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?

  • 3
    $\begingroup$ 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

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 $4\cdot 7777\cdot \frac{1}{12}=\frac{7777}{3}$.


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

Then we can derive this equation.


Multiply out the fractions and factor out x, you get:

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

Thus :

Multiply both sides by $4$ to get:

Divide this by 12, for each month,

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

  • 1
    $\begingroup$ Please check $-(\frac{1}{2}x(\frac{1}{3}(\frac{1}{4})))$ in your equation. $\endgroup$ – Américo Tavares Feb 9 '11 at 20:51
  • 1
    $\begingroup$ ... 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.