Difference between revisions of "2022 AMC 10B Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
| − | Since the donkey hiccupped the 1st hiccup at <math>4:00</math>, | + | Since the donkey hiccupped the 1st hiccup at <math>4:00</math>, it hiccupped for <math>5 \cdot (700-1) = 3495</math> seconds, which is <math>58</math> minutes and <math>15</math> seconds, so the answer is <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>. |
~MrThinker | ~MrThinker | ||
| Line 22: | Line 22: | ||
We see that the minute has already been determined. | We see that the minute has already been determined. | ||
| − | The donkey hiccups once every 5 seconds, or 12 times a minute. <math>700\ | + | The donkey hiccups once every 5 seconds, or 12 times a minute. <math>700\equiv 4 \pmod{12}</math>, 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 and HIPHOPFROG1 | ~not_slay and HIPHOPFROG1 | ||
| − | ==Video Solution | + | ==Solution 3 (Another Fast Way)== |
| − | https://youtu.be/ | + | |
| + | We can add <math>7\cdot500=3500</math> and then minus <math>5</math> seconds. | ||
| + | <cmath>\begin{align*} | ||
| + | 4:00+3500\text{ seconds} &= 4:00+3600\text{ seconds }-100\text{ seconds} \\ | ||
| + | &=4:00+1\text{ hour }-100\text{ seconds} \\ | ||
| + | &=5:00-100\text{ seconds} \\ | ||
| + | &=20 \text{ seconds after } 4:58. | ||
| + | \end{align*}</cmath> | ||
| + | Finally, we minus <math>5</math> seconds giving <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>. | ||
| + | |||
| + | ~Pancakerunner2 | ||
| + | |||
| + | ==Video Solution (🔥Fast and Easy🔥)== | ||
| + | https://youtu.be/HLuGbW2P_tA | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ==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}} | ||
Latest revision as of 00:14, 9 November 2024
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 , it 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 (🔥Fast and Easy🔥)
~Education, the Study of Everything
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. 
