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Difference between revisions of "2001 AMC 10 Problems/Problem 8"

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== Problem ==
 
== Problem ==
  
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Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
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A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells every 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together? (Answer in the form AB:CD without am or pm, such as 08:00)
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<math> \textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }168\qquad\textbf{(E) }252 </math>
 
  
 
== Solution ==
 
== Solution ==

Revision as of 10:31, 11 November 2022

Problem

A church rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells every 25 minutes. If they all ring their bells at noon on the same day, at what time will they next all ring their bells together? (Answer in the form AB:CD without am or pm, such as 08:00)

Solution

We need to find the least common multiple of the four numbers given. That is, the next time they will be together. First, find the least common multiple of $3$ and $4$.

$3 \times 4 = 12$.

Find the least common multiple of $12$ and $6$.

Since $12$ is a multiple of $6$, the least common multiple is $12$.

Lastly, the least common multiple of $12$ and $7$ is $12 \times\ 7 = \boxed{\textbf{(B) }84}$.

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 AMC 10 Problems and Solutions

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