Difference between revisions of "2022 AMC 10B Problems/Problem 4"
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The donkey hiccups once every 5 seconds, or 12 times a minute. <math>700\equiv4</math> (mod 12), so the 700th hiccup happened on the same second as the 4th, which occurred on the <math>5(4-1)=15</math>th second. <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>. | The donkey hiccups once every 5 seconds, or 12 times a minute. <math>700\equiv4</math> (mod 12), so the 700th hiccup happened on the same second as the 4th, which occurred on the <math>5(4-1)=15</math>th second. <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>. | ||
− | ~not_slay | + | ~not_slay and HIPHOPFROG1 |
==Video Solution 1== | ==Video Solution 1== |
Revision as of 15:10, 20 November 2022
Contents
[hide]Problem
A donkey suffers an attack of hiccups and the first hiccup happens at one afternoon. Suppose that the donkey hiccups regularly every seconds. At what time does the donkey’s th hiccup occur?
Troll Solution
Obviously, the donkey will have its th hiccup seconds after the moment it started. This is .
This is not correct, though. Hiccup number occurred at , so actually the time of hiccup is .
Solution 1
Since the donkey hiccupped the 1st hiccup at , he hiccupped for seconds, which is minutes and seconds, so the answer is .
~MrThinker
Solution 2 (Faster)
We see that the minute has already been determined. The donkey hiccups once every 5 seconds, or 12 times a minute. (mod 12), so the 700th hiccup happened on the same second as the 4th, which occurred on the th second. .
~not_slay and HIPHOPFROG1
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 AMC 10 Problems and Solutions |
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