2022 AMC 10B Problems/Problem 4

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?

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

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

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

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

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

Troll 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 $4\text{:}00 + 5(n-1)~\text{seconds}$ .

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. $700\equiv4$ (mod 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

Video Solution 1

https://youtu.be/7q45hNtIelU

~Education, the Study of Everything

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.

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