Time and Distance - 01

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{distance}} = x\,km\,{\text{and}}\,{\text{usual}}\,{\text{rate}} = y\,kmph \cr & {\text{Then}},\,\frac{x}{y} - \frac{x}{{y + 3}} = \frac{{40}}{{60}} \cr & \Rightarrow 2y\left( {y + 3} \right) = 9x\,..........\left( i \right) \cr & {\text{and}},\,\frac{x}{{y - 2}} - \frac{x}{y} = \frac{{40}}{{60}} \cr & \Rightarrow y(y - 2) = 3x\,.............\left( {ii} \right) \cr & {\text{On}}\,{\text{dividing}}\,\left( {\text{i}} \right)\,{\text{by}}\,\left( {{\text{ii}}} \right){\text{,}}\,{\text{we}}\,{\text{get}}\,x = 40 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{foot}}\,{\text{be}}\,x\,km \cr & {\text{Then,}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{bicycle}} = \left( {61 - x} \right)km \cr & {\text{So}},\,\frac{x}{4} + \frac{{\left( {61 - x} \right)}}{9} = 9 \cr & \Rightarrow 9x + 4\left( {61 - x} \right) = 9 \times 36 \cr & \Rightarrow 5x = 80 \cr & \Rightarrow x = 16\,km \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}} \cr & {\text{and}}\,{\text{that}}\,{\text{of}}\,{\text{the}}\,{\text{car}}\,{\text{be}}\,y\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{{120}}{x} + \frac{{480}}{y} = 8 \cr & \Rightarrow \frac{1}{x} + \frac{4}{y} = \frac{1}{{15}}\,.........\left( i \right) \cr & {\text{and}},\,\frac{{200}}{x} + \frac{{400}}{y} = \frac{{25}}{3} \cr & \Rightarrow \frac{1}{x} + \frac{2}{y} = \frac{1}{{24}}\,.........\left( {ii} \right) \cr & {\text{Solving}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right){\text{,}}\, \cr & {\text{we}}\,{\text{get}},\,x = 60\,{\text{and}}\,y = 80 \cr & \therefore {\text{Ratio}}\,{\text{of}}\,{\text{speeds}} \cr & = 60:80 = 3:4 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{by}}\,x\,{\text{km}} \cr & {\text{Then}},\,\frac{x}{{10}} - \frac{x}{{15}} = 2 \cr & \Rightarrow 3x - 2x = 60 \cr & \Rightarrow x = 60\,km \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,60\,km\,{\text{at}}\,10\,{\text{km/hr}} \cr & = \left( {\frac{{60}}{{10}}} \right)\,hrs = 6\,hrs \cr & {\text{So,}}\,{\text{Robert}}\,{\text{started}}\,{\text{6}}\,{\text{hours}}\,{\text{before}}\,2\,P.M.\,i.e.,\,at\,A.M. \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = \left( {\frac{{60}}{5}} \right)\,kmph = 12\,kmph \cr}$$

Explanation:-

$$\eqalign{ & {\text{Time}}\,{\text{taken = 1}}\,{\text{hr}}\,{\text{40}}\,{\text{min}}\,{\text{48}}\,{\text{sec}} \cr & = 1\,{\text{hr}}\,40\frac{4}{5}\,{\text{min}} = 1\frac{{51}}{{75}}\,hrs = \frac{{126}}{{75}}\,hrs \cr & {\text{Let}}\,{\text{the}}\,{\text{actual}}\,{\text{speed}}\,{\text{be}}\,x\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{5}{7}x \times \frac{{126}}{{75}} = 42 \cr & \Rightarrow x = \left( {\frac{{42 \times 7 \times 75}}{{5 \times 126}}} \right) \cr & \Rightarrow x = 35\,{\text{km/hr}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Total}}\,{\text{time}}\,{\text{taken}} \cr & = \left( {\frac{{160}}{{64}} + \frac{{160}}{{80}}} \right)hrs \cr & = \frac{9}{2}\,hrs \cr & \therefore {\text{Average}}\,{\text{speed}} \cr & = \left( {320 \times \frac{2}{9}} \right)\,{\text{km/hr}} \cr & = 71.11\,{\text{km/hr}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,7x\,{\text{and}}\,8x\,{\text{km/hr}} \cr & {\text{Then}},\,8x = \left( {\frac{{400}}{4}} \right) = 100 \cr & \Rightarrow x = \left( {\frac{{100}}{8}} \right) = 12.5 \cr & \therefore {\text{Speed}}\,{\text{of}}\,{\text{first}}\,{\text{train}} \cr & = \left( {7 \times 12.5} \right){\text{km/hr}} \cr & = 87.5\,{\text{km/hr}} \cr}$$

Explanation:-

$$\eqalign{ & \frac{{\left( {1/2} \right)x}}{{21}} + \frac{{\left( {1/2} \right)x}}{{24}} = 10 \cr & \Rightarrow \frac{x}{{21}} + \frac{x}{{24}} = 20 \cr & \Rightarrow 15x = 168 \times 20 \cr & \Rightarrow x = \left( {\frac{{168 \times 20}}{{15}}} \right) \cr & \Rightarrow x = 224\,km \cr}$$

Explanation:-

$$\eqalign{ & {\text{Due}}\,{\text{to}}\,{\text{stoppages,}}\,{\text{it}}\,{\text{covers}}\,{\text{9}}\,{\text{km}}\,{\text{less}} \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{cover}}\,{\text{9}}\,{\text{km}} \cr & = \left( {\frac{9}{{54}} \times 60} \right)\min \cr & = 10\min \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}}\,{\text{be}}\,x\,kmph \cr & {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \frac{{150}}{{100}}x = \left( {\frac{3}{2}x} \right)kmph \cr & \therefore \frac{{75}}{x} - \frac{{75}}{{\left( {3/2} \right)x}} = \frac{{125}}{{10 \times 60}} \cr & \Rightarrow \frac{{75}}{x} - \frac{{50}}{x} = \frac{5}{{24}} \cr & \Rightarrow x = \left( {\frac{{25 \times 24}}{5}} \right) \cr & \Rightarrow x = 120\,kmph \cr}$$