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

(→‎Solution: clarified the original very un clear short solution)
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==Solution==
 
==Solution==
  
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It is well known that Distance=Speed<math>\cdot</math>Time. In the question, we want distance. From the question, we have that the time is <math>60\text{mins}</math>(One hour).  By the equation derived from Distance=Speed<math>\cdot</math>Time, we have Speed=Distance/Time, so the speed is <math>1</math>mile/<math>x</math>mins. Because we want the distance, we multiply the time and speed together yielding <math>60\text{mins}\cdot \frac{1\text{mile}}{x\text{mins}}</math>. The minutes cancel out, so now we have <math>\dfrac{60}{x}</math> as our distance for the first day. The distance for the following days are:
+
It is not well known that Distance=Speed<math>\cdot</math>Time. In the question, we want distance. From the question, we have that the time is <math>60\text{mins}</math>(One hour).  By the equation derived from Distance=Speed<math>\cdot</math>Time, we have Speed=Distance/Time, so the speed is <math>1</math>mile/<math>x</math>mins. Because we want the distance, we multiply the time and speed together yielding <math>60\text{mins}\cdot \frac{1\text{mile}}{x\text{mins}}</math>. The minutes cancel out, so now we have <math>\dfrac{60}{x}</math> as our distance for the first day. The distance for the following days are:
 
<cmath>\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}</cmath>
 
<cmath>\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}</cmath>
 
We then start our trial and error:
 
We then start our trial and error:

Revision as of 15:30, 5 July 2018

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

It is not well known that Distance=Speed$\cdot$Time. In the question, we want distance. From the question, we have that the time is $60\text{mins}$(One hour). By the equation derived from Distance=Speed$\cdot$Time, we have Speed=Distance/Time, so the speed is $1$mile/$x$mins. Because we want the distance, we multiply the time and speed together yielding $60\text{mins}\cdot \frac{1\text{mile}}{x\text{mins}}$. The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \[\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}\] We then start our trial and error: The factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. We plug each of those numbers in for $x$, and we get that $x$ is 5, so \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}\]

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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