Answer:
[tex]\frac{-100(100+x)}{(x-100)}[/tex]
Step-by-step explanation:
Let the original price of the item = p
Let there was q% of increase on the original value, so
the printed price of the item, [tex]C= p+p \frac{q}{100}=p(1+\frac{q}{100})[/tex]
After a price reduction of x%, the has its price increased to its original value, so,
[tex]C-C\times \frac {x}{100} =[/tex] original price + increased price
[tex]p(1+\frac{q}{100})-p(1+\frac{q}{100})\times \frac {x}{100} = p+p[/tex]
[tex](1+\frac{q}{100})-(1+\frac{q}{100})\times \frac{x}{100} =2[/tex]
[tex]-(1+\frac{q}{100})\times \frac{x}{100} =2-1-\frac{q}{100}[/tex]
[tex]-(1+\frac{q}{100})\times \frac{x}{100} =1-\frac{q}{100}[/tex]
[tex](1+\frac{q}{100})\times \frac{x}{100} =\frac{q}{100}-1[/tex]
[tex]\frac{x}{100}+\frac{q}{100}\times\frac{x}{100}-\frac{q}{100}=-1[/tex]
[tex]x+x \times \frac{q}{100}-q=-100[/tex]
[tex]q(\frac{x}{100}-1)=-100-x[/tex]
[tex]q=\frac{-100(100+x)}{(x-100)}[/tex]
Hence, the percent of the increase is [tex]\frac{-100(100+x)}{(x-100)}[/tex]
