Question 357·Hard·Math

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A laboratory has a 50-liter cleaning solution that is 20% bleach by volume. The technician drains off some of this solution and replaces it with an equal volume of a 40% bleach solution. After thorough mixing, the resulting solution is 28% bleach. How many liters of the original solution were drained and replaced?

Your answer

A mixture equation tracks the amount of one ingredient, not just the total liquid. In a drain-and-replace problem, the removed solution takes its share of bleach with it while the total volume stays fixed. Write the bleach amounts as concentration times volume, then use a Desmos regression to solve the equation. Counting the removed bleach as still present gives the wrong balance.

Hints

Hint 1
Let $x$ be the liters drained. Removing $x$ liters leaves $50-x$ liters of the original solution. Which portion has the $20\%$ concentration, meaning the percent of its volume that is bleach?
Hint 2
Use an ingredient balance: bleach in the original solution that remains plus bleach in the replacement equals bleach in the final mixture. Don't count the drained portion as remaining. What volume should you multiply by $20\%$ ?
Hint 3
The final volume is still $50$ liters because the technician adds exactly as much solution as was drained. Take $28\%$ of that volume to find the final amount of bleach.

Step-by-step

Balance the liters of bleach
Step 1
Find how much original solution remains
Let $x$ be the liters drained and replaced. Removing $x$ liters from the $50$ -liter solution leaves $50-x$ liters of the original solution still in the container. The replacement does not bring that original solution back.
Step 2
Find each portion's bleach amount
Percent means per hundred, so $20\%=0.20$ and $40\%=0.40$ . Multiply each concentration by its portion's volume: the original solution that remains contains $0.20(50-x)$ liters of bleach, and the replacement brings in $0.40x$ liters. Using $0.20(50)$ would count bleach that was drained.
Step 3
Write the bleach balance
The technician adds the same volume that was drained, so the final volume is $50$ liters. Since $28\%=0.28$ , the final mixture contains $0.28(50)$ liters of bleach. Set the bleach remaining plus the bleach added equal to that amount: $0.20(50-x)+0.40x=0.28(50)$ . The right side is not $0.28(50+x)$ ; the new liquid replaces the drained liquid.
Step 4
Solve for the volume replaced
Use a regression to find the value that makes the bleach amounts equal. Type $0.20(50-x_1)+0.40x_1\sim0.28(50)$ : Desmos uses $x_1$ as the value to find, and $\sim$ tells it to fit the two sides. Under PARAMETERS, it shows $x_1=20$ . Because $x$ measures the volume drained and replaced, the technician replaced 20 liters. Grid in 20.

Question 357·Hard·Math

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