Boats & Streams Questions and Answers

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.

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

Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.

Now 30/(a + b) + 20/(a - b) = 4 and 45/(a + b) + 40/(a - b) = 7

Solving the equation, the speed of man in still water is 12.5 kmph.

Answer: D

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

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

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

\[ b = \frac{18 + 10}{2} = 14 \text{ kmph} \]

\[ r = \frac{18 - 10}{2} = 4 \text{ kmph} \]

Every listed pair sums to 18 except one, so the discriminating check is the difference: only 14 and 4 give an upstream speed of 10 kmph.

Answer: D

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

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

\[ b + r = 10 \text{ kmph}, \qquad b - r = 8 \text{ kmph} \]

\[ \frac{D}{10} + \frac{D}{8} = 4.5 \]

\[ D\left(\frac{4 + 5}{40}\right) = 4.5 \quad\Longrightarrow\quad \frac{9D}{40} = 4.5 \]

\[ D = 20 \text{ km} \]

20 km appears nowhere among the choices, so the answer is none of these. Checking: 2 hours out and 2½ back, which is the stated 4½ hours.

Answer: D

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

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

\[ b + r = 15 + 3 = 18 \text{ km/hr} \]

The distance is in metres and the answer in seconds, so convert the speed:

\[ 18 \text{ km/hr} = 18 \times \frac{5}{18} = 5 \text{ m/s} \]

\[ t = \frac{60}{5} = 12 \text{ seconds} \]

The factor \( \tfrac{5}{18} \) converts km/hr to m/s; skipping it would leave the units mismatched and the answer meaningless.

Answer: A

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

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

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

\[ r = \frac{15.5 - 8.5}{2} = \frac{7}{2} = 3.5 \text{ km/hr} \]

The boat does 12 km/hr in still water. The 7 km/hr option is the raw difference, which is twice the current.

Answer: D

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

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

\[ b + r = \frac{24}{4} = 6 \text{ km/hr}, \qquad b - r = \frac{24}{6} = 4 \text{ km/hr} \]

\[ b = \frac{6 + 4}{2} = 5 \text{ km/hr} \]

5 km/hr is not among the listed values, so the answer is none of these. The nearest option, 5.5, is the sort of near-miss these questions use to punish rounding.

Answer: D

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

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

The upstream trip takes 2 hours more, so 12 hours in all.

\[ b + r = \frac{24}{10} = 2.4 \text{ km/hr}, \qquad b - r = \frac{24}{12} = 2 \text{ km/hr} \]

\[ b = \frac{2.4 + 2}{2} = 2.2 \text{ km/hr} \]

2.2 km/hr is not offered, so the answer is none of these. The current works out at a modest 0.2 km/hr.

Answer: D

Let the distance between the two parts be x km. 

Then speed downstream = x/4 km/hr. 

Speed Upstream = x/5 km/hr 

Speed of the stream = ½ (x/4 -x/5) 

Therefore, ½(x/4 -x/5) = 2. => x/4 -x/5 = 4 => x = 80. 

Hence, the distance between the two ports is 80km.