21. A was travelling in his boat, when the wind blew his hat off and the hat started floating back downstream. The boat continued to travel upstream for 15 more minutes before A realized that his hat had fallen off and turned back downstream. He caught up with that as soom as it reached the starting point. Find the speed of the river if A’s hat flew off exactly 5 km from where he started.
Sol:
In 15 minutes distance covered by boat upstream = (B – W) × \(\frac{{15}}{{60}}\)
Since hat does not have its own speed, it flows with the speed of current.
∴ in 15 min distance travelled by hat = \(\frac{{15}}{{60}} \times W = \frac{W}{4}\)
∴ in 15 min distance between boat and hat = \(\frac{{B – W}}{4} + \frac{W}{4}\) = \(\frac{B}{4}\)
= \(\frac{1}{4}\)
∴ total time = \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}h\)
according to question in \(\frac{1}{2}h\) hat covers a distance of 5 km.
∴ Speed of hat = \(\frac{5}{{\frac{1}{2}}}\) = 10 km/h
which is same as speed of river
∴ Speed of river (W) = 10 km/h Answer
22. A boat goes 40 km and back to starting point in 8 h. The time taken by the boat to row 4 km downstream is equal to the time taken by the boat to row 3 km upstream. Find the speed of boat in still water and the rate of current.
Sol:
\(40 = \frac{{{B^2} – {W^2}}}{{2B}} \times 8\)
B² – W² = 10B ……………….(i)
& \(\frac{4}{{B + W}} = \frac{3}{{B – W}}\)
4B – 4W = 3B + 3W
B = 7W ………………(ii)
put this value in equation (i)
(7W)² – W² = 10 × (7W)
48W² = 70W
W = \(\frac{{70}}{{48}}\) = \( \boldsymbol{\frac{{35}}{{24}}} \) km/h Answer
& B = 7W = \(\frac{{7 \times 35}}{{24}}\) = \( \boldsymbol {10\frac{5}{{24}}}\) km/h Answer
23. The speed of 30 m long motor boat in still water is 32 km/h. This motorboat is moving in the river whose speed is 5 km/h. Travelling upstream it crosses a temple situated at the bank of river in 10 seconds. Find the length of the temple.
Sol:
Upstream speed of boat = 32 – 5 = 27 km/h = 27 × \(\frac{5}{{18}}\) m/sec
= \(\frac{{15}}{2}\) m/sec
∴ \(\frac{{30 + x}}{{\frac{{15}}{2}}}\) = 10
x = 45 m ⟵ Length of Temple Answer
24. A boat rows downstream 68 km and upstream 45 km in 9 hours. The same boat rows 51 km downstream & 72 km upstream in 2 hours more. Find the rate of current and the speed of the boat in still water.
Sol:
\(\frac{{68}}{{B + W}} + \frac{{45}}{{B – W}} = 9\) ……………(i)
& \(\frac{{51}}{{B + W}} + \frac{{72}}{{B – W}} = 11\) ……………….(ii)
Let \(\frac{1}{{B + W}} = x\) & \(\frac{1}{{B – W}} = y\)
∴ 68x + 45y = 9 …………..(iii)
51x + 72y = 11 …………(iv)
(72×68 – 51×45)x = 9×72 – 11×45
2601x = 153
x = \(\frac{1}{{17}} = \frac{1}{{B + W}}\) ⟹ B + W = 17
put this value in equation (iii)
68×\(\frac{1}{{17}}\) + 45y = 9
y = \(\frac{1}{9}\) ⟹ \(\frac{1}{{B – W}} = \frac{1}{9}\) ⟹ B – W = 9
∴ B + W = 17
& B – W = 9
B = 13 km/h & W = 4 km/h Answer
25. A ship is 126 km from the shore, springs a leak which admits 3.5 ton of water every 7 minute. An outlet tank can through out 18 tons of water per hour. Find at what speed it should move such that when it begins to sink a rescue ship moves with 8 km/h escapes the passengers of the ship, if 84 ton of water is enough to sink ?
Sol:
26. Two boats go downstream from point A to point B. The faster boat covers the distance from A to B \(1\frac{1}{4}\) times as fast as the slower boat. It is known that for every hour the slower boat lags behind the faster boat by 10 km. However if they go upstream then the faster boat covers the distance from y to x in half the time as the slower boat. Find the speed of the slower boat in still water.
Sol: