A milk vendor mixes water with milk and sells the mixture at the same price per liter as if it...
GMAT Data Sufficiency : (DS) Questions
A milk vendor mixes water with milk and sells the mixture at the same price per liter as if it were undiluted milk. The selling price per liter of the mixture is the vendor's cost per liter of the milk plus a markup of \(\mathrm{x}\%\). The water costs the vendor nothing. If the vendor gets a \(50\%\) profit on the sale of the mixture, what is the value of x ?
- If the vendor mixes half the intended quantity of water and sells every liter of the mixture at the cost price per liter of the undiluted milk, the vendor will get a \(10\%\) profit.
- The concentration of milk in the mixture after adding water is \(\frac{5}{6}\).
Understanding the Question
Let's break down this profit puzzle step by step.
What we need to find: The value of x (the markup percentage on milk cost)
Given information:
- A vendor mixes free water with milk
- The mixture is sold at the same price as pure milk
- Selling price = Cost of milk + \(x\%\) markup
- The vendor makes \(50\%\) total profit on the mixture
The key insight: The vendor has two sources of profit:
- The markup (\(x\%\)) on the milk cost
- The dilution profit from selling free water at milk price
These two profit sources must combine to yield exactly \(50\%\) total profit. Once we know how much profit comes from dilution, we can determine how much must come from markup - and this will uniquely determine x.
What "sufficient" means here: We can determine the exact value of x.
Analyzing Statement 1
What Statement 1 tells us: When the vendor uses only half the intended water amount and sells at cost price (no markup), the profit is \(10\%\).
Let's think through this logically:
- With half the water and \(0\%\) markup → \(10\%\) profit
- This \(10\%\) profit comes entirely from selling water at milk price
- So with the full intended water amount and \(0\%\) markup → \(20\%\) profit (doubling the water doubles this profit source)
Now here's the crucial insight: Since we need \(50\%\) total profit and dilution provides \(20\%\), the markup must provide exactly \(30\%\) additional profit.
To visualize this: If milk costs \(\$100\) per liter, a \(30\%\) profit contribution means the markup must result in selling at \(\$130\), which requires \(x = 30\).
[STOP - Sufficient!]
Conclusion: There's only one markup percentage that contributes exactly \(30\%\) to the profit equation. Statement 1 is sufficient.
This eliminates choices B, C, and E.
Analyzing Statement 2
Now let's forget Statement 1 completely and analyze Statement 2 independently.
What Statement 2 provides: The milk concentration in the mixture is \(\frac{5}{6}\) (meaning water is \(\frac{1}{6}\)).
Let's reason through this:
- For every 5 liters of milk costing 5C, the vendor adds 1 liter of free water
- The vendor sells 6 liters at the milk price but only paid for 5 liters
- This dilution alone gives: \(\frac{6}{5} - 1 = \frac{1}{5} = 20\%\) profit
Again, we arrive at the same critical point: dilution provides \(20\%\) profit, so markup must provide \(30\%\) to reach the required \(50\%\) total.
[STOP - Sufficient!]
Conclusion: This uniquely determines x. Statement 2 is sufficient.
This eliminates choices A, C, and E.
The Answer: D
Both statements independently lead us to the same conclusion - the dilution provides \(20\%\) profit, so the markup must provide exactly \(30\%\) profit. This uniquely determines the value of x.
Answer Choice D: "Each statement alone is sufficient."