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Word Problem Solving Strategies
相關課程
- Let's do some more word problems. And this first one's
- a bit of a classic.
- Brit has $2.25 in nickels and dimes.
- If she has 40 coins in total, how many of each
- coin does she have?
- So let's define some variables here.
- Let's let n equal the number of nickels she has.
- And let d is equal to the number of dimes she has.
- So let's see, the first thing we know is that she has $2.25
- in nickels and dimes.
- So how much money does she have?
- Well, if you take the number of nickels, each nickel is
- worth $0.05.
- So the number of nickels times 0.05, that's how much money
- her nickels represent, plus the amount of
- money her dimes represent.
- So she's got d dimes.
- They're each worth 0.1 of a dollar, or .10, or 0.10 if we
- want to put the 0 in front of the decimal.
- So this is the total amount of money she has.
- If she has 5 nickels, you'd do 5 times $0.05, it'd be $0.25.
- If she had 4 dimes, that'd be $0.40, and you would add it
- all together.
- Now, the problem tells us that the total amount of money she
- has is $2.25.
- So this the first statement right there can be translated
- into this equation right here.
- She has a certain number of nickels and a
- certain number of dimes.
- You multiply $0.05 times the nickels, plus $0.10 times the
- number of dimes, you end up with 225.
- Now, this second statement, she has 40 coins in total.
- And I'm assuming that she only has nickels and dimes, that
- there aren't any other coins.
- So 40 coins in total, that means that the number of
- nickels plus the number of dimes has to be equal to 40.
- So the nickels plus the dimes has got to be equal to 40.
- That's what that information tells us.
- Now, we have two equations with two unknowns, and there's
- two ways we can do this.
- We can do this as a system, and we can subtract one
- equation from the other.
- Or we can do substitution.
- Let's do substitution.
- That's often a more intuitive way to do it.
- So what I'm going to do is, I'm going to solve for-- I
- could solve for either one-- but I want to solve for-- I'll
- solve for d and then I'll substitute whatever I get for
- the d in this equation right here.
- So if we subtract n from both sides of this equation-- so
- let me do that.
- Negative n plus this, minus n.
- Obviously, these cancel out and you're just left with d is
- equal to 40 minus n.
- Now, we can take this.
- We just established using the second equation, using this
- second statement, that d is equal to 40 minus the number
- of nickels.
- And we can use that to substitute back in for d.
- So this first equation will now become 0.05 times our
- number of nickels, plus 0.-- I'll just write 0.1-- times
- our number of dimes.
- We already know that the number of dimes is
- equal to 40 minus n.
- That's what that second statement told us.
- So instead of d, I'm going to substitute it with 40 minus n.
- 40-- that's the number of dimes we have. It's 40 minus
- the number of nickels, and that is equal to $2.25.
- And now we can just try to solve for n.
- So we get a 0.05n plus .1 times 40, that is 4.
- Let me show you that I'm distributing it.
- So .1 times 40, that is 4 minus 0.1n-- I just distribute
- this times each of these terms-- is equal to $2.25.
- And then what can we do?
- Well, we can take .05 minus .1-- or your .10, depending on
- how you want to do it-- if you take those two terms, you're
- actually going to get a negative 0.05n minus .10--
- this is twice as big as that, so it's going to get a
- negative number-- plus 4-- so I still have that
- 4-- is equal to $2.25.
- Now I can subtract 4 from both sides of this equation.
- I get negative 0.05n is equal to-- if you subtract 4 from
- the left, that 4's going to go away-- you subtract 4 over
- here; 2.25 minus 4-- that's negative 1.75.
- All right.
- Now we can multiply both sides of this equation by negative
- 1, just to make it a little bit cleaner, so that becomes a
- plus and a plus.
- And then divide both sides by .05.
- I could have just divided both sides by negative .05 to begin
- with, but I like getting everything positive.
- So if you divided by 0.05, you get n is equal to $1.75
- divided by 0.05.
- So let's do this division.
- Let me do it in a different color.
- 0.05 divided into $1.75-- let's multiply both the
- divisor and the dividend by 100.
- So we're going to go over two decimal spaces, so this just
- becomes a 5.
- This becomes 175.
- This is the same thing as 175 divided by 5.
- 5 goes into 17 three times.
- 3 times 5 is 15.
- 17 minus 15 is 2.
- Bring down the 5.
- 25.
- 5 goes into 25 five times.
- 5 times 5 is 25.
- You have no remainder.
- So this is equal to 35.
- So n is equal to 35, and now we can solve for d. d is equal
- to 40 minus n.
