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28 March, 02:56

You have one type of nut that sells for $2.80/lb and another type of nut that sells for $9.60/lb. You would like to have 20.4 lbs of a nut mixture that sells for $6.60/lb. How much of each nut will you need to obtain the desired mixture?

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  1. 28 March, 04:01
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    Answer:11.396Ibs of nuts that cost $9.60/Ib and 9.014Ibs that cost $2.80/Ib

    Step-by-step explanation:

    First we find the cost of the supposed mixture we are to get by selling it $6.60/Ib which weighs 20.41Ibs

    Which is 6.6 x 20.41 = $134.64

    Now we label the amount of mixture we want to get with x and y

    x = amount of nuts that cost $2.8/Ib

    y = amount of nuts that cost $9.6/Ib

    Now we know the amount of mixture needed is 20.41Ibs

    So x + y = 20.41Ibs

    And then since the price of the mixture to be gotten overall is $134.64

    We develop an equation with x and y for that same amount

    We know the first type of nut is $2.8/Ib

    So for x amount we have 2.8x

    For the second type of nut that is $9.6/Ib

    For y amount we have 9.6y

    So adding these to equate to $134.64

    2.8x + 9.6y = 134.64

    So we have two simultaneous equations

    x + y = 20.41 (1)

    2.8x + 9.6y = 57.148 (2)

    We can solve either using elimination or factorization method

    I'm using elimination method

    Multiplying the first equation by 2.8 so that the coefficient of x for both equations will be the same

    2.8x + 2.8y = 57.148

    2.8x + 9.6y = 134.64

    Subtracting both equations

    -6.8y = - 77.492

    Dividing both sides by - 6.8

    y = - 77.492/-6.8 = 11.396

    y = 11.396Ibs which is the amount of nuts that cost $9.6/Ib

    Putting y = 11.396 in (1)

    x + y = 20.41 (1)

    x + 11.396 = 20.41

    Subtract 11.396 from both sides

    x + 11.396-11.396 = 20.41-11.396

    x = 9.014Ibs which is the amount of nuts that cost $2.8/Ibs
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