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>
  
==Solution==
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==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>, 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>.
  
 
~MrThinker
 
~MrThinker
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==Solution 2==
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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:21, 17 November 2022

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 $4:00$, he 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

We see that the minute has already been determined. The donkey hiccups once every 5 seconds, or 12 times a minute. $700\equiv4$ (mod 12), so the 700th hiccup happened on the same second as the 4th, which occurred on the $3(4-1)=15$th second. $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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
2022 AMC 12B (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

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