Boats & Streams Questions and Answers

Answer: D

The distance is the same both ways, so the speeds are inversely proportional to the times. Upstream takes 10 minutes and downstream 5 minutes, so the downstream speed is twice the upstream speed. With the boat's speed b and the stream's speed 2 m/s, the equation is as shown.

\[ b + 2 = 2(b - 2) \]

\[ b = 6 \]

The speed of the boat in still water is 6 m/s. Check: upstream 4 m/s and downstream 8 m/s, which is double.

Answer: B

Let the speed of the boat in still water be v km/hr. Going down the stream the speed is v + 4 and coming back it is v - 4, and the total time is 40 hours.

\[ \frac{120}{v+4} + \frac{120}{v-4} = 40 \]

\[ \frac{240v}{v^{2}-16} = 40 \Rightarrow v^{2} - 6v - 16 = 0 \]

This gives v = 8, so the speed in still water is 8 km/hr, option (B). Speeds of 15 and 10 would take much less than 40 hours. Check: 120 / 12 = 10 hours downstream and 120 / 4 = 30 hours upstream, and 10 + 30 = 40 hours, which agrees with the given time.

Answer: C

The boat takes 4 hours to go 28 km downstream. Equal times for 21 km downstream and 9 km upstream mean the speeds are in the ratio 21 : 9 = 7 : 3, so downstream speed is \(7k\) and upstream speed is \(3k\). \[ \begin{aligned} \frac{12}{7k} + \frac{12}{3k} &= 5\frac{5}{7} \\ &= \frac{40}{7} \end{aligned} \] \[ \begin{aligned} \frac{12 + 28}{7k} &= \frac{40}{7k} \\ &= \frac{40}{7} \Rightarrow k \\ &= 1 \end{aligned} \] Downstream speed is 7 km/h, so \[ \frac{28}{7} = 4\text{ hours} \] The other options do not match this speed.

Answer: B

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

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

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

\[ b = \frac{6 + 4}{2} = 5 \text{ kmph} \]

The stream's own rate is \( \tfrac{6-4}{2} = 1 \) kmph, which is offered as a distractor. The man's rate must lie between his two observed speeds, so it has to be 5.

Answer: C

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

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

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

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

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

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

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

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

\[ b + r = 10.5 \text{ kmph}, \qquad b - r = 7.5 \text{ kmph} \]

\[ t = \frac{105}{10.5} + \frac{105}{7.5} = 10 + 14 = 24 \text{ hours} \]

Using the still-water speed for both legs would give \( \tfrac{210}{9} = 23.3 \) hours - less than the truth, because the extra time lost upstream outweighs the time saved downstream.

Answer: C

Please remember,

If a is rate downstream and b is rate upstream

Rate in still water = 1/2(a+b)

Rate of current = 1/2(a-b)

=> Rate in still water = 1/2(20+10) = 15 kmph

=> Rate of current = 1/2(20-10) = 5 kmph

Answer: D

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 over the same distance:

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad 4.5 + r = 2(4.5 - r) = 9 - 2r \]

\[ 3r = 4.5 \quad\Longrightarrow\quad r = 1.5 \text{ km/h} \]

Checking: 6 km/h downstream and 3 km/h upstream. In general \( r = \tfrac{b}{3} \) whenever the upstream leg takes twice as long.

Answer: C

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

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

\[ b + r = 20 + 5 = 25 \text{ km/h} \]

\[ 12 \text{ min} = \frac{12}{60} = 0.2 \text{ h} \]

\[ \text{distance} = 25 \times 0.2 = 5 \text{ km} \]

Using the still-water speed of 20 km/h instead would give 4 km - the current is helping, so the distance must exceed that.

Answer: C

Convert both quantities to the units the answer is wanted in.

\[ 11.25 \text{ min} = \frac{11.25}{60} = 0.1875 \text{ h} \]

\[ \text{speed} = \frac{0.75}{0.1875} = 4 \text{ km/h} \]

Equivalently, three quarters of a kilometre in \( \tfrac{3}{16} \) of an hour means \( \tfrac{3}{4} \times \tfrac{16}{3} = 4 \) km/h. No current is mentioned, so this speed is the still-water speed.