Boats & Streams Questions and Answers

Answer: C

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

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

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: B

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

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

\[ b + r = \frac{20}{2} = 10 \text{ km/h}, \qquad b - r = \frac{20}{5} = 4 \text{ km/h} \]

\[ b = \frac{10 + 4}{2} = 7 \text{ km/h} \]

The data are entirely sufficient, so "data inadequate" is a trap: two journey times over a known distance always determine both speeds.

Answer: C

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

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

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

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

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

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: B

x = 7 km/hr ; y = 3 km/hr
Ds = 10 km/hr ; Us = 4 km/hr
Distance (d) is same. Therefore, if time taken for downstream is t hours, the time for upstream is (t + 6) hours.
10*t = 4*(t + 6)
6t = 24 ; t = 4 hours
d = 10*4 = 40 km

Answer: A

If x: speed of boats man in still water
y: speed of the river
Downstream speed (Ds) = x + y
Upstream speed (Us) = x - y
x = (Ds + Us) / 2
y = (Ds - Us) / 2
In the above problem Ds = 10 ; Us = 4
x = (10 + 4) / 2 = 14/2 = 7 km/hr
y = (10 - 4)/2 = 6/2 = 3 km/hr

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

Speed = distance/time

Speed downstream: 96 / 8 km/hr = 12kmph

Speed of stream = 4kmph

Effective speed of boat = (12-4) kmph
= 8kmph

Distance to be travelled upstream= 8 km

Speed upstream = boat speed-current speed

= 8-4 kmph

= 4 kmph

Time taken = distance/speed = 8/ 4 hours
=2 hours

Thus it will take 2 hours to go upstream.

Answer: C

Speed of the boat in still water: 15 km/hr

Speed of boat downstream: (15+x)km/hr where x is speed of the current.

The boat travels 30 km downstream and then 30 km upstream and takes 9/2 hours.

Total time= Time taken to travel downstream + Time taken to travel upstream

=> 4/5= (30/(15+x))+(30/(15-x))

=>x= 5

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: C

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

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

\[ b + r = 13 + 4 = 17 \text{ km/h} \]

\[ t = \frac{68}{17} = 4 \text{ hours} \]

Using the still-water speed alone would give \( \tfrac{68}{13} = 5.2 \) hours - the stream is pushing, so the journey must be quicker than that.