Question

A person sells 3 article, frist at a profit of 20% second at a profit of 16(2/3)%. At what profit or loss % he will sell the third article so there will be no profit no loss in the whole transaction. Selling price of all the article are in ration of 1:2:3.

Answer 1

Let me derive the net profit / loss of whole transaction of selling $n$ articles of cost prices $C_1,\ C_2,\ C_3,\,\dots\,,\ C_n$ at profit $r_1\%,\ r_2\%,\ r_3\%,\,\dots\,,\ r_n\%$ sold at selling price $S_1,\ S_2,\ S_3,\,\dots\,,\ S_n$ each.

[Note:- For loss we can take negative sign for the respective percentage.]

Then the profit percentage of $i^{th}$ article for $i\in\{1,\ 2,\ 3,\,\dots\,,\ n\}$ is given by,

$\longrightarrow r_i=\dfrac{S_i-C_i}{C_i}\times100$

$\longrightarrow \dfrac{C_ir_i}{100}=S_i-C_i$

$\longrightarrow S_i=C_i\left[1+\dfrac{r_i}{100}\right]$

$\longrightarrow S_i=C_i\left[\dfrac{100+r_i}{100}\right]$

$\longrightarrow C_i=\dfrac{100S_i}{100+r_i}$

Then, net cost price of the whole transaction will be,

$\displaystyle\longrightarrow \sum_{i=1}^nC_i=\sum_{i=1}^n\dfrac{100S_i}{100+r_i}$

$\displaystyle\longrightarrow \sum_{i=1}^nC_i=100\,\sum_{i=1}^n\dfrac{S_i}{100+r_i}$

Let $r$ be the profit percentage of whole transaction. Then,

$\longrightarrow r=\dfrac{\displaystyle\sum_{i=1}^nS_i-\sum_{i=1}^nC_i}{\displaystyle\sum_{i=1}^nC_i}\times100$

$\longrightarrow r=\dfrac{\displaystyle\sum_{i=1}^nS_i-100\sum_{i=1}^n\dfrac{S_i}{100+r_i}}{\displaystyle100\sum_{i=1}^n\dfrac{S_i}{100+r_i}}\times100$

$\longrightarrow r=\dfrac{\displaystyle\sum_{i=1}^nS_i}{\displaystyle\sum_{i=1}^n\dfrac{S_i}{100+r_i}}-100$

$\longrightarrow\large\boxed{r=\displaystyle\sum_{i=1}^nS_i\cdot\left[\displaystyle\sum_{i=1}^n\dfrac{S_i}{100+r_i}\right]^{-1}-100}$

According to the question, $n=3.$ Hence net profit is,

$\longrightarrow r=\left[S_1+S_2+S_3\right]\left[\dfrac{S_1}{100+r_1}+\dfrac{S_2}{100+r_2}+\dfrac{S_3}{100+r_3}\right]^{-1}-100$

which is zero as per the question.

$\longrightarrow\left[S_1+S_2+S_3\right]\left[\dfrac{S_1}{100+r_1}+\dfrac{S_2}{100+r_2}+\dfrac{S_3}{100+r_3}\right]^{-1}-100=0$

Since $r_1=20$ and $r_2=16\,\dfrac{2}{3}=\dfrac{50}{3},$

$\longrightarrow\left[S_1+S_2+S_3\right]\left[\dfrac{S_1}{100+20}+\dfrac{S_2}{100+\frac{50}{3}}+\dfrac{S_3}{100+r_3}\right]^{-1}-100=0$

Given that, $S_1:S_2:S_3=1:2:3.$ Hence let,

• $S_1=S;\quad S_2=2S;\quad S_3=3S$

Then,

$\longrightarrow\left[S+2S+3S\right]\left[\dfrac{S}{100+20}+\dfrac{2S}{100+\frac{50}{3}}+\dfrac{3S}{100+r_3}\right]^{-1}=100$

$\longrightarrow6S\cdot S^{-1}\left[\dfrac{1}{120}+\dfrac{6}{350}+\dfrac{3}{100+r_3}\right]^{-1}=100$

$\longrightarrow\dfrac{350(100+r_3)+120\times6(100+r_3)+120\times350\times3}{120\times350(100+r_3)}=\dfrac{6}{100}$

$\longrightarrow\dfrac{350(100+r_3)+720(100+r_3)+126000}{420(100+r_3)}=6$

$\longrightarrow1070(100+r_3)+126000=2520(100+r_3)$

$\longrightarrow145(100+r_3)=12600$

$\longrightarrow100+r_3=\dfrac{12600}{145}$

$\longrightarrow r_3=\dfrac{12600}{145}-100$

$\longrightarrow\underline{\underline{r_3=-13\ \dfrac{3}{29}}}$

I.e., the third article should be sold at a loss of $\bf{13\ \dfrac{3}{29}\%.}$