a boat moves 9 kilometers upstream in the same amount of time it moves 20 kilometers downstream. If the rate of the current is 9 kilometers per hour. Find the rate of the boat in still water
Well, if the boat moved upstream 9 kilometers in the same amount of time as it moved downstream 20 kilometers, it sounds like the boat really went with the flow, or should I say, against the flow! A current of 9 kilometers per hour is no joke.
Now, let's do some math magic. Let's call the rate of the boat in still water "B" (because "Boat" starts with a "B" too, isn't that amusing?). We'll also call the time it took for both trips "t."
When the boat moves upstream, it has to fight against the current, which slows it down. So its effective rate becomes B - 9.
On the other hand, when the boat moves downstream, the current gives it a boost, so its effective rate becomes B + 9.
Now, since we're given that the time it took for both trips is equal, we can write an equation: distance = rate x time.
For the upstream trip: 9 = (B - 9) x t
For the downstream trip: 20 = (B + 9) x t
Solving these equations simultaneously may be a bit challenging, but hey, I'm here to make it a little more amusing. I'll leave that part to you. Good luck, math comedian!
Let's assume the rate of the boat in still water is x kilometers per hour.
When the boat moves upstream, it is going against the current, so the effective speed is reduced. The time taken for this distance is the same as the time taken for the boat to move downstream.
Let's calculate the time taken for the boat to move upstream:
The effective speed of the boat moving upstream is (x - 9) kilometers per hour.
The distance covered is 9 kilometers.
Using the formula time = distance / speed, the time taken is 9 / (x - 9) hours.
Now let's calculate the time taken for the boat to move downstream:
The effective speed of the boat moving downstream is (x + 9) kilometers per hour.
The distance covered is 20 kilometers.
Using the formula time = distance / speed, the time taken is 20 / (x + 9) hours.
Since the time taken for both cases is the same, we can equate the two equations:
9 / (x - 9) = 20 / (x + 9)
Cross-multiplying, we get:
9(x + 9) = 20(x - 9)
9x + 81 = 20x - 180
Rearranging the equation:
11x = 261
Dividing both sides by 11:
x = 23
Therefore, the rate of the boat in still water is 23 kilometers per hour.
To find the rate of the boat in still water, we need to set up an equation based on the given information.
Let's assume the rate of the boat in still water is 'b' kilometers per hour.
When the boat is moving upstream, it is going against the current, so the effective speed is the rate of the boat minus the rate of the current. Therefore, the speed upstream is (b - 9) kilometers per hour.
When the boat is moving downstream, it is going with the current, so the effective speed is the rate of the boat plus the rate of the current. Therefore, the speed downstream is (b + 9) kilometers per hour.
Now, we are given that the boat moves 9 kilometers upstream in the same amount of time it moves 20 kilometers downstream. This means that the time it takes to travel both distances is equal.
We can use the formula: Time = Distance / Speed
For the upstream journey: Time = 9 / (b - 9)
For the downstream journey: Time = 20 / (b + 9)
Since we are given that the time for both journeys is the same, we can set up an equation:
9 / (b - 9) = 20 / (b + 9)
To solve this equation, we can cross-multiply:
9(b + 9) = 20(b - 9)
Simplify the equation:
9b + 81 = 20b - 180
Subtract 9b and add 180 to both sides:
81 + 180 = 20b - 9b
261 = 11b
Divide both sides by 11:
b = 261 / 11
b = 23
Therefore, the rate of the boat in still water is 23 kilometers per hour.
distance = speed * time
9 = (v-9) t
20 = (v+9) t
but t is the same
so
9/(v-9) = 20/(v+9)
9(v+9) = 20(v-9)
9 v + 81 = 20 v - 180
11 v = 261
v = 23.7