Quantitative Aptitude
Boats and Streams
369 Questions
Boats and Streams Questions
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$6\ h$
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$7.5\ h$
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$10\ h$
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$15\ h$
B
Correct answer
Explanation
Let v be boat speed, u be stream speed. Downstream: v+u = D/6. Upstream: v-u = D/10. Adding: 2v = D/6 + D/10 = (5D+3D)/30 = 8D/30 = 4D/15. v = 2D/15. Time in still water = D/v = D / (2D/15) = 15/2 = 7.5 hours.
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$\dfrac{m + n}{2\, m\, n}$
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$\dfrac{m - n}{2\, m\, n}$
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$\dfrac{n - m}{2\, m\, n}$
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$\dfrac{2(m + n)}{ m\, n}$
A
Correct answer
Explanation
Speed upstream = 1/m km/min. Speed downstream = 1/n km/min. Let boat speed be B and current speed be C. B - C = 1/m and B + C = 1/n. Adding the two equations: 2B = 1/m + 1/n = (n+m)/(mn). So B = (m+n)/(2mn).
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$2\ hr$
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$2\ hr\ 40\ min$
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$1\ hr\ 20\ min$
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$2\ hr\ 30\ min$
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$1-\dfrac { v^ 2 }{ V^ 2 } $
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$\dfrac{1}{1-\dfrac { v^ 2 }{ V^ 2 } }$
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$1+\dfrac { v^ 2 }{ V^ 2 } $
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$\dfrac{1}{1+\dfrac { v^ 2 }{ V^ 2 } }$
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120 min
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160 min
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200 min
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None of these
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$=(2u+2v\cos \theta )\dfrac{d}{v\sin\theta}$
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$=(2u+v\cos \theta )\dfrac{d}{v\sin\theta}$
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$=(u+3v\cos \theta )\dfrac{d}{v\sin\theta}$
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$=(u+v\cos \theta )\dfrac{d}{v\sin\theta}$
D
Correct answer
Explanation
The swimmer's velocity along the river is v cos(theta), while the river adds velocity u in the same direction. The time to cross is d/(v sin(theta)), so the downstream displacement is (u + v cos(theta))d/(v sin(theta)). The listed expression answers part c, but parts a and b are not represented separately.
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$d=765m$
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$d=865m$
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$d=965m$
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$d=1065m$
A
Correct answer
Explanation
Let d be the width, v=17, u=8. Quickest path time t1 = d / sqrt(v^2 - u^2) = d / sqrt(289 - 64) = d / sqrt(225) = d / 15. Shortest path time t2 = d / v = d / 17. The difference t2 - t1 = 6. So d/15 - d/17 = 6. (17d - 15d) / 255 = 6, 2d = 1530, d = 765 m.
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$3km{h}^{-1}$
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$km{h}^{-1}$
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$5km{h}^{-1}$
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$6km{h}^{-1}$
A
Correct answer
Explanation
Let v be boat speed and u be stream speed. The raft moves at speed u. The boat travels downstream for 1 hour (60 min), then turns back. The relative speed of the boat and raft is (v+u) - u = v downstream and (v-u) + u = v upstream. The distance covered by the raft in the time the boat returns is 6km. This setup implies the stream velocity is 3km/h.
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$\dfrac{t_1+t_2}{2}$
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$2(t_2=t_1)$
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$\dfrac{2 t_1 t_2}{t_1+t_2}$
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$\sqrt{t_1 t_2}$
C
Correct answer
Explanation
Let d be the distance, v be boat speed, u be stream speed. t1 = d/(v+u) and t2 = d/(v-u). Then 1/t1 = (v+u)/d and 1/t2 = (v-u)/d. Adding these: 1/t1 + 1/t2 = 2v/d. So v/d = (t1+t2)/(2*t1*t2). The time in still water is d/v = (2*t1*t2)/(t1+t2).
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11 $kmh^{-1}$
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22 $kmh^{-1}$
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33 $kmh^{-1}$
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44 $kmh^{-1}$
B
Correct answer
Explanation
Boat velocity vB = 25 (North). Current vC = 10 (60 deg east of south). East of south is 120 deg from North. Use vector addition: Resultant^2 = 25^2 + 10^2 + 2 * 25 * 10 * cos(120). Resultant^2 = 625 + 100 + 500 * (-0.5) = 725 - 250 = 475. Resultant = sqrt(475) approx 21.8, which is 22.
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$6\ km\ h^{-1}$
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$3\ km\ h^{-1}$
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$4\ km\ h^{-1}$
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$2\ km\ h^{-1}$
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$30^\circ$
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$37^\circ$
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$53^\circ$
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$63.4^\circ$
D
Correct answer
Explanation
The river contributes 3 m/s eastward and the boat contributes 6 m/s northward. Therefore, tan(theta) = 6/3 = 2, so theta is approximately 63.4 degrees from the river flow.
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$v_{0} = 10\ ms^{-1}, \theta_{0} = 30^{\circ}$
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$v_{0} = 10\ ms^{-1}, \theta_{0} = 40^{\circ}$
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$v_{0} = 10\ ms^{-1}, \theta_{0} = 60^{\circ}$
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Same for all the above
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$ 2\sqrt10$
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$ 4\sqrt10$
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$ 5\sqrt10$
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$ 6\sqrt10$
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$\dfrac{1}{8}$
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$\dfrac{1}{4}$
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$\dfrac{1}{2}$
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$1$
A
Correct answer
Explanation
For the shortest path, the boat must head upstream such that its resultant velocity is perpendicular to the river flow. The resultant speed across the river is sqrt(10^2 - 6^2) = 8 km/h. Time = distance / speed = 1 / 8 hours.