A tank contains x gallons of antifreeze that is, by volume, y% of propylene glycol and \((100-\mathrm{y})\%\) water, where y...

Understanding the Question

We need to find how many gallons of pure propylene glycol Shilah must add to transform a \(\mathrm{y}\%\) propylene glycol solution into a \(60\%\) propylene glycol solution.

Given Information

• Tank contains x gallons of antifreeze
• Current mixture: \(\mathrm{y}\%\) propylene glycol and \((100-\mathrm{y})\%\) water
• Constraint: \(\mathrm{y} < 60\)
• Target mixture: \(60\%\) propylene glycol and \(40\%\) water
• We're adding pure propylene glycol (\(100\%\) concentration)

What We Need to Determine

We need a specific numerical value - the exact number of gallons to add. This is a value question.

Setting Up Our Approach

When we add p gallons of pure propylene glycol:

• Initial propylene glycol: \(\frac{\mathrm{xy}}{100}\) gallons
• After adding p gallons: \(\left(\frac{\mathrm{xy}}{100} + \mathrm{p}\right)\) gallons of propylene glycol in \((\mathrm{x} + \mathrm{p})\) total gallons
• Target equation: \(\frac{\frac{\mathrm{xy}}{100} + \mathrm{p}}{\mathrm{x} + \mathrm{p}} = 0.6\)

Solving this equation algebraically:

• \(\frac{\mathrm{xy}}{100} + \mathrm{p} = 0.6(\mathrm{x} + \mathrm{p})\)
• \(\frac{\mathrm{xy}}{100} + \mathrm{p} = 0.6\mathrm{x} + 0.6\mathrm{p}\)
• \(\frac{\mathrm{xy}}{100} + 0.4\mathrm{p} = 0.6\mathrm{x}\)
• \(0.4\mathrm{p} = 0.6\mathrm{x} - \frac{\mathrm{xy}}{100}\)
• \(\mathrm{p} = \frac{0.6\mathrm{x} - \frac{\mathrm{xy}}{100}}{0.4}\)

Key Insight: To find a unique value for p, we need to determine the value of the expression \(\left(0.6\mathrm{x} - \frac{\mathrm{xy}}{100}\right)\).

Analyzing Statement 1

Statement 1 tells us: \(\mathrm{xy} = 3200\)

What Statement 1 Provides

Substituting \(\mathrm{xy} = 3200\) into our formula:

\(\mathrm{p} = \frac{0.6\mathrm{x} - \frac{3200}{100}}{0.4} = \frac{0.6\mathrm{x} - 32}{0.4}\)

Testing Different Values

Since we still have x as an unknown variable, let's test what happens with different values:

• If \(\mathrm{x} = 80\) gallons: then \(\mathrm{y} = 40\%\), and \(\mathrm{p} = \frac{48 - 32}{0.4} = 40\) gallons
• If \(\mathrm{x} = 100\) gallons: then \(\mathrm{y} = 32\%\), and \(\mathrm{p} = \frac{60 - 32}{0.4} = 70\) gallons

Conclusion

Different values of x that satisfy \(\mathrm{xy} = 3200\) give us different amounts of propylene glycol to add. Without knowing x specifically, we cannot determine a unique value for p.

Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Important: We now analyze Statement 2 independently, forgetting Statement 1 completely.

Statement 2 tells us: \(0.6\mathrm{x} - \frac{\mathrm{xy}}{100} = 16\)

The Critical Recognition

Look at what Statement 2 gives us - it's exactly the numerator in our formula!

Recall: \(\mathrm{p} = \frac{0.6\mathrm{x} - \frac{\mathrm{xy}}{100}}{0.4}\)

Direct Calculation

Since Statement 2 tells us that \(0.6\mathrm{x} - \frac{\mathrm{xy}}{100} = 16\), we can substitute directly:

\(\mathrm{p} = \frac{16}{0.4} = 40\) gallons

[STOP - Sufficient!] We have found a unique value for p.

Why This Works

Statement 2 provides the exact expression we identified as critical. We don't need to know x and y individually - Statement 2 directly gives us their key combination that determines p uniquely.

Statement 2 is sufficient.

This eliminates choices C and E.

The Answer: B

Statement 2 alone provides the exact expression needed to calculate the gallons of propylene glycol to add (\(40\) gallons), while Statement 1 leaves us with multiple possible values depending on x.

Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."