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This question showed up in an exam for 11-year olds:

Alice bought some chocolate and vanilla muffins. $\frac{3}{4}$ of her muffins are chocolate and the rest are vanilla. She then bought another 60 vanilla muffins. In the end, $\frac{1}{3}$ of her muffins are chocolate. How many muffins did Alice have at first?

I'm sure everyone attacks this problem with algebra, but the exam is aimed at 11 years old, and my understanding is at that age, students use more elementary methods involving drawing figures. However, one student came up with the following method that nonetheless gives the correct answer:

  • Originally $\frac{1}{4}$ of Alice's muffins are vanilla. Finally $\frac{2}{3}$ of her muffins are vanilla.
  • The difference is $\frac{2}{3} - \frac{1}{4} = \frac{5}{12}$
  • 60 muffins were added, so we divide 60 by 5 (the numerator above) and multiply by 4 (the denominator in Alice's original muffins) = 48

The reasoning looks clearly bogus to me, but the answer is correct. I'm trying to figure out why it is correct, and possibly its domain of validity. What I've figured out so far:

  • The method appears to be robust to changing $\frac{3}{4} \rightarrow \frac{3}{5}$.
  • It appears to be robust to changing $\frac{3}{4} \rightarrow \frac{3}{7}$.
  • It appears to not be robust to changing $\frac{1}{3} \rightarrow \frac{2}{3}$
  • It also appears to not be robust to simultaneously changing $\frac{1}{3} \rightarrow \frac{3}{5}$ and $\frac{3}{4} \rightarrow \frac{7}{10}$.

It seems like changing the second number breaks the method, but not the first number, which makes it seem like the method has partial validity. That's a surprise, since bogus methods ought to fail almost all the time. Why?

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    $\begingroup$ Please explain what happens when you do a change that you don't think the method is robust to. $\endgroup$ Commented Nov 18 at 8:30
  • $\begingroup$ @Henriksupportsthecommunity It just fails. The answer is wrong. $\endgroup$ Commented Nov 18 at 13:24
  • $\begingroup$ There is no such thing as "just fails", something must happen. And the answer (which wasn't there when I wrote the initial comment) looks like what I though - and probably right. $\endgroup$ Commented Nov 18 at 14:02
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    $\begingroup$ Never forget: $\require{cancel}$ $\frac {\cancel{6}4}{1\cancel{6}} = 4$. Sometimes doing something completely wrong yields the correct answer entirely by coincidence. $\endgroup$ Commented Nov 18 at 20:34
  • $\begingroup$ When I imagine a procedure for doing this without algebra, I just try numbers in sequence until I find the answer. It doesn't take very long, honestly. $\endgroup$ Commented Nov 18 at 20:41

2 Answers 2

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Why, just do the algebra and see what happens. Say, your unknown is $n$, and your $60$ is $m$, and your $\frac34$ is $a\over b$, and your $1\over3$ is $\frac{c}d$. Then the actual answer is $$n=\frac{m\cdot{c\over d}}{{a\over b}-{c\over d}}$$ While the answer produced by the bogus method is $$n=\frac{m\cdot b}{\text{numerator}\left({a\over b}-{c\over d}\right)}=\frac{m\cdot\dfrac{b}{\text{LCM}(b,d)}}{{a\over b}-{c\over d}}$$

Obviously, the two are equal when $c=1$ and $(b,\,d)$ are coprime, which is relatively often true if we pick small numbers at random, and in particular, indeed true in all cases where you say the bogus method worked, and not true where you say it didn't.

So it goes.

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    $\begingroup$ In fact, there is easily seen to be equality precisely when $c=\gcd(b,d)$ (use the identity that relates lcm to gcd). Of course, this means that $c$ divides $d$, which means we can simplify the fraction, so $c=\gcd(b,d)=1$ is the only essential case. $\endgroup$ Commented Nov 18 at 12:22
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An answer has already been given as to why the bogus method doesn't always work,

but how could $11$ year old Alice properly solve it, "drawing figures"

Let us take a case where the bogus method is said to not work, raising Vanilla from $\frac13$ to $\frac23$

With Vanilla/Chocolate muffins = V/C respectively,
initially, she had $\boxed{V C C}$
To raise vanilla to $\frac23$, add $3$ vanilla, $\boxed{V C C\mid V V V}$

So as the $3$ added vanillas represent $60$ muffins, the original mix also had $60$ muffins

or for the original problem, $\boxed{VCCC}$ to $\boxed{VCCC\mid VVVVV}$, the $5$ added muffins represent $60$ so the original mix had $48$ muffins

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