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A boat goes 10kms an hour in still water, but takes twice as much time in going the same distance against the current. The speed of the current (in km/hr) is –
- a)2 km/hr
- b)4 km/hr
- c)3 km/hr
- d)5 km/hr
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A boat goes 10kms an hour in still water, but takes twice as much time...
Speed of Boat in Current = Speed of Boat in still water – Speed of Current
A boat goes 10 kms an hour in still water
∴ Speed of Boat in still water = 10km/hr
Let it travels for 1 hr and covered 10 km
It takes twice as much time in going the same distance against the current
Thus, it will take 2 hr to cover the same distance of 10 km
∴ Speed of Boat against Current = 10km / 2hrs = 5km/hr
Speed of Current = 10km/hr – 5km/hr = 5km/hr
A boat goes 10 kms an hour in still water
∴ Speed of Boat in still water = 10km/hr
Let it travels for 1 hr and covered 10 km
It takes twice as much time in going the same distance against the current
Thus, it will take 2 hr to cover the same distance of 10 km
∴ Speed of Boat against Current = 10km / 2hrs = 5km/hr
Speed of Current = 10km/hr – 5km/hr = 5km/hr
Most Upvoted Answer
A boat goes 10kms an hour in still water, but takes twice as much time...
Let's assume the speed of the current is x km/hr.
When the boat is going with the current, its effective speed is increased by x km/hr. Therefore, the speed of the boat with the current is 10 + x km/hr.
When the boat is going against the current, its effective speed is decreased by x km/hr. Therefore, the speed of the boat against the current is 10 - x km/hr.
Let's say the distance traveled by the boat is D km.
The time taken to travel D km with the current is D / (10 + x) hours.
The time taken to travel D km against the current is D / (10 - x) hours.
According to the problem, the time taken to travel D km against the current is twice the time taken to travel D km with the current. So we can write the equation:
D / (10 - x) = 2 * (D / (10 + x))
Simplifying this equation:
D(10 + x) = 2D(10 - x)
10x + x^2 + 20D = 20D - 2xD
x^2 + 10x - 2xD = 0
Factoring out D:
D(x^2 + 10x - 2x) = 0
x^2 + 10x - 2x = 0
x(x + 10 - 2) = 0
x(x + 8) = 0
So either x = 0 or x = -8.
Since the speed of the current cannot be negative, the speed of the current is 0 km/hr.
When the boat is going with the current, its effective speed is increased by x km/hr. Therefore, the speed of the boat with the current is 10 + x km/hr.
When the boat is going against the current, its effective speed is decreased by x km/hr. Therefore, the speed of the boat against the current is 10 - x km/hr.
Let's say the distance traveled by the boat is D km.
The time taken to travel D km with the current is D / (10 + x) hours.
The time taken to travel D km against the current is D / (10 - x) hours.
According to the problem, the time taken to travel D km against the current is twice the time taken to travel D km with the current. So we can write the equation:
D / (10 - x) = 2 * (D / (10 + x))
Simplifying this equation:
D(10 + x) = 2D(10 - x)
10x + x^2 + 20D = 20D - 2xD
x^2 + 10x - 2xD = 0
Factoring out D:
D(x^2 + 10x - 2x) = 0
x^2 + 10x - 2x = 0
x(x + 10 - 2) = 0
x(x + 8) = 0
So either x = 0 or x = -8.
Since the speed of the current cannot be negative, the speed of the current is 0 km/hr.
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A boat goes 10kms an hour in still water, but takes twice as much time in going the same distance against the current. The speed of the current (in km/hr) is –a)2 km/hrb)4 km/hrc)3 km/hrd)5 km/hrCorrect answer is option 'D'. Can you explain this answer?
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