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11 February, 16:25

An alloy weighing 40 lbs. is 10% tin. The alloy was made by mixing a 15% tin alloy and a 8% tin alloy. How many pounds of each alloy (to the nearest tenth) were used to make the 10% alloy?

lbs. of the 15% alloy and lbs. of the 8% alloy.

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Answers (2)
  1. 11 February, 16:26
    0
    Let the weight of 15% tin be x.

    So the weight of the 8% tin will be: (40 - x), since the total weight is 40 lbs

    Using the mixture formula:

    w1*p1 + w2*p2 = w*p

    Where w1 = weight of the first percentage, p1 = First Percentage

    w2 = weight of the second percentage, p2 = Second Percentage

    w = Final weight, p1 = Final Percentage

    0.15*x + 0.08 (40 - x) = 40*0.1

    0.15x + 3.2 - 0.08x = 4

    0.15x - 0.08x = 4 - 3.2

    0.07x = 0.8

    x = 0.8/0.07

    x ≈ 11.429

    Recall the 15% was made with x, which is ≈ 11.429

    Recall the 8% was made with (40-x) ≈ (40 - 11.429) ≈ 28.571

    So the 10% tin of the 40 lbs was made with ≈11.4 lbs of the 15% tin and ≈28.6 lbs of the 8% tin
  2. 11 February, 16:47
    0
    X=15% alloy

    y=8% alloy

    x+y=40

    y=40-x

    0.15x+0.08y=40*0.10

    0.15x+0.08 (40-x) = 40*0.1

    times 100 both sides for easy

    15x+8 (40-x) = 400

    15x+320-8x=400

    7x+320=400

    minus 320 from both sides

    7x=80

    divide both sides by 7

    x=80/7

    x+y=40

    80/7+y=280/7

    y=200/7

    aprox

    x=11.4lb

    y=28.6lb

    11.4lb of 15%

    28.6lb of the 8%

    what?
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