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.
Answers
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
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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g=number of generall
r+g=1787
4r+3g=5792
multiply first equation by -3 and add to second
-3r-3g=-5361
4r+3g=5792 +
1r+0g=431
r=431
sub
r+g=1787
431+g=1787
minus 431 both sides
g=1356
1356 general sold
421 reserved seat sold