Q1. (i) A car covers a distance of 60 km in 40 minutes. Find the speed of the car in km/hr.
(ii) A distance of 800 m is covered in 30 minutes, find the speed in km/hr.
(iii) A boy runs a distance of 16 km in $1\dfrac{1}{3}$ hrs; find his speed m/sec.
Show Answer
(i) Given Distance = 60 kms
$\Rightarrow$ Given Time = 40 minutes
$\Rightarrow$ Converting minutes into hours = $\dfrac{40}{60} = \dfrac{2}{3}$
$\therefore$ Required Speed = $\dfrac{60}{\dfrac{2}{3}} = \dfrac{60 \times 3}{2} = 30 \times 3 = 90$ Km/hr Ans
(ii) Given Distance = 800 metres
$\Rightarrow$ Converting metres into Kms = $\dfrac{800}{1000} = \dfrac{4}{5}$
$\Rightarrow$ Given Time = 30 minutes
$\Rightarrow$ Converting minutes into hours = $\dfrac{30}{60} = \dfrac{1}{2}$
$\therefore$ Required Speed = $\dfrac{\dfrac{4}{5}}{\dfrac{1}{2}} = \dfrac{4 \times 2}{5} = \dfrac{8}{5} = 1.6$ Km/hr Ans
(iii) Given Distance = 16 kms
$\Rightarrow$ Given Time = $\dfrac{4}{3}$ hrs
$\therefore$ Required Speed = $\dfrac{16}{\dfrac{4}{3}} = \dfrac{16 \times 3}{4} = 4 \times 3 = 12$ Km/hr
$\Rightarrow$ Converting Km/hr into m/s = $12 \times \dfrac{5}{18} = 2 \times \dfrac{5}{3} = \dfrac{10}{3} = 3\dfrac{1}{3}$ m/s Ans
$\Rightarrow$ Given Time = 40 minutes
$\Rightarrow$ Converting minutes into hours = $\dfrac{40}{60} = \dfrac{2}{3}$
$\therefore$ Required Speed = $\dfrac{60}{\dfrac{2}{3}} = \dfrac{60 \times 3}{2} = 30 \times 3 = 90$ Km/hr Ans
(ii) Given Distance = 800 metres
$\Rightarrow$ Converting metres into Kms = $\dfrac{800}{1000} = \dfrac{4}{5}$
$\Rightarrow$ Given Time = 30 minutes
$\Rightarrow$ Converting minutes into hours = $\dfrac{30}{60} = \dfrac{1}{2}$
$\therefore$ Required Speed = $\dfrac{\dfrac{4}{5}}{\dfrac{1}{2}} = \dfrac{4 \times 2}{5} = \dfrac{8}{5} = 1.6$ Km/hr Ans
(iii) Given Distance = 16 kms
$\Rightarrow$ Given Time = $\dfrac{4}{3}$ hrs
$\therefore$ Required Speed = $\dfrac{16}{\dfrac{4}{3}} = \dfrac{16 \times 3}{4} = 4 \times 3 = 12$ Km/hr
$\Rightarrow$ Converting Km/hr into m/s = $12 \times \dfrac{5}{18} = 2 \times \dfrac{5}{3} = \dfrac{10}{3} = 3\dfrac{1}{3}$ m/s Ans
Q2. (i) Find the distance covered by a bus in 50 minutes at the speed of 42 km/hr.
(ii) Find the distance covered by an object in 2 minutes at the speed of 13.5 m/sec.
Show Answer
(i) Given time = 50 mins
$\Rightarrow$ Converting time = $\dfrac{50}{60} = \dfrac{5}{6}$ hrs
$\Rightarrow$ Given speed = 42 km/hr
$\therefore$ Required distance = $42 \times \dfrac{5}{6} = 7 \times 5 = 35$ kms Ans
(ii) Given time = 2 mins
$\Rightarrow$ Given speed = 13.5 m/s
$\Rightarrow$ Converting minutes into seconds = $2 \times 60 = 120$ sec
$\therefore$ Required distance = $120 \times 13.5 = 120 \times \dfrac{135}{10} = 12 \times 135 = 1620$ metres
$\Rightarrow$ Converting metres into kms = $\dfrac{1620}{1000} = 1.620$ kms Ans
$\Rightarrow$ Converting time = $\dfrac{50}{60} = \dfrac{5}{6}$ hrs
$\Rightarrow$ Given speed = 42 km/hr
$\therefore$ Required distance = $42 \times \dfrac{5}{6} = 7 \times 5 = 35$ kms Ans
(ii) Given time = 2 mins
$\Rightarrow$ Given speed = 13.5 m/s
$\Rightarrow$ Converting minutes into seconds = $2 \times 60 = 120$ sec
$\therefore$ Required distance = $120 \times 13.5 = 120 \times \dfrac{135}{10} = 12 \times 135 = 1620$ metres
$\Rightarrow$ Converting metres into kms = $\dfrac{1620}{1000} = 1.620$ kms Ans
Q3. (i) Find the time taken to cover a distance of 120 km at the speed of 10 m/sec.
