A Jar contains a mixture of A & B in the ratio . When L of Mixture is replaced with liquid B, ratio becomes . How many liters of liquid A was present in mixture initially?
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This one took me way too long. I kept setting up the replacement wrong until I realized you have to subtract 10 L from BOTH parts before adding the pure B back. That was the key for me.
Same here, that was my exact mistake the first time through.
Yeah once I saw it as remove 2 L of A and 8 L of B it got much more manageable.
Got it in under a minute but I've drilled this exact pattern before. For anyone stuck, think about what 10 L of the original mixture actually contains — it's not a 50/50 split.
That hint is perfect. Took me a second to realize the 4:1 matters for the removal too.
Is this really 655-705? Felt more like a 600 level once you set it up algebraically. Maybe I'm missing a trap?
I think the trap is exactly what the first comment said — people forget the removed mixture isn't pure A or pure B.
Can someone explain why we can't just do 4x - 10 and x + 10? I keep getting an answer that isn't even in the choices.
Because the 10 L removed is 4:1 mixture, not 10 L of pure A. Subtract 8 L from B and 2 L from A, then add 10 L pure B.
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.