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2022 AMC 10B Problem 4

2022 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Clocks, Calendars & Time · difficulty 1 of 5 · about 2 min · position P1-10

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?

A)
$15 \text{ seconds after } 4:58$
B)
$20 \text{ seconds after } 4:58$
C)
$25 \text{ seconds after } 4:58$
D)
$30 \text{ seconds after } 4:58$
E)
$35 \text{ seconds after } 4:58$

Answer

Key insight

The 700th hiccup is 699 gaps after the first, so it occurs 699 × 5 = 3495 seconds, i.e. 58 minutes 15 seconds, after 4:00.

Solution

The first hiccup is at $4{:}00$ exactly. Between the first and the $700$ th hiccup there are $700 - 1 = 699$ intervals of $5$ seconds each, so the $700$ th hiccup happens
$699 \cdot 5 = 3495 \text{ seconds}$
after $4{:}00$ .

Convert: $3495 = 58 \cdot 60 + 15$ , so that is $58$ minutes and $15$ seconds. The time is $15$ seconds after $4{:}58$ .

The answer is $\boxed{\textbf{(A)}\ 15 \text{ seconds after } 4{:}58}$ .

Why this works

Regularly spaced events form an arithmetic sequence of times, $t_n = t_1 + (n-1)d$ . The " $n-1$ " is the fencepost principle: $n$ posts enclose $n-1$ gaps. The answer choices are all within a few seconds of each other precisely to punish an off-by-one error.

The trap

Counting 700 gaps instead of 699, which gives 3500 seconds and choice (B).

Common mistakes

Counting 700 gaps instead of 699, which gives 3500 seconds and choice (B).
Slipping in the seconds-to-minutes conversion, e.g. dividing 3495 by 60 and misreading the remainder.

Techniques

Set up the equation/formula and compute; no special trick needed