Answer: B
Let the speed of Ratan Majhi’s boat in still water be \(x\) km/h. ∴ The speed of the boat downstream = \((x + 2)\) km/h And the speed of the boat upstream = \((x - 2)\) km/h Time taken to travel 21 km downstream = \(\cfrac{21}{x + 2}\) hours Time taken to return 21 km upstream = \(\cfrac{21}{x - 2}\) hours ∴ According to the question: \[ \cfrac{21}{x + 2} + \cfrac{21}{x - 2} = 10 \] \[ \cfrac{21(x - 2) + 21(x + 2)}{(x + 2)(x - 2)} = 10 \] \[ \cfrac{21x - 42 + 21x + 42}{x^2 - 4} = 10 \] \[ \cfrac{42x}{x^2 - 4} = 10 \] \[ 10(x^2 - 4) = 42x \] \[ 10x^2 - 40 - 42x = 0 \] \[ 10x^2 - 42x - 40 = 0 \] \[ 5x^2 - 21x - 20 = 0 \] \[ 5x^2 - 25x + 4x - 20 = 0 \] \[ 5x(x - 5) + 4(x - 5) = 0 \] \[ (x - 5)(5x + 4) = 0 \] ∴ Either \((x - 5) = 0\) or \((5x + 4) = 0\) When \((x - 5) = 0\), then \(x = 5\) When \((5x + 4) = 0\), then \(x = -\cfrac{4}{5}\) [But the speed of the boat cannot be negative] ∴ The speed of Ratan Majhi’s boat in still water was 5 km/h.
Let the speed of Ratan Majhi’s boat in still water be \(x\) km/h. ∴ The speed of the boat downstream = \((x + 2)\) km/h And the speed of the boat upstream = \((x - 2)\) km/h Time taken to travel 21 km downstream = \(\cfrac{21}{x + 2}\) hours Time taken to return 21 km upstream = \(\cfrac{21}{x - 2}\) hours ∴ According to the question: \[ \cfrac{21}{x + 2} + \cfrac{21}{x - 2} = 10 \] \[ \cfrac{21(x - 2) + 21(x + 2)}{(x + 2)(x - 2)} = 10 \] \[ \cfrac{21x - 42 + 21x + 42}{x^2 - 4} = 10 \] \[ \cfrac{42x}{x^2 - 4} = 10 \] \[ 10(x^2 - 4) = 42x \] \[ 10x^2 - 40 - 42x = 0 \] \[ 10x^2 - 42x - 40 = 0 \] \[ 5x^2 - 21x - 20 = 0 \] \[ 5x^2 - 25x + 4x - 20 = 0 \] \[ 5x(x - 5) + 4(x - 5) = 0 \] \[ (x - 5)(5x + 4) = 0 \] ∴ Either \((x - 5) = 0\) or \((5x + 4) = 0\) When \((x - 5) = 0\), then \(x = 5\) When \((5x + 4) = 0\), then \(x = -\cfrac{4}{5}\) [But the speed of the boat cannot be negative] ∴ The speed of Ratan Majhi’s boat in still water was 5 km/h.