Boats and Streams

Upstream speed = 20/6 = 10/3 kmph (boat - current). Downstream speed = 18/4 = 9/2 kmph (boat + current). Solving: (b+c) - (b-c) = 2c, so 9/2 - 10/3 = 7/6, giving c = 7/12 kmph. The water current speed is 7/12 kmph, which is approximately 0.58 kmph.

Time = distance/speed. Downstream: 12/(9+v) hours. Upstream: 12/(9-v) hours. Total: 12/(9+v) + 12/(9-v) = 3. Solving gives 3v² = 9, so v = 3 km/hr. Wait, let me solve: 12(9-v+9+v)/(81-v²) = 3, so 216 = 3(81-v²), 72 = 81-v², v² = 9, v = 3. But answer says 4.5. Let me re-verify the claimed answer.

The speed of the stream is calculated as (Downstream Velocity - Upstream Velocity) / 2. Thus, (40 - 14) / 2 = 26 / 2 = 13 km/hr. Note: the option says 'hm/hr' which is a typo for 'km/hr'.

Let boat speed in still water = b, stream speed = s. Upstream: b - s = 10. Downstream: b + s = 14. Adding: 2b = 24, so b = 12 mph. Then s = 14 - 12 = 2 mph. The stream flows at 2 mph. This is a classic boats and streams problem - the key is recognizing that downstream speed is sum and upstream is difference.

Speed downstream = (30 + 6) = 36 kmph Time = 12 min = (12/60) hours = 1/5 hours Distance = 36/5 = 7.2 km Hence, this option is correct.

The speed while going downstream = (15 + 3 - 2) kmph = 16 kmph

Distance = 16/4 = 4 km

Time taken in downstream $= \displaystyle \frac{d}{15 + 5} = \frac{d}{20}$
Time taken in upstream $= \displaystyle \frac{d}{15 - 5} = \frac{d}{10}$
$\therefore$ Average speed $ = \displaystyle \frac{2d}{\displaystyle \frac{d}{20} + \frac{d}{10}} = \frac{40}{3} km/hr$
So, $\displaystyle \frac{\text{Average speed}}{\text{Speed in still water}} = \frac{40/3}{15} = \frac{40}{45} = \frac{8}{9}$.

Speed of the man in still water $\cfrac{2}{15/60}$ = $8$ km/hr
Speed of the man in downstream = $\cfrac{4}{20/60}$ = $12$ km/hr
Speed of stream = $4$ km/hr
Distance covered by paper boat in $2$ $\frac{1}{2}$ hours = $\cfrac{5}{2} \, \times$  $4$ = $10$ km 

Boat speed = 4, current = 2. Upstream speed = 4-2 = 2 km/hr. Distance = 2 * 9 = 18 km. Downstream speed = 4+2 = 6 km/hr. Time = 18 / 6 = 3 hours.

The resultant speed is the vector sum of the two perpendicular velocities. Resultant = sqrt(10^2 + 6^2) = sqrt(100 + 36) = sqrt(136). sqrt(136) is approximately 11.66, which rounds to 11.7.

The boat velocity vector is (10, 6) relative to the shore. The angle theta with the shoreline is tan(theta) = 10 / 6 = 1.666. theta = arctan(1.666) = 59 degrees.

Let upstream speed be U and downstream speed be D. Time taken to row 7 km downstream = 7/D. Time taken to row 3 km upstream = 3/U. Given 7/D = 3/U, so U/D = 3/7.

Let the still-water speed be u and the original stream speed be v. From (u + v)/(u - v) = 5/2, u = 7v/3; using the changed ratio after the stream slows by 1 gives v = 15/8 and u = 35/8 km/hr.

The speed of a boat in still water is the average of its downstream and upstream speeds. Speed = (Downstream + Upstream) / 2 = (9 + 6) / 2 = 15 / 2 = 7.5 km/hr.