Writing a Percent as a Fraction and Reducing the ResultDate: 05/03/2004 at 22:38:52 From: Mike Subject: algebra I need help. I just can't seem to figure out how to convert a percent into a fraction, such as 20% = 20/100 = 1/5. All I know is that you put it into a fraction like 20/100 but after that I get stuck and I just don't know how to get the answer. I thought you did 20 divided by 100 but it does not work. Date: 05/04/2004 at 17:54:54 From: Doctor Ian Subject: Re: algebra Hi Mike, You're right--you start by putting the percentage over 100, because that's what we _mean_ when we write something like 20%--it's just another name for 20/100. So 20 20% = --- 100 Now what? Well, the easiest thing to do is break the numerator and denominator into prime factors: 20 2 * 2 * 5 20% = --- = ------------- 100 2 * 2 * 5 * 5 Since we're multiplying by 2 and dividing by 2, we can cancel a pair of 2's: x 20 2 * 2 * 5 20% = --- = ------------- 100 2 * 2 * 5 * 5 x We can do the same thing for the other pair of 2's, and a pair of 5's: x x x 20 2 * 2 * 5 20% = --- = ------------- 100 2 * 2 * 5 * 5 x x x Since we're left with nothing in the numerator, we can include a factor of 1 up there, as a placeholder: x x x 20 2 * 2 * 5 * 1 20% = --- = ------------- 100 2 * 2 * 5 * 5 x x x 1 = - 5 Now, we could have approached it slightly differently. We could have looked at 20 --- = ? 100 and asked ourselves: "What divides both 20 and 100 evenly?" If we realize that 5 does, we could divide both the numerator and denominator by 5 to get 20 4 5 --- = -- * - 100 20 5 4 = -- 20 Now we could do the same thing again: "What divides both 4 and 20 evenly?" They're both evenly divided by 2, so 4 2 2 -- = -- * - 20 10 2 2 = -- 10 And again: 2 1 2 -- = - * - 10 5 2 1 = - 5 If you use prime factors instead of going step by step like in the second method, you're guaranteed never to miss any common factors. So sometimes it seems like more work to do the prime factor approach, but there's something to be said for an approach that you _know_ will always work, isn't there? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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