Finding Original Amount before a Percent Was Added

Date: 04/24/2007 at 11:36:13
From: Sytolli
Subject: tax percentage

I paid $25.00 for gas.  The cost included 42% in tax.  What is the 
cost of the gas without the tax?

I multiplied 42% by $25.00 and subtracted the result from 25, but this 
answer is not appropriate.  I think I am missing a step. 

Date: 04/24/2007 at 12:56:18
From: Doctor Ian
Subject: Re: tax percentage

Hi Sytolli,

Here's one way to think about it.  For every $1 you pay for gas, you
pay $0.42 in tax, right?  That is, $1 worth of gas costs you $1.42.

Let's call that one 'unit' of gas.  How many units can you buy for
$25?  That would be

   25
  ---- = 17.6
  1.42 

That is, you can buy 17.6 units of gas for $25.  Each of those units
contains $1 in gas, and $0.42 in tax.  So the price without tax would
be 17.6 dollars, or $17.60. 

Does that make sense?  Let me know if you need more help.

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/24/2007 at 14:28:31
From: Doctor Riz
Subject: Re: tax percentage

Hi Sytolli -

Just a quick follow up on Dr. Ian's comments to show you a slightly
more formal algebraic way to think about this problem.

Let g be the cost of the gasoline before the tax.  Then, since you pay
42% of that amount in tax, the amount of tax is .42g.  That means the
total amount you paid is g + .42g or 1.42g.  Since the total was $25,
we can say that:

  1.42g = 25

Dividing both sides by 1.42, we get g = 17.6, so the cost of the gas
before taxes was $17.60.

Note that this approach works exactly the same as Dr. Ian's, with the
1.42 representing the "actual cost" of each dollar of gasoline.

- Doctor Riz, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/25/2007 at 13:14:26
From: Sytolli
Subject: Thank you (tax percentage)

Doctors Ian and Riz,

Thank you for your replies.  I knew there was a simpler way to look at
it and just couldn't figure it out.  It makes a lot of sense.  Your
help is much appreciated.