Question 102426: I looked over my notes several times, but I still have trouble setting up the problem and solving it. I'm not even sure I have this under the right category. This is being solved using a Quantity-Value Table.
Juan has coffee that sells for $9 a pound and $4 a pound. How many pounds of each must be mixed to get 20 pounds of coffee worth $8.25 a pound?
I know the table is set up as Q|V|Q*V. What I tried using was x*4, x*9, and 20*8.25. I'm not sure I have it set up correctly. Then I got to 9x+4x=165. Solve and get x = 165/13. I know I'm supposed to apply that answer to get the resulting answer for both the $9 a pound and $4 a pound, but I'm not sure how.
Please and thank you for any help you can give!
Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Juan has coffee that sells for $9 a pound and $4 a pound. How many pounds of each must be mixed to get 20 pounds of coffee worth $8.25 a pound?
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Let amt. of $9 coffee be "x" ; Value of this is 9x dollars
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Amt. of $4 coffee is "20-x" ; value of this is 4(20-x)= 80-4x dollars
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Amt of mixture = 20 lbs ; value of this is 8.25*20 = 165 dollars
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EQUATION:
value + value = value
9x + 80-4x = 165
5x = 85
x =17 lbs (amount of $9 coffee in the mixture)
20-x = 3 lbs (amount of $4 coffee in the misture)
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Cheers,
Stan H.
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! Let's pick this problem apart. We'll start by recognizing that we need to find the weight of
two different coffees. There is $4 a pound coffee, so let's call its unknown weight F
(standing for 4). There is also $9 a pound coffee, so let's call its unknown weight N
(for 9).
.
Common sense will tell us that the most we can sell coffee for is $9 per pound. Since the
coffee we want to mix up is going to cost $8.25 per pound, we're going to need more of the
$9 per pound stuff in the mix than the $4 a pound cheap stuff. In fact if we mixed equal weights
of the $9 coffee and $4 coffee, we should get coffee that sells for $6.50 a pound -- halfway
between the two prices. I'm throwing this info in just as a mental check we can use later
to make sure our answer makes sense.
.
We want to get 20 pounds of mixed coffee. Therefore, we know that the sum of the two weights
F and N must equal 20. So we can write our first equation as:
.
F + N = 20
.
That's pretty straightforward. No magic there.
.
Now let's start looking at the cost. If we have F pounds of coffee at $4 per pound, then
the value of the $4 coffee we have in the mix is $4 times the number of pounds F.
.
Similarly, if we have N pounds of coffee in the mix at $9 per pound, then the total value of
the $9 per pound coffee in the mix is $9 times the number of pounds which is N.
.
Now how much is the mix worth. We know there are 20 pounds of mix and it sells for $8.25
per pound. So we multiply 20 times $8.25 and get $165 dollars.
.
Now let's add up the values of the two ingredients and we can set that equal to the value
of the mix. The values of the two ingredients is $4*F and $9*N and these combined amounts
must equal the $165 that the mix is worth. In equation form this is:
.
4F + 9N = 165
.
So we have two equations to work with:
.
F + N = 20 and
4F + 9N = 165
.
We can solve this set a number of ways. Let's use variable elimination. Suppose we multiply
the top equation [both sides and all terms] by -4. If we do that the equation set becomes:
.
-4F - 4N = -80 and
+4F + 9N = 165
.
Now vertically in columns let's add the two equations together. Notice that the -4F in the
top equation cancels the +4F in the bottom equation. [We deliberately made that happen by
multiplying the top equation by -4.]
.
The -4N and the +9N add together to give +5N and the -80 and +165 add together to give +85.
So after adding these two equations we are left with the single equation:
.
5N = 85
.
Solve for N by dividing both sides by 5 to get:
.
N = 85/5 = 17
.
So we need 17 pounds of the $9 a pound coffee. Since we need a total of 20 pounds, this
means that we must have 3 pounds of the $4 per pound coffee.
.
Just as we "guessed" earlier, we needed a lot more of the $9 coffee in the mix than we did
of the $4 coffee.
.
Hope this helps you to see the general way that you can approach problems such as these.
You need two equations to solve for two unknowns, so you can start on that basis.
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