Mike needs 10 lbs of metal with 36% silver. If mike combines one metal with 28% silver, and another with 61% silver, how much of each metal does Mike need?

Question

I have the answer, just need to know how to work it.

Answer 1

M1C1=M2C2+(M1-M2)C3

We know M1; C1; C2; C3; Then we can calculate M2 and M3

M3=M1-M2

3.6=0.28x+ 6.1- 0.61x

x=M2= 7.572 lbs and M3 = 2.428

M2 is the one with 28% and M3 is the one with 61%

 

Answer 2

Morris,

 

This is similar to the previous problem, except we now have 2 unknowns.

 

Let x be the quantity of silver at 28% and y the quantity at 61%

 

Then x*28% + y*61% = 10*36%

 

and x + y = 10

 

write x (or y) in terms of the other variable using the second equation, then substitute back into the 1st equation and solve.

 

Morris:  for some reason it's not wanting me to save a comment to your comment so I'll ad the additional work here.

 

Let y = 10-x then

 

28*x + (10-x)*61 = 360

 

28*x + 610 - 61*x = 360

 

collecting like terms:

 

33*x = 250 and x = 250/33 = 7.58 lbs.

 

so y = 10 - x = 2.42 lbs.

 

check:  7.58*28 + 2.42*61 = 359.8, so our work is correct.