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The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction is increased by 1/24. What is the difference between the numerator and denominator of the fraction?

Do we have any shortcuts for questions like these?

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  • $\begingroup$ The difference is...what? $\endgroup$ – ndh Oct 22 '13 at 4:40
  • $\begingroup$ we have to find it $\endgroup$ – gandhigcpp Oct 22 '13 at 4:40
  • $\begingroup$ Ok, makes sense. You might want to rephrase to "What is the difference between the num and the den?" $\endgroup$ – ndh Oct 22 '13 at 4:42
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    $\begingroup$ We know a+b=11, (a+2)/(b+2)=a/b+1/24. Can you do the rest from here? $\endgroup$ – user98602 Oct 22 '13 at 4:55
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    $\begingroup$ Posting your current solution may help to know what is considered as "shortcutted" and what's not. $\endgroup$ – chubakueno Oct 22 '13 at 5:00
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The quickest solution that I’ve found is to write the fraction as $\frac{a}b$ and note that

$$\frac1{24}=\frac{a+2}{b+2}-\frac{a}b=\frac{2(b-a)}{b(b+2)}\;,$$

so

$$\frac{b-a}{b(b+2)}=\frac1{48}\;.$$

$6\cdot8=48$, so try $b=6$; then $a=5$, which works.

Replacing $a$ by $11-b$ to get

$$\frac{13-b}{b+2}=\frac{11-b}b+\frac1{24}=\frac{264-23b}{24b}$$

and solving the resulting quadratic $24b(13-b)=(b+2)(264-23b)$ for $b$ confirms that this is the only solution. However, it’s hardly a general technique, unlike the solution of the quadratic.

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If the numerator and denominator are supposed to be positive integers that sum to $11$, there are only $10$ possibilities, and it's pretty easy to check them one by one.

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With the fraction $\ds{p \over q}$ we get $\ds{47q - 48 p = q\pars{q + 1}}$. Since $q\pars{q + 1}$ and $48p$ are even, $q$ must be even. We only have to check $q = 2, 4, 6, 8 , 10$ such that $$ p = {q\pars{46 - q} \over 48} = {s\pars{23 - s} \over 12} \quad \mbox{is an integer}\,, \quad q= 2s\,,\quad s = 1,2,3,4,5 $$ That yields the allowed $s = 3\quad\imp\quad q = 6,\quad p = 5\imp\quad$ $\ds{\large{p \over q} = {5 \over 6}}$

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  • $\begingroup$ We can also infer that $q>p$ from the fact that both are positive and that adding a positive value to both increases the fraction, so we only have to check $q=6,8,10$. $\endgroup$ – Jaycob Coleman Oct 22 '13 at 6:25
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    $\begingroup$ @JaycobColeman Fine. That's true. Since the "range" was quite small we didn't pay attention to that. Thanks. $\endgroup$ – Felix Marin Oct 22 '13 at 6:30
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$\dfrac{x}{y}=\dfrac{x+a}{y+a}+b=\dfrac{x+a+by+ba}{y+a}$

$xa=ya+by^2+bay$

$x=y+by\dfrac{y+a}{a}$

$x+y=n$

$n=y(b\dfrac{y+a}{a}+2)=\dfrac{b}ay^2+(b +2)y$,

so

$y=\dfrac{-(b+2)\pm \sqrt{(b+2)^2+4bn/a}}{2b/a}$

Yours is the particular case $a=2,b=-1/24,n=11$. This gives two solutions, but one can be cast out as it would require one of $x,y$ to be negative, which would make the fraction negative as well.

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