Fraction: Algebra without the EquationsDate: 8/23/96 at 23:36:49 From: Donna J Brewer Subject: Solving Fraction Problems in a Different Way Dear Dr. Math, I am no wizard at math, but I discovered a cool way to solve certain fraction problems. The only trouble is, I don't know why it works! It is bugging the heck out of me. Here is an example. 2/3 of ? = 66 Now, I know that I have 3 equal groups, with a certain number in each group. But I don't know what that number is. I found that if I divide the given sum (in this case 66) by the number in the top of the fraction (in this case 2), I will ALWAYS come up with the number in each group (or a single portion of whatever the fraction is, one third in this instance). Example: I divide 66 by the top number in the fraction (2) and get the answer of 33. Since I now know how much comprises one portion, I simply multiply by the denominator to find out the unknown number. In the above case, I multiply 33 by 3 and get 99. My solved problem now reads 2/3 of 99 = 66 This solution ALWAYS works. 3/4 of ? = 75 75 divided by 3 = 25 25 x 4 = 100 3/4 of 100 = 75. Date: 8/26/96 at 13:25:34 From: Doctor Leigh Subject: Re: Solving Fraction Problems in a Different Way Thank you for your problem. Your solution works because you are doing the correct algebra without needing to write out the equations. Let's take your sample problem. 3/4 of ? = 75 To do this algebraically you would set up an equation: 3/4 * x = 75 From there you need to get x alone on one side of the equation. Normally you would just divide by 3/4, but you can also divide by three and multiply by four - which is the same thing as dividing by 3/4. This is why your solution works. You are really just dividing by the fraction, but you have separated the division. Good work in taking something and changing it into what you understand! -Doctor Leigh Ann, 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/