Boats & Streams Questions and Answers

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

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

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

\[ 1 \text{ h } 20 \text{ min} = \frac{4}{3} \text{ h} \]

\[ b + r = \frac{80}{4/3} = 60 \text{ km/h} \]

\[ r = 60 - 45 = 15 \text{ km/h} \quad\Longrightarrow\quad b - r = 45 - 15 = 30 \text{ km/h} \]

\[ t = \frac{80}{30} = \frac{8}{3} \text{ h} = 2 \text{ h } 40 \text{ min} \]

The upstream speed is half the downstream speed here, so the return leg takes exactly twice as long.

Answer: A

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

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

Swimming back against the current is the upstream leg:

\[ b - r = 4 - 2 = 2 \text{ km/h} \]

\[ t = \frac{6}{2} = 3 \text{ hours} \]

All the data needed is present, so "insufficient data" is a trap. Using the still-water speed of 4 km/h would give 1.5 hours - too optimistic, since the current is working against him.

Answer: A

A boat travels at the speed of 14 km/hr in still water.

=> Speed of stream = 1.2 kmph

=> Speed of boat downstream becomes 14 + 1.2 = 15.2 kmph

=> Speed of boat upstream becomes 14 - 1.2 = 12.8 kmph

Total time taken in the journey = Journey downstream + Journey upstream

Therefore, 4864(1/12.8 + 1/15.2) = 700 hours

Answer: A

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

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

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

\[ \frac{b + 6}{b - 6} = \frac{18}{12} = \frac{3}{2} \]

\[ 2(b + 6) = 3(b - 6) \quad\Longrightarrow\quad 2b + 12 = 3b - 18 \]

\[ b = 30 \text{ km/h} \]

Checking: downstream 36 km/h for 12 h and upstream 24 km/h for 18 h both cover 432 km.

Answer: B

Correct answer: 3 km/hr

Downstream speed is \(b+s\) and upstream is \(b-s\), so their difference is twice the stream speed:

\[s = \frac{18 - 12}{2} = 3 \text{ km/hr}\]

Their sum gives the boat’s own speed, \(\tfrac{18+12}{2} = 15\) km/hr.

Answer: C

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

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

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

\[ b = \frac{35 + 25}{2} = 30 \text{ kmph} \]

The still-water speed is the average of the two, so it must lie between 25 and 35 - ruling out 60 and 10 on sight. The stream itself runs at \( \tfrac{35-25}{2} = 5 \) kmph.

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: C

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

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

Equal times mean the distances are in proportion to the speeds:

\[ \frac{26}{20 + r} = \frac{14}{20 - r} \]

\[ 26(20 - r) = 14(20 + r) \quad\Longrightarrow\quad 520 - 26r = 280 + 14r \]

\[ 240 = 40r \quad\Longrightarrow\quad r = 6 \text{ km/h} \]

Checking: 26 km at 26 km/h and 14 km at 14 km/h both take exactly one hour.

Answer: D

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

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

\[ b + r = 20 + 6 = 26 \text{ kmph} \]

\[ t = \frac{60}{26} = \frac{30}{13} \text{ hours} \]

That is about 2 hours 18 minutes. Using the still-water speed would give 3 hours - the current is helping, so the trip must be faster than that.