Ametallurgist has one alloy containing 49%49% copper and another containing 62%62% copper. how many pounds of each alloy must he use to make 5151 pounds of a third alloy containing 56%56% copper? (round to two decimal places if necessary.)

2377.38 pounds of the first alloy

2773.62 pounds of the second alloy

Step-by-step explanation:

First, let Alloy1 represent the alloy with 49% copper, Alloy2 the alloy with 62%, and Alloy3 the alloy with 56%.

The third alloy is made from a mixture of the first two, so:

MassAlloy1 + MassAlloy2 = MassAlloy3        

MassAlloy1 + MassAlloy2 = 5151 lb         eq.1

MassCopper1 = MassAlloy1 * 0.49        eq.2

MassCopper2 = MassAlloy2 * 0.62       eq.3

MassCopper3 = MassAlloy3 * 0.56        eq.4

The problem tells us that MassAlloy3 = 5151 lb, thus the copper mass in the third alloy is:

MassCopper3 =  5151 lb * 0.56 = 2884.56 lb

MassCopper1 + MassCopper2 = MassCopper3

MassCopper1 + MassCopper2 = 2884.56 lb         eq.5

MassAlloy1 * 0.49 + MassAlloy2 * 0.62 = 2884. 56 lb     eq.6

Eq. 1 and Eq. 6 give us a system of two equations and two unknowns, so we can solve it:

MassAlloy1 + MassAlloy2 = 5151 lb  

MA1 = 5151 - MA2

MassAlloy1 * 0.49 + MassAlloy2 * 0.62 = 2884. 56 lb

(5151-MA2) * 0.49 + MA2 * 0.62 = 2884.56

2523.99 - 0.49MA2 + 0.62MA2 = 2884.56

0.13MA2 = 360.57

MA2 = 2773.62

MassAlloy1 + MassAlloy2 = 5151 lb

MassAlloy1 + 2773.62 lb = 5151 lb

MassAlloy1 =2377.38 lb

26

step-by-step explanation:

-1/2 +2

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