Hi Gmatiseasyforall,Great instinct spotting the pattern, and you're right that the answer here is
140·
60/(
140+
60) =
8400/
200 =
42, which is exactly half the harmonic mean of the two volumes.
But here's the key idea:
the harmonic mean isn't actually the reason the problem works - it's just the
shape the answer happens to take once you do the algebra. There's no special "harmonic mean property" being invoked. Let me show you where the formula really comes from.
The real structureLook back at KarishmaB's step in the thread - the one that reads x/
140 = (
60−x)/
60. That single equation is the heart of it. Two facts force it:
- After you draw x and swap,
each container's total volume is unchanged (
140 stays
140,
60 stays
60).
- "Equal price per liter" is the same as
equal concentration of either oil in both containers.
Set the concentration of A's oil equal in both containers, using general volumes a and b:
(a-x)/a = x/b
Cross-multiply:
- b(a - x) = x·a
- ab - xb = xa
- ab = x(a + b)
- x = ab/(a + b)
That's it. The formula ab/(a+b)
falls straight out of the equal-concentration condition - no harmonic mean needed to derive it. It only
looks like half the harmonic mean because ab/(a+b) and
2ab/(a+b) share the same algebraic form.
When the shortcut is safe to useThe formula x = ab/(a+b) applies whenever the problem has this exact structure:
- Equal amounts drawn from each container and
swapped,
- Volumes end up unchanged,
- The final condition is
equal price/concentration,
- The two starting prices are different.
Strip away any of those and the derivation breaks - so lean on the
structure, not on remembering "harmonic mean."
Answer: DGmatiseasyforall
Hi Bunuel! Thanks for posting this question! This is one of those mixture problems that confuse me a lot!
The Answer provided by andreogon is right on the dots! The answer to this type of mixture problems is always
\( \frac{ ab}{a+b } \) where
a and
b are the respective volumes of the two containers. Notice that this is the HALF of the Harmonic Mean formula i.e \(\frac{ Harmonic Mean}{2 } \). My question is why is Harmonic mean is applicable to this type of mixture problems? Is there any specific structure of the question that make Harmonic mean appropriate for this type of question? Or is there any specific property of Harmonic Mean which makes it a fit for this type of question?
Will much appreciate your insights in this regard! Thanks in advance!