Based on a reduced price, what was the original price?

[algabre y78y]

What is the first price if it was reduced by 20% and the new price is 8
Help me please I'm completely lost

 

[Steven G]

Let p = original price.
So 80% of the original price is $8
That is (80/100)*p = $8.

Can you solve for p?

 

[MarkFL]

I've been giving your threads descriptive titles hoping you'd catch on and take the hint. A thread posted in the "Arithmetic" forum with the title "Arithmetic" offers no addition information about the nature of the question being asked. Good thread titles make the site more useful for everyone.

 

[algabre y78y]

I still don't get it.

 

[HallsofIvy]

Exactly which part do you not get? First, Do you understand what it means to "reduce a price by 20%"? If the original price was $30 what is 20% of that? What is the price after you take away that 20%? Do you understand that if you "reduce" a price by 20% you are leaving 100- 20= 80%? Do you understand that "80%" means 80/100? So reducing a price by 20% the new price is 80% of the original price? That is how Jomo got "\(\displaystyle \left(\frac{80}{100}\right)p= 8\)" where "p" is, as Jomo said, the unknown old price and "8" is the given new price.

Or is your problem solving that equation? Do you know that \(\displaystyle \frac{80}{100}= \frac{8}{10}= \frac{4}{5}\)? So you were asked to solve the equation \(\displaystyle \frac{4}{5}p= 8\). What do you get if you multiply both sides of the equation by \(\displaystyle \frac{5}{4}\)?

 

[TheJason]

Let p = original price.
So 80% of the original price is $8
That is (80/100)*p = $8.

Can you solve for p?

(0.20)p = 8

is another way to solve - as @HallsofIvy suggested.