2022 AMC 10B Problem 4

Below is the professionally curated solution for Problem 4 of the 2022 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

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Concepts:arithmetic sequenceunit conversion

Difficulty rating: 870

4.

A donkey suffers an attack of hiccups and the first hiccup happens at 4:004:00 one afternoon. Suppose that the donkey hiccups regularly every 55 seconds. At what time does the donkey’s 700700th hiccup occur?

15 seconds after 4:5815 \text{ seconds after } 4:58

20 seconds after 4:5820 \text{ seconds after } 4:58

25 seconds after 4:5825 \text{ seconds after } 4:58

30 seconds after 4:5830 \text{ seconds after } 4:58

35 seconds after 4:5835 \text{ seconds after } 4:58

Solution:

Since we want to look at the 700700th hiccup, we need to look at time that is 699699 hiccups after the first one.

This would be 6995=3495699\cdot 5 = 3495 seconds. Note that 3495=6058+15,3495 = 60\cdot 58+15, so the time would be 5858 minutes and 1515 seconds after the first hiccup. This would therefore be 4:584:58 and 1515 seconds.

Thus, the answer is A .

Problem 4 in Other Years