(ii) Find the time taken to cover a distance of 660 m at the speed of 72 km/hr.
Show Answer
(i) Given distance = 120 kms
$\Rightarrow$ Converting kms into metres = $120 \times 1000$ m
$\Rightarrow$ Given speed = 10 m/s
$\therefore$ Required time = $\dfrac{120 \times 1000}{10} = 12000$ seconds
$\Rightarrow$ Converting seconds into hours = $\dfrac{12000}{60 \times 60} = \dfrac{120}{36} = \dfrac{20}{6} = \dfrac{10}{3} = 3\dfrac{1}{3}$ hrs Ans
Also, $3\dfrac{1}{3}$ hrs = 3 hrs and $\dfrac{1}{3} \times 60$ = 3 hrs and 20 minutes
(ii) Given distance = 660 metres
$\Rightarrow$ Given speed = 72 km/hr
$\therefore$ Required time = $\dfrac{660}{72} = \dfrac{55}{6}$ hrs
$\Rightarrow$ Converting km/hr into m/s = $\dfrac{55}{6} \times \dfrac{18}{5} = 11 \times 3$ seconds Ans
$\Rightarrow$ Converting kms into metres = $120 \times 1000$ m
$\Rightarrow$ Given speed = 10 m/s
$\therefore$ Required time = $\dfrac{120 \times 1000}{10} = 12000$ seconds
$\Rightarrow$ Converting seconds into hours = $\dfrac{12000}{60 \times 60} = \dfrac{120}{36} = \dfrac{20}{6} = \dfrac{10}{3} = 3\dfrac{1}{3}$ hrs Ans
Also, $3\dfrac{1}{3}$ hrs = 3 hrs and $\dfrac{1}{3} \times 60$ = 3 hrs and 20 minutes
(ii) Given distance = 660 metres
$\Rightarrow$ Given speed = 72 km/hr
$\therefore$ Required time = $\dfrac{660}{72} = \dfrac{55}{6}$ hrs
$\Rightarrow$ Converting km/hr into m/s = $\dfrac{55}{6} \times \dfrac{18}{5} = 11 \times 3$ seconds Ans
Q4. An aeroplane flies 750 km in 40 min. Find:
(i) Its speed in km/hr.
(ii) Time taken by it for covering a distance of 900 km.
Show Answer
(i) Given distance = 750 kms
$\Rightarrow$ Given time = 40 minutes
$\Rightarrow$ Converting time = $\dfrac{40}{60} = \dfrac{2}{3}$ hrs
$\therefore$ Required Speed = $\dfrac{750}{\dfrac{2}{3}} = \dfrac{750 \times 3}{2} = 375 \times 3 = 1125$ Km/hr Ans
(ii) Given distance = 900 metres
$\Rightarrow$ Given speed = 1125 km/hr
$\therefore$ Required time = $\dfrac{900}{1125} = \dfrac{180}{225} = \dfrac{36}{45} = \dfrac{4}{5}$ hrs
$\Rightarrow$ Converting hours into minutes = $\dfrac{4}{5} \times 60 = 4 \times 12 = 48$ minutes Ans
$\Rightarrow$ Given time = 40 minutes
$\Rightarrow$ Converting time = $\dfrac{40}{60} = \dfrac{2}{3}$ hrs
$\therefore$ Required Speed = $\dfrac{750}{\dfrac{2}{3}} = \dfrac{750 \times 3}{2} = 375 \times 3 = 1125$ Km/hr Ans
(ii) Given distance = 900 metres
$\Rightarrow$ Given speed = 1125 km/hr
$\therefore$ Required time = $\dfrac{900}{1125} = \dfrac{180}{225} = \dfrac{36}{45} = \dfrac{4}{5}$ hrs
$\Rightarrow$ Converting hours into minutes = $\dfrac{4}{5} \times 60 = 4 \times 12 = 48$ minutes Ans
Q5. A man takes 150 steps in walking 120 metres. Find his dpeed in:
(i) m/sec, and
(ii) km/hr, if he takes 3 steps in one second.
