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

Revision as of 18:23, 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 $\boxed{C}$.

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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