Boats & Streams Questions and Answers

Answer: A

The time taken is 4 hours.

The speed in still water is the average of the upstream and downstream speeds. With the ratio 3 : 5, write them as 3k and 5k:

\[ \begin{aligned} \frac{3k + 5k}{2} &= 24,\ k = 6 \\ \text{up} &= 18,\ \text{down} = 30\ \text{km/h} \\ t &= \frac{36}{18} + \frac{60}{30} = 2 + 2 = 4 \end{aligned} \]

5 hours and 3 hours do not follow from these speeds, and since 4 hours is offered, 'None of these' is wrong.

Answer: D

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

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

\[ b + r = 30 + 6 = 36 \text{ km/h} \]

\[ 24 \text{ min} = \frac{24}{60} = 0.4 \text{ h} \]

\[ \text{distance} = 36 \times 0.4 = 14.4 \text{ km} \]

Converting the minutes to hours before multiplying is the whole trick; leaving 24 in place would give 864 km - which is why 864 m appears as a distractor.

Answer: C

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

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

For a fixed distance, time is inversely proportional to speed, so the time ratio inverts to give the speed ratio:

\[ \frac{t_{\text{up}}}{t_{\text{down}}} = \frac{11}{5} \quad\Longrightarrow\quad \frac{b + r}{b - r} = \frac{11}{5} \]

\[ 5(b + r) = 11(b - r) \quad\Longrightarrow\quad 5b + 5r = 11b - 11r \]

\[ 16r = 6b \quad\Longrightarrow\quad \frac{b}{r} = \frac{16}{6} = \frac{8}{3} \]

So the ratio is \( 8 : 3 \). The 10 km cancels out, so the distance is never needed.

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: B

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

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

Time is inversely proportional to speed over the same distance, so taking twice as long upstream means going half as fast:

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

\[ 6 + r = 2(6 - r) = 12 - 2r \]

\[ 3r = 6 \quad\Longrightarrow\quad r = 2 \text{ km/h} \]

Checking: downstream 8 km/h and upstream 4 km/h - exactly double.

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

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

Equal times mean the two distances are in the same ratio as the two speeds.

\[ \frac{26}{b + 3} = \frac{14}{b - 3} \]

\[ 26(b - 3) = 14(b + 3) \quad\Longrightarrow\quad 26b - 78 = 14b + 42 \]

\[ 12b = 120 \quad\Longrightarrow\quad b = 10 \text{ km/h} \]

Checking: 26 km at 13 km/h and 14 km at 7 km/h both take exactly 2 hours.

Answer: A

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

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

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: B

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

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

The upstream trip already gives one speed outright:

\[ b - r = \frac{48}{6} = 8 \text{ km/h} \]

With statement A (\( b = 10 \)): \( r = 2 \), so \( b + r = 12 \) and the downstream time is \( \tfrac{48}{12} = 4 \) hours.

With statement B (\( r = 2 \)): \( b = 10 \), giving the same \( b + r = 12 \) and the same 4 hours.

Each statement alone completes the pair, so either one is sufficient. Once \( b - r \) is known, any single further fact about \( b \) or \( r \) determines the other - which is exactly the situation this data-sufficiency format is testing.

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 \( d \):

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

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