A man went downstream for 28 km in a motor boat and immediately returned. It too

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Question
A man went downstream for 28 km in a motor boat and immediately returned. It took the man twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.
Options
A. 12 km/hr, 3 km/hr
B. 9 km/hr, 3 km/hr
C. 8 km/hr, 2 km/hr
D. 9 km/hr, 6 km/hr

Correct Answer

9 km/hr, 3 km/hr

Explanation
Let's call the speed of the boat in still water "b" and the speed of the river flow "r".
When the man goes downstream, he travels with the current, so his effective speed is b + r. When he goes upstream, he travels against the current, so his effective speed is b - r.
We know that the man traveled 28 km downstream and then immediately returned, so his total distance traveled is 56 km. We also know that it took him twice as long to make the return trip, so his time going upstream was twice his time going downstream.
Using the formula distance = rate x time, we can set up two equations:
28 = (b + r) * t1
28 = (b - r) * 2t1
where t1 is the time it took the man to go downstream.
Simplifying the second equation, we get:
14 = (b - r) * t1
Now let's consider what happens if the speed of the river flow is doubled. The new effective speed downstream would be b + 2r, and the new effective speed upstream would be b - 2r. The total distance traveled is still 56 km. We can set up two more equations:
56 = (b + 2r) * t2
56 = (b - 2r) * 2t2
where t2 is the new total time it takes to make the trip downstream and back.
Simplifying the second equation, we get:
28 = (b - 2r) * t2
Now we have four equations and four unknowns (b, r, t1, and t2). We can solve for them using substitution and elimination.
From the first equation, we can solve for t1:
t1 = 28 / (b + r)
From the second equation, we can solve for t1 in terms of t2:
t1 = 14 / (b - r)
Setting these two expressions for t1 equal to each other and simplifying, we get:
(b + r) / (b - r) = 2
Solving for r in terms of b, we get:
r = b / 3
Now we can substitute this expression for r into any of the four equations to solve for the other variables. Let's use the first equation:
28 = (b + r) * t1
28 = (4b/3) * t1
t1 = 21 / (2b)
Now we can use this expression for t1 to solve for t2 in terms of b:
t2 = 42 / (2b - 4r)
t2 = 42 / (2b - 4b/3)
t2 = 126 / b
Finally, we can use the third equation to solve for b:
56 = (b + 2r) * t2
56 = (4b/3 + 2b/3) * 126 / b
56 = 2 * 126
b = 9
So the speed of the boat in still water is 9 km/h, and the speed of the river flow is b/3 = 3 km/h.

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