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Difference between revisions of "2022 AMC 10B Problems/Problem 4"

(→‎Video Solution by Interstigation)
(→‎Solution 3 (Another fast way))
 
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==Solution 1==
 
==Solution 1==
  
−
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>.
+
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 26: Line 26:
 
~not_slay and HIPHOPFROG1
 
~not_slay and HIPHOPFROG1
  
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==Solution 3 (Another fast way)==
+
==Solution 3 (Another Fast Way)==
  
 
We can add <math>7\cdot500=3500</math> and then minus <math>5</math> seconds.
 
We can add <math>7\cdot500=3500</math> and then minus <math>5</math> seconds.
−
<cmath>4:00+3500\text{ seconds}</cmath>
+
<cmath>\begin{align*}
−
<cmath>=4:00+3600\text{ seconds }-100\text{ seconds}</cmath>
+
4:00+3500\text{ seconds} &= 4:00+3600\text{ seconds }-100\text{ seconds} \\
−
<cmath>=4:00+1\text{ hour }-100\text{ seconds}</cmath>
+
&=4:00+1\text{ hour }-100\text{ seconds} \\
−
<cmath>=5:00-100\text{ seconds}</cmath>
+
&=5:00-100\text{ seconds} \\
−
<cmath>=20 \text{ seconds after } 4:58</cmath>
+
&=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>.
 
Finally, we minus <math>5</math> seconds giving <math>\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}</math>.
  

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)
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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