Boats & Streams Questions and Answers

Answer: C

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

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

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

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: A

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

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

\[ 3 \text{ h } 45 \text{ min} = 3.75 \text{ h}, \qquad 2 \text{ h } 30 \text{ min} = 2.5 \text{ h} \]

\[ b + r = \frac{15}{3.75} = 4 \text{ km/hr}, \qquad b - r = \frac{5}{2.5} = 2 \text{ km/hr} \]

\[ r = \frac{4 - 2}{2} = 1 \text{ km/hr} \]

The boat itself manages 3 km/hr. Converting the mixed times to decimal hours before dividing keeps the arithmetic clean.

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: A

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

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

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

\[ b = \frac{80 + 60}{2} = 70 \text{ kmph} \]

\[ r = \frac{80 - 60}{2} = 10 \text{ kmph} \]

Checking: \( 70 + 10 = 80 \) and \( 70 - 10 = 60 \). The still-water speed must lie between the two given speeds, which alone eliminates every other option.

Answer: C

Correct answer: 5 hours

Let the speeds be \(4k\) and \(k\), so downstream is \(5k\) and upstream \(3k\).

\[5k = \frac{30}{3} = 10 \quad\Longrightarrow\quad k = 2\]

The upstream speed is \(3k = 6\) km/hr, so

\[\text{Time} = \frac{30}{6} = 5 \text{ hours}\]

Answer: C

Correct answer: 12 hours

\[\text{Upstream speed} = 3 - 2 = 1 \text{ km/hr}, \qquad \text{Downstream speed} = 3 + 2 = 5 \text{ km/hr}\]

\[\text{Total time} = \frac{10}{1} + \frac{10}{5} = 10 + 2 = 12 \text{ hours}\]

Almost the whole journey is spent on the upstream leg, because the swimmer barely outpaces the current.

Answer: A

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

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

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

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

\[ b = 3r \quad\Longrightarrow\quad b : r = 3 : 1 \]

The boat must be faster than the stream, so the first term has to be the larger - which alone rules out the reversed 1 : 3.

Answer: D

Correct answer: 8 km

Downstream he moves at 8 km/hr and upstream at 4 km/hr. Letting the distance be \(x\):

\[\frac{x}{8} + \frac{x}{4} = 3 \quad\Longrightarrow\quad \frac{x + 2x}{8} = 3 \quad\Longrightarrow\quad 3x = 24\]

So \(x = 8\) km, taking 1 hour out and 2 hours back.

Answer: D

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

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

\[ b + r = 10 \text{ kmph}, \qquad b - r = 8 \text{ kmph} \]

\[ \frac{D}{10} + \frac{D}{8} = 4.5 \]

\[ D\left(\frac{4 + 5}{40}\right) = 4.5 \quad\Longrightarrow\quad \frac{9D}{40} = 4.5 \]

\[ D = 20 \text{ km} \]

20 km appears nowhere among the choices, so the answer is none of these. Checking: 2 hours out and 2½ back, which is the stated 4½ hours.

Answer: C

Speed upstream = 2km/hr.

Speed downstream = 6 km/hr 

Speed in stationary water = ½ (6 + 2) km/hr = 4 km/hr 

Time is taken to cover 5 km in stationary water 

= (1/4 x 5) 1 hr 15 min