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 | ||
+ | |||
+ | ==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== | ||
+ | |||
+ | https://youtu.be/-rjeTcs3lGA | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2022|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:35, 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.