Question

Can someone explain this problem to me?? After a price reduction of x%, an item has its price increased to its original value. What was the percent of increase? Express your answer as a common fraction in terms of x.

Answer 1

Answer:

$\frac{-100(100+x)}{(x-100)}$

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, $C= p+p \frac{q}{100}=p(1+\frac{q}{100})$

After a price reduction of x%, the has its price increased to its original value, so,

$C-C\times \frac {x}{100} =$ original price + increased price

$p(1+\frac{q}{100})-p(1+\frac{q}{100})\times \frac {x}{100} = p+p$

$(1+\frac{q}{100})-(1+\frac{q}{100})\times \frac{x}{100} =2$

$-(1+\frac{q}{100})\times \frac{x}{100} =2-1-\frac{q}{100}$

$-(1+\frac{q}{100})\times \frac{x}{100} =1-\frac{q}{100}$

$(1+\frac{q}{100})\times \frac{x}{100} =\frac{q}{100}-1$

$\frac{x}{100}+\frac{q}{100}\times\frac{x}{100}-\frac{q}{100}=-1$

$x+x \times \frac{q}{100}-q=-100$

$q(\frac{x}{100}-1)=-100-x$

$q=\frac{-100(100+x)}{(x-100)}$

Hence, the percent of the increase is $\frac{-100(100+x)}{(x-100)}$