Difference between revisions of "2006 AMC 8 Problems/Problem 16"
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==Solution== | ==Solution== | ||
| − | \boxed{\textbf{(C)} 6800}$ The least common multiple of 20, 45 and 30 is 2 | + | \boxed{\textbf{(C)} 6800}$ |
| + | The least common multiple of 20, 45 and 30 is 2 | ||
2 | 2 | ||
| − | ¢ 3 | + | ¢ 3 |
2 | 2 | ||
¢ 5 = 180. Using the | ¢ 5 = 180. Using the | ||
Revision as of 13:49, 3 November 2012
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?
Solution
\boxed{\textbf{(C)} 6800}$ The least common multiple of 20, 45 and 30 is 2 2 ¢ 3 2 ¢ 5 = 180. Using the LCM, in 180 seconds Alice reads 18 20 = 9 pages, Chandra reads 180 30 = 6 pages and Bob reads 180 45 = 4 pages. Together they read a total of 19 pages in 180 seconds. The total number of seconds each reads is 760 19 ¢ 180 = 7200.}}
| 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 | ||
