Difference between revisions of "2010 AMC 8 Problems/Problem 21"

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==Problem==
 
==Problem==
Hui is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book?  
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Hu is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book?  
  
 
<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>

Revision as of 18:32, 10 November 2016

Problem

Hu is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?

$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$

Solution

Let $x$ be the number of pages in the book. After the first day, Hui had $\frac{4x}{5}-12$ pages left to read. After the second, she had $(\frac{3}{4})(\frac{4x}{5}-12)-15 = \frac{3x}{5}-24$ left. After the third, she had $(\frac{2}{3})(\frac{3x}{5}-24)-18=\frac{2x}{5}-34$ left. This is equivalent to $62.$

\begin{align*} \frac{2x}{5}-34&=62\\ 2x - 170 &= 310\\ 2x &= 480\\ x &= \boxed{\textbf{(C)}\ 240} \end{align*}

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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