Boats & Streams Questions and Answers

Answer: C

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

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

\[ b - r = \frac{48}{12} = 4 \text{ km/h}, \qquad b + r = \frac{48}{8} = 6 \text{ km/h} \]

\[ b = \frac{6 + 4}{2} = 5 \text{ km/h}, \qquad r = \frac{6 - 4}{2} = 1 \text{ km/h} \]

Half the sum gives the boat and half the difference gives the river - the standard pair of formulas, and the reason the longer time always belongs to the upstream leg.

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: C

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

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

\[ 1 \text{ h } 20 \text{ min} = \frac{4}{3} \text{ h} \]

\[ b + r = \frac{80}{4/3} = 60 \text{ km/h} \]

\[ r = 60 - 45 = 15 \text{ km/h} \quad\Longrightarrow\quad b - r = 45 - 15 = 30 \text{ km/h} \]

\[ t = \frac{80}{30} = \frac{8}{3} \text{ h} = 2 \text{ h } 40 \text{ min} \]

The upstream speed is half the downstream speed here, so the return leg takes exactly twice as long.

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: D

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

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

\[ b + r = \frac{20}{1} = 20 \text{ km/h}, \qquad b - r = \frac{20}{2} = 10 \text{ km/h} \]

\[ b = \frac{20 + 10}{2} = 15 \text{ km/h} \]

\[ r = \frac{20 - 10}{2} = 5 \text{ km/h} \]

Every option lists a pair summing to 20, so the discriminator is the difference: only 15 and 5 give an upstream speed of 10 km/h.

Answer: A

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

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

\[ b + r = 12 \text{ km/h}, \qquad b - r = 8 \text{ km/h} \]

The upstream trip takes 4 hours longer over the same distance \( D \):

\[ \frac{D}{8} - \frac{D}{12} = 4 \]

\[ D\left(\frac{3 - 2}{24}\right) = 4 \quad\Longrightarrow\quad \frac{D}{24} = 4 \]

\[ D = 96 \text{ km} \]

Checking: 12 hours upstream against 8 hours downstream - a difference of exactly 4.

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: A

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

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

The distance is the same both ways, so the speeds are inversely proportional to the times:

\[ \frac{b + 6}{b - 6} = \frac{18}{12} = \frac{3}{2} \]

\[ 2(b + 6) = 3(b - 6) \quad\Longrightarrow\quad 2b + 12 = 3b - 18 \]

\[ b = 30 \text{ km/h} \]

Checking: downstream 36 km/h for 12 h and upstream 24 km/h for 18 h both cover 432 km.

Answer: D

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

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

\[ b + r = 14.4 \text{ km/h}, \qquad b - r = 9.6 \text{ km/h} \]

The round trip to a place \( D \) away takes 1 hour:

\[ \frac{D}{14.4} + \frac{D}{9.6} = 1 \]

\[ D\left(\frac{1}{14.4} + \frac{1}{9.6}\right) = D\left(\frac{2 + 3}{28.8}\right) = 1 \]

\[ \frac{5D}{28.8} = 1 \quad\Longrightarrow\quad D = 5.76 \text{ km} \]

The 4.8 km on offer is what you get by using the still-water speed of 12 km/h for both legs - but the current does not cancel out over a round trip.

Answer: D

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

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

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

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad 12 + r = 2(12 - r) \]

\[ 12 + r = 24 - 2r \quad\Longrightarrow\quad 3r = 12 \quad\Longrightarrow\quad r = 4 \text{ km/h} \]

Checking: 16 km/h downstream and 8 km/h upstream. In general, when the upstream time is \( n \) times the downstream time, \( r = b \times \tfrac{n-1}{n+1} \) - here \( 12 \times \tfrac{1}{3} = 4 \).