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?


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