A vessel contains L milk. The milkman delivers L to the first house and adds an equal quantity of water. He does exactly the same at the second and third house. What is the ratio of milk and water when he has finished delivering at the third house?
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Classic removal-and-replacement trap. The key is that you're always removing 20 L of the *current* mixture, not 20 L of pure milk. Took me a second to see why the answer isn't just 20:60 or something.
Same here — I kept subtracting 20 from the milk each time and got a totally different number. The fraction remaining after each step is what matters.
Got 27:37 after working through it, but let me double check the third step. After two operations milk is 80*(3/4)^2 = 45, then remove 20 of the 80 L mixture... yeah that lines up with choice D.
Nice, the (3/4) pattern makes it way faster than tracking actual liters each round.
Is there a shortcut for this? I feel like doing it step by step takes too long for a 700-level question.
Yeah — milk left = 80 * (1 - 20/80)^3 = 80 * (3/4)^3. Then water is just 80 minus that. Saves a lot of time.
That formula only works because the vessel stays at 80 L the whole time. If the total volume changed you'd have to go step by step.
The wording threw me — 'adds an equal quantity of water' means he adds 20 L each time, so total stays 80 L. Once I got that, it was manageable. Solid question.
What it tests
Your ability to reason about weighted averages and the composition of mixtures.
Common trap
Averaging the percentages of two solutions without weighting by their volumes.