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

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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>
  
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==Solution==
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==Solution 1(algebra solution)==
  
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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>(\frac{3}{4})(\frac{4x}{5}-12)-15 = \frac{3x}{5}-24</math> left. After the third, she had <math>(\frac{2}{3})(\frac{3x}{5}-24)-18=\frac{2x}{5}-34</math> left. This is equivalent to <math>62.</math>
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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>
  
 
<cmath>\begin{align*} \frac{2x}{5}-34&=62\\
 
<cmath>\begin{align*} \frac{2x}{5}-34&=62\\
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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.
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==Video Solution by OmegaLearn==
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https://youtu.be/rQUwNC0gqdg?t=1739
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 +
~pi_is_3.14
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==Video by MathTalks==
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https://youtu.be/mSCQzmfdX-g
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 +
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==Video Solution by WhyMath==
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https://youtu.be/p2wtDPepQhY
 +
 +
~savannahsolver
  
 
==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}}

Latest revision as of 20:54, 23 October 2024

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(algebra 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 $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ 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 (working backwards)

On the last day Hui read $62$ pages. Working backwards, she read $18$ pages more ($80$ total). $80$ is $\frac{2}{3}$ of the remaining pages, so the third morning Hui had $120$ pages to read. Add back the $15$ and you find $135$ must be $\frac{3}{4}$ of the remaining pages, so there were $180$ pages on the second morning. Add $12$ more, then $192$ is $\frac{4}{5}$ of the total book, which must be $\boxed{\textbf{(C)}\ 240}$ pages long.

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=1739

~pi_is_3.14

Video by MathTalks

https://youtu.be/mSCQzmfdX-g


Video Solution by WhyMath

https://youtu.be/p2wtDPepQhY

~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

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