Question 775847: In 20 oz of one alloy, there are 6 oz of copper, 4 oz of zinc, and 10 oz of lead. In
20 oz of a second alloy, there are 12 oz of copper, 5 oz of zinc and 3 oz of lead. In
20 oz of a third alloy, there are 8 oz of copper, 6 oz of zinc, and 6 oz of lead. How
many ounces of each alloy should be combined to make a new alloy containing
34 oz of copper, 17 oz of zinc, and 19 oz of lead?
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! In 20 oz of one alloy, there are 6 oz of copper, 4 oz of zinc, and 10 oz of lead. In
20 oz of a second alloy, there are 12 oz of copper, 5 oz of zinc and 3 oz of lead. In
20 oz of a third alloy, there are 8 oz of copper, 6 oz of zinc, and 6 oz of lead. How
many ounces of each alloy should be combined to make a new alloy containing
34 oz of copper, 17 oz of zinc, and 19 oz of lead?
Alloy A ---6 oz of copper, 4 oz of zinc, and 10 oz of lead
proportion of copper = 6/20 =3/10
proportion of zinc = 4/20 =1/5
proportion of lead = 10/20 = 1/2
Alloy B----------- 12 oz of copper, 5 oz of zinc and 3 oz of lead.
proportion of copper = 12/20 =3/5
proportion of zinc = 5/20 =1/4
proportion of lead = 3/20 = 3/20
Alloy C----------- 8 oz of copper, 6 oz of zinc, and 6 oz of lead.
proportion of copper = 4/10 =2/5
proportion of zinc = 6/20 =3/10
proportion of lead = 6/20 = 3/10
Let x be the Alloy A required in the mixture, y --alloy B, z---alloy C
copper
3/10 x +3/5 y +2/5 z =34
multiply by 10
3x+6y+4z=340..................(1)
Zinc
1/5x+1/4y+3/10 z=17
multiply by 20
4x+5y+6z=340.......................(2)
1/2 x+3/20y+3/10z=19
multiply by 20
10x+3y+6z=380......................(3)
Solve the three equations
3 x + 6 y + 4 z = 340 -------------- 1
4 x + 5 y 6 z = 340 -------------- 2
10 x + 3 y + 6 z 380 -------------- 3
consider equation 1 &2 Eliminate y
Multiply 1 by -5 -5
Multiply 2 by 6 4
we get
-15 x + -30 y + -20 z = -1700
24 x + 30 y + 36 z = 2040
Add the two
9 x + 0 y + 16 z = 340 ------------- 4
consider equation 2 & 3 Eliminate y
Multiply 2 by -3
Multiply 3 by 5
we get
-12 x + -15 y + -18 z = -1020
50 x + 15 y + 30 z = 1900
Add the two
38 x + 0 y + 12 z = 880 -------------5 5
Consider (4) & (5) Eliminate x
Multiply 4 by -38
Multiply (5) by 9
we get
-342 x + -608 z = -12920
342 x + 108 z = 7920
Add the two
0 x + -500 z = -5000
/ -69840
z = 10
Plug the value of z in (5)
38 x + 12 z = 880
38 x = 760
x = 20
plug value of x & z in 1
60 6 y + 40 = 340
6 y = 340 + -60 + -40
6y = 240
y= 40
Alloy A = 20 Oz
Alloy B = 40 Oz
Alloy C = 10 Oz
m.ananth@hotmail.ca
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