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Difference between revisions of "2022 AMC 10B Problems/Problem 4"
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<math>\textbf{(E) }35 \text{ seconds after } 4:58</math>
 
<math>\textbf{(E) }35 \text{ seconds after } 4:58</math>
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==Troll Solution==
 
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Obviously, the donkey will have its <math>700</math>th hiccup <math>700\cdot 5 = 3500</math> seconds after the moment it started. This is <math>4\text{:}00 + 3500~\text{seconds}=\boxed{\textbf{(B)}~20 \text { seconds after } 4\text{:}58}</math>.
 
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This is not correct, though. Hiccup number <math>1</math> occurred at <math>4\text{:}00</math>, so actually the time of hiccup <math>n</math> is <math>4\text{:}00 + 5(n-1)~\text{seconds}</math>.
 
  
 
==Solution 1==
 
==Solution 1==
  
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Since the donkey hiccupped the 1st hiccup at <math>4:00</math>, he 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>.
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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
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We see that the minute has already been determined.
 
We see that the minute has already been determined.
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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>3(4-1)=15</math>th second. <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>.
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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>.
  
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~not_slay
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~not_slay and HIPHOPFROG1
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 +
==Solution 3 (Another Fast Way)==
 +
 
 +
We can add <math>7\cdot500=3500</math> and then minus <math>5</math> seconds.
 +
<cmath>\begin{align*}
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4:00+3500\text{ seconds} &= 4:00+3600\text{ seconds }-100\text{ seconds} \\
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&=4:00+1\text{ hour }-100\text{ seconds} \\
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&=5:00-100\text{ seconds} \\
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&=20 \text{ seconds after } 4:58.
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\end{align*}</cmath>
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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
 +
 
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/_KNR0JV5rdI?t=310
 +
 
 +
==Video Solution by Math4All999==
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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}}
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{{AMC12 box|year=2022|ab=B|num-b=3|num-a=5}}
 
−
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:14, 9 November 2024

Problem

A donkey suffers an attack of hiccups and the first hiccup happens at $4:00$ one afternoon. Suppose that the donkey hiccups regularly every $5$ seconds. At what time does the donkey’s $700$th hiccup occur?

$\textbf{(A) }15 \text{ seconds after } 4:58$

$\textbf{(B) }20 \text{ seconds after } 4:58$

$\textbf{(C) }25 \text{ seconds after } 4:58$

$\textbf{(D) }30 \text{ seconds after } 4:58$

$\textbf{(E) }35 \text{ seconds after } 4:58$

Solution 1

Since the donkey hiccupped the 1st hiccup at $4:00$, it hiccupped for $5 \cdot (700-1) = 3495$ seconds, which is $58$ minutes and $15$ seconds, so the answer is $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~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. $700\equiv 4 \pmod{12}$, so the 700th hiccup happened on the same second as the 4th, which occurred on the $5(4-1)=15$th second. $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~not_slay and HIPHOPFROG1

Solution 3 (Another Fast Way)

We can add $7\cdot500=3500$ and then minus $5$ seconds. \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*} Finally, we minus $5$ seconds giving $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~Pancakerunner2

Video Solution (🔥Fast and Easy🔥)

https://youtu.be/HLuGbW2P_tA

~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

2022 AMC 10B (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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All AMC 10 Problems and Solutions

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