Difficult math question- not from a textbook either..

Question

A metallurgist has one alloy containing 40%
aluminum and another containing 61% aluminum. How many pounds of each alloy must he use to make 57 pounds of a third alloy containing 46% aluminum?

Answer 1

So we have to add x pounds of 40% Al to y pounds of 61% Al and we will get 57 pounds of 46% Al.

 

.40x + .61y = .46(x+y)

x + y = 57

x = 57 - y

 

Substitute:

 

.40(57 - y) + .61y = .46(57 - y + y)

22.8 - .40y + .61y = 26.22

.21y = 3.42

y = 16.286

x = 57 - 16.268 = 40.732

 

Using significant figures, he will need to use 41 pounds of the 40% alloy and 16 pounds of the 61% alloy.

 

Answer 2

     X is the amount of the 40% aluminum

    Y is the amount of the 61% aluminum

 

  X+Y = 57

 

 The amount of aluminum is 0.4X + 0.61Y = 57 * 0.46 = 26.22

 

    Solving the first equation for Y and multiplying the second equation by 100 to clear away the decimals:

 

 Y = 57 -x

 

 40X + 61Y = 2622

 

 40x + 61( 57 - x) = 2622

 

 40x + 3477 - 61x = 2622

 

 0 = 2622 - 3477 + 61x - 40x

 

0 = -855 + 21x

 

855 = 21x

 

x = 855/21 

 

y = 57 - 855/21 = (1197 - 855)/21 = 342/21 = 114/7

 

Check: 

                40% of  855/21  + 61% of 342/21 =

 

                40/100 * 855/21 + 61/100 * 342/21 =

 

                     34200/2100 + 20862/2100 =

 

                       55062 / 2100 = 

                 

                      2622/100 = <-- divides by 21

 

                      (57*46)/100 = 

 

                         57 * 46/100 = 46% of 57

 

 

The answers are in bold above, verified,tested, and proven to be correct