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

(Solution 2 (Faster))
(Solution 2 (Faster))
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\equiv4</math> (mod 12), 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>.
+
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

Revision as of 19:34, 7 December 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 (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

Bogus Solution

Obviously, the donkey will have its $700$th hiccup $700\cdot 5 = 3500$ seconds after the moment it started. This is $4\text{:}00 + 3500~\text{seconds}=\boxed{\textbf{(B)}~20 \text { seconds after } 4\text{:}58}$.

This is not correct, though. Hiccup number $1$ occurred at $4\text{:}00$, so actually the time of hiccup $n$ is $4\text{:}00 + 5(n-1)~\text{seconds}$.

Video Solution 1

https://youtu.be/7q45hNtIelU

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png