Show Answer
Given, a man takes 150 steps to cover 120 metres
$\therefore$ metres covered per step = $\dfrac{120}{150} = \dfrac{4}{5} = 0.8$ m/step
(i) Given, the man takes 3 steps per second
$\therefore$ His speed in m/sec = $3 \times 0.8 = 2.4$ m/sec Ans
(ii) His speed in km/hr = $2.4 \times \dfrac{18}{5}$
$\Rightarrow \dfrac{24}{10} \times \dfrac{18}{5} = \dfrac{12}{5} \times \dfrac{18}{5} = \dfrac{216}{25} = 8.64$ km/hr Ans.
$\therefore$ metres covered per step = $\dfrac{120}{150} = \dfrac{4}{5} = 0.8$ m/step
(i) Given, the man takes 3 steps per second
$\therefore$ His speed in m/sec = $3 \times 0.8 = 2.4$ m/sec Ans
(ii) His speed in km/hr = $2.4 \times \dfrac{18}{5}$
$\Rightarrow \dfrac{24}{10} \times \dfrac{18}{5} = \dfrac{12}{5} \times \dfrac{18}{5} = \dfrac{216}{25} = 8.64$ km/hr Ans.
Q6. Out of a total distance covered of 200 km, first half is covered at the speed of 60 km/hr and remaining half is covered at 75 km/hr. Find:
(i) Total time taken to cover the whole journey.
(ii) Average speed for the whole journey.
Show Answer
(i) Given, total distance = 200 kms
$\Rightarrow$ speed for covering 1st half od journey = 60 km/hr
$\therefore$ time taken to cover 1st half od journey = $\dfrac{100}{60} = \dfrac{5}{3}$
$\Rightarrow$ speed for 2nd half of the journey = 75 km/hr
$\therefore$ time taken to cover 2nd half of the journey = $\dfrac{100}{75} = \dfrac{4}{3}$
$\Rightarrow$ Total time taken = $\dfrac{5}{3} + \dfrac{4}{3} = \dfrac{5 + 4}{3} = \dfrac{9}{3} = 3$ hrs. Ans.
(ii) Total distance = 200 kms
$\Rightarrow$ Total time = 3 hrs
$\therefore$ Average speed = $\dfrac{200}{3} = 66\dfrac{2}{3}$ km/hr Ans
$\Rightarrow$ speed for covering 1st half od journey = 60 km/hr
$\therefore$ time taken to cover 1st half od journey = $\dfrac{100}{60} = \dfrac{5}{3}$
$\Rightarrow$ speed for 2nd half of the journey = 75 km/hr
$\therefore$ time taken to cover 2nd half of the journey = $\dfrac{100}{75} = \dfrac{4}{3}$
$\Rightarrow$ Total time taken = $\dfrac{5}{3} + \dfrac{4}{3} = \dfrac{5 + 4}{3} = \dfrac{9}{3} = 3$ hrs. Ans.
(ii) Total distance = 200 kms
$\Rightarrow$ Total time = 3 hrs
$\therefore$ Average speed = $\dfrac{200}{3} = 66\dfrac{2}{3}$ km/hr Ans
Q7. A part of a journey is covered in 1 hour 40 minutes at 75 km/hr and the remaining part of it is covered in 2 hrs and 50 minutes at 60 km/hr. Find:
(i) The total lenght of the journey.
(ii) The average speed during the whole journey.
