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/
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