Boats & Streams Questions and Answers

Answer: B

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

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

\[ b + r = 6 \text{ km/h}, \qquad b - r = 3 \text{ km/h} \]

Average speed over a round trip is total distance over total time, which gives the harmonic mean, not the arithmetic one:

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

The arithmetic mean of 6 and 3 is 4.5, which is too high: more time is spent at the slower speed, so the slow leg carries more weight. Notice the answer is below the shark's still-water speed of 4.5 - a current always costs time on a round trip.

Answer: C

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

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

"With the current" is the downstream speed, so

\[ b + r = 15, \qquad r = 2.5 \quad\Longrightarrow\quad b = 12.5 \text{ km/h} \]

\[ b - r = 12.5 - 2.5 = 10 \text{ km/h} \]

Equivalently, the current is subtracted twice in going from downstream to upstream: \( 15 - 2 \times 2.5 = 10 \). Subtracting it only once gives 12.5 - the still-water speed, offered as the last distractor.

Answer: D

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

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

\[ b - r = \frac{15}{3} = 5 \text{ km/hr}, \qquad b + r = \frac{21}{3} = 7 \text{ km/hr} \]

\[ r = \frac{7 - 5}{2} = 1 \text{ km/hr} \]

Both trips take 3 hours, so the distances are proportional to the speeds. The 2 km/hr option is the raw difference, twice the current.

Answer: C

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

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

\[ b + r = 13 + 4 = 17 \text{ km/h} \]

\[ t = \frac{68}{17} = 4 \text{ hours} \]

Using the still-water speed alone would give \( \tfrac{68}{13} = 5.2 \) hours - the stream is pushing, so the journey must be quicker than that.

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

Correct answer: 3 km/hr

Write \(p = \tfrac{1}{u}\) and \(q = \tfrac{1}{d}\), which makes both conditions linear.

\[30p + 44q = 10, \qquad 40p + 55q = 13\]

Multiplying the first by 4 and the second by 3 and subtracting gives \(11q = 1\), so \(d = 11\) km/hr. Substituting back, \(30p = 6\), so \(u = 5\) km/hr.

\[s = \frac{11 - 5}{2} = 3 \text{ km/hr}\]

Answer: D

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

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

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

\[ b = \frac{18 + 10}{2} = 14 \text{ kmph} \]

\[ r = \frac{18 - 10}{2} = 4 \text{ kmph} \]

Every listed pair sums to 18 except one, so the discriminating check is the difference: only 14 and 4 give an upstream speed of 10 kmph.

Answer: D

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

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

The upstream trip takes 2 hours more, so 12 hours in all.

\[ b + r = \frac{24}{10} = 2.4 \text{ km/hr}, \qquad b - r = \frac{24}{12} = 2 \text{ km/hr} \]

\[ b = \frac{2.4 + 2}{2} = 2.2 \text{ km/hr} \]

2.2 km/hr is not offered, so the answer is none of these. The current works out at a modest 0.2 km/hr.

Answer: A

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

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

\[ \frac{30}{15 - r} + \frac{30}{15 + r} = 4.5 \]

Over the common denominator \( 225 - r^{2} \):

\[ \frac{30(15 + r) + 30(15 - r)}{225 - r^{2}} = \frac{900}{225 - r^{2}} = 4.5 \]

\[ 225 - r^{2} = 200 \quad\Longrightarrow\quad r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/hr} \]

The \( r \) terms cancel in the numerator, leaving a simple equation. Checking: 30 km at 10 km/hr takes 3 hours and at 20 km/hr takes 1½.

Answer: B

Correct answer: 1 km/hr

Equal times over 4 km downstream and 3 km upstream give \(\tfrac{4}{d} = \tfrac{3}{u}\), so \(u = \tfrac{3d}{4}\).

\[\frac{48}{d} + \frac{48}{u} = 14 \quad\Longrightarrow\quad \frac{48}{d} + \frac{64}{d} = 14 \quad\Longrightarrow\quad \frac{112}{d} = 14\]

So \(d = 8\) and \(u = 6\), giving \(s = \tfrac{8-6}{2} = 1\) km/hr.