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In a box, there are red and blue marbles. If you take away one red marble, 1/4 of all marbles are red. If, on the other hand, you add a red one, 1/3 of all marbles are red. How many blue marbles are in the box?

If we say:

  • x = Amount of red marbles
  • y = Amount of blue marbles
  • n = Total amount of marbles

Then, what I get from the statement above is:

  • x + y = n
  • (x - 1) = n/4
  • (x + 1) = n/3

But I dont get any further from there.

I do not want the solution, but a hint would be helpful.

Source: http://monoid.mathematik.uni-mainz.de/monoid127.pdf (Page 21, Exercise 1163)

Arno Nymo
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  • it's three equations in three unknowns...just eliminate one at a time. – lulu Sep 07 '17 at 21:35
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    Note: your equations are not correct. Well, $x+y=n$ is correct. But the next should be $x-1=\frac {n-1}4$ and similarly for the third. – lulu Sep 07 '17 at 21:36
  • if you wanted a limited search chinese remainder theorem might be helpful. –  Sep 07 '17 at 22:21

2 Answers2

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Let $\color{red}{r}$ be the number of red marbles and $\color{blue}{b}$ be the number of blue marbles. If we remove a red marble ... \begin{eqnarray*} \frac{\color{red}{r}-1}{\color{blue}{b}+\color{red}{r}-1} =\frac{1}{4} \\ \color{blue}{b}+3=3\color{red}{r} \end{eqnarray*} If we add a red marble ... \begin{eqnarray*} \frac{\color{red}{r}+1}{\color{blue}{b}+\color{red}{r}+1} =\frac{1}{3} \\ 2\color{red}{r}+2=\color{blue}{b} \end{eqnarray*} So $ 2\color{red}{r}+5=\color{blue}{b}+3=3\color{red}{r}$ etc...

$b=12$

Donald Splutterwit
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Guide:

$$x-1 = \frac{n-1}{4}$$

$$x+1 = \frac{n+1}{3}$$

Solve for $x$ and $n$. After which, solve for $y$.

Siong Thye Goh
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