Difference between revisions of "2010 AMC 8 Problems/Problem 21"
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x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | ||
| − | ==Solution 2== | + | ==Solution 2(working backwards)== |
If Hui has <math>62</math> pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had <math>80</math> pages left to read. <math>80</math> pages is <math>\frac{2}{3}</math> of the pages she had left at the end of Day 2. So at the end of Day 2, she had <math>80 \cdot \frac{3}{2} = 120</math> pages left to read. | If Hui has <math>62</math> pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had <math>80</math> pages left to read. <math>80</math> pages is <math>\frac{2}{3}</math> of the pages she had left at the end of Day 2. So at the end of Day 2, she had <math>80 \cdot \frac{3}{2} = 120</math> pages left to read. | ||
Revision as of 11:20, 28 April 2021
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 1(algebra 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 
Solution 2(working backwards)
If Hui has pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had 
pages left to read. pages is 
of the pages she had left at the end of Day 2. So at the end of Day 2, she had 
pages left to read.
Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had 
pages left to read. Again working backwards on Day 1 we find that Hui originally had 
pages to read.
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. 
