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.
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