Question

a metalworker has a metal alloy that is 15% copper and another alloy that is 55% copper. how many kilograms of each alloy should the metalworker combine to create 50kg of a 39% copper alloy?The metal worker should use ____ kilograms of the metal alloy that is 15% copper and _____ kilograms of the metal alloy that is 55% copper.

Answer 1

Given that:

- The metal alloy is 15% copper and another alloy is 55% copper.

- The metalworker needs to create 50 kilograms of a 39% copper alloy.

Let be "x" the kilograms of metal alloy that is 15% copper and "y" the kilograms of metal alloy that is 55% copper.

Knowing that the metalworker needs to create a total of 50 kilograms, you can set up this first equation:

$x+y=50$

Knowing the percents that are given in the exercise, you can set up the second equation (remember that a percent can be converted to a decimal number by dividing it by 100):

$\begin{gathered} \frac{15}{100}x+\frac{55}{100}y=(\frac{39}{100})(50) \\ \\ 0.15x+0.55y=19.5 \end{gathered}$

Now you can set up this System of Equations:

$\begin{cases}x+y=50 \\ \\ 0.15x+0.55y=19.5\end{cases}$

In order to solve it, you can use the Elimination Method:

1. Multiply the first equation by -0.15:

$\begin{cases}-0.15x-0.15y=-7.5 \\ \\ 0.15x+0.55y=19.5\end{cases}$

2. Add the equations:

$\begin{gathered} \begin{cases}-0.15x-0.15y=7.5 \\ \\ 0.15x+0.55y=19.5\end{cases} \\ --------------- \\ 0+0.40y=12 \\ 0.40y=12 \end{gathered}$

3. Solve for "y":

$\begin{gathered} y=\frac{12}{0.40} \\ \\ y=30 \end{gathered}$

4. You can substitute the value of "y" into the first equation:

$\begin{gathered} x+y=50 \\ x+(30)=50 \end{gathered}$

5. Solve for "x":

$\begin{gathered} x=50-30 \\ x=20 \end{gathered}$

Hence, the answer is:

The metalworker should use 20 kilograms of the metal alloy that is 15% copper and 30 kilograms of the metal alloy that is 55% copper.