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    notcoolenough's Avatar
    notcoolenough Posts: 95, Reputation: 7
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    #1

    Jun 25, 2015, 01:46 PM
    Elementary Algebra (Math 100) question
    On a practice test, a word problem has confused me. I don't actually understand the question on the surface. The word problem was a variation of the one from the link below:

    Jimmy is a partner in an Internet-based coffee supplier.The company offers gourmet coffee beans for $11 per pound and regular coffee beans for $5 per pound. Jimmy is creating a product that will sell for $7 per pound.The first thing to go into the mixing bin was 10 pounds of the gourmet beans. How many pounds of the less expensive regular beans should be added? - http://www.algebra.com/algebra/homew...ion.296831.htm

    I know the answer. I know how to do it. We went over it in class. But I still don't get the actual question. This type of question is bound to show up on the test and if I don't understand what is being asked, I'm not going to be able to figure out its equation. What is the mixing bin? What does it do? Why does 20lbs need to be in the mixture? Why not 5 or 70 lbs? What is jimmy trying to accomplish?

    I'm reading it like 11$ per apple, 7$ per orange, and 9$ per lemon. If jimmy puts ten apples in the mixing bowl, how many oranges should be added? That seems arbitrary. Just put in whatever.

    Thank you.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jun 25, 2015, 02:17 PM
    A mixing bin is simply a large bowl. I don't like the wording of the question, because it doesn't make it clear that the objective is to have a mixture that has an average cost of $7/pound.

    Think about a similar situation regarding mixing water of different temperatures. Suppose you want to take a bath in water that is 70 degrees F. And suppose that the hot water tap delivers water at 110 degrees while the cold water tap delivers water at 50 degrees. Let's say you first put 10 gallons of hot water in the tub - how much cold water must you add to get the temp down to 70? You can't put in any old arbitrary amount of cold water - too little and the temp will be above 70, too much and it will be below 70. The final temp is the weighted average of the water you put in. Given Q_hot gallons of hot water at temp T_hot, and Q_cold gallons of cold water at temp T_cold, so that the total number of gallons is Q_total = Q_hot + Q_cold, you have:

    T_total x Q_total = (T_hot x Q_hot)+(T_cold x Q_cold).

    Substituting in the numbers:

    70 x (10 + Q_cold) = (110 x 10)+(50 x Q_cold)

    Solve for Q_cold (which turns out to be 20 gallons). Hope this analogy helps.
    notcoolenough's Avatar
    notcoolenough Posts: 95, Reputation: 7
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    #3

    Jun 26, 2015, 06:35 PM
    THANK YOU! That makes sense now! Yeah the wording definitely tripped me up. A huge weight has been lifted off my shoulders.

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