Boats & Streams Questions and Answers

Answer: B

Correct answer: 1 km/hr

Equal times over 4 km downstream and 3 km upstream give \(\tfrac{4}{d} = \tfrac{3}{u}\), so \(u = \tfrac{3d}{4}\).

\[\frac{48}{d} + \frac{48}{u} = 14 \quad\Longrightarrow\quad \frac{48}{d} + \frac{64}{d} = 14 \quad\Longrightarrow\quad \frac{112}{d} = 14\]

So \(d = 8\) and \(u = 6\), giving \(s = \tfrac{8-6}{2} = 1\) km/hr.

Answer: B

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

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

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

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

\[ 3r = b \quad\Longrightarrow\quad \frac{b}{r} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Checking with \( b = 3 \) and \( r = 1 \): downstream 4 and upstream 2, so the upstream leg does take twice as long. Neither distance nor actual speed is needed - only the ratio.

Answer: C

Correct answer: 6 hours

Against the stream the speeds subtract:

\[\text{Upstream speed} = 15 - 5 = 10 \text{ km/hr}\]

\[\text{Time} = \frac{60}{10} = 6 \text{ hours}\]

The same 60 km downstream, at 20 km/hr, would take only 3 hours.

Answer: B

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

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

\[ 42 \text{ min} = \frac{42}{60} = 0.7 \text{ h} \]

\[ b - r = \frac{7}{0.7} = 10 \text{ kmph} \]

\[ b = (b - r) + r = 10 + 3 = 13 \text{ kmph} \]

The current is added back once, not twice: subtracting it gave the upstream speed, so restoring it gives the still-water speed. Adding it a second time would give the downstream speed of 16 kmph.

Answer: B

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

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

\[ b + r = \frac{72}{9} = 8 \text{ kmph}, \qquad b - r = \frac{45}{9} = 5 \text{ kmph} \]

\[ r = \frac{8 - 5}{2} = 1.5 \text{ kmph} \]

The 3 kmph option is the full difference; halving it is the step that separates the current from the gap between the two speeds.

Answer: B

Let the speed of the boat in still water be 4 km/hr and speed of the current be u km/hr. 

Rate downstream = (u + v) km/hr. 

Rate upstream = (u- v)km/hr 

Let the distance covered in each case be x km. 

Then 2x/(u + v) = x / (u - v) 

=> 2 (u - v) = ( u + v)

=> u =3v 

=> u/v =3/1

Answer: B

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

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

Let the distance \( PQ = QR = d \), with Q and R downstream of P.

P to R is \( 2d \) downstream in 4 hours:

\[ \frac{2d}{b + r} = 4 \quad\Longrightarrow\quad \frac{d}{b + r} = 2 \]

The round trip P–Q–P takes 10 hours:

\[ \frac{d}{b + r} + \frac{d}{b - r} = 10 \quad\Longrightarrow\quad 2 + \frac{d}{b - r} = 10 \]

\[ \frac{d}{b - r} = 8 \]

Dividing the two results eliminates \( d \):

\[ \frac{b + r}{b - r} = \frac{8}{2} = 4 \quad\Longrightarrow\quad b + r = 4b - 4r \]

\[ 5r = 3b \quad\Longrightarrow\quad \frac{b}{r} = \frac{5}{3} \]

Answer: D

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

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

\[ b + r = \frac{20}{1} = 20 \text{ km/h}, \qquad b - r = \frac{20}{2} = 10 \text{ km/h} \]

\[ b = \frac{20 + 10}{2} = 15 \text{ km/h} \]

\[ r = \frac{20 - 10}{2} = 5 \text{ km/h} \]

Every option lists a pair summing to 20, so the discriminator is the difference: only 15 and 5 give an upstream speed of 10 km/h.

Answer: D

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

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

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

\[ \frac{b + r}{b - r} = 2 \quad\Longrightarrow\quad 12 + r = 2(12 - r) \]

\[ 12 + r = 24 - 2r \quad\Longrightarrow\quad 3r = 12 \quad\Longrightarrow\quad r = 4 \text{ km/h} \]

Checking: 16 km/h downstream and 8 km/h upstream. In general, when the upstream time is \( n \) times the downstream time, \( r = b \times \tfrac{n-1}{n+1} \) - here \( 12 \times \tfrac{1}{3} = 4 \).

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