Boats & Streams Questions and Answers

Answer: D

With \( b \) the still-water speed and \( r \) the stream's,

\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]

\[ b + r = 20 + 6 = 26 \text{ kmph} \]

\[ t = \frac{60}{26} = \frac{30}{13} \text{ hours} \]

That is about 2 hours 18 minutes. Using the still-water speed would give 3 hours - the current is helping, so the trip must be faster than that.

Answer: C

Correct answer: 1 km/hr

\[\text{Downstream} = \frac{24}{4} = 6 \text{ km/hr}, \qquad \text{Upstream} = \frac{24}{6} = 4 \text{ km/hr}\]

\[s = \frac{6 - 4}{2} = 1 \text{ km/hr}\]

The boat itself moves at \(\tfrac{6+4}{2} = 5\) km/hr in still water.

Answer: C

Let the speed of motor boat be 36x km/hr and that of current of water be 5x km/hr 

Speed downstream = (36x +5x)km/hr = 41x km/hr 

Speed upstream = (36 -5x)km/hr = 31x km/hr 

Distance covered downstream = (41x × 31/6)km 

Distance upstream = [1271x/6 × 1/31x]hrs 

= 41/6 hrs = 6hrs 50 min

Answer: C

With \( b \) the still-water speed and \( r \) the stream's,

\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]

\[ \frac{6}{b + 4} + \frac{6}{b - 4} = 2 \]

Over the common denominator \( b^{2} - 16 \):

\[ \frac{6(b - 4) + 6(b + 4)}{b^{2} - 16} = \frac{12b}{b^{2} - 16} = 2 \]

\[ 12b = 2b^{2} - 32 \quad\Longrightarrow\quad b^{2} - 6b - 16 = 0 \]

\[ (b - 8)(b + 2) = 0 \quad\Longrightarrow\quad b = 8 \text{ kmph} \]

The negative root is discarded. Checking: 6 km at 12 kmph takes half an hour and 6 km at 4 kmph takes one and a half - two hours in total.

Answer: C

Write \( b \) for the boat's speed in still water and \( r \) for the stream's. Then

\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]

For a fixed distance, time is inversely proportional to speed, so the time ratio inverts to give the speed ratio:

\[ \frac{t_{\text{up}}}{t_{\text{down}}} = \frac{11}{5} \quad\Longrightarrow\quad \frac{b + r}{b - r} = \frac{11}{5} \]

\[ 5(b + r) = 11(b - r) \quad\Longrightarrow\quad 5b + 5r = 11b - 11r \]

\[ 16r = 6b \quad\Longrightarrow\quad \frac{b}{r} = \frac{16}{6} = \frac{8}{3} \]

So the ratio is \( 8 : 3 \). The 10 km cancels out, so the distance is never needed.

Answer: D

Write \( b \) for the boat's speed in still water and \( r \) for the stream's. Then

\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]

\[ b + r = \frac{100}{10} = 10 \text{ km/h}, \qquad b - r = \frac{75}{15} = 5 \text{ km/h} \]

\[ r = \frac{10 - 5}{2} = 2.5 = 2\tfrac{1}{2} \text{ km/h} \]

The stream's speed is half the difference of the two speeds, not the difference itself - the 5 km/h option is exactly that omission.

Answer: C

With \( b \) the still-water speed and \( r \) the stream's,

\[ \text{downstream} = b + r, \qquad \text{upstream} = b - r \]

\[ b + r = 15 + 3 = 18 \text{ km/hr} \]

\[ 12 \text{ min} = \frac{12}{60} = 0.2 \text{ h} \]

\[ \text{distance} = 18 \times 0.2 = 3.6 \text{ km} \]

Converting the minutes to hours is the only real step; using 15 km/hr instead of the downstream 18 would give 3 km.

Answer: C

Correct answer: 7 km/hr

The upstream speed is \(b - s\), so

\[5 - s = 3 \quad\Longrightarrow\quad s = 2 \text{ km/hr}\]

\[\text{Downstream speed} = b + s = 5 + 2 = 7 \text{ km/hr}\]

The still-water speed always lies exactly midway between the two.