Show Answer
(i) Time taken to cover 1st part of journey = 1 hr 40 min
$\Rightarrow$ Converting time = $1 + \dfrac{40}{60} = \dfrac{60 + 40}{60} = \dfrac{100}{60} = \dfrac{5}{3}$
$\Rightarrow$ speed for 1st part of journey = 75 km/hr
$\therefore$ distance of 1st part = $\dfrac{5}{3} \times 75 = 5 \times 25 = 125$ kms
$\Rightarrow$ Time taken to cover 1st part of journey = 2 hr 50 min
$\Rightarrow$ Converting time = $2 + \dfrac{50}{60} = \dfrac{120 + 50}{60} = \dfrac{170}{60} = \dfrac{17}{6}$
$\Rightarrow$ speed for 1st part of journey = 60 km/hr
$\Rightarrow$ distance of 1st part = $\dfrac{17}{6} \times 60 = 17 \times 10 = 170$ kms
$\therefore$ total distance = $125 + 170 = 295$ kms Ans
(ii) total time taken = $\dfrac{5}{3} + \dfrac{17}{6} = \dfrac{10 + 17}{6} = \dfrac{27}{6} = \dfrac{9}{2}$
$\therefore$ Average speed = $\dfrac{295}{\dfrac{9}{2}}$
$\Rightarrow \dfrac{295 \times 2}{9} = \dfrac{590}{9} = 65\dfrac{5}{9}$ km/hr Ans
$\Rightarrow$ Converting time = $1 + \dfrac{40}{60} = \dfrac{60 + 40}{60} = \dfrac{100}{60} = \dfrac{5}{3}$
$\Rightarrow$ speed for 1st part of journey = 75 km/hr
$\therefore$ distance of 1st part = $\dfrac{5}{3} \times 75 = 5 \times 25 = 125$ kms
$\Rightarrow$ Time taken to cover 1st part of journey = 2 hr 50 min
$\Rightarrow$ Converting time = $2 + \dfrac{50}{60} = \dfrac{120 + 50}{60} = \dfrac{170}{60} = \dfrac{17}{6}$
$\Rightarrow$ speed for 1st part of journey = 60 km/hr
$\Rightarrow$ distance of 1st part = $\dfrac{17}{6} \times 60 = 17 \times 10 = 170$ kms
$\therefore$ total distance = $125 + 170 = 295$ kms Ans
(ii) total time taken = $\dfrac{5}{3} + \dfrac{17}{6} = \dfrac{10 + 17}{6} = \dfrac{27}{6} = \dfrac{9}{2}$
$\therefore$ Average speed = $\dfrac{295}{\dfrac{9}{2}}$
$\Rightarrow \dfrac{295 \times 2}{9} = \dfrac{590}{9} = 65\dfrac{5}{9}$ km/hr Ans
Q8. Ramesh goes from station A to station B at the speed of 40 km/hr and returns from station B to station A at the speed of 60 km/hr. Find average speed for the whole journey.
Show Answer
$\Rightarrow$ Let distance from station A to B be 100 kms
$\Rightarrow$ Given speed for station A to B = 40 km/hr
$\therefore$ time taken for station A to B = $\dfrac{100}{40} = \dfrac{10}{4} = \dfrac{5}{2}$
$\Rightarrow$ Given speed for station B to A = 60 km/hr
$\therefore$ time taken for station B to A = $\dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$
$\Rightarrow$ total distance = $100 + 100 = 200$ kms
$\Rightarrow$ total time = $\dfrac{5}{2} + \dfrac{5}{3} = \dfrac{15 + 10}{6} = \dfrac{25}{6}$
$\therefore$ Average speed = $\dfrac{200 \times 6}{25} = 8 \times 6 = 48$ km/hr Ans
$\Rightarrow$ Given speed for station A to B = 40 km/hr
$\therefore$ time taken for station A to B = $\dfrac{100}{40} = \dfrac{10}{4} = \dfrac{5}{2}$
$\Rightarrow$ Given speed for station B to A = 60 km/hr
$\therefore$ time taken for station B to A = $\dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$
$\Rightarrow$ total distance = $100 + 100 = 200$ kms
$\Rightarrow$ total time = $\dfrac{5}{2} + \dfrac{5}{3} = \dfrac{15 + 10}{6} = \dfrac{25}{6}$
$\therefore$ Average speed = $\dfrac{200 \times 6}{25} = 8 \times 6 = 48$ km/hr Ans
Q9. A journey of 540 kms is completed in 7 hrs and 30 mins. If $\dfrac{2}{3}$ of the journey is completed at 80 km/hr, at what speed is the remaining journey completed?
