Boats & Streams Questions and Answers

Answer: C

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

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

\[ b + r = 20 + 5 = 25 \text{ km/h} \]

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

\[ \text{distance} = 25 \times 0.2 = 5 \text{ km} \]

Using the still-water speed of 20 km/h instead would give 4 km - the current is helping, so the distance must exceed that.

Answer: C

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

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

Set \( x = \dfrac{1}{b+r} \) and \( y = \dfrac{1}{b-r} \):

\[ 24x + 36y = 6.5, \qquad 36x + 24y = 6 \]

Adding: \( 60x + 60y = 12.5 \), so \( x + y = 0.20833 \).

Subtracting: \( -12x + 12y = 0.5 \), so \( y - x = 0.04167 \).

\[ y = 0.125 \Rightarrow b - r = 8, \qquad x = 0.08333 \Rightarrow b + r = 12 \]

\[ r = \frac{12 - 8}{2} = 2 \text{ km/h} \]

Adding and subtracting the two equations is quicker than eliminating one variable at a time, because the coefficients 24 and 36 appear in both.

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.

Answer: B

Correct answer: 7.5 km/hr

\[\text{Downstream time} = \frac{30}{15} = 2 \text{ h}, \qquad \text{Upstream time} = \frac{30}{5} = 6 \text{ h}\]

\[\text{Average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{60}{8} = 7.5 \text{ km/hr}\]

Note that this is well below the still-water speed of 10 km/hr: the slow return leg occupies most of the time, so it dominates the average.

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: 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: B

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

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

\[ b = \frac{18 + 10}{2} = 14 \text{ km/h} \]

\[ r = \frac{18 - 10}{2} = 4 \text{ km/h} \]

Checking: \( 14 + 4 = 18 \) downstream and \( 14 - 4 = 10 \) upstream. The 28 and 8 option forgets to halve; the still-water speed must always lie between the two given speeds.

Answer: B

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

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

The upstream trip already gives one speed outright:

\[ b - r = \frac{48}{6} = 8 \text{ km/h} \]

With statement A (\( b = 10 \)): \( r = 2 \), so \( b + r = 12 \) and the downstream time is \( \tfrac{48}{12} = 4 \) hours.

With statement B (\( r = 2 \)): \( b = 10 \), giving the same \( b + r = 12 \) and the same 4 hours.

Each statement alone completes the pair, so either one is sufficient. Once \( b - r \) is known, any single further fact about \( b \) or \( r \) determines the other - which is exactly the situation this data-sufficiency format is testing.

Answer: D

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

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

\[ b + r = \frac{24}{4} = 6 \text{ km/hr}, \qquad b - r = \frac{24}{6} = 4 \text{ km/hr} \]

\[ b = \frac{6 + 4}{2} = 5 \text{ km/hr} \]

5 km/hr is not among the listed values, so the answer is none of these. The nearest option, 5.5, is the sort of near-miss these questions use to punish rounding.

Answer: B

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

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

Substituting \( x = \dfrac{1}{b-r} \) and \( y = \dfrac{1}{b+r} \) makes both conditions linear:

\[ 30x + 44y = 10, \qquad 40x + 55y = 13 \]

Multiply the first by 4 and the second by 3:

\[ 120x + 176y = 40, \qquad 120x + 165y = 39 \]

\[ 11y = 1 \quad\Longrightarrow\quad y = \frac{1}{11} \quad\Longrightarrow\quad b + r = 11 \]

\[ 30x + 4 = 10 \quad\Longrightarrow\quad x = \frac{1}{5} \quad\Longrightarrow\quad b - r = 5 \]

\[ b = \frac{11 + 5}{2} = 8 \text{ km/h} \]

The reciprocal substitution is the standard move for these paired-journey problems: without it the equations are non-linear and far harder to solve.