The fraction $\frac{m}{n}$, where $m$ and $n$ are positive integers is given. If we increase the numerator of the fraction by 5%, then by what percent should we increase the denominator in order to decrease the fraction $\frac{m}{n}$ by 10%?

(answer is $50/3$)

I did this, $\frac{105}{n}=90$ seems wrong?

  • $\begingroup$ You're correct. The new $n'=(105/90)n$. $\endgroup$ – poyea Jun 8 '18 at 18:35
  • $\begingroup$ answer says 50/3 though $\endgroup$ – user567775 Jun 8 '18 at 18:37

1. Let $x=\frac{m}{n}$ and $x'=\frac{m'}{n'}$.


You know how $m$ changes:

$$\frac{1.05m}{n'}=0.9\frac{m}{n}\implies n'=\frac{1.05}{0.9}n=\frac{7}{6}n=(1+\frac{1}{6})n.$$

That much.

2. Say $x=\frac{4}{9}$. $x'=\frac{4.2}{9k}$. We want $$\frac{x'-x}{x}=-0.1$$ $$x'=0.9x$$ $$\frac{4.2}{9k}=0.9(4/9)\implies k=\frac{4.2}{9\cdot0.9(4/9)}=\frac{7}{6}.$$

3. Note that $n'=(1+\frac{50}{3}\color{blue}{\%})n=(1+\frac{1}{6})n$. (Credit to Daniel)

  • $\begingroup$ I got that too, but test's answer says it's $50/3$% $\endgroup$ – user567775 Jun 8 '18 at 18:55
  • 1
    $\begingroup$ @user567775 $\frac{50}{3}\% = \frac{50}{300} = \frac{1}{6}$, so that agrees with poyea's result. $\endgroup$ – Daniel Fischer Jun 8 '18 at 19:04
  • $\begingroup$ Ah, it's $(50/3)\%$. I thought it was just a fraction, without the percentage sign. @user567775 $\endgroup$ – poyea Jun 8 '18 at 19:07

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