Show Answer
$\Rightarrow$ Total distance = 540 kms
$\Rightarrow \dfrac{2}{3}$ of total distance = $\dfrac{2}{3} \times 540 = 2 \times 180 = 360$ kms
$\Rightarrow$ speed for covering 360 kms = 80 km/hr
$\Rightarrow$ time taken for covering 360 kms = $\dfrac{360}{80} = \dfrac{36}{8} = \dfrac{9}{2}$
$\Rightarrow$ Given, total time taken to complete the journey = 7 hr 30 min
$\Rightarrow$ Converting time = $7 + \dfrac{30}{60} = \dfrac{420 + 30}{60} = \dfrac{450}{60} = \dfrac{45}{6} = \dfrac{15}{2}$
$\therefore$ Remaining time = $\dfrac{15}{2} - \dfrac{9}{2} = \dfrac{6}{2} = 3$ hrs
$\Rightarrow$ Remaining distance = $540 - 360 = 180$ kms
$\therefore$ speed for remaining distance = $\dfrac{180}{3} = 60$ km/hr Ans
$\Rightarrow \dfrac{2}{3}$ of total distance = $\dfrac{2}{3} \times 540 = 2 \times 180 = 360$ kms
$\Rightarrow$ speed for covering 360 kms = 80 km/hr
$\Rightarrow$ time taken for covering 360 kms = $\dfrac{360}{80} = \dfrac{36}{8} = \dfrac{9}{2}$
$\Rightarrow$ Given, total time taken to complete the journey = 7 hr 30 min
$\Rightarrow$ Converting time = $7 + \dfrac{30}{60} = \dfrac{420 + 30}{60} = \dfrac{450}{60} = \dfrac{45}{6} = \dfrac{15}{2}$
$\therefore$ Remaining time = $\dfrac{15}{2} - \dfrac{9}{2} = \dfrac{6}{2} = 3$ hrs
$\Rightarrow$ Remaining distance = $540 - 360 = 180$ kms
$\therefore$ speed for remaining distance = $\dfrac{180}{3} = 60$ km/hr Ans
Q10. The total distance between two stations is 570 kms out of which, 224 km is covered at 56 km/hr, 210 kms is covered 70 km/hr and the remaining is covered in 2 hrs. Find the average speed during the whole journey.
Show Answer
$\Rightarrow$ Total distance = 540 kms
$\Rightarrow$ 1st part of journey = 224 kms
$\Rightarrow$ speed to cover 224 kms = 56 km/hr
$\therefore$ time to cover 224 kms = $\dfrac{224}{56} = 4$ hrs
$\Rightarrow$ 2nd part of journey = 210 kms
$\Rightarrow$ speed to cover 210 kms = 70 km/hr
$\therefore$ time to cover 224 kms = $\dfrac{210}{70} = 3$ hrs
$\Rightarrow$ Remaining distance = $570 - (224 + 210) = 570 - 434 = 146$
$\Rightarrow$ time taken to cover remaining distance = 2 hrs
$\Rightarrow$ Total time = $4 + 3 \times 2 = 9$ hrs
$\therefore$ Average speed = $\dfrac{570}{9} = \dfrac{190}{3} = 63\dfrac{1}{3}$ km/hr Ans
$\Rightarrow$ 1st part of journey = 224 kms
$\Rightarrow$ speed to cover 224 kms = 56 km/hr
$\therefore$ time to cover 224 kms = $\dfrac{224}{56} = 4$ hrs
$\Rightarrow$ 2nd part of journey = 210 kms
$\Rightarrow$ speed to cover 210 kms = 70 km/hr
$\therefore$ time to cover 224 kms = $\dfrac{210}{70} = 3$ hrs
$\Rightarrow$ Remaining distance = $570 - (224 + 210) = 570 - 434 = 146$
$\Rightarrow$ time taken to cover remaining distance = 2 hrs
$\Rightarrow$ Total time = $4 + 3 \times 2 = 9$ hrs
$\therefore$ Average speed = $\dfrac{570}{9} = \dfrac{190}{3} = 63\dfrac{1}{3}$ km/hr Ans
Q11. A distance of 33 kms is covered in 1 hr and 12 minutes. if the first 15 kms is covered at the speed of 25 km/hr, find the speed during the remaining distance.
