Question 26·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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A $40$ -liter tank is initially full of a solution in which the ratio of dye to water is $3:5$ . The solution is thoroughly mixed, and $x$ liters of the solution are removed. Then $x$ liters of pure water are added to the tank. The ratio of dye to water in the tank is now $1:3$ .

Which choice is the value of $x$ ?

In a mixture-ratio question, dye-to-water compares two parts, not dye with the whole tank. Add the ratio parts to find the dye fraction. Then track the dye through the removal, because pure water adds no dye. A Desmos regression can find how much mixture remains; subtract that from the original volume to answer how much was removed.

Hints

Hint 1
A part-to-part ratio compares dye with water. To find dye's fraction of the entire solution, count the dye parts and water parts together. What fraction of the original mixture is dye?
Hint 2
Because the solution is thoroughly mixed, the portion left after removal has the original dye fraction. Call its volume $R$ liters. How much dye does that portion contain before any water is added?
Hint 3
Adding back the same volume restores the tank to $40$ liters, but adds no dye. Use the new ratio to find the dye in that full tank, then equate it to the dye that remained.

Step-by-step

Track the dye that remains
Step 1
Find the original dye fraction
A part-to-part ratio compares dye with water, not dye with the whole solution. The $3:5$ ratio has $3+5=8$ parts in all, so dye is $\frac{3}{8}$ of any thoroughly mixed portion.
Step 2
Write the dye left after removal
Let $R$ be the liters of solution left after removal. Since the solution was thoroughly mixed, those $R$ liters contain $\frac{3}{8}R$ liters of dye, and the added pure water contributes no dye.
Step 3
Find the final dye amount
Adding back the same number of liters restores the full tank to $40$ liters. The new $1:3$ ratio has $1+3=4$ parts in all, so dye makes up $\frac14(40)$ liters of the final solution.
Step 4
Equate the two dye amounts
The water added doesn't change the dye amount, so the dye left after removal must be the dye in the final tank: $\frac38R=\frac14(40)$ .
Step 5
Solve for the mixture left
Type $\frac38x_{1}\sim\frac14(40)$ in Desmos, using $x_{1}$ for $R$ . The $\sim$ asks Desmos to find the value that makes both sides equal. Under PARAMETERS, it shows $x_{1}\approx26.6667$ . That's liters remaining, not liters removed.
Step 6
Subtract to find what was removed
Add $40-x_{1}$ on the next Desmos line: the tank started with $40$ liters, so this takes away the volume remaining. Desmos shows $13.333\ldots$ ; tap its fraction button to see $\frac{40}{3}$ . $\frac{40}{3}$ liters of solution were removed. Choice A.

Question 26·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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