The speed of boat A in still water is 2 km/h less than the speed of the boat B in still water. The time taken by boat A to travel a distance of 20km downstream is 30 minutes more than time taken by boat B to travel the same distance downstream. If the speed of the current is \( \Large \frac{1}{3} \)rd of the speed of the boat A in still water, what is the speed of boat B ? (km/h)
Correct Answer: Description for Correct answer:
Let the speed of boat A in still water be x kmph
\( \Large \therefore \) Speed of boat B = (x + 2) kmph
Rate downstream of boat A
= \( \Large (x + \frac{x}{3} ) = \frac{4x}{3} \ kmph \)
Rate downstream of boat B
= \( \Large (x + 2 + \frac{x}{3}) \ kmph \)
= \( \Large (\frac{4x}{3} + 2 ) \ kmph \)
According to the question,
\( \Large \frac{20}{\frac{4x}{3}} - \frac{20}{\frac{4x}{3} + 2} = \frac{30}{60} = \frac{1}{2} \)
=> \( \Large \frac{60}{4x} - \frac{60}{4x + 6} = \frac{1}{2} \)
=> \( \Large 60 (\frac{4x + 6 - 4x}{4x(4x+ 6)}) = \frac{1}{2} \)
=> \( \Large 4x(4x + 6) = 360 \times 2 \)
=> \( \Large x (2x + 3) = \frac{360 \times 2}{8} = 90 \)
=> \( \Large 2x^{2} + 3x - 90 = 0 \)
=> \( \Large 2x^{2} + 15x - 12x - 90 = 0 \)
=> \( \Large x (2x + 15) - 6 (2x + 15) = 0 \)
=> (x - 6)(2x + 15) = 0
=> \( \Large x = 6 \ because \ x \neq \frac{-15}{2} \)
\( \Large \therefore \) Speed of boat B = 6 + 2
= 8 kmph
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