Boats & Streams Questions and Answers

Answer: B

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

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

\[ b + r = \frac{30}{3} = 10 \text{ kmph}, \qquad b - r = \frac{18}{3} = 6 \text{ kmph} \]

\[ b = \frac{10 + 6}{2} = 8 \text{ kmph} \]

Because both legs take the same 3 hours, the distances are directly proportional to the speeds - which is what makes the two speeds so easy to read off.

Answer: B

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

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

\[ b + r = \frac{72}{9} = 8 \text{ kmph}, \qquad b - r = \frac{45}{9} = 5 \text{ kmph} \]

\[ r = \frac{8 - 5}{2} = 1.5 \text{ kmph} \]

The 3 kmph option is the full difference; halving it is the step that separates the current from the gap between the two speeds.

Answer: B

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

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

\[ b + r = 7.2 \text{ kmph}, \qquad b - r = 4.8 \text{ kmph} \]

\[ \frac{D}{7.2} + \frac{D}{4.8} = 1 \]

\[ D\left(\frac{1}{7.2} + \frac{1}{4.8}\right) = D\left(\frac{2 + 3}{14.4}\right) = 1 \]

\[ \frac{5D}{14.4} = 1 \quad\Longrightarrow\quad D = 2.88 \text{ km} \]

Using the still-water speed of 6 kmph for both legs would give 3 km - the current always costs time overall, so the real distance is a little less.

Answer: C

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

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

\[ b + r = 7.2 \text{ kmph}, \qquad b - r = 4.8 \text{ kmph} \]

\[ \frac{D}{7.2} + \frac{D}{4.8} = 1 \quad\Longrightarrow\quad \frac{5D}{14.4} = 1 \quad\Longrightarrow\quad D = 2.88 \text{ km} \]

That is the one-way distance. The question asks for the total journey, so it must be doubled:

\[ 2 \times 2.88 = 5.76 \text{ km} \]

2.88 km is the correct answer to a different question - how far away the place is.

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: B

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

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

\[ b + r = 6 \text{ kmph}, \qquad b - r = 3 \text{ kmph} \]

Average speed is total distance over total time, which is the harmonic mean of the two speeds:

\[ \bar{v} = \frac{2 \times 6 \times 3}{6 + 3} = \frac{36}{9} = 4 \text{ kmph} \]

The arithmetic mean would be 4.5 - the still-water speed - but more time is spent on the slow upstream leg, so the true average falls below it. A round trip in a current is always slower on average than the same trip in still water.

Answer: A

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

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

\[ \frac{35}{b + 1} + \frac{35}{b - 1} = 12 \]

\[ \frac{35(b - 1) + 35(b + 1)}{b^{2} - 1} = \frac{70b}{b^{2} - 1} = 12 \]

\[ 70b = 12b^{2} - 12 \quad\Longrightarrow\quad 6b^{2} - 35b - 6 = 0 \]

\[ b = \frac{35 + \sqrt{1225 + 144}}{12} = \frac{35 + 37}{12} = 6 \text{ kmph} \]

Checking: 35 km at 7 kmph takes 5 hours and at 5 kmph takes 7 - twelve hours in total.

Answer: B

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

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

\[ \text{downstream} = 60 + 20 = 80 \text{ kmph} \]

\[ \text{upstream} = 60 - 20 = 40 \text{ kmph} \]

The two speeds always straddle the still-water speed, one above and one below by the same amount - here 20 kmph either side of 60.

Answer: A

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

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

\[ b = \frac{(b+r) + (b-r)}{2}, \qquad r = \frac{(b+r) - (b-r)}{2} \]

\[ b = \frac{80 + 60}{2} = 70 \text{ kmph} \]

\[ r = \frac{80 - 60}{2} = 10 \text{ kmph} \]

Checking: \( 70 + 10 = 80 \) and \( 70 - 10 = 60 \). The still-water speed must lie between the two given speeds, which alone eliminates every other option.

Answer: D

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

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

\[ b + r = \frac{85}{2.5} = 34 \text{ kmph}, \qquad b - r = \frac{45}{2.5} = 18 \text{ kmph} \]

\[ r = \frac{34 - 18}{2} = 8 \text{ kmph} \]

The boat itself does 26 kmph in still water. Halving the difference is the step that matters - the full 16 kmph gap is twice the current.