Percentage of IncreaseDate: 7/23/96 at 8:57:37 From: Sue Subject: Percentage of Increase Dear Dr. Math, My math question is about finding the percent of increase between two numbers. We added memory to our computer system. We had 96 MB of main memory and now with our new addition, we have 256 MB of main memory. I would like to figure out what percent increase this represents. Where should I begin? Sue Date: 7/23/96 at 13:21:6 From: Doctor Paul Subject: Re: Percentage of Increase One way to begin would be to estimate the percent increase, and then solve for it exactly. If you go from 100 MB of memory to 200 MB then you've increased it by 100 percent, because the amount of the increase (100 MB) is 100% of the original amount (100 MB). That is... if you double your memory then you've increased it by 100 percent. If you add another 100 MB, you're adding another 100% of the original amount, so you have a 200% increase, from 100 MB to 300 MB. In your case, you have gone from about 100 to about 250. Since 250 is halfway between 200 MB and 300 MB, you could guess that the answer is about 150 percent. Does this make sense? Now let's find the actual value. I'm going to do a simple example first so you see how percentages work. If I go from 100 MB to 105 MB, what is the percent increase? In this case, the numbers are straightforward: the increase (5 MB) is 5 percent of the original amount (100 MB). But we can use a method that will work even when the numbers aren't this tidy: I ask: 100 times what number will give me 105? 100 * x = 105 x = 105 / 100 x = 1.05 Then I ask: What increase is that over 100%? x - 1 = 1.05 - 1 = 0.05 = 5/100 = 5% So I have an increase of 5%. Now let's do the same thing with your numbers: 1) 96 * x = 256 x = 256 / 96 x = 2.67 2) x - 1 = 2.67 - 1 = 1.67 = 167/100 = 167% which is pretty close to the original estimate of 150%. That gives us some confidence that we have the right answer. -Doctor Paul, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/