Difference between revisions of "2022 AMC 10B Problems/Problem 4"
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~not_slay and HIPHOPFROG1 | ~not_slay and HIPHOPFROG1 | ||
− | == | + | ==Solution 3 (Another fast way)== |
− | + | ||
+ | We can add <math>7\cdot500=3500</math> and then minus <math>5</math> seconds. | ||
+ | <cmath>4:00+3500\text{ seconds}</cmath> | ||
+ | <cmath>=4:00+3600\text{ seconds }-100\text{ seconds}</cmath> | ||
+ | <cmath>=4:00+1\text{ hour }-100\text{ seconds}</cmath> | ||
+ | <cmath>=5:00-100\text{ seconds}</cmath> | ||
+ | <cmath>=20 \text{ seconds after } 4:58</cmath> | ||
+ | Finally, we minus <math>5</math> seconds giving <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>. | ||
+ | |||
+ | ~Pancakerunner2 | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/_KNR0JV5rdI?t=310 | https://youtu.be/_KNR0JV5rdI?t=310 |
Revision as of 11:35, 23 November 2023
Contents
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?
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. , so the 700th hiccup happened on the same second as the 4th, which occurred on the th second. .
~not_slay and HIPHOPFROG1
Solution 3 (Another fast way)
We can add and then minus seconds. Finally, we minus seconds giving .
~Pancakerunner2
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=310
Video Solution by Math4All999
https://youtu.be/Jaybq_YT4Pk?feature=shared
Video Solution by paixiao
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.