A chemist has milliliters of a acid solution. The chemist will add x milliliters of water (which contains acid) to obtain a new solution whose acid concentration is between and , inclusive. Which compound inequality represents all possible values of x?

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SAT Linear Inequalities Question 42 | Free SAT Prep | Aniko

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Question 42·Hard·Linear Inequalities in One or Two Variables

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A chemist has $50$ milliliters of a $36\%$ acid solution. The chemist will add x milliliters of water (which contains $0\%$ acid) to obtain a new solution whose acid concentration is between $8\%$ and $10\%$ , inclusive. Which compound inequality represents all possible values of x?

For mixture and concentration inequality problems, first compute the actual amount of the substance (here, acid) so you can keep that number fixed. Then express the new concentration as a fraction using the fixed amount over the new total volume, translate any verbal range like "between A% and B%, inclusive" into a compound inequality, and solve each side carefully. Always check that your final solution set respects both inequalities and matches the context (for example, volumes must be nonnegative).

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Track the amount of acid

First figure out how many milliliters of acid are in the original 50 milliliters of 36% solution. That amount of acid does not change when water is added.

Write the new concentration as a fraction

After adding $x$ milliliters of water, what is the new total volume? Use that and the fixed amount of acid to write the concentration as $\text{acid} / \text{total volume}$ .

Turn the word condition into a math inequality

"Between 8% and 10%, inclusive" should become a compound inequality involving the expression you wrote for the concentration and the decimals 0.08 and 0.10.

Solve carefully and combine conditions

Solve each side of the compound inequality step by step, and remember that the final $x$ values must satisfy both resulting inequalities, not just one of them.

Desmos Guide

Enter the concentration function

In Desmos, type y = 18/(50 + x) to represent the acid concentration (as a decimal) after adding $x$ milliliters of water.

Graph the boundary concentrations

Add two more expressions: y = 0.08 and y = 0.10. These are horizontal lines showing the 8% and 10% concentration limits.

Find where the concentration is within the desired range

Look for the $x$ -values where the curve $y = 18/(50 + x)$ lies between the lines $y = 0.08$ and $y = 0.10$ (including the intersection points). Use the intersection tool or tap the intersection points to read off the $x$ -values that mark the lower and upper bounds of this interval.

Step-by-step Explanation

Find the amount of acid in the original solution

The chemist starts with 50 milliliters of a 36% acid solution.

Amount of acid:

0.36 \times 50 = 18 \text{ milliliters of acid}

This 18 milliliters of acid will stay the same after water is added, because water has 0% acid.

Write the concentration of the new solution in terms of x

The chemist adds $x$ milliliters of water.

New total volume: $50 + x$ milliliters
Amount of acid: still 18 milliliters

So the concentration (as a decimal) of the new solution is

\text{concentration} = \frac{18}{50 + x}.

Translate the phrase "between 8% and 10%, inclusive" into an inequality

"Between 8% and 10%, inclusive" means the concentration can be 8%, 10%, or any value in between.

Write this in decimal form:

8% $= 0.08$
10% $= 0.10$

So we need

0.08 \le \frac{18}{50 + x} \le 0.10.

This is a compound inequality that we will solve for $x$ .

Solve each part of the compound inequality

First solve the left part:

0.08 \le \frac{18}{50 + x}.

Since $50 + x > 0$ , we can multiply both sides by $50 + x$ without flipping the inequality:

\begin{aligned} 0.08(50 + x) &\le 18 \\ 4 + 0.08x &\le 18 \\ 0.08x &\le 14 \\ x &\le 175 \end{aligned}

This gives an upper bound on $x$ .

Now solve the right part:

\frac{18}{50 + x} \le 0.10.

Multiply both sides by $50 + x$ :

\begin{aligned} 18 &\le 0.10(50 + x) \\ 18 &\le 5 + 0.10x \\ 13 &\le 0.10x \\ x &\ge 130 \end{aligned}

This gives a lower bound on $x$ .

Combine the bounds to get the full solution for x

From the two parts, we found:

$x \ge 130$ from the right inequality
$x \le 175$ from the left inequality

Both conditions must be true at the same time, so $x$ must be between 130 and 175, inclusive:

130 \le x \le 175.

This matches answer choice C.

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