← SIG (Susquehanna) Interview Insights
I set up variables pretty quickly: let p be paddling speed relative to water, so upstream speed is p-2 and downstream speed is p+2.
First, use the first day's travel to determine the paddling speed relative to the water by setting up an equation based on the distances traveled upstream and downstream. Then, use that speed to calculate the time needed for the return trip (23 miles upstream) and subtract from the arrival time to find the departure time.
Pro tip: Clearly define your variables and state the assumptions (e.g., constant paddling speed, no rest stops) before diving into calculations. This demonstrates structured thinking and prevents errors.
Let p be the paddling speed in still water (mph). Assume constant paddling speed and no breaks. The current is 2 mph, so upstream speed is p-2 and downstream speed is p+2.
On day 1, they paddle upstream for 4 hours and downstream for 5 hours. The net displacement is 23 miles upstream. So, 4(p-2) - 5(p+2) = 23. Solve for p.
Expand and solve: 4p - 8 - 5p - 10 = 23 → -p - 18 = 23 → -p = 41 → p = -41? Wait, check sign: Actually, net upstream displacement means upstream distance minus downstream distance equals 23. So 4(p-2) - 5(p+2) = 23. That gives 4p-8 -5p-10 =23 → -p -18 =23 → -p=41 → p=-41, impossible. So re-evaluate: The net displacement is 23 miles upstream, meaning they ended 23 miles upstream from start. So upstream distance (4(p-2)) minus downstream distance (5(p+2)) equals 23. But if p is positive, upstream speed is less than downstream, so upstream distance might be less. Let's solve correctly: 4(p-2) - 5(p+2) = 23 → 4p-8 -5p-10 =23 → -p -18 =23 → -p=41 → p=-41. Negative speed is impossible. So maybe the net displacement is downstream? The problem says: 'The next day they canoe back to their starting point, which is 23 miles upstream'. That implies the starting point of the first day is 23 miles upstream from where they ended? Actually, re-read: 'Two friends paddle upstream for 4 hours, then downstream for 5 hours. The next day they canoe back to their starting point, which is 23 miles upstream, and arrive at 4pm.' This is ambiguous. Let's parse: They paddle upstream for 4 hours, then downstream for 5 hours. The next day they canoe back to their starting point, which is 23 miles upstream. So their starting point is 23 miles upstream from where they are at the end of day 1? That would mean after day 1, they are 23 miles downstream from their starting point? But they paddled upstream first, then downstream. If they end up downstream of start, then net displacement is downstream. But the problem says 'canoe back to their starting point, which is 23 miles upstream' meaning from their current location, the starting point is 23 miles upstream. So they need to go upstream 23 miles to return. So on day 1, they ended 23 miles downstream from start. So net displacement downstream = 23 miles. So downstream distance minus upstream distance = 23. So 5(p+2) - 4(p-2) = 23. Solve: 5p+10 -4p+8 =23 → p+18=23 → p=5 mph. That works. So p=5 mph.
Return trip is 23 miles upstream. Upstream speed = p-2 = 3 mph. Time = distance/speed = 23/3 ≈ 7.666... hours = 7 hours 40 minutes.
They arrive at 4pm. Subtract 7 hours 40 minutes: 4pm - 7h40m = 8:20am. So they left at 8:20am.
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