Boats & Streams Questions and Answers

Answer: D

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

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

\[ b + r = 10.5 \text{ kmph}, \qquad b - r = 7.5 \text{ kmph} \]

\[ t = \frac{105}{10.5} + \frac{105}{7.5} = 10 + 14 = 24 \text{ hours} \]

Using the still-water speed for both legs would give \( \tfrac{210}{9} = 23.3 \) hours - less than the truth, because the extra time lost upstream outweighs the time saved downstream.

Answer: B

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

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

Twice the time means half the speed over the same distance:

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad b + r = 2b - 2r \]

\[ 3r = b \quad\Longrightarrow\quad \frac{b}{r} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Checking with \( b = 3 \) and \( r = 1 \): downstream 4 and upstream 2, so the upstream leg does take twice as long. Neither distance nor actual speed is needed - only the ratio.

Answer: A

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

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

Equal times over 4 km downstream and 3 km upstream give the speed ratio:

\[ \frac{4}{b + r} = \frac{3}{b - r} \quad\Longrightarrow\quad 4b - 4r = 3b + 3r \]

\[ b = 7r \quad\Longrightarrow\quad b + r = 8r, \qquad b - r = 6r \]

Now use the 48 km round trip in 14 hours:

\[ \frac{48}{8r} + \frac{48}{6r} = 14 \quad\Longrightarrow\quad \frac{6}{r} + \frac{8}{r} = 14 \]

\[ \frac{14}{r} = 14 \quad\Longrightarrow\quad r = 1 \text{ km/h} \]

So the boat does 7 km/h in still water: 8 km/h downstream for 6 hours and 6 km/h upstream for 8 hours.

Answer: A

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

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

Swimming back against the current is the upstream leg:

\[ b - r = 4 - 2 = 2 \text{ km/h} \]

\[ t = \frac{6}{2} = 3 \text{ hours} \]

All the data needed is present, so "insufficient data" is a trap. Using the still-water speed of 4 km/h would give 1.5 hours - too optimistic, since the current is working against him.

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: B

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

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

\[ b + r = 6 \text{ km/h}, \qquad b - r = 4 \text{ km/h} \]

\[ b = \frac{6 + 4}{2} = 5 \text{ kmph} \]

The stream's own rate is \( \tfrac{6-4}{2} = 1 \) kmph, which is offered as a distractor. The man's rate must lie between his two observed speeds, so it has to be 5.

Answer: D

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

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

Twice the time means half the speed over the same distance:

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad 4.5 + r = 2(4.5 - r) = 9 - 2r \]

\[ 3r = 4.5 \quad\Longrightarrow\quad r = 1.5 \text{ km/h} \]

Checking: 6 km/h downstream and 3 km/h upstream. In general \( r = \tfrac{b}{3} \) whenever the upstream leg takes twice as long.

Answer: C

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

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

\[ b + r = 15 + 5 = 20 \text{ kmph} \]

Downstream the current helps, so the speeds add. Upstream they would subtract, giving 10 kmph - which is why 10 is offered as a distractor.

Answer: C

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

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

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

\[ b = \frac{35 + 25}{2} = 30 \text{ kmph} \]

The still-water speed is the average of the two, so it must lie between 25 and 35 - ruling out 60 and 10 on sight. The stream itself runs at \( \tfrac{35-25}{2} = 5 \) kmph.

Answer: A

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

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

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

\[ r = \frac{35 - 25}{2} = 5 \text{ kmph} \]

The current's speed is half the difference, not the difference itself: it helps by \( r \) one way and hinders by \( r \) the other, so the gap between the two speeds is \( 2r \). That is why 10 kmph is on the list.