Show Answer
$\Rightarrow$ 1st part of journey = 15 kms
$\Rightarrow$ speed to cover 15 kms = 25 km/hr
$\therefore$ time to cover 1st part of journey = $\dfrac{15}{25} = \dfrac{3}{5}$
$\Rightarrow$ Given, total time = 1 hr 12 min
$\Rightarrow$ Converting time = $1 + \dfrac{12}{60} = \dfrac{60 + 12}{60} = \dfrac{72}{60} = \dfrac{6}{5}$
$\Rightarrow$ Remaining distance = $33 - 15 = 18$ kms
$\Rightarrow$ Remaining time = $\dfrac{6}{5} - \dfrac{3}{5} = \dfrac{3}{5}$
$\therefore$ speed to cover 18 kms = $\dfrac{18}{\dfrac{3}{5}} = \dfrac{18 \times 5}{3} = 6 \times 5 = 30$ kms Ans
$\Rightarrow$ speed to cover 15 kms = 25 km/hr
$\therefore$ time to cover 1st part of journey = $\dfrac{15}{25} = \dfrac{3}{5}$
$\Rightarrow$ Given, total time = 1 hr 12 min
$\Rightarrow$ Converting time = $1 + \dfrac{12}{60} = \dfrac{60 + 12}{60} = \dfrac{72}{60} = \dfrac{6}{5}$
$\Rightarrow$ Remaining distance = $33 - 15 = 18$ kms
$\Rightarrow$ Remaining time = $\dfrac{6}{5} - \dfrac{3}{5} = \dfrac{3}{5}$
$\therefore$ speed to cover 18 kms = $\dfrac{18}{\dfrac{3}{5}} = \dfrac{18 \times 5}{3} = 6 \times 5 = 30$ kms Ans
Q12. X and Y are two towns. A amn goes from town X to town Y at an average speed of 30 km/hr and returns at the average speed of 20 km/hr. Find the distance between the two towns if the man takes 10 hrs on the whole.
Show Answer
$\Rightarrow$ Let distance between towns be = $x$ kms
$\Rightarrow$ Given, speed to go from town $x$ to $y$ = 30 km/hr
$\therefore$ time taken to cover distance town $x$ to $y$ = $\dfrac{x}{30}$
$\Rightarrow$ speed to go from town $y$ to $x$ = 20 km/hr
$\therefore$ time taken to cover distance town $y$ to $x$ = $\dfrac{x}{20}$
$\Rightarrow$ Given, time taken to cover distance town $y$ to $x$ = 10 hrs
So, we have the equation:
$\Rightarrow \dfrac{x}{30} + \dfrac{x}{20} = 10$
$\Rightarrow \dfrac{2x + 3x}{60} = 10$
$\Rightarrow \dfrac{5x}{60} = 10$
$\Rightarrow x = \dfrac{10 \times 60}{5} = 2 \times 60 = 120$ kms Ans
$\Rightarrow$ Given, speed to go from town $x$ to $y$ = 30 km/hr
$\therefore$ time taken to cover distance town $x$ to $y$ = $\dfrac{x}{30}$
$\Rightarrow$ speed to go from town $y$ to $x$ = 20 km/hr
$\therefore$ time taken to cover distance town $y$ to $x$ = $\dfrac{x}{20}$
$\Rightarrow$ Given, time taken to cover distance town $y$ to $x$ = 10 hrs
So, we have the equation:
$\Rightarrow \dfrac{x}{30} + \dfrac{x}{20} = 10$
$\Rightarrow \dfrac{2x + 3x}{60} = 10$
$\Rightarrow \dfrac{5x}{60} = 10$
$\Rightarrow x = \dfrac{10 \times 60}{5} = 2 \times 60 = 120$ kms Ans
Q13. A man takes 40 mins to cover a certain distance at the speed of 6 km/hr. How much should he increase his speed to cover the same distance in 30 mins?
