Boats & Streams Questions and Answers

Answer: B

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

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

Substituting \( x = \dfrac{1}{b-r} \) and \( y = \dfrac{1}{b+r} \) makes both conditions linear:

\[ 30x + 44y = 10, \qquad 40x + 55y = 13 \]

Multiply the first by 4 and the second by 3:

\[ 120x + 176y = 40, \qquad 120x + 165y = 39 \]

\[ 11y = 1 \quad\Longrightarrow\quad y = \frac{1}{11} \quad\Longrightarrow\quad b + r = 11 \]

\[ 30x + 4 = 10 \quad\Longrightarrow\quad x = \frac{1}{5} \quad\Longrightarrow\quad b - r = 5 \]

\[ b = \frac{11 + 5}{2} = 8 \text{ km/h} \]

The reciprocal substitution is the standard move for these paired-journey problems: without it the equations are non-linear and far harder to solve.

Answer: A

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

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

Adding the two eliminates \( r \):

\[ (b + r) + (b - r) = 34 + 22 = 56 \quad\Longrightarrow\quad 2b = 56 \]

\[ b = 28 \text{ km/h} \]

The still-water speed is always the average of the two, and the stream's speed is half their difference: \( r = \tfrac{34-22}{2} = 6 \) km/h. The 56 on offer is the sum, one step short.

Answer: A

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

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

\[ b + r = 16 + 5 = 21 \text{ km/hr} \]

\[ t = \frac{84}{21} = 4 \text{ hours} \]

Using the still-water speed of 16 km/hr would give 5.25 hours - the stream is helping, so the trip must take less than that.

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

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

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

\[ r = b - (b - r) = 5 - 3.5 = 1.5 \text{ kmph} \]

\[ b + r = 5 + 1.5 = 6.5 \text{ kmph} \]

The still-water speed sits exactly midway between the two, so the downstream speed is as far above 5 as the upstream speed is below it - 1.5 kmph either way.

Answer: A

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

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

Equal times over 4 km downstream and 3 km upstream give the speed ratio:

\[ \frac{4}{b + r} = \frac{3}{b - r} \quad\Longrightarrow\quad 4b - 4r = 3b + 3r \]

\[ b = 7r \quad\Longrightarrow\quad b + r = 8r, \qquad b - r = 6r \]

Now use the 48 km round trip in 14 hours:

\[ \frac{48}{8r} + \frac{48}{6r} = 14 \quad\Longrightarrow\quad \frac{6}{r} + \frac{8}{r} = 14 \]

\[ \frac{14}{r} = 14 \quad\Longrightarrow\quad r = 1 \text{ km/h} \]

So the boat does 7 km/h in still water: 8 km/h downstream for 6 hours and 6 km/h upstream for 8 hours.

Answer: A

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

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

Put \( b = 7k \) and \( r = k \), so

\[ b - r = 6k, \qquad b + r = 8k \]

From the upstream trip, with 14 min = \( \tfrac{7}{30} \) h:

\[ 6k = \frac{4.2}{7/30} = 18 \quad\Longrightarrow\quad k = 3 \]

\[ b + r = 8 \times 3 = 24 \text{ km/h} \]

\[ t = \frac{18.4}{24} = \frac{23}{30} \text{ h} = 46 \text{ minutes} \]

Setting the ratio up with a single unknown \( k \) is what turns two unknowns into one.

Answer: A

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

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

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

\[ \frac{D}{6} + \frac{D}{4} = 1 \]

\[ D\left(\frac{2 + 3}{12}\right) = 1 \quad\Longrightarrow\quad \frac{5D}{12} = 1 \]

\[ D = 2.4 \text{ km} \]

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: 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

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

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

\[ r = b - (b - r) = 2 - 1 = 1 \text{ km/hr} \]

The current slows the boat from 2 km/hr to 1 km/hr, so it must itself be 1 km/hr. Downstream the boat would manage \( 2 + 1 = 3 \) km/hr, which is why 3 appears as a distractor.