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 a lloy (to the nearest tenth) were used to make the 10% alloy? lbs. of the 15% alloy and lbs. of the 8% alloy.
2 answers:
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
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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