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.

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

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

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