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Difference between revisions of "2017 AMC 8 Problems/Problem 23"

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==Solution==
 
==Solution==
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We know that some multiples of 5 go into 60 evenly. Let's check some because the problem specifies 5 minute increase. 5 does, 10 does, 15 does, and 20 does. This is 4. This seems to satisfy the problem! In the first day, she travels 1 mile in 5 minutes, or 12 miles in 60 minutes. In the second day, she travels 1 mile in 10 minutes, or 6 miles in 60 minutes. In the third day, she travels 1 mile in 15 minutes, or 4 miles in 60 minutes. In the fourth day, she travels 1 mile in 20 minutes, or 3 miles in 60 minutes. Adding these up, we get 12+6+4+3, which is 25. So our answer is 25, \boxed{\textbf{(C)} 25}$
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We know that some multiples of 5 go into 60 evenly. Let's check some because the problem specifies 5 minute increase. 5 does, 10 does, 15 does, and 20 does. This is 4. This seems to satisfy the problem! In the first day, she travels 1 mile in 5 minutes, or 12 miles in 60 minutes. In the second day, she travels 1 mile in 10 minutes, or 6 miles in 60 minutes. In the third day, she travels 1 mile in 15 minutes, or 4 miles in 60 minutes. In the fourth day, she travels 1 mile in 20 minutes, or 3 miles in 60 minutes. Adding these up, we get 12+6+4+3, which is 25. So our answer is 25, or <math>C</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:07, 10 November 2019

Problem 23

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips? $\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution

We know that some multiples of 5 go into 60 evenly. Let's check some because the problem specifies 5 minute increase. 5 does, 10 does, 15 does, and 20 does. This is 4. This seems to satisfy the problem! In the first day, she travels 1 mile in 5 minutes, or 12 miles in 60 minutes. In the second day, she travels 1 mile in 10 minutes, or 6 miles in 60 minutes. In the third day, she travels 1 mile in 15 minutes, or 4 miles in 60 minutes. In the fourth day, she travels 1 mile in 20 minutes, or 3 miles in 60 minutes. Adding these up, we get 12+6+4+3, which is 25. So our answer is 25, or $C$.

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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