I am preparing for a competitive examination which consists of quantitative aptitude. We came across a question on Ratios as follows:

Two numbers are in the ratio 4 : 5. If 7 is added to each, the ratio becomes 5 : 6. Find the numbers.

Now to solve such questions, which have a given ratio, a number added to both the numbers in the ratio and the final ratio, our teacher suggested a simple short cut to save some time in the examination. It is as follows:

If the difference between the antecedent and the consequent of the ratio of the original numbers (here, ratio of original numbers = 4 : 5, difference = 1) is same as the difference between the antecedent and the consequent of the ratio of the final numbers (here, 5 : 6, difference = 1), then compare the difference (1) with the number added to the original numbers(here, 7).

Using that, proportionately find the number corresponding to the original ratio points.

E.g. here, difference = 1 and number added = 7. So "if 7 corresponds to 1 ratio point" says he, "4 ratio points (the antecedent of original ratio) must correspond to 28 and 5 to 35. And in this way we get the original numbers as 28 and 35."

Can somebody please explain why/how this short cut works? Also would this work only in this specific case, or could it be applied to some other cases also, albeit with some modification? Also, he used the term "ratio point" to denote both the difference between the ante. and conse. of the ratio as well as the ante. and conse. of the ratios. Is the term legit?


So if your first numbers are $N$ and $M$ with $N \ne M$ and $\frac NM = \frac nm$.

An if you add $w\ne 0$ to each number you get the ration $\frac {N+w}{M+w} =\frac pq$.

The shortcut says if $d= p-n= q-m$ then $N = n*\frac {w}{d}$ and $M = m*\frac {w}{d}$.


Well $\frac NM = \frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.

And $\frac {nk + w}{mk + w} = \frac pq$

So $q(nk+w) = p(mk + w)$ so

$k(qn - pm) = w(p-q)$

Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$

$k(q(m+v) - (q+v)m) = wv$

$k(qm + qv - qm - mv) = wv$

$k*v(q-m) = wv$

$k*v*d = wv$

$k= \frac wd$.

The only assumption is $v\ne 0$ and $d\ne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $\frac NM = \frac nm = \frac {N+w}{M+w} $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $w\ne 0$ then $\frac NM = \frac ww = 1$ and $N=M$ which we assumed was not the case.

  • $\begingroup$ Perfect. This is it. $\endgroup$ – Neil Jul 26 '18 at 21:00

Your numbers are $4k$ and $5k$

When you add $7$ to them you get $4k+7$ and $5k+7$

The new ration is $$\frac { 4k+7}{5k+7}=\frac { 5}{6}$$

Solve for $k$ and you get $k=7$

Thus your numbers are $28$ and $35$


Because you pass from $\displaystyle{4x\over 5x}$ to $\displaystyle{5x\over 6x}$ by adding $x$ to both numerator and denominator.


Yes, your statement is true.

$$\frac {nk+d}{(n+1)k+d}=\frac {(n+1)k}{(n+2)k} \implies d=k $$

  • $\begingroup$ @fleablood offers a more generalized solution $\endgroup$ – Neil Jul 26 '18 at 21:01
  • $\begingroup$ Thanks for the comment, my solution is also correct but shorter in details. $\endgroup$ – Mohammad Riazi-Kermani Jul 26 '18 at 21:11

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