A container holds a liquid mixture that is vinegar and water by volume. If milliliters of only vinegar are removed and replaced with milliliters of water, the resulting mixture is vinegar. How many milliliters of the mixture were in the container before any liquid was removed?

Solution available with step-by-step explanation

SAT Percentages Question 36 | Free SAT Prep | Aniko

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Question 36·Hard·Percentages

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A container holds a liquid mixture that is $40\%$ vinegar and $60\%$ water by volume. If $20$ milliliters of only vinegar are removed and replaced with $20$ milliliters of water, the resulting mixture is $25\%$ vinegar.

How many milliliters of the mixture were in the container before any liquid was removed?

For mixture and percentage problems where something is removed and replaced, first assign a variable to the total starting amount, then write expressions for the amount of the key ingredient (here, vinegar) before and after the change. Use the given final percentage to form an equation: (final amount of ingredient)/(final total amount) = given percent as a decimal. Solve the resulting linear equation carefully, converting decimals like 0.4, 0.25, and 0.15 to fractions if that makes the arithmetic easier, and double-check that your denominator uses the final total amount, not an intermediate amount after removal.

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Introduce a variable

Let $V$ be the total volume of the mixture (in milliliters) before any liquid is removed. How much vinegar is that in terms of $V$ if vinegar is $40\%$ of the mixture?

Carefully track the vinegar amount

When you remove 20 mL, you are taking out only vinegar, not the mixture. What does that do to the vinegar amount? After you add 20 mL of water, does the vinegar amount change again?

Translate the 25% information into an equation

After the remove-and-replace step, you know the final total volume and the final amount of vinegar (both in terms of $V$ ). How can you express "the mixture is $25\%$ vinegar" as a fraction or equation using those two quantities?

Solve carefully with decimals or fractions

Your equation should end up with a term like $0.15V = 20$ or an equivalent fraction equation. Make sure you divide correctly to solve for $V$ ; you may find it easier to convert the decimals to fractions.

Desmos Guide

Set up the equation in Desmos

In one expression line, type (0.4x - 20)/x to represent the final fraction of vinegar. In a second line, type 0.25 to represent 25% vinegar as a decimal.

Find the x-value that makes the fractions match

Look for the intersection point of the graph of y = (0.4x - 20)/x and the horizontal line y = 0.25. The x-coordinate of this intersection is the original volume of the mixture in milliliters.

Step-by-step Explanation

Define a variable and the starting amounts

Let $V$ be the total number of milliliters of mixture in the container at the start.

Vinegar at the start: $40\%$ of $V$ , which is $0.4V$ .
Water at the start: $60\%$ of $V$ , which is $0.6V$ .

Track what happens when 20 mL of vinegar is removed and water is added

We remove 20 mL of pure vinegar and then add 20 mL of water.

After removing vinegar, vinegar amount becomes $0.4V - 20$ .
Total volume after removal is $V - 20$ .
Then we add $20$ mL of water, so the total volume goes back to $V$ .
Adding water does not change the amount of vinegar, so the final vinegar amount is still $0.4V - 20$ .

So in the final mixture:

Total volume is $V$ .
Vinegar amount is $0.4V - 20$ .

Use the final percentage to write an equation

We are told the resulting mixture is $25\%$ vinegar.

That means:

\frac{\text{final vinegar amount}}{\text{final total volume}} = 0.25

Substitute the expressions from Step 2:

\frac{0.4V - 20}{V} = 0.25

This equation relates the original volume $V$ to the given percentages and the 20 mL change.

Solve the equation for V

Start with the equation:

\frac{0.4V - 20}{V} = 0.25

Multiply both sides by $V$ :

0.4V - 20 = 0.25V

Subtract $0.25V$ from both sides:

0.4V - 0.25V = 20 \\[0.7em] 0.15V = 20

Now solve for $V$ :

V = \frac{20}{0.15}

Write $0.15$ as a fraction: $0.15 = \tfrac{15}{100} = \tfrac{3}{20}$ .

V = 20 \div \frac{3}{20} = 20 \cdot \frac{20}{3} = \frac{400}{3}

Convert $\tfrac{400}{3}$ to a mixed number:

$3$ goes into $400$ a total of $133$ times (since $133 \cdot 3 = 399$ ), with a remainder of $1$ .
So $\tfrac{400}{3} = 133\tfrac{1}{3}$ .

Therefore, the container originally held $133\tfrac{1}{3}$ milliliters of mixture.

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Step-by-step Explanation

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