Boats & Streams Questions and Answers

Answer: A

The time taken is 4 hours.

The speed in still water is the average of the upstream and downstream speeds. With the ratio 3 : 5, write them as 3k and 5k:

\[ \begin{aligned} \frac{3k + 5k}{2} &= 24,\ k = 6 \\ \text{up} &= 18,\ \text{down} = 30\ \text{km/h} \\ t &= \frac{36}{18} + \frac{60}{30} = 2 + 2 = 4 \end{aligned} \]

5 hours and 3 hours do not follow from these speeds, and since 4 hours is offered, 'None of these' is wrong.

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.

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.