Boats & Streams Questions and Answers

Answer: D

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

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

\[ b + r = \frac{30}{2.5} = 12 \text{ km/hr}, \qquad b - r = \frac{30}{3.75} = 8 \text{ km/hr} \]

\[ b = \frac{12 + 8}{2} = 10 \text{ km/hr} \]

10 km/hr is absent from the options, so the answer is none of these. The 12 and 8 on the list are the downstream and upstream speeds - intermediate values, not the answer.

Answer: A

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

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

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

Over the common denominator \( 225 - r^{2} \):

\[ \frac{30(15 + r) + 30(15 - r)}{225 - r^{2}} = \frac{900}{225 - r^{2}} = 4.5 \]

\[ 225 - r^{2} = 200 \quad\Longrightarrow\quad r^{2} = 25 \quad\Longrightarrow\quad r = 5 \text{ km/hr} \]

The \( r \) terms cancel in the numerator, leaving a simple equation. Checking: 30 km at 10 km/hr takes 3 hours and at 20 km/hr takes 1½.

Answer: C

Let the speed of motor boat be 36x km/hr and that of current of water be 5x km/hr 

Speed downstream = (36x +5x)km/hr = 41x km/hr 

Speed upstream = (36 -5x)km/hr = 31x km/hr 

Distance covered downstream = (41x × 31/6)km 

Distance upstream = [1271x/6 × 1/31x]hrs 

= 41/6 hrs = 6hrs 50 min

Answer: B

Speed downstream = (5+1)km/hr = 6 km/hr 

Speed upstream = (5-1)km/hr = 4 km/hr 

Let the required distance be x km. 

Then x/6 + x/4 = 75/60 = 5/4 => (2x + 3x) =15 => x =3. 

Required distance = 3 km

Answer: A

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

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

Both legs take 3 hours, so read each speed straight off its distance.

\[ b - r = \frac{3}{3} = 1 \text{ km/hr}, \qquad b + r = \frac{15}{3} = 5 \text{ km/hr} \]

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

The boat does 3 km/hr in still water - only slightly faster than the current, which is why the upstream progress is so slow.

Answer: C

Let the speed of boat be x km/hr and speed of stream be ykm/hr. 

44/5 × (x -y) = 4x (x +y) 

=> 44(x-y) = 20 (x+y) 

 =>11 (x-y) = 5 (x +y) 

 => 6x =16 y 

 => x/y = 16/6 = 8/3 

 => x : y = 8 : 3

Answer: C

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

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

Still water gives \( b = 6 \) km/hr. Taking three times as long against the current means going at one third the speed:

\[ b - r = \frac{6}{3} = 2 \text{ km/hr} \]

\[ r = 6 - 2 = 4 \text{ km/hr} \]

The comparison here is with the still-water time, not the downstream time - so \( b - r \) is compared with \( b \), giving \( r = \tfrac{2b}{3} \).

Answer: D

Distance covered upstream in 45/4min =3/4 km 

Distance covered upstream in 1 hr =(3/4 × 2/15 × 60) km/hr = 4 km/hr 

Distance covered downstream in 15/2 min = ¾ km 

Distance covered downstream in 1hr 

= (3/4 × 2/15 × 60)km/hr = 6 km/hr 

Speed of the man in still water 

= ½ (6 + 4)km/hr = 5 km/hr

Answer: B

Let the rate of the stream be x km/hr. 

Then Speed downstream = (15/2 + x)km/hr. 

Speed upstream = (15/2 -x)km/hr 

(15/2 + x) = 2 (15/2 - x) 

=> 3x =15/2 

=> x = 15/6 = 5/2 = 2.5 

Therefore, Rate of stream = 2.5 km/hr

Answer: C

Speed upstream = 2km/hr.

Speed downstream = 6 km/hr 

Speed in stationary water = ½ (6 + 2) km/hr = 4 km/hr 

Time is taken to cover 5 km in stationary water 

= (1/4 x 5) 1 hr 15 min