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Question 110049: A coffee merchant has coffee beans that sell for $9 per pound and $12 per pound. The two types are to be mixed to create 100 lb of a mixture that will sell for $11.25 per pound. How much of each type of bean should be used in the mixture?
Thank you for your help!
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let's pick this problem apart one step at a time. We'll begin by letting N represent the
the number of pounds of coffee at Nine dollars a pound and T represent the number of pounds
of coffee at Twelve dollars a pound.
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As a rough check we can tell that T must be more than N. Why is that? because the final mixture
must sell at $11.25 a pound ... and this price is closer to $12 a pound than it is to $9.00 a pound.
[If N equaled T, we would expect the coffee mix to sell for $10.50 a pound, which is the price
that is halfway between $9.00 a pound and $12.00 a pound.]
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Now to the problem. The total weight of the mix must be 100 pounds. Therefore, we know that
N + T must equal 100. In equation form this is:
.
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Next, we can tell that at $11.25 per pound the entire 100 pounds of mix must be worth
$11.25 times 100 which equals $1125.00.
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Of that total amount, the value of the $9.00 per pound coffee is $9 times the number of
pounds of that coffee (which is N). So 9*N is the dollar amount of the $9 coffee in the mix.
And the value of the of the $12.00 per pound coffee in the mix is 12*T. The combined
value of the two coffees in the mix must add to $1125.00 and in equation form this is:
.
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By itself, this equation cannot be solved because it is a single equation and there are
two unknowns ... N and T. But we can go back to the equation:
.
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and solve for one of the variables in terms of the other. For example, let's solve for
N by subtracting T from both sides of the equation to get:
.
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The right side of this equation is equal to N so it can be substituted for N in the dollar
equation. Start with:
.
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and replace N with 100 - T to get:
.
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Multiply out the left side to make it:
.
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Get rid of the 900 on the left side by subtracting 900 from both sides to reduce the equation to:
.
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Combine the two terms on the left side:
.
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Solve for T by dividing both sides by 3 and you have:
.
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This tells you that the 100 pounds of mix contains 75 pounds of the coffee that costs $12 per pound.
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The remainder of the 100 pounds of mix must therefore be 25 pounds of the coffee that costs
$9 per pound.
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Check ... 25 pounds + 75 pounds = 100 pounds of mix.
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And at $9 per pound, the 25 pounds of it in the mix is worth $9 * 25 = $225. Plus at
$12 per pound, the 75 pounds of it in the mix is worth $12 * 75 = $900. So the total worth
of the mix is $225 + $900 = $1125. Finally there are 100 lbs of the mix and at a total worth
of $1125, the cost per pound should be $1125/100 = $11.25 and this is what the problem said
it should be.
.
Everything checks. Therefore, you can say that the required mix consists of 25 pounds of
the $9 per pound coffee and 75 pounds of the $12 per pound coffee.
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Finally, note that there is more of the $12 per pound coffee in the mix than there is
$9 per pound coffee just as we originally thought.
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Hope this helps you to understand the problem.
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