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Word Problem Solving Strategies

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Word Problem Solving Strategies

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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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Word Problem Solving Strategies

Word Problem Solving Strategies