Difference between revisions of "2018 AMC 8 Problems/Problem 6"
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− | Since Anh drives 3 times as fast on the highway, it takes him <math>\frac{1}{3}</math> of the time to drive 10 miles on the highway than on the coastal road. 1/ | + | Since Anh drives <math>3</math> times as fast on the highway, it takes him <math>\frac{1}{3}</math> of the time to drive <math>10</math> miles on the highway than on the coastal road. <math>\frac{1}{3}</math> of <math>30</math> is <math>10</math>, and since he drives <math>50</math> miles on the highway, we multiply <math>10</math> by <math>5</math> to get <math>50</math>. This means it took him <math>50</math> minutes to drive on the highway, and if we add the <math>30</math> minutes it took for him to drive on the coastal road, we would get <math>80</math>. |
-UnstoppableGoddess | -UnstoppableGoddess |
Revision as of 20:14, 11 November 2019
Contents
Problem 6
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
Solution 1
Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is mph. His speed on the highway then is mph. He drives miles, so he also drives minutes. The total amount of minutes spent on his trip is
Solution 2
Since Anh drives times as fast on the highway, it takes him of the time to drive miles on the highway than on the coastal road. of is , and since he drives miles on the highway, we multiply by to get . This means it took him minutes to drive on the highway, and if we add the minutes it took for him to drive on the coastal road, we would get .
-UnstoppableGoddess (helped by qkddud)
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.