A motorboat running upstream takes 8 hours 48 minutes to cover a certain distance and 4 hours to cover the same distance downstream. What is the ratio of the speed of the motorboat in still water to the speed of the water current?

Introduction / Context:
This problem is structurally identical to a typical boats and streams question in which we are given upstream and downstream times for the same distance. The aim is to find the ratio of the speed of the boat in still water to the speed of the current. It illustrates a useful shortcut that expresses this ratio directly in terms of the upstream and downstream times.

Given Data / Assumptions:

• Upstream time = 8 hours 48 minutes.
• Downstream time = 4 hours.
• Distance upstream and downstream is the same.
• Speeds are constant and current is uniform.

Concept / Approach:
For a fixed distance D:

• Upstream speed = D / T_up.
• Downstream speed = D / T_down.

Let b be the speed of the boat in still water and c be the speed of the current. Then b = (downstream speed + upstream speed) / 2 and c = (downstream speed - upstream speed) / 2. There is a known result that b : c equals (T_up + T_down) : (T_up - T_down). We can use this time based ratio directly to avoid computing D.

Step-by-Step Solution:
Convert 8 hours 48 minutes into hours: 48 minutes = 48 / 60 = 0.8 hours. So T_up = 8.8 hours. T_down = 4 hours. Ratio b : c = (T_up + T_down) : (T_up - T_down). Compute T_up + T_down = 8.8 + 4 = 12.8. Compute T_up - T_down = 8.8 - 4 = 4.8. So b : c = 12.8 : 4.8. Divide both numbers by 0.8 to simplify: 12.8 / 0.8 = 16 and 4.8 / 0.8 = 6, giving 16 : 6. Divide by 2 to get 8 : 3.

Verification / Alternative check:
We can also confirm by assuming a convenient distance, say D = 1 unit, computing upstream and downstream speeds as 1 / 8.8 and 1 / 4, then finding b and c from (v_up + v_down) / 2 and (v_down - v_up) / 2. Simplifying b : c from those values leads to the same ratio 8 : 3. This double checks the shortcut formula and validates the answer.

Why Other Options Are Wrong:
Ratios like 3:7 or 7:9 do not match the relationship between times, and if we try to back compute speeds from them we do not obtain the given upstream and downstream times. The ratio 5:4 is also inconsistent with the time difference. Only 8:3 is consistent with the given travel times for the same distance.

Common Pitfalls:
A frequent mistake is to use T_up : T_down directly as the ratio of speeds, which is incorrect because speed is inversely proportional to time. Another issue is failing to convert 8 hours 48 minutes correctly to 8.8 hours, which leads to wrong numbers in the ratio formula. Proper unit conversion and careful use of the shortcut formula are essential.

Final Answer:
The ratio of the speed of the motorboat in still water to the speed of the current is 8:3.