Boats And Streams — Part 2 MCQs with Explanations

Speed of boat in downstream $$\eqalign{ & = \frac{{56}}{{3.5}} \cr & = \frac{{56}}{7} \times 2 \cr & = 16{\text{ km/hr}} \cr} $$ ∴ Required time $$ = \frac{{41.6}}{{16}} = 2.6{\text{ hrs}}$$

$$\eqalign{ & {\text{Speed of boat}} = x \cr & {\text{Speed of stream}} = y \cr & {\text{Distance}} = 9\left( {x + y} \right) = 18\left( {x - y} \right) \cr & x + y = 2x - 2y \cr & x = 3y \cr & {\text{Distance}} = 9\left( {x + y} \right) \cr & = 9\left( {3y + y} \right) \cr & = 36y \cr & 36y \times \frac{3}{5} = t \times x \cr & 36y \times \frac{3}{5} = t \times 3y \cr & t = \frac{{36}}{5} = 7.2 \cr} $$

Distance Relative Speed = Time Distance = 36 + 54 × 1 = 90 × 5 18 × 1 km/hr → × 5 18 m/sec = 25 metre

$$\eqalign{ & {\text{Motor boat}} \cr & \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = 10\,.\,.\,.\,.\,.\,\left( 1 \right) \cr & \frac{{80}}{{x + y}} + \frac{{40}}{{x - y}} = 15\,.\,.\,.\,.\,.\,\left( 2 \right) \cr & \frac{{20}}{{x - y}} = 5 \cr & x - y = 4\,.\,.\,.\,.\,.\,\left( 3 \right) \cr & {\text{From equation }}\left( 2 \right),\,\frac{{80}}{{x + y}} + 10 = 15 \cr & x + y = 16\,.\,.\,.\,.\,.\,\left( 4 \right) \cr & {\text{From equation }}\left( 3 \right)\,{\text{and }}\left( 4 \right) \cr & x = 10,\,y = 6 \cr & {\text{Speed of motor boat}} = 10\,{\text{Answer}} \cr & \cr & {\bf{Alternate\, Solution}} \cr & \left( {40 + 30} \right) - \left( {80 + 40} \right) = \left( {10 - 15} \right){\text{hr}} \cr & 5\,{\text{hr}} \to {\text{50 km}} \cr & 1\,{\text{hr}} \to 1{\text{0 km}} \cr & {\text{Speed }} = 1{\text{0 km/hr}} \cr} $$

Speed of boat in still water = 15 km/hr Speed of stream = 5 km/hr Total time of upstream and downstream $$\eqalign{ & = \frac{{60}}{{15 - 5}} + \frac{{60}}{{15 + 5}} \cr & = 6 + 3 \cr & = 9{\text{ hr}} \cr} $$

$$\eqalign{ & {\text{Speed of boat}} = x \cr & {\text{Speed of current}} = y \cr & {\text{According to the question,}} \cr & \frac{{40}}{{x + y}} + \frac{{25}}{{x - y}} = \frac{{15}}{2} \cr & {\text{Multiplying both sides by}}\frac{6}{5}{\text{ and}} \cr & \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = \frac{{18}}{2} \cr & \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = 9\,.\,.\,.\,.\,.\,.\,.\,\left( 1 \right) \cr & \frac{{48}}{{x + y}} + \frac{{36}}{{x - y}} = 10\,.\,.\,.\,.\,.\,.\,.\,\left( 2 \right) \cr & \underline {\,\, - \,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{ - 6}}{{x - y}} = - 1 \cr & x - y = 6\,.\,.\,.\,.\,.\,.\,.\,\left( 3 \right) \cr & {\text{From equation }}\left( 2 \right), \cr & \frac{{48}}{{x + y}} + \frac{{36}}{6} = 10 \cr & \frac{{48}}{{x + y}} = 4 \cr & x + y = 12\,.\,.\,.\,.\,.\,.\,.\,\left( 4 \right) \cr & {\text{From equation }}\left( 3 \right){\text{ and }}\left( 4 \right) \cr & x - y = 6 \cr & x + y = 12 \cr & \overline {\,x = 9,\,y = 3\,} \cr & {\text{Speed of boat}} = 9 \cr} $$

Understanding the Problem: Imagine a swimmer swimming in a river. The river's current pushes the swimmer, affecting their speed. The swimmer swims against the current (meaning the current slows them down) for 6 minutes, then turns around and swims with the current (the current speeds them up) for another 6 minutes. The total distance covered during this time is 120 meters. We need to find the speed of the river's current. Breaking it Down: Let's say the swimmer's speed in still water is 'x' m/min and the speed of the current is 'y' m/min. When swimming against the current, the effective speed is (x - y) m/min (because the current slows them down). When swimming with the current, the effective speed is (x + y) m/min (because the current helps them). Calculations: Distance covered against the current = (x - y) * 6 minutes Distance covered with the current = (x + y) * 6 minutes The difference between these two distances is 120 meters (the distance between points P and Q): [(x + y) * 6] - [(x - y) * 6] = 120 Simplifying this equation, we get: 6x + 6y - 6x + 6y = 120 12y = 120 y = 10 m/min Converting to km/h: We have the speed of the current as 10 m/min. To convert this to km/h, we need to multiply by 60 (minutes in an hour) and divide by 1000 (meters in a kilometer): (10 m/min) * (60 min/hour) / (1000 m/km) = 0.6 km/h Therefore, the speed of the current is 0.6 km/h. The correct option is D.

$$\eqalign{ & \frac{{90}}{{{\text{Boat}} - {\text{Stream}}}} + \frac{{90}}{{{\text{Boat}} + {\text{Stream}}}} = {\text{Time}} \cr & \frac{{90}}{{12}} + \frac{{90}}{{24}} = {\text{Time}} \cr & {\text{Time}} = \frac{{270}}{{24}} = \frac{{45}}{4} = 11\frac{1}{4} \cr} $$

$$\eqalign{ & \frac{5}{{x - y}} + \frac{{15}}{{2\left( {x + y} \right)}} = \frac{3}{4}{\text{hr }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{{2.5 \times 2}}{{x - y}} + \frac{{5 \times 2}}{{x + y}} = \frac{{5 \times 2}}{{12}}{\text{hr }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{5}{{2\left( {x + y} \right)}} = \frac{1}{{12}} \cr & x + y = 30 \cr & {\text{Put }}x + y = 30{\text{ in equation }}\left( {\text{i}} \right) \cr & \frac{5}{{x - y}} + \frac{{15}}{{2 \times 30}} = \frac{3}{4} \cr & \frac{5}{{x - y}} = \frac{1}{2} \cr & x - y = 10 \cr & {\text{Now,}} \cr & \,x + y = 30 \cr & \underline {\,x - y = 10\,} \cr & x = 20 \cr & y = 10 \cr & \frac{{6\,{\text{km}}}}{{x - y}} = \frac{6}{{10}} \times 60 = 36\min \cr} $$