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Difference between revisions of "2006 AMC 8 Problems/Problem 16"

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==Solution==
 
==Solution==
  
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{(E)} The least common multiple of 20, 45 and 30 is 2
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The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply <math>160 \cdot 45 = \boxed{\textbf{(E)}\ 7200}</math>.
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2
 
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¢ 3
 
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2
 
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¢ 5 = 180. Using the
 
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LCM, in 180 seconds Alice reads
 
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180
 
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20 = 9 pages, Chandra reads
 
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180
 
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30 = 6 pages
 
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and Bob reads
 
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180
 
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45 = 4 pages. Together they read a total of 19 pages in 180
 
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seconds. The total number of seconds each reads is
 
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760
 
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19
 
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¢ 180 = 7200.}}
 
  
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==Solution 2==
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We can solve this using equations. Let Bob read <math>x</math> pages and Chandra read <math>y</math> pages. Then Alice would read <math>760-x-y</math> pages. Then we can set up the equations using the time it takes them to read a page: <math>45x=30y=20(760-x-y)</math>. Since we only have two variables, we only need two equations. Let's use <math>45x=30y</math> and <math>45x=20(760-x-y)</math>. We can simplify <math>45x=30y</math> down to <math>3x=2y</math> and <math>45x=20(760-x-y)</math> down to <math>9x=4(760-x-y)</math>. We can then solve the equations accordingly:
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<cmath>9x=3040-4x-4y</cmath>
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<cmath>13x=3040-4y</cmath>
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From the equation <math>3x=2y</math>, we know that <math>4y=6x</math>.
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<cmath>13x=3040-6x</cmath>
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<cmath>19x=3040</cmath>
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<cmath>x=160</cmath>
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Now that we know <math>x</math>, we can multiply it by <math>45</math> (Bob's speed) to get the answer: <math>160\times45</math>=<math>\boxed{\textbf{(E)}\ 7200}</math>
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~Trex226
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==See Also==
 
{{AMC8 box|year=2006|num-b=15|num-a=17}}
 
{{AMC8 box|year=2006|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 18:12, 11 January 2023

Problem

Problems 14, 15 and 16 involve Mrs. Reed's English assignment.

A Novel Assignment

The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.

Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?

$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$

Solution

The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply $160 \cdot 45 = \boxed{\textbf{(E)}\ 7200}$.

Solution 2

We can solve this using equations. Let Bob read $x$ pages and Chandra read $y$ pages. Then Alice would read $760-x-y$ pages. Then we can set up the equations using the time it takes them to read a page: $45x=30y=20(760-x-y)$. Since we only have two variables, we only need two equations. Let's use $45x=30y$ and $45x=20(760-x-y)$. We can simplify $45x=30y$ down to $3x=2y$ and $45x=20(760-x-y)$ down to $9x=4(760-x-y)$. We can then solve the equations accordingly:

\[9x=3040-4x-4y\] \[13x=3040-4y\]

From the equation $3x=2y$, we know that $4y=6x$.

\[13x=3040-6x\] \[19x=3040\] \[x=160\]

Now that we know $x$, we can multiply it by $45$ (Bob's speed) to get the answer: $160\times45$=$\boxed{\textbf{(E)}\ 7200}$

~Trex226

See Also

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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