When Jake stops drinking alcohol at 10:30 pm, he has a blood alcohol content (BAC) of 0.08375 - HSC - SSCE Mathematics Standard - Question 13 - 2020 - Paper 1

Since Jake has a BAC of 0.08375 and needs to reach 0.05, we find the difference:

$Difference = 0.08375 - 0.05 = 0.03375$

Now, we can calculate the time to reduce from 0.08375 to 0.05:

$Time = \frac{0.03375}{0.015} = 2.25 \text{ hours}$

The total time taken to reach 0.05 BAC will be the sum of the time taken to reach zero BAC and the time taken to reach 0.05:

$Total Time = 5.5833 + 2.25 = 7.8333 \text{ hours}$

Starting from 10:30 pm:

$10:30 \text{ pm} + 7.8333 \text{ hours} \approx 6:19:59 \text{ am}$

Jake's BAC will reach 0.05 around 6:20 am the next day. Therefore, choosing the closest option from the available choices would lead us to:

Answer: 12:45 am (Option A)