Question 1190205: Bags of coffee worth $90 a bag must be mixed with coffee worth $75 a bag to get 50 bags worth $87 a bag. How many bags of each are needed? Found 2 solutions by math_tutor2020, greenestamps:Answer by math_tutor2020(3817) (Show Source):
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Let's define two brand labels such that
Brand A is worth $90 per bag
Brand B is worth $75 per bag
Let
x = number of bags of brand A
y = number of bags of brand B
The final mix has 50 bags total, so,
x+y = 50
y = 50-x
which we'll use later in substitution
90x = total cost of buying x bags of brand A
75y = total cost of buying y bags of brand B
90x+75y = total cost buying all bags of coffee
We divide this total cost by the number of bags overall (50) to get the final cost per bag ($87)
(total cost)/(total number of bags) = final cost per bag
(90x+75y)/(50) = 87
90x+75y = 87*(50)
90x+75y = 4350
Now we plug in y = 50-x and solve for x.
90x+75y = 4350
90x+75( y ) = 4350
90x+75( 50-x ) = 4350
90x+75*50-75x = 4350
90x+3750-75x = 4350
15x+3750 = 4350
15x = 4350-3750
15x = 600
x = 600/15
x = 40
Then compute the value of y
y = 50-x
y = 50-40
y = 10
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Answers:
40 bags of the $90 per bag brand.
10 bags of the $75 per bag brand.
The other tutor showed one of many ways to solve the problem using formal algebra.
If a formal algebraic solution is not required, you can solve the problem quickly with a little mental arithmetic.
(1) Look at the three prices per bag on a number line -- 75, 87, and 90 -- and observe/calculate that 87 is 12/15 = 4/5 of the way from 75 to 90
(2) That means 4/5 of the mixture should be the more expensive coffee
ANSWER: 4/5 of 50 bags, or 40 bags, of the coffee worth $90 per bag; the other 10 bags of coffee worth $75 per bag
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