Difference between revisions of "2010 AMC 8 Problems/Problem 21"
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2x &= 480\\ | 2x &= 480\\ | ||
x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | x &= \boxed{\textbf{(C)}\ 240} \end{align*}</cmath> | ||
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| + | ==Solution 2 (working backwards)== | ||
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| + | On the last day Hui read <math>62</math> pages. Working backwards, she read <math>18</math> pages more (<math>80</math> total). <math>80</math> is <math>\frac{2}{3}</math> of the remaining pages, so the third morning Hui had <math>120</math> pages to read. Add back the <math>15</math> and you find <math>135</math> must be <math>\frac{3}{4}</math> of the remaining pages, so there were <math>180</math> pages on the second morning. Add <math>12</math> more, then <math>192</math> is <math>\frac{4}{5}</math> of the total book, which must be <math>\boxed{\textbf{(C)}\ 240}</math> pages long. | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== | ||
Latest revision as of 20:54, 23 October 2024
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(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)
On the last day Hui read pages. Working backwards, she read pages more (
total). is 
of the remaining pages, so the third morning Hui had 
pages to read. Add back the and you find 
must be 
of the remaining pages, so there were 
pages on the second morning. Add more, then 
is 
of the total book, which must be 
pages long.
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=1739
~pi_is_3.14
Video by MathTalks
Video Solution by WhyMath
~savannahsolver
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. 
