A boat goes 15 km upstream and 10 ½ km downstream in 3 hours 15...
Let speed of the boat in still water be x km/h and speed of current be y km/h. Then, upstream speed = (x – y) km/h and downstream speed = (x + y) km/h Now,
From Equation (i) and (ii) x = 10 km/hr and y = 4 km/hr.
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A boat goes 15 km upstream and 10 ½ km downstream in 3 hours 15...
Km downstream in a total of 5 hours. If the speed of the boat in still water is 30 km/h, what is the speed of the current?
Let's assume the speed of the current is x km/h.
When the boat goes upstream, its effective speed is reduced by the speed of the current, so it is (30 - x) km/h. The distance traveled upstream is 15 km. Therefore, the time taken to go upstream is 15 / (30 - x) hours.
When the boat goes downstream, its effective speed is increased by the speed of the current, so it is (30 + x) km/h. The distance traveled downstream is 10 km. Therefore, the time taken to go downstream is 10 / (30 + x) hours.
The total time taken for the entire trip is 5 hours, so we can write the equation:
15 / (30 - x) + 10 / (30 + x) = 5
To solve this equation, we can first get rid of the denominators by multiplying every term by (30 - x)(30 + x):
15(30 + x) + 10(30 - x) = 5(30 - x)(30 + x)
Expanding and simplifying:
450 + 15x + 300 - 10x = 5(900 - x²)
750 + 5x = 4500 - 5x²
Rearranging the terms:
5x² + 5x - 3750 = 0
Dividing every term by 5:
x² + x - 750 = 0
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. By factoring, we find:
(x - 25)(x + 30) = 0
This gives us two possible values for x: x = 25 or x = -30. Since the speed cannot be negative, the speed of the current is 25 km/h.