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
Officer on Horseback Not quite a brain teaser but fun nonetheless!
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- I was sent some problems by one of the viewers out there.
- I believe his name is Cortaggio or Cortajio, I apologize.
- I'm sure I'm mispronouncing it.
- But they are really interesting problems.
- What's interesting about them is that they don't involve
- super-fancy mathematics.
- They just involve an elegant way to apply fairly
- simple mathematics.
- So, without me talking too much about the problems, let's
- try to solve one of them.
- Actually, I don't know where I'm going to categorize this.
- Either in algebra or in the brain-teaser playlist,
- or maybe both.
- So, an officer on horseback starts at the back of a column
- of marching soldiers and rides to the front of the column.
- Then turns around and rides to the rear of the column.
- If the rider travels three times as fast as the column
- moves, and the column is 100 meters long -- OK, I think we
- have enough information to start drawing this.
- Let me draw the column of soldiers.
- I'll just draw this as a big fat line.
- So, that's the column of soldiers.
- I'm about to cough.
- Excuse me.
- I've learned not to cough directly in the microphone.
- Don't want blow your speakers out.
- So that's the column of soldiers right there.
- And the problem tells us that it's 100 meters long.
- So this distance right here is 100 meters.
- And it's moving with some velocity.
- Let's say, to the right.
- So I'm going to just call it, it's moving with
- some velocity, v.
- We'll have to stay abstract there because it doesn't
- tell us the velocity.
- And then we have, what is this, the officer. Is that what they...right.
- I wanted to make sure I wasn't giving the incorrect title.
- So, an officer on horseback stars at the back of a column.
- So he starts here.
- He's on his horse.
- He's on his horse.
- That's my rendition of the horse for these purposes, it
- has an officer on on the back.
- And he is going to, while the whole column is going forward
- with the velocity, v, he's going to go to the front of it.
- Obviously he can go faster because he's on a horse.
- And then he's going to go to the back and they want to
- know how far did this whole column move.
- And how fast is he going?
- Let's see, if the rider travels 3 times as fast as the column,
- if the column is this traveling with the velocity v, he's going
- to be traveling with the velocity of 3v.
- So let's think about two things.
- Let's think about the time it takes him to go to the front
- of the column, and the time it takes to go back.
- So let's say t1 is equal to time to front.
- And then t2 would be time to back.
- There might be other ways to solve this, but this is the
- way that's jumping out into my brain.
- So let's figure out what the time to the front
- of the column is.
- So, over some period of time -- so, how far is the column
- going to move over to t1?
- The column is going to move, well, actually, let me write
- a little formula here.
- Although I'm sure you know this formula.
- Distance is equal to rate times time.
- Right?
- So, over time 1, how far does the column move?
- It's going to move v.
- v times time 1.
- And how far is this guy going to move?
- Well, we're saying over time, when he moves to
- the front of the column.
- So, this whole column is moving to the right.
- And at the same time, this guy's moving to
- the right faster.
- So at the end of time 1, which I've defined as the time
- it takes him to get to the front, what's true about the
- officer on the horseback?
- He will have had to travel 100 meters further than
- the column, right?
- In order to catch up to the front of the column, he would
- have had to go 100 meters further than the column.
- So, the distance that the column travels, plus 100 meters
- is going to be the distance that the officer travels in
- the same amount of time.
- And what's the distance that he travels in that
- same amount of time?
- Well, distance is equal to rate, 3v, times time, t1.
- Time to the front.
- So this is a relationship between velocity and
- the time to the front.
- And let's see if we can simplify this a little bit.
- So, if we subtract v t1 from both sides, we get 100 is equal to
- 3v t1 minus v t1, that's 2v t1, and divide both sides
- by 2, you get 50.
- v t1 is equal to 50.
- The velocity of this column of soldiers times the time it
- takes this officer to get to the front is equal to 50.
- Well, that doesn't solve our problem yet.
- We want to know how far does the column move?
- We have two variables with one equation.
- Not helpful yet.
- Let's see if t2 can help us a little bit?
- All right.
- I'll switch colors to ease the monotony.
- Time to back.
- So, now we're in the opposite situation, where the
- guy's gotten here.
- He turns around.
- I'd argue, immediately.
- He turns around immediately.
- And he goes back.
- With the velocity of 3v.
- So, my question to you as he starts out here, and relative
- to him, he's going this direction at a velocity of 3v.
- And the back of the column is moving towards him with
- the velocity of v, right?
- So if you think about it, the back of the column is going to
- be approaching the rider with the velocity of 4v.
- When you have two velocities that are moving in opposite
- directions -- if I move in this direction, at 60 miles per
- hour, and you're moving in that direction at 60 miles per hour.
- Relative to me, if I assume that I'm stationary, you would
- look like you're coming at me at 120 miles per hour.
- So that same idea -- this officer is going to be
- approaching the end of the column, the back of the column,
- with the velocity of 4v.
- So how long does it take them to get to the back?
- Well, let's see.
- His velocity is 4v.
- I'll do that in green.
- So his velocity is 4v.
- That's how fast he's approaching the back
- of the column.
- And it's going to take him time 2, times time 2.
- And his distance, he's going to travel 100 meters, because
- that's the length of the column.
- Is equal to 100.
- And let's see, if we divide both sides of this by 4,
- we got velocity times time 2 is equal to 25.
- And, once again, we have 1 equation with two unknowns.
- It doesn't help a lot.
- Let's review the problem again to see if somehow we can use
- this information and this information to solve what
- they're asking for.
- So, they want to know how far does the column move by the
- time the officer arrives back at the rear of the column.
- So, they want to know how far did this whole thing move over
- the entire time of this problem happening?
- What was the entire time?
- It was t1 plus t2.
- That's the entire time. t1 to go to the front, and then
- t2 to go back to the back.
- So how far did the column move?
- Well, the column, distance is equal to rate times time.
- So the column will move its distance -- is equal to the
- column's rate, velocity.
- And then what's the time that this whole little
- problem occurs on?
- Well t1 is how long it takes the officer to
- get to the front.
- Plus t2.
- So this is what we're solving for.
- This is what we need to know.
- We need to know the distance traveled by the column.
- And, once again, we have all these variables.
- But maybe we can do something interesting.
- Let's look at this.
- So what is -- so if we just distribute this, we have
- distance is equal to v times t1 plus v times t2.
- And do we know what these things are?
- Well, sure.
- We were able to kind of stumble our way into into what
- these values are.
- And that's what's interesting about this problem.
- We never figured out v.
- We never figured out t1 or t2.
- But we can figure out this whole thing.
- Because v times t1 -- they tell us, is 50.
- So we can substitute it back here.
- So the distance is equal to 50.
- Plus, what's v times t2?
- Well, we solved it here when we figured out how long it would
- take to go back to the front of the column.
- So that's 25.
- So that, we'll put over here.
- So the total distance that the column traveled is 50 plus
- 25 meters, or 75 meters.
- And this is a neat problem, because they didn't tell us
- how fast the column is actually going.
- They're just saying it's v.
- They just said that the officer is going 3 times
- as fast, but we don't know absolute velocities.
- But we were still able to figure out, even without even
- knowing the absolute times, we were still able to figure out
- the total distance that the column travelled.
- Anyway, I hope you enjoyed that.
- And if you did, you can thank Cortagio for the 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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