Boats & Streams Questions and Answers

Answer: C

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

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

"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: C

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

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

\[ 8 \text{ h } 48 \text{ min} = 8.8 \text{ h} \]

Speeds are inversely proportional to times over the same distance:

\[ \frac{b + r}{b - r} = \frac{8.8}{4} = \frac{11}{5} \]

\[ 5(b + r) = 11(b - r) \quad\Longrightarrow\quad 16r = 6b \]

\[ \frac{b}{r} = \frac{8}{3} \quad\Longrightarrow\quad 8 : 3 \]

The distance cancels, so "cannot be determined" is wrong - a ratio of times is enough to fix a ratio of speeds.

Answer: B

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

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

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

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

Combining the fractions over the common denominator \( (15+r)(15-r) = 225 - r^{2} \):

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

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

The \( r \) terms in the numerator cancel, which is what turns a messy equation into a simple one.

Answer: C

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

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

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

The stream is \( \tfrac{11-5}{2} = 3 \) km/h. The still-water speed is always the average of the two given speeds, so it must fall between 5 and 11 - which alone rules out two of the options.

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: 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: A

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

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

\[ 90 \text{ min} = 1.5 \text{ h} \]

\[ \frac{36}{10 - r} - \frac{36}{10 + r} = 1.5 \]

\[ \frac{36(10 + r) - 36(10 - r)}{100 - r^{2}} = \frac{72r}{100 - r^{2}} = 1.5 \]

\[ 72r = 150 - 1.5r^{2} \quad\Longrightarrow\quad r^{2} + 48r - 100 = 0 \]

\[ r = \frac{-48 + \sqrt{2304 + 400}}{2} = \frac{-48 + 52}{2} = 2 \text{ mph} \]

The negative root is discarded as a speed. Checking: 36 miles at 8 mph takes 4.5 hours and at 12 mph takes 3 - exactly 90 minutes apart.

Answer: A

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

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

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

\[ \frac{D}{6} + \frac{D}{4} = 1 \]

\[ D\left(\frac{2 + 3}{12}\right) = 1 \quad\Longrightarrow\quad \frac{5D}{12} = 1 \]

\[ D = 2.4 \text{ km} \]

Using 5 km/h both ways would give 2.5 km. The current always costs time overall, so the true distance covered in the hour is a little less.

Answer: D

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

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

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

\[ \frac{b + 3}{b - 3} = \frac{1.5}{1} = \frac{3}{2} \]

\[ 2(b + 3) = 3(b - 3) \quad\Longrightarrow\quad 2b + 6 = 3b - 9 \]

\[ b = 15 \text{ kmph} \]

Checking: 18 kmph for 1 hour and 12 kmph for 1½ hours both cover 18 km.

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.