Difference between revisions of "2006 AMC 8 Problems/Problem 15"
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<math> \textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506 </math> | <math> \textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506 </math> | ||
| − | + | ==Solution 1== | |
| − | |||
Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve. | Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve. | ||
Let <math>x</math> be the number of pages that Chandra reads. | Let <math>x</math> be the number of pages that Chandra reads. | ||
| − | <math>30x = 45(760-x)</math> Distribute the 45 | + | <math>30x = 45(760-x)</math> Distribute the <math>45</math> |
| − | <math>30x = 45(760) - 45x</math> Add 45x to both sides | + | <math>30x = 45(760) - 45x</math> Add <math>45x</math> to both sides |
| − | <math>75x = 45(760)</math> Divide both sides by 15 to make it easier to solve | + | <math>75x = 45(760)</math> Divide both sides by <math>15</math> to make it easier to solve |
| − | <math>5x = 3(760)</math> Divide both sides by 5 | + | <math>5x = 3(760)</math> Divide both sides by <math>5</math> |
<math>x = 3(152) = \boxed{\textbf{(C)} 456}</math> | <math>x = 3(152) = \boxed{\textbf{(C)} 456}</math> | ||
| − | + | ==Solution 2== | |
| − | Bob and Chandra read at a rate of <math>30:45</math> seconds per page, respectively. Simplifying that gets us Bob reads 2 pages for every 3 pages that Chandra reads. Therefore Chandra should read <math>\frac{3}{2+3}=\frac{3}{5}</math> of the book. <math>\frac{3}{5}\cdot760</math>=<math>\boxed{\textbf{(C)} 456}</math> | + | Bob and Chandra read at a rate of <math>30:45</math> seconds per page, respectively. Simplifying that gets us Bob reads <math>2</math> pages for every <math>3</math> pages that Chandra reads. Therefore Chandra should read <math>\frac{3}{2+3}=\frac{3}{5}</math> of the book. <math>\frac{3}{5}\cdot760</math>=<math>\boxed{\textbf{(C)} 456}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=14|num-a=16}} | {{AMC8 box|year=2006|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:33, 25 April 2016
Contents
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.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
Solution 1
Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.
Let 
be the number of pages that Chandra reads.
Distribute the 
Add 
to both sides
Divide both sides by 
to make it easier to solve
Divide both sides by 
Solution 2
Bob and Chandra read at a rate of 
seconds per page, respectively. Simplifying that gets us Bob reads 
pages for every 
pages that Chandra reads. Therefore Chandra should read 
of the book. 
=
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
