The article **Boats and Streams Practice Quiz** provides information about Boats and Streams, a important topic of **Quantitative Aptitude section.** Consists of different types Boats and Streams questions with solutions useful for candidates preparing for different competitive examinations like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.**

**Answer –** Option B

**Explanation –**

Let \( {v}_{1}\) be the speed of boat in still water and \( {v}_{2}\) be the speed of current

\( {v}_{1} + {v}_{2}\) = \(\frac {20}{1}\) = 20 ……(1)

\({v}_{1} – {v}_{2}\) = \(\frac {20}{2}\) = 10 ……(2)

From equations (i) and (ii) we get

\({v}_{1}\) = 15 km/hr

\(\frac {d}{{v}_{1}} \) = \(\frac {30}{15}\) = 2 hrs

**2. In the above question, the speed at which the stream is flowing is**

**Answer –** Option D

**Explanation –**

From the above two equation, we get \(\frac {d}{{v}_{2}} \) = 5 km/hr

**3. A boat travels 10 km in 1 hr downstream and 14 km in 2 hrs upstream. How much time this boat will take to travel 17 km in still water?**

**Answer –** Option C

**Explanation –**

\( {v}_{1} + {v}_{2}\) = \(\frac {10}{1}\) = 10 ……(1)

\( {v}_{1} – {v}_{2}\) = \(\frac {14}{2}\) = 7 ……(2)

Adding equations (i) and (ii), we get

\({v}_{1}\) = \(\frac {17}{2}\) km/hr

\(\frac {d}{{v}_{1}} \) = \(\frac {17}{\frac{17}{2}}\) = 2 hrs

**4. A man goes by motor boat a certain distance up stream at 15 km/hr and return the same
downstream at 20 km/hr. The total time taken for the journey was 7 hrs. Find how far did he go.**

**Answer –** Option A

**Explanation –**

\(\frac {d}{20} + \frac {d}{10}\) = 7

d = 60 km

**5. A man can row upstream a distance of \(\frac {2} {3}\) km in 10 minutes and returns the same distance downstream in 5 minutes. Ratio of manâ€™s speed in still water and that of the stream will be**

**Answer –** Option A

**Explanation –**

\( {v}_{1} – {v}_{2}\) = \(\frac {\frac{2}{3}km}{10 min}\)

i.e, \( {v}_{1} – {v}_{2}\) = \(\frac {2}{30}\) km/min ……(1)

and \( {v}_{1} + {v}_{2}\) = \(\frac {\frac{2}{3} km}{5 min}\) = 10

\( {v}_{1} + {v}_{2}\) = \(\frac {2}{15}\) = 7 ……(2)

Solving equation (i) and (ii), we get

\( {v}_{1}\) = \(\frac {2 + 4}{60}\) km/min = \(\frac {6}{60}\) km/min = \(\frac {1}{10}\) km/min

\( {v}_{2}\) = \(\frac {4 – 2}{60}\) km/min = \(\frac {2}{60}\) km/min = \(\frac {1}{30}\) km/min

\( \frac{{v}_{1}}{{V}_{2}}\) = \(\frac {1}{10}\) * \(\frac {30}{1}\) = 3 : 1

**Answer –** Option A

**Explanation –**

\(( {v}_{1} + {v}_{2}) {t}_{1}\) = \(( {v}_{1} – {v}_{2}) {t}_{2}\)

i.e, \(( {v}_{1} + {v}_{2}) * 6\) = \(( {v}_{1} – {v}_{2}) * 9\)

\(( {v}_{1}\) = 10 km/hr

**2. A boat against the current of water goes 9 km/hr and in the direction of the current 12 km/hr. The boat takes 4 hours and 12 minute es to move upward and downward direction from A to B. What is the distance between A and B?**

**Answer –** Option A

**Explanation –**

\(\frac {d}{9} + \frac {d}{12}\) = \(4 \frac {12}{60}\)

d = 21.6 km

**3. A man takes 3 hours and 45 minutes to boat 15 km with the current in a river and 2 hours 30 minutes to cover a distance of 5 km against the current. Speed of the boat in still water and speed of the current respectively will be**

**Answer –** Option A

**Explanation –**

\(({v}_{1} + {v}_{2})\) = \( \frac {15}{3 \frac {3}{4}}\) = 4 km/hr

\( ({v}_{1} – {v}_{2})\) = \( \frac {5}{2 \frac {1}{2}}\) = 2 km/hr

Solving equations (i) and (ii), we get

\( {v}_{1}\) = 3 km/hr and \({v}_{2}\) = 1 km/hr

**4. A boat can be rowed 6 km/hr along the current and 4 km/hr against the current. Speed of the current and speed of the boat in still water, respectively will be**

**Answer –** Option B

**Explanation –**

\(({v}_{1} + {v}_{2})\) = 6 …..(1)

\(( {v}_{1} – {v}_{2})\) = 4 …..(2)

From equations (i) and (ii), we get

\( {v}_{1}\) = 5 km/hr and \({v}_{2}\) = 1 km/hr

**5. A boat moves down the stream at the rate of 1 km in 6 minutes and up the stream at the rate of
1 km in 10 minutes. The speed of the current is**

**Answer –** Option A

**Explanation –**

\(({v}_{1} + {v}_{2})\) = \( \frac {1 km}{6 min}\) = 10 km/hr …(1)

\(({v}_{1} – {v}_{2})\) = \( \frac {1 km}{10 min}\) = 6 km/hr ….(2)

subtracting equation (2) from (1), we get

\({v}_{1}\) = 2 km/hr

**Answer –** Option A

**Explanation –**

\(\frac {d}{5 + 1} + \frac {d}{5 – 1}\) = \(\frac {75}{60}\)

\(\frac {2d + 3d}{12}\) = \(\frac {5}{4}\)

d = 3 km

**2. Speed of a boat in still water is 7 km/hr and speed of the stream is 1.5 km/hr . How much time will it take to move up is stream of a distance 7.7 km?**

**Answer –** Option B

**Explanation –**

t = \(\frac {d}{{v}_{1} + {v}_{2}}\) = \(\frac {7.7}{7 + 1.5}\) = \(\frac {7.7}{5.5}\)

= \(\frac {7}{5}\) hrs = \(\frac {7}{5} * 60 min\) = 84 min

**3. A motorboat takes 2 hours to travel a distance of 9 km down the current and it takes 6 hours to
travel the same distance against the current. What is the speed of the boat in still water in kmph?**

**Answer –** Option A

**Explanation –**

Let the speed of boat in still water and speed of current are x and y km/h respectively.

i.e, Downward speed of boat = (x + y) km/h.

According to question,

x + y = \(\frac {9}{2}\)

2x + 2y = 9 …(i)

x – y = \(\frac {9}{6}\)

2x â€“ 2y = 3 …(ii)

On solving equations (i) and (ii), we get

x = 3, y = \(\frac {3}{2}\)