Question

18. By reducing the selling price of an article by
Rs 100, a gain of 10% turns into a loss of 10%.
Find the original selling price of the article.
​

Answer 1

Solution :

$\underline{\bf{Given\::}}}}$

By reducing the selling price of an article by Rs.100, a gain of 10% turns into a loss of 10%.

$\underline{\bf{Explanation\::}}}}$

Let the cost price of article be Rs.r

A/q

$\longrightarrow\sf{Original\:Selling\:price=10\% \:of\:r\:+\:r}\\\\\longrightarrow\sf{Original\:Selling\:price=\dfrac{10}{100} \times r + r}\\\\\longrightarrow\sf{Original\:Selling\:price=\dfrac{10r}{100} +r}\\\\\longrightarrow\sf{Original\:Selling\:price=\dfrac{10r+100r}{100} }\\\\\longrightarrow\sf{Original\:Selling\:price=\dfrac{110r}{100}......................(1)}$

∴ S.P. = 110r/ 100 - 100

Using formula of the loss % :

$\longrightarrow\tt{Loss\%=\dfrac{C.P- S.P}{C.P} \times 100}\\\\\\\longrightarrow\tt{\cancel{10}\%=\dfrac{r-\bigg(\dfrac{110r}{100}-100\bigg) }{r} \times 10\cancel{0}}}\\\\\longrightarrow\tt{\dfrac{10r}{100} =\dfrac{100r}{100} -\dfrac{110r}{100} +100}\\\\\longrightarrow\tt{\dfrac{10r}{100} =\dfrac{100r-110r}{100} +100}\\\\\longrightarrow\tt{\dfrac{10r}{\cancel{100}} =\dfrac{-10r}{\cancel{100}} +100}\\\\\longrightarrow\tt{10r=-10r+100}\\\\\longrightarrow\tt{10r+10r=100}$

$\longrightarrow\tt{20r=100}\\\\\longrightarrow\tt{r=\cancel{100/20}}\\\\\longrightarrow\bf{r=Rs.5}$

∴ Putting the value of r in equation (1),we get;

$\longrightarrow\tt{\dfrac{11\cancel{0}\times 5}{10\cancel{0}}}\\ \\\longrightarrow\tt{\cancel{\dfrac{55}{10} }}\\\\\longrightarrow\bf{Rs.5.5}$

Thus;

The Original selling price of the article will be Rs.5.5 .