Boats & Streams Questions and Answers

Answer: C

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

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

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

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

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

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

"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.

Answer: C

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

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

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

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

The current runs at \( \tfrac{11-5}{2} = 3 \) km/hr. Since the still-water speed is the average, it must fall between 5 and 11 - which eliminates 6 as too close to the slow end without any calculation.

Answer: D

Speed downstream = (9+3) km/hr = 12km/hr 

Speed upstream = (9-3) km/hr = 6 km/hr 

Distance AB = x km 

 x/6 + x/12 = 3 

=> 2x + x = 36 

=> x= 12

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 mean the two distances are in the same ratio as the two speeds.

\[ \frac{26}{b + 3} = \frac{14}{b - 3} \]

\[ 26(b - 3) = 14(b + 3) \quad\Longrightarrow\quad 26b - 78 = 14b + 42 \]

\[ 12b = 120 \quad\Longrightarrow\quad b = 10 \text{ km/h} \]

Checking: 26 km at 13 km/h and 14 km at 7 km/h both take exactly 2 hours.