Difference between revisions of "2010 AMC 8 Problems/Problem 21"
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==Solution== | ==Solution== | ||
− | Let <math>x</math> be the number of pages in the book. After the first day, Hui had <math>\frac{4x}{5}-12</math> pages left to read. After the second, she had <math>(\frac{3}{4})(\frac{4x}{5}-12)-15 = \frac{3x}{5}-24</math> left. After the third, she had <math>(\frac{2}{3})(\frac{3x}{5}-24)-18=\frac{2x}{5}-34</math> left. This is equivalent to <math>62.</math> | + | Let <math>x</math> be the number of pages in the book. After the first day, Hui had <math>\frac{4x}{5}-12</math> pages left to read. After the second, she had <math>\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24</math> left. After the third, she had <math>\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34</math> left. This is equivalent to <math>62.</math> |
<cmath>\begin{align*} \frac{2x}{5}-34&=62\\ | <cmath>\begin{align*} \frac{2x}{5}-34&=62\\ |
Revision as of 22:38, 16 September 2018
Problem
Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read of the pages plus more, and on the second day she read of the remaining pages plus pages. On the third day she read of the remaining pages plus pages. She then realized that there were only pages left to read, which she read the next day. How many pages are in this book?
Solution
Let be the number of pages in the book. After the first day, Hui had pages left to read. After the second, she had left. After the third, she had left. This is equivalent to
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.