Boats & Streams Questions and Answers

Answer: C

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:

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad 42 + r = 2(42 - r) = 84 - 2r \]

\[ 3r = 42 \quad\Longrightarrow\quad r = 14 \text{ kmph} \]

Checking: 56 kmph downstream and 28 kmph upstream - exactly double. The rule \( r = \tfrac{b}{3} \) holds for any "twice as long" problem.

Answer: D

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

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

Convert each time to hours and divide into the three-quarter kilometre.

\[ b - r = \frac{0.75}{11.25/60} = \frac{0.75 \times 60}{11.25} = 4 \text{ km/hr} \]

\[ b + r = \frac{0.75}{7.5/60} = \frac{0.75 \times 60}{7.5} = 6 \text{ km/hr} \]

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

The current runs at 1 km/hr. Both times must be converted before averaging - averaging the times instead of the speeds gives a different and wrong figure.

Answer: D

Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.

Now 30/(a + b) + 20/(a - b) = 4 and 45/(a + b) + 40/(a - b) = 7

Solving the equation, the speed of man in still water is 12.5 kmph.

Answer: C

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

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

\[ b - r = \frac{2}{1} = 2 \text{ km/h} \]

\[ b + r = \frac{1}{10/60} = 6 \text{ km/h} \]

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

\[ t = \frac{5}{4} = 1.25 \text{ h} = 1 \text{ hr } 15 \text{ min} \]

"Stationary water" means no current, so the still-water speed is what applies - the 10 minutes must be converted to hours before it can be used as a speed.

Answer: D

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

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

\[ 4 \text{ h } 30 \text{ min} = 4.5 \text{ h} \]

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

\[ \frac{900}{225 - r^{2}} = 4.5 \quad\Longrightarrow\quad 225 - r^{2} = 200 \]

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

The \( r \) terms cancel when the two fractions are combined, which is what makes the quadratic so simple. Checking: 1½ hours down at 20 km/hr and 3 hours back at 10.

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: 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: D

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

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

\[ b + r = 15 + 3 = 18 \text{ km/h} \]

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

\[ \text{distance} = 18 \times 0.2 = 3.6 \text{ km} \]

Using 15 km/h gives 3 km and using 3 km/h gives 0.6 km; the downstream speed is the sum, so the answer must exceed 3 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: B

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

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

\[ b + r = \frac{16}{2} = 8 \text{ km/h}, \qquad b - r = \frac{16}{4} = 4 \text{ km/h} \]

\[ b = \frac{8 + 4}{2} = 6 \text{ km/h} \]

Two times over a known distance always determine both speeds, so "data inadequate" is never right in this configuration.