Two numbers are in the ratio $3:5.$ If $9$ is subtracted from each, they become in the ratio $12:23.$ Find the smaller number.


Let's denote the two numbers as $3x$ and $5x$.

According to the problem: $ \dfrac{3x - 9}{5x - 9} = \dfrac{12}{23}$

Cross-multiplying: $$23(3x - 9) = 12(5x - 9)\\69x - 207 = 60x - 108\\x = 11$$

Therefore, the smaller number $3x$ is $33$.

There is a trick to solving this problem. The difference between $12$ and $23$ is $11.$ If you multiply $11$ to $3$ and $5$ you get $33$ and $55.$ So, the smaller number is $33.$

I understand the above solution. How does the trick work, though?

  • $\begingroup$ Think about it for a moment. How can this "trick" be correct advice when it does not consider the fact that the number we subtracted was $9$ and not some completely different number? $\endgroup$
    – David K
    Commented May 26 at 15:01

2 Answers 2


The trick is fake: it claims that $ \dfrac{3x - 18}{5x - 18} = \dfrac{12}{23}$ has solution $23-12=11$ while its actual solution is $22.$


hint: try solving for $\frac{ax-c}{bx-c}$=$\frac{d}{e}$
when you will try to solve it you'll get $x=\frac{c(e-d)}{ae-bd}$ and when you'll do that you'll find that the trick is just a coincidence because the numbers in the original ratio will be $ax:bx$ and in your case $x=|d-e|$ which is not true for all cases

  • $\begingroup$ as a final nod try this trick for other numbers,this trick will be false where $\frac{c}{ae-bd}$ =1 $\endgroup$ Commented May 26 at 6:07
  • 1
    $\begingroup$ Do you mean the trick will be false unless $\frac{c}{ae-bd}=1$? That equation happens to be true in the original question. $\endgroup$
    – David K
    Commented May 26 at 14:59
  • $\begingroup$ Oh yeah thanks for telling me $\endgroup$ Commented May 26 at 17:23

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