- So d is equal to 40 minus 35, which is equal to 5.
- So we have 5 dimes and 35 nickels.
- And just to verify that that actually does make sense, 35
- nickels are going to be worth how much?
- Well 35 times 5 cents is going to get us to
- 175 cents or $1.75.
- So this times 35, that is going to be $1.75 in nickels.
- 35 times 5 or times .05, depending on how you
- want to view it.
- And then this is going to be 5.
- So this right here is going to be $0.50 in dimes.
- $1.75 plus $0.50, that's $2.25.
- Let's do one more.
- A pattern of squares is out together-- oh.
- A pattern of squares is out together as shown.
- I guess they're hanging out.
- How many squares are in the 12th diagram?
- So let's see what's happening.
- So this is the first, that is the second, that is the third.
- Let's see here, each side, this is a 1 by 1 square, and
- then here it's 2 by 2, but then they take out this block
- right over there.
- So let's see.
- So one way you could think about it is this is 2 by 2,
- and this is a 3 by 3, but they took out the 2 by 2.
- And so the next one is going to be a 4 by 4 where they took
- out the 3 by 3.
- So I guess one way that we can describe the number of
- squares-- actually, the easiest way, you don't have to
- think about that is, look, I have 3 here, and then I have 3
- minus 1 over here.
- Let me do that in a different color.
- 3 minus 1 over here.
- Here I have 2, and here I have 2 minus 1.
- And then here I have 1, and then here I have 1 minus 1.
- I have nothing.
- So in general, the number of squares for any n is going to
- be n plus, then we have n minus 1.
- Plus n minus 1, which is the same thing as 2n minus 1.
- And it works, right.
- 2n minus 1.
- 2 times 3 minus 1 is 5.
- 1, 2, 3, 4, 5.
- So works for these three.
- So the 12th square, so when n is equal to 12, it's going to
- be 2 times 12 minus 1, which is equal to 24 minus 1, which
- is equal to 23 squares in the 12th diagram.
- Actually, I said lets do only one, let's do another one.
- We're in the zone.
- Grace starts biking at 12 miles per hour.
- 1 hour later, Dan starts biking at 15 miles per hour.
- Following the same route, how long will it take him to catch
- up with Grace?
- So let t equal the time Grace is biking.
- Actually, let me do it this way.
- Let me do it the time Dan is biking.
- Then how long will Grace be biking?
- Well, he starts an hour-- let's see, did I say, yeah--
- an hour later, he starts.
- So whatever number this is, Grace is going to be biking
- for 1 more hour.
- So t plus 1 is equal to time Grace is biking.
- When this is 0 this is going to be 1.
- She will have already been biking for 1 hour.
- And we just have to remember that distance is equal to rate
- times time.
- So Dan's distance is going to be equal to 15 miles an hour
- times the time in hours.
- This is Dan's distance.
- And then Grace's distance-- maybe I'll call that g, or
- let's say D for Grace, that's her distance as well-- is
- going to be equal to 12 miles an hour.
- And her time in terms of t is going to be t plus 1.
- She's going to be always be traveling for 1
- more hour than Dan.
- 12 times t plus 1.
- And the point at which they catch up, that's the point at
- which these two things are going to be equal to.
- That Dan's distance is going to be
- equal to Grace's distance.
- So we just have to set these two to be equal to each other.
- So you get 15t is going to be equal to-- that is how far Dan
- has been riding his bike-- and then it's going to be equal to
- 12 times t plus 1.
- That is how far Grace has ridden her bike.
- And remember, her time is always going to be 1 hour more
- than Dan's.
- So let's just solve for time.
- So we get 15t is equal to 12t plus 1.
- 12t plus 1.
- And then if we subtract 12-- I'm sorry, 12t plus 12-- I
- don't want to make a mistake there.
- If we subtract 12t from both sides, you get
- 3t is equal to 12.
- Divide both sides by 3-- let me scroll down a little bit--
- you get t is equal to 4.
- So after 4 hours, after Dan has been riding for 4 hours,
- he will catch up with Grace.
- This is Dan, and if Dan's ridden for 4 hours, Grace has
- been riding for 6 hours.
- 6 hours is how far Grace has ridden for.
- And Dan after 4 hours, 4 times 15, he would
- have gone 60 miles.
- And Grace after-- sorry, there's 1 more, this isn't a
- six-- she goes 1 more hour than Dan.
- So it's 5, she's traveled for 5 hours.
- So Grace would have gone 5 hours at 12 miles an hour.
- And that, once again, is also 60 miles.
- So they meet at exactly the 60 mile mark.
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