Question 80386: 60. Business and finance. A coffee merchant has coffee beans that sell for $9 perpound 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?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let N equal the number of pounds of $9 per pound nuts.
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Let T equal the number of pounds of $12 per pound nuts.
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From the problem you can infer that when you mix these two amounts together, the total
weight of the mixture will be 100 lbs. In equation form this becomes:
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N + T = 100
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If you multiply $9 times N you get the total dollar amount of N lbs of nuts in the mixture.
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If you then multiply $12 times T you get the total dollar amount of the T lbs of nuts in
the mixture.
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The price of the mixture is given as $11.25 a pound and there are 100 lbs of the mixture.
So the mixture is worth $11.25 times the 100 lbs of mixture ... a total of $1125.
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So the dollar amount for each type of nut must add up to be $1125. In equation form this
is:
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$9*N + $12*T = $1125.
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So we have two equations:
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N + T = 100 and
9N + 12T = 1125
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Let's solve these two equations by elimination of the variable N. Multiply all the terms
in the top equation by 9 so that the term involving N in the top equation equals the term
that involves N in the bottom equation. The multiplication of the top equation by 9
results in:
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9N + 9T = 900
9N + 12T = 1125
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Now subtract the bottom equation from the top equation and you get:
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0*N - 3T = -225
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and the term 0*N = zero so the equation is just
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-3T = - 225
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Solve for T by dividing both sides by -3 to get:
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(-3T)/(-3) = -225/-3
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The division leads to:
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T = 75
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So the mixture contains 75 lbs of nuts that cost $12 per pound.
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Since there are 100 lbs of the mixture, and we have accounted for 75 lbs of it, the
remaining 25 lbs must be the nuts that cost $9 per lb.
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This is how you do this problem. You could have used other methods to solve the pair of
equations (methods such as substitution or determinants) but variable elimination
used above works just as well. Hope this all makes sense to you and you can see how you
need to find two equations to solve this problem.
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