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A boat can cover 10 km up the stream and 5 km down the stream in 6 hours . If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.

Quadratic Equations

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Answer

Let the speed of boat in still water be x km/h

Speed of stream = 1.5 km/h

∴ Speed of boat upstream = (x - 1.5) km/h and Speed of boat downstream = (x + 1.5) km/h

According to given,

⇒10x−1.5+5x+1.5=6⇒10(x+1.5)+5(x−1.5)(x+1.5)(x−1.5)=6⇒10x+15+5x−7.5x2−1.5x+1.5x−2.25=6⇒15x+7.5x2−2.25=6⇒15x+7.5=6(x2−2.25)⇒15x+7.5=6x2−13.50⇒6x2−13.5−7.5−15x=0⇒6x2−15x−21=0⇒3(2x2−5x−7)=0⇒2x2−5x−7=0⇒2x2−7x+2x−7=0⇒x(2x−7)+1(2x−7)=0⇒(x+1)(2x−7)=0⇒x=−1 or 2x−7=0⇒x+1=0 or x=72⇒x=−1 or x=3.5\Rightarrow \dfrac{10}{x - 1.5} + \dfrac{5}{x + 1.5} = 6 \\[1em] \Rightarrow \dfrac{10(x + 1.5) + 5(x - 1.5)}{(x + 1.5)(x - 1.5)} = 6 \\[1em] \Rightarrow \dfrac{10x + 15 + 5x - 7.5}{x^2 - 1.5x + 1.5x - 2.25} = 6 \\[1em] \Rightarrow \dfrac{15x + 7.5}{x^2 - 2.25} = 6 \\[1em] \Rightarrow 15x + 7.5 = 6(x^2 - 2.25) \\[1em] \Rightarrow 15x + 7.5 = 6x^2 - 13.50 \\[1em] \Rightarrow 6x^2 - 13.5 - 7.5 - 15x = 0 \\[1em] \Rightarrow 6x^2 - 15x - 21 = 0 \\[1em] \Rightarrow 3(2x^2 - 5x - 7) = 0 \\[1em] \Rightarrow 2x^2 - 5x - 7 = 0 \\[1em] \Rightarrow 2x^2 - 7x + 2x - 7 = 0 \\[1em] \Rightarrow x(2x - 7) + 1(2x - 7) = 0 \\[1em] \Rightarrow (x + 1)(2x - 7) = 0 \\[1em] \Rightarrow x = -1 \text{ or } 2x - 7 = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x = \dfrac{7}{2} \\[1em] \Rightarrow x = -1 \text{ or } x = 3.5

Since speed of bus cannot be negative hence, x ≠ -1

Hence speed of boat in still water is 3.5 km/h.

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