Show Answer
$\Rightarrow$ Time taken = 40 mins
$\Rightarrow$ Converting time = $\dfrac{40}{60} = \dfrac{2}{3}$
$\Rightarrow$ speed = 6 km/hr
$\therefore$ distance = $\dfrac{2}{3} \times 6 = 4$ kms
$\Rightarrow$ Given time = 30 mins
$\Rightarrow$ Converting time = $\dfrac{30}{60} = \dfrac{1}{2}$
$\therefore$ speed to cover 4 kms in 30 mins = $\dfrac{4}{\dfrac{1}{2}} = 4 \times 2 = 8$ km/hr
So, speed should be increased by = $8 - 6 = 2$ km/hr Ans
$\Rightarrow$ Converting time = $\dfrac{40}{60} = \dfrac{2}{3}$
$\Rightarrow$ speed = 6 km/hr
$\therefore$ distance = $\dfrac{2}{3} \times 6 = 4$ kms
$\Rightarrow$ Given time = 30 mins
$\Rightarrow$ Converting time = $\dfrac{30}{60} = \dfrac{1}{2}$
$\therefore$ speed to cover 4 kms in 30 mins = $\dfrac{4}{\dfrac{1}{2}} = 4 \times 2 = 8$ km/hr
So, speed should be increased by = $8 - 6 = 2$ km/hr Ans
Q14. In how much time a car, wunning at 60 km/hr, cover a distance of 600 km if it stops for $\dfrac{1}{2}$ hour after every 100 kms?
Show Answer
Given, the car halts after every 100 kms for $\dfrac{1}{2}$ hour
$\Rightarrow$ Now, no. of halts in 600 kms = 5
$\therefore$ total halt time = $\dfrac{1}{2} \times 5 = \dfrac{5}{2} = 2\dfrac{1}{2}$ hrs
$\Rightarrow$ speed of car = 6- km/hr
$\Rightarrow$ distance covered = 600 kms
$\therefore$ time taken = $\dfrac{600}{60} = 10$ hrs
$\Rightarrow$ time taken including halt time = $10 + 2\dfrac{1}{2} = 12\dfrac{1}{2}$ hrs Ans
$\Rightarrow$ Now, no. of halts in 600 kms = 5
$\therefore$ total halt time = $\dfrac{1}{2} \times 5 = \dfrac{5}{2} = 2\dfrac{1}{2}$ hrs
$\Rightarrow$ speed of car = 6- km/hr
$\Rightarrow$ distance covered = 600 kms
$\therefore$ time taken = $\dfrac{600}{60} = 10$ hrs
$\Rightarrow$ time taken including halt time = $10 + 2\dfrac{1}{2} = 12\dfrac{1}{2}$ hrs Ans
Q15. A policeman follows a thief who is 600 metres ahead of him. If they run at the speeds of 6 km/hr and 5 km/hr, how long will the policeman take to catch the thief?
Show Answer
$\Rightarrow$ speed for policeman = 6 kms
$\Rightarrow$ Converting speed into m/s = $6 \times \dfrac{5}{18} = \dfrac{5}{3}$ m/s
$\Rightarrow$ speed for thief = 5 kms
$\Rightarrow$ Converting speed into m/s = $5 \times \dfrac{5}{18} = \dfrac{25}{18}$ m/s
$\because$ the policeman and thief are running in the same direction
$\therefore$ Relative speed = $\dfrac{5}{3} - \dfrac{25}{18} = \dfrac{30 - 25}{18} = \dfrac{5}{18}$ m/s
$\Rightarrow$ Converting seconds into minutes = $\dfrac{5}{18} \times 60 = \dfrac{50}{3}$
$\therefore$ Time taken by policeman to catch the thief:
$\Rightarrow \dfrac{600}{\dfrac{50}{3}} = \dfrac{600 \times 3}{50} = 12 \times 3 = 36$ Ans
$\Rightarrow$ Converting speed into m/s = $6 \times \dfrac{5}{18} = \dfrac{5}{3}$ m/s
$\Rightarrow$ speed for thief = 5 kms
$\Rightarrow$ Converting speed into m/s = $5 \times \dfrac{5}{18} = \dfrac{25}{18}$ m/s
$\because$ the policeman and thief are running in the same direction
$\therefore$ Relative speed = $\dfrac{5}{3} - \dfrac{25}{18} = \dfrac{30 - 25}{18} = \dfrac{5}{18}$ m/s
$\Rightarrow$ Converting seconds into minutes = $\dfrac{5}{18} \times 60 = \dfrac{50}{3}$
$\therefore$ Time taken by policeman to catch the thief:
$\Rightarrow \dfrac{600}{\dfrac{50}{3}} = \dfrac{600 \times 3}{50} = 12 \times 3 = 36$ Ans