Boats & Streams Questions and Answers

Answer: A

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

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

\[ 3 \text{ h } 45 \text{ min} = 3.75 \text{ h}, \qquad 2 \text{ h } 30 \text{ min} = 2.5 \text{ h} \]

\[ b + r = \frac{15}{3.75} = 4 \text{ km/hr}, \qquad b - r = \frac{5}{2.5} = 2 \text{ km/hr} \]

\[ r = \frac{4 - 2}{2} = 1 \text{ km/hr} \]

The boat itself manages 3 km/hr. Converting the mixed times to decimal hours before dividing keeps the arithmetic clean.

Answer: C

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

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

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

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

The current runs at \( \tfrac{11-5}{2} = 3 \) km/hr. Since the still-water speed is the average, it must fall between 5 and 11 - which eliminates 6 as too close to the slow end without any calculation.

Answer: D

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

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

\[ 4 \text{ h } 30 \text{ min} = 4.5 \text{ h} \]

\[ \frac{30}{15 + r} + \frac{30}{15 - r} = 4.5 \]

\[ \frac{900}{225 - r^{2}} = 4.5 \quad\Longrightarrow\quad 225 - r^{2} = 200 \]

\[ r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/hr} \]

The \( r \) terms cancel when the two fractions are combined, which is what makes the quadratic so simple. Checking: 1½ hours down at 20 km/hr and 3 hours back at 10.

Answer: D

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

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

\[ b - r = \frac{15}{3} = 5 \text{ km/hr}, \qquad b + r = \frac{21}{3} = 7 \text{ km/hr} \]

\[ r = \frac{7 - 5}{2} = 1 \text{ km/hr} \]

Both trips take 3 hours, so the distances are proportional to the speeds. The 2 km/hr option is the raw difference, twice the current.

Answer: B

Rate upstream = (750/675) = 10/9 m/sec

Rate downstream (750/450) m/sec = 5/3 m/sec

Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.

= 25/18 m/sec

= (25/18) × (18/5) kmph

= 5 kmph

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

Correct answer: 3 km/hr

Downstream speed is \(b+s\) and upstream is \(b-s\), so their difference is twice the stream speed:

\[s = \frac{18 - 12}{2} = 3 \text{ km/hr}\]

Their sum gives the boat’s own speed, \(\tfrac{18+12}{2} = 15\) km/hr.

Answer: A

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

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

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

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

\[ b = 3r \quad\Longrightarrow\quad b : r = 3 : 1 \]

The boat must be faster than the stream, so the first term has to be the larger - which alone rules out the reversed 1 : 3.

Answer: C

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

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

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

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

\[ \text{distance} = 18 \times 0.2 = 3.6 \text{ km} \]

Converting the minutes to hours is the only real step; using 15 km/hr instead of the downstream 18 would give 3 km.

Answer: A

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

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

Convert each time to hours and divide into the 3 km.

\[ b - r = \frac{3}{20/60} = 9 \text{ km/hr} \]

\[ b + r = \frac{3}{18/60} = 10 \text{ km/hr} \]

\[ r = \frac{10 - 9}{2} = \frac{1}{2} \text{ km/hr} \]

The current's speed is half the difference of the two, not the difference itself. Sahil rows at 9.5 km/hr in still water.