A boat goes 4 km upstream and 4 km downstream in a total time of 1 hour. The same boat goes 5 km downstream and 3 km upstream in 55 minutes. What is the speed (in km/h) of the boat in still water?

Introduction / Context:
This is a more involved boats and streams problem. We are told about two different trips made by the same boat, each involving a mix of upstream and downstream travel over different distances and times. From this combined information, we must determine the speed of the boat in still water. This requires setting up a pair of equations in two unknowns and solving them systematically.

Given Data / Assumptions:

• Let the speed of the boat in still water be b km/h.
• Let the speed of the current be c km/h.
• Upstream speed = b - c.
• Downstream speed = b + c.
• First trip: 4 km upstream + 4 km downstream in 1 hour.
• Second trip: 5 km downstream + 3 km upstream in 55 minutes (which is 11/12 hour).
• We must find b.

Concept / Approach:
For each trip, total time equals the sum of individual times. Time is given by distance divided by speed. We write one time equation for the first trip and another for the second trip, each involving b and c. To simplify, we introduce variables x = b - c and y = b + c so that the equations involve x and y directly. After solving for x and y, we recover b as (x + y) / 2. This method turns a boats and streams word problem into a clean algebra exercise.

Step-by-Step Solution:
Step 1: Define x and y for convenience. Let x = b - c (upstream speed) and y = b + c (downstream speed). Step 2: Form the equation for the first trip. Time upstream = 4 / x hours. Time downstream = 4 / y hours. Total time = 1 hour, so 4 / x + 4 / y = 1. Step 3: Form the equation for the second trip. Time downstream = 5 / y hours. Time upstream = 3 / x hours. Total time = 55 minutes = 11 / 12 hours. So 5 / y + 3 / x = 11 / 12. Step 4: Multiply the first equation by xy. 4y + 4x = xy. Rearrange as xy - 4x - 4y = 0. Step 5: Multiply the second equation by 12xy to clear denominators. 12(5 / y + 3 / x) = 11. This becomes 60x + 36y = 11xy. Rearrange to 11xy - 60x - 36y = 0. Step 6: Use the first equation to express xy. From xy - 4x - 4y = 0, we get xy = 4x + 4y. Substitute into 11xy - 60x - 36y = 0. 11(4x + 4y) - 60x - 36y = 0. 44x + 44y - 60x - 36y = 0. (-16x) + 8y = 0, so 8y = 16x and y = 2x. Step 7: Substitute y = 2x back into xy = 4x + 4y. x(2x) = 4x + 4(2x). 2x^2 = 4x + 8x = 12x. If x is not zero, divide by 2x: x = 6. Then y = 2x = 12. Step 8: Recover the boat speed in still water. b = (x + y) / 2 = (6 + 12) / 2 = 18 / 2 = 9 km/h.
Verification / Alternative check:
With b = 9 and c = (y - x) / 2 = (12 - 6) / 2 = 3 km/h, upstream speed is 6 km/h and downstream speed is 12 km/h. First trip: 4 / 6 + 4 / 12 = 2/3 + 1/3 = 1 hour, which matches. Second trip: 5 / 12 + 3 / 6 = 5/12 + 1/2 = 5/12 + 6/12 = 11/12 hour = 55 minutes, also correct.
Why Other Options Are Wrong:
If b were 6.5, 7.75, 10.5 or 8.5 km/h, it would be impossible to choose a single current speed c that satisfies both time equations simultaneously. Quick substitution of these values leads to inconsistencies between the calculated trip times and the given total times.
Common Pitfalls:
A common pitfall is to try to treat each trip independently as if they give direct upstream and downstream speeds, without setting up proper equations. Another is algebraic error when working with fractions, especially when clearing denominators; careful multiplication is essential to avoid mistakes.
Final Answer:
The speed of the boat in still water is 9 km/h.