Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Figuring a Tax

Date: 05/13/2003 at 12:07:08
From: Annette
Subject: Percentage increase of an amount

I work in insurance. We have a premium of $75,329. This premium needs 
to be increased by a tax of 3.75%.  

We have been doing the following: $75,329 / .9625 =  $78,263.90  
Why doesn't this work compared to the example below? If I follow what 
I read in the archives I should be using the following:

(1.00 * Y) + (.0375 * Y) = X
(1.00 + .0375) * Y = X
1.0375 * Y = X

So 1.0375 * $75,329  =  $78,153.84


Date: 05/13/2003 at 14:09:03
From: Doctor Ian
Subject: Re: Percentage increase of an amount

Hi Annette, 

If the premium before tax is P, and the tax is 3.75%, then the final
charge should be 

    P + 0.0375P

  = P(1 + 0.0375)

  = 1.0375P

So if the premium is $75,329, the amount charged should be

  1.0375 * $75,329 = 78,153.84

which is what you got at the end of your question.

What are you doing when you divide by 0.9625, instead of multiplying 
by 1.0375?  What amount are you ending up with?  

To make this easier to write, I'm going to use r = 0.0375 to represent
the tax rate:
 
                 something = P/0.9625

                           = P/(1 - r)

          something(1 - r) = P

   something - r*something = P

                 something = P + r*something
  
Now, since we can get 'something' by dividing P by a number that is
less than 1, it has to be the case that 

  something > P

So what you're doing is computing the tax on an amount _greater_ than
the premium, and adding it to the premium. Which is to say, you've
been charging too much tax.  

Another way to look at it is this: Since

  something(1 - r) = P

the 'something' that you're computing is the amount that you'd have to
_reduce_ by 3.75% to end up with your original premium. 

An example with simpler numbers might make this clearer. Suppose the 
premium is $100, and the tax is 10%.  The final amount should be 

  $100 + (10% of $100) = $100 + $10 = $110

But suppose we ask instead: What number would we have to _reduce_ by
10%, in order to end up with $100? It's not $110. If you reduce $110
by 10%, you end up with 

  $110 - (10% of $110) = $110 - $11 = $99

In fact, the amount A that you'd have to start with would be given by

  A(1 - 10/100) = 100

              A = 100/(1 - 10/100)

                = 111.11

So what you're doing is kind of like this:

  premium with tax = $100 + (10% of 111.11)

                   = $100 + (10% of (100 + 11.11))

                   = $100 + (10% of 100) + (10% of 11.11)
                       ^         ^              ^
                       |         |              |
                     premium    tax          overcharge

Note that dividing by (1-rate) instead of multiplying by (1+rate)
gives you the same answer only when

     P 
  ------- = P(1 + r)
  (1 - r)

     1 
  ------- = (1 + r)
  (1 - r)

        1 = (1 + r)(1 - r)

        1 = 1 - r^2

      r^2 = 0

Which is to say, it works only when the rate is zero. For any other
rate, you end up adding too much tax. How much too much?  We can 
compute that this way:

        overcharge = P/(1-r) - P(1+r)  

                   = P[1/(1-r) - (1+r)]

                   = P[r^2/(1 - r)]

Let's check that on our $100 premium:

        overcharge = $100 * [(0.1)^2 / (1 - 0.1)]

                   = $100 * [0.01 / 0.9]

                   = $1.11

which agrees with what we got before. In the actual case, this works
out to 

        overcharge = $75,329 * [(0.0375)^2 / (1 - 0.0375)]

                   = $100 * [0.00140625 / 0.9625]

                   = $110.06

which is the difference between $78,263.90 and $78,153.84.  

I hope this helps. Write back if you'd like to talk more about this,
or anything else. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Middle School Fractions

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/