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Boats and Streams Questions Answers

Boats and Streams Questions Answers (MCQ) listings with explanations are important for BANK PO, Clerk, IBPS, SBI-PO, RBI, MBA, MAT, CAT, IIFT, IGNOU, SSC CGL, CBI, CPO, CLAT, CTET, NDA, CDS, Specialist Officers and other competitive exams.

1) A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the strea

  • 1) 0,5
  • 2) 5,5
  • 3) 15,5
  • 4) 10,5
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Ans.   C
Explanation :
Please remember, If a is rate downstream and b is rate upstream Rate in still water = 1/2(a+b) Rate of current = 1/2(a-b) => Rate in still water = 1/2(20+10) = 15 kmph => Rate of current = 1/2(20-10) = 5 kmph

2) A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the curren

  • 1) 1 km/hr
  • 2) 2 km/hr
  • 3) 3 km/hr
  • 4) 4 km/hr
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Ans.   A
Explanation :
First of all, we know that speed of current = 1/2(speed downstream - speed upstream) [important] So we need to calculate speed downstream and speed upstream first. Speed = Distance / Time [important] \begin{aligned} \text {Speed upstream =}\\ (\frac{15}{3\frac{3}{4}}) km/hr \ = 15 \times \frac{4}{15} = 4 km/hr \ \text{Speed Downstream = } (\frac{5}{2\frac{1}{2}}) km/hr \ = 5 \times \frac{2}{5} = 2 km/hr \\text {So speed of current = } \frac{1}{2}(4-2) \ = 1 km/hr \end{aligned}

3) In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still wat

  • 1) 6
  • 2) 7
  • 3) 8
  • 4) 9
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Ans.   C
Explanation :
We know we can calculate it by 1/2(a+b) => 1/2(11+5) = 1/2(16) = 8 km/hr

4) If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream

  • 1) 5 km/hr
  • 2) 4 km/hr
  • 3) 2 km/hr
  • 4) 1 km/hr
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Ans.   D
Explanation :
Rate upstream = (15/3) kmph Rate downstream (21/3) kmph = 7 kmph. Speed of stream (1/2)(7 - 5)kmph = 1 kmph

5) A man can row \begin{aligned} 9\frac{1}{3} \end{aligned} kmph in still water and finds that it takes him thrice as much time to row up than as to row, down the same distance in the river. The speed of the current i

  • 1) \begin{aligned} 3\frac{2}{3}kmph \end{aligned}
  • 2) \begin{aligned} 4\frac{2}{3}kmph \end{aligned}
  • 3) \begin{aligned} 5\frac{2}{3}kmph \end{aligned}
  • 4) \begin{aligned} 6\frac{2}{3}kmph \end{aligned}
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Ans.   B
Explanation :
Friends first we should analyse quickly that what we need to calculate and what values we require to get it. So here we need to get speed of current, for that we will need speed downstream and speed upstream, because we know Speed of current = 1/2(a-b) [important] Let the speed upstream = x kmph Then speed downstream is = 3x kmph [as per question] \begin{aligned} \text{speed in still water = } \frac{1}{2}(a+b) \=> \frac{1}{2}(3x+x) \=> 2x \\text{ as per question we know, }\2x = 9\frac{1}{3} \=> 2x = \frac{28}{3} => x = \frac{14}{3} \\end{aligned} So, Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr. Speed of the current \begin{aligned} =\frac{1}{2}[14 - \frac{14}{3}]\= \frac{14}{3} = 4 \frac{2}{3} kmph \end{aligned}

6) A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water

  • 1) 4 kmph
  • 2) 5 kmph
  • 3) 6 kmph
  • 4) 7 kmph
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Ans.   B
Explanation :
Rate upstream = (750/675) = 10/9 m/sec Rate downstream (750/450) m/sec = 5/3 m/sec Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec. = 25/18 m/sec = (25/18)*(18/5) kmph = 5 kmph

7) A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr)

  • 1) 2 km/hr
  • 2) 3 km/hr
  • 3) 4 km/hr
  • 4) 5 km/hr
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Ans.   D
Explanation :
Let the speed of the stream be x km/hr. Then, Speed downstream = (15 + x) km/hr, Speed upstream = (15 - x) km/hr So we know from question that it took 4(1/2)hrs to travel back to same point. So, \begin{aligned} \frac{30}{15+x} - \frac{30}{15-x} = 4\frac{1}{2} \=> \frac{900}{225 - x^2} = \frac{9}{2} \=> 9x^2 = 225 \=> x = 5 km/hr \end{aligned}
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