Difference between revisions of "2017 AMC 8 Problems/Problem 23"
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− | ==Problem | + | ==Problem== |
− | 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? | + | 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 <math>5</math> 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? |
<math>\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82</math> | <math>\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82</math> | ||
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==Solution== | ==Solution== | ||
− | From the question | + | It is well known that <math>\text{Distance}=\text{Speed} \cdot \text{Time}</math>. In the question, we want distance. From the question, we have that the time is <math>60</math> minutes or <math>1</math> hour. By the equation derived from <math>\text{Distance}=\text{Speed} \cdot \text{Time}</math>, we have <math>\text{Speed}=\frac{\text{Distance}}{\text{Time}}</math>, so the speed is <math>1</math> mile per <math>x</math> minutes. 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},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15 | + | <cmath>\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}</cmath> |
− | + | We know that <math>x,x+5,x+10,x+15</math> are all factors of <math>60</math>, therefore, <math>x=5</math> because the factors have to be in an arithmetic sequence with the common difference being <math>5</math> and <math>x=5</math> is the only solution. | |
− | <cmath>\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}</cmath> | + | <cmath>\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}</cmath> |
− | - | + | |
+ | ==Video Solution== | ||
+ | https://youtu.be/EiN5oMJRL-8 - Happytwin | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2017|num-b=22|num-a=24}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 13:11, 18 January 2021
Contents
Problem
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 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?
Solution
It is well known that . In the question, we want distance. From the question, we have that the time is minutes or hour. By the equation derived from , we have , so the speed is mile per minutes. Because we want the distance, we multiply the time and speed together yielding . The minutes cancel out, so now we have as our distance for the first day. The distance for the following days are: We know that are all factors of , therefore, because the factors have to be in an arithmetic sequence with the common difference being and is the only solution.
Video Solution
https://youtu.be/EiN5oMJRL-8 - Happytwin
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.