Difference between revisions of "2010 AMC 8 Problems/Problem 21"
Thebeast5520 (talk | contribs) m (→Solution) |
(→Solution) |
||
Line 4: | Line 4: | ||
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math> | <math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math> | ||
− | ==Solution== | + | ==Solution 1== |
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> | 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> | ||
Line 12: | Line 12: | ||
2x &= 480\\ | 2x &= 480\\ | ||
x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had <math>180</math> pages left to read. Again working backwards on Day 1 we find that Hui originally had <math>\boxed{\textbf{(C)}\ 240}</math> pages to read. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=20|num-a=22}} | {{AMC8 box|year=2010|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:46, 9 November 2018
Contents
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
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
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.