Pre-Algebra Word Problem

[lovelyshannon]

A metallurgist has one alloy containing 26% copper and another containing 53% copper. How many pounds of each alloy must he use to make 58 pounds of a third alloy containing 48% copper? (Round off the answers to the nearest hundredth.) Please help and explain how to solve this problem.

 

[soroban]

Hello, lovelyshannon!

A metallurgist has one alloy containing 26% copper and another containing 53% copper.
How many pounds of each alloy must he use to make 58 pounds of a third alloy containing 48% copper?
(Round off the answers to the nearest hundredth.)

Let \(\displaystyle x\) = pounds of 26% alloy.
Then \(\displaystyle 58-x\) = pounds of 53% alloy.

We have \(\displaystyle x\) pounds which is 26% copper.
. . It contains: .\(\displaystyle 0.26x\) pounds of copper.

We have \(\displaystyle 58-x\) pounds whch is 53% copper.
. . It contains: .\(\displaystyle 0.53(58-x)\) pounds of copper.

The mixture will contain: .\(\displaystyle 0.26x + 0.53(58-x)\) pounds of copper. [1]


But we know that the mixture will be 58 pounds which is 48% copper.
. . It contains: .\(\displaystyle 0.48(58) \,=\,27.84\) pounds of copper. [2]


We just described the amount of copper in the mixture in two ways.

There is our equation! . . . . \(\displaystyle 0.26x + 0.53(58-x) \:=\:27.84\)

Solve for \(\displaystyle x\!:\;\;0.26x + 30.74 - 0.53x \:=\:27.84\)

. . . . . . . . . . . . . . . . . . . . .\(\displaystyle -0.27x \:=\:-2.9\)

. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle x \:=\:\dfrac{-2.9}{-0.27} \:=\:10.740740...\)

Therefore: .\(\displaystyle \begin{Bmatrix} 10.74\text{ pounds of 26\% alloy} \\ 47.26\text{ pounds of 53\% alloy}\end{Bmatrix}\)

 

[lovelyshannon]

Thank you for explaining how to solve the problem!!