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

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(Solution)
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<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==
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==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>
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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==
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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.
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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

Problem

Hui 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 1

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 $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ left. After the third, she had $\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34$ 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*}

Solution 2

If Hui has $62$ pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had $80$ pages left to read. $80$ pages is $\frac{2}{3}$ of the pages she had left at the end of Day 2. So at the end of Day 2, she had $80 \cdot \frac{3}{2} = 120$ pages left to read. Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had $180$ pages left to read. Again working backwards on Day 1 we find that Hui originally had $\boxed{\textbf{(C)}\ 240}$ pages to read.

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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