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Every month, a girl gets An allowance. Assume last year she had no money, and kept all the money she has earned up to now. Then she spends $\frac{1}{2}$ of her money on clothes, then $\frac{1}{3}$ of the remaining money on games, and then $\frac{1}{4}$ of the remaining money on toys. After she bought all of that, she had $7777$ left. Assuming she only gets money by allowance, how much money does she earn every month?

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    $\begingroup$ work your way backwards. how much money did she have before spending on toys? before spending on games? before spending on clothes? now what is her monthly allowance? $\endgroup$ – Yuval Filmus Feb 9 '11 at 7:14
  • $\begingroup$ why "Interests" in the title? $\endgroup$ – Américo Tavares Feb 10 '11 at 23:09
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Solution without equations, only fractions. If we represent by $1$ the total money earned by the girl before spending on cloths, games and toys, we can split this unit according to the fractions she spent:

$$1=\overset{3/4}{\overbrace{\underset{2/3}{\underbrace{\frac{1}{2}+\frac{1}{3} \cdot \frac{1}{2}}}+\frac{1}{4}\cdot \frac{1}{3}}}+\frac{1}{4}$$

The fraction $\frac{1}{4}$ represents the money left, which we know is $7777 $. So the money she earned in a year is $4\cdot 7777$. An in a month $4\cdot 7777\cdot \frac{1}{12}=\frac{7777}{3}$.

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Let $x$ be the amount of money the girl started with.

Then we can derive this equation.

$x-\frac{1}{2}x-(\frac{1}{2}x(\frac{1}{3}))-(\frac{1}{2}x(\frac{2}{3}(\frac{1}{4})))=7777$

Multiply out the fractions and factor out x, you get:
$x(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12})=7777$

So $x(\frac{12}{12}-\frac{6}{12}-\frac{2}{12}-\frac{1}{12})=7777$

Thus :
$\frac{1}{4}x=7777$

Multiply both sides by $4$ to get:
$x=31108$

Divide this by 12, for each month,
$31108/12\approx2592.33$

So the girl gets an allowance of $2592.33$$/month

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    $\begingroup$ Please check $-(\frac{1}{2}x(\frac{1}{3}(\frac{1}{4})))$ in your equation. $\endgroup$ – Américo Tavares Feb 9 '11 at 20:51
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    $\begingroup$ ... I've got $-\frac{1}{12}x$ instead. And $x=31108$, $x/12=2592.3$. $\endgroup$ – Américo Tavares Feb 9 '11 at 20:55
  • $\begingroup$ You are correct, thanks. $\endgroup$ – Justin Feb 9 '11 at 21:47

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