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

Revision as of 01:45, 8 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

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
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All AMC 10 Problems and Solutions

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

@@ Line 25: / Line 25: @@
 ~not_slay and HIPHOPFROG1
-===Bogus Solution===
-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>.
-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>.
 ==Video Solution 1==

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