pass each other in 30 minutes.
- OK, so they're going to pass each other someplace
- in between A and B.
- I'll just arbitrarily draw that.
- They're going to pass each other there.
- And what are these distances?
- It took them 30 minutes to both to get this point, right?
- They pass each other in 30 minutes.
- And Ann, headed from point B, arrives there in
- another 20 minutes.
- OK, so this distance takes Ann 30 minutes.
- So what is the actual distance?
- Well, it's always useful to write distance is equal
- to rate times time.
- So what's the actual distance she traveled?
- Well, the rate is va.
- She traveled a rate of va and her time -- they tell us -- it
- took her 30 minutes to get there.
- So va times 30.
- Let's say this velocity is expressed in minutes.
- Fair enough.
- And then Ann, headed for point B, arrives there
- in another 20 minutes.
- So this is going to take her another 20 minutes.
- So this distance right here is what?
- She's going at the same velocity, but it
- takes her 20 minutes.
- So the distance is velocity times time.
- So it's va times 20, 20va.
- Fair enough.
- How many minutes after Ann arrives at point B will
- Betty arrive at point A?
- So Betty was going in this direction, they meet right
- here, and then she keeps going and hits point A.
- So the key realization here is that this took Ann 30 minutes
- to get here, but it also took Betty 30 minutes to get here.
- So what's this distance in terms of Betty's velocity?
- What's this distance?
- Well, distance is equal to rate times time.
- It took Betty 30 minutes to go this distance
- at a velocity of vb.
- So it took her 30vb meters, or whatever, actually.
- We don't even have to specify the units.
- Whatever the units the velocity is in.
- If it's in meters per minute, then this is
- the distance in meters.
- This distance is 30vb.
- Right?
- So essentially, we just have to figure out how long it takes
- her to travel this remaining distance and we would
- have solved the problem.
- So how can we do that?
- Well, we can already set up a relationship between
- vb and va, because this distance is the same.
- It doesn't matter how fast you travel it, the
- distance doesn't change.
- So we know that 30 times Betty's velocity is equal to 20
- times Ann's velocity, 20va.
- And so if we wanted to express Ann's velocity in terms of
- Betty's velocity -- let's just do it that way -- if we divide
- both sides by 20 we get 3/2 of Betty's velocity is equal
- to Ann's velocity.
- Fair enough.
- So what is this distance in terms of Betty's velocity?
- We know it in terms of Ann's velocity.
- In terms of Ann's velocity, it's 30 times Ann's velocity.
- Right?
- So what is 30 times Ann's velocity in terms of
- Betty's velocity?
- Well, if we multiply this times 30, that's 30 times Ann's
- velocity, so we can multiply both sides of this times 30.
- So if you multiply 30 times here and times thirty,
- this is the distance.
- This distance right here.
- Ann's velocity times 30 minutes.
- It took her 30 minutes to travel that far.
- So this is the same distance expressed in Betty's velocity.
- So 30 times 3 is 90 divided by 2 is 45.
- So it's 45vb.
- So this distance up here is 45vb.
- So distance is equal to rate times -- so this
- is rate -- times time.
- So this is how many minutes it took Betty to
- travel this distance.
- So what is the question asking us?
- How many minutes after Ann reaches point B will
- Betty arrive at point A.
- So they both reach right here.
- In another 20 minutes, Ann gets here.
- And in another 45 minutes, Betty will get over there.
- Another 45 minutes.
- So it takes Ann 20 minutes to get here.
- It takes Betty 45 minutes to get here, after their meeting
- right at the center.
- So it's going to take Betty another 25 minutes.
- Right?
- After 20 minutes Ann is here and then we have to wait
- another 25 minutes for Betty to get here, 45 minutes
- after this meeting.
- So the answer is 25 minutes.
- How many minutes after Ann arrives at point B will
- Betty arrive at point A.
- Yes, 25 minutes.
- They could have just asked how many minutes after the meeting
- will it take Betty to reach point A.
- It would have been 45.
- But they want to know how many more minutes
- after Ann gets there.
- Anyway, I thought you would find that useful and, a final
- time, thanks to Kortaggio for sending that problem.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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