Evaporation

Date: 12/08/2003 at 22:00:52
From: Olo
Subject: Mixture and percent

A substance is 99% water.  Some water eveporates, leaving a substance 
that is 98% water.  How much of the water evaporated?

Date: 12/09/2003 at 01:52:54
From: Doctor Greenie
Subject: Re: Mixture and percent

Hi, Olo --

The original subtance is 99% water.  That means we have

              # parts water
  ----------------------------------- = 99/100
  total # parts of (water plus other)  

So we can say we start with 99 parts water and 1 part other.

We take away x parts of water.  We then have (99-x) parts of water 
and still 1 part other, giving us (100-x) parts of (water plus other).  
The percentage of water in the new substance is 98%:

   99-x
  ----- = .98
  100-x

We solve this to determine x, the number of parts of water that 
evaporate:

  99-x = .98(100-x)

  99-x = 98 - .98x

  99-98 = -.98x + 1x

      1 = .02x
  
     50 = x

So 50 parts of water evaporated.  In terms of percentages, we started 
with 99 parts water, and 50 of these parts evaporated, so the 
percentage of water that evaporated is

  50
  -- * 100 = 50.5 (approximately)
  99


You can find a different approach to the problem here:

  How Much Water Evaporated?
    http://mathforum.org/library/drmath/view/63108.html 

I hope this helps.  Please write back if you have any further 
questions about any of this.

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