Q31. Walking at $\dfrac{3}{5}$ of his usual speed, a man is 4 hrs too late. Find his usual time.
Show Answer
$\because$ The man walks $\dfrac{3}{5}$ of his usual speed, time taken is $\dfrac{5}{3}$ of his usual time. As such, we have the following equation:
$\Rightarrow (\dfrac{5}{3})$ of usual time = usual time $+ 4$ hrs
$\Rightarrow (\dfrac{5}{3} - 1)$ of usual time = $4$ hrs
$\Rightarrow (\dfrac{2}{3})$ of usual time = $4$ hrs
$\therefore$ Usual time = $\dfrac{3}{2} \times 4 = 3 \times 2 = 6$ hrs Ans
$\Rightarrow (\dfrac{5}{3})$ of usual time = usual time $+ 4$ hrs
$\Rightarrow (\dfrac{5}{3} - 1)$ of usual time = $4$ hrs
$\Rightarrow (\dfrac{2}{3})$ of usual time = $4$ hrs
$\therefore$ Usual time = $\dfrac{3}{2} \times 4 = 3 \times 2 = 6$ hrs Ans
Q32. A man covers a distance of 40 km in 6 hours, partly at 6 km/hr and partly at 8 km/hr. Find the distance covered 8 km/hr.
Show Answer
$\Rightarrow$ Let distance covered at 8 km/hr be = $x$
$\Rightarrow$ As such, distance covered at 6 km/hr be = $40 - x$
$\Rightarrow$ Time taken to travel $x$ kms = $\dfrac{x}{8}$
$\Rightarrow$ Time taken to travel $40 - x$ kms = $\dfrac{40 - x}{6}$
$\Rightarrow$ So, we have the following equation:
$\Rightarrow \dfrac{x}{8} + \dfrac{40 - x}{6} = 6$
$\Rightarrow 3x + 160 - 4x = 6 \times 24$
$\Rightarrow x + 160 = 144$
$\Rightarrow x = 160 - 144$
$\Rightarrow x = 16$
Hence, distance covered at 8 km/hr = 16 kms
And, distance covered at 6 km/hr = $40 - 16 = 24$ kms
$\Rightarrow$ As such, distance covered at 6 km/hr be = $40 - x$
$\Rightarrow$ Time taken to travel $x$ kms = $\dfrac{x}{8}$
$\Rightarrow$ Time taken to travel $40 - x$ kms = $\dfrac{40 - x}{6}$
$\Rightarrow$ So, we have the following equation:
$\Rightarrow \dfrac{x}{8} + \dfrac{40 - x}{6} = 6$
$\Rightarrow 3x + 160 - 4x = 6 \times 24$
$\Rightarrow x + 160 = 144$
$\Rightarrow x = 160 - 144$
$\Rightarrow x = 16$
Hence, distance covered at 8 km/hr = 16 kms
And, distance covered at 6 km/hr = $40 - 16 = 24$ kms
Q33. A train covers a distance of 72 kms between two stations in 1 hr and 48 mins. Calculate the speed of train?
Show Answer
$\Rightarrow$ Given, distance = 72 kms
$\Rightarrow$ Time = 1 hr. 48 mins.
$\Rightarrow$ Converting Hours into Minutes = $\dfrac{60 + 48}{60}$
$\Rightarrow \dfrac{108}{60} = \dfrac{54}{30} = \dfrac{27}{15} = \dfrac{9}{5}$
$\Rightarrow$ Now, speed = $\dfrac{72}{\dfrac{9}{5}}$
$\Rightarrow \dfrac{72 \times 5}{9} = 8 \times 5 = 40$ km/hr Ans
$\Rightarrow$ Time = 1 hr. 48 mins.
$\Rightarrow$ Converting Hours into Minutes = $\dfrac{60 + 48}{60}$
$\Rightarrow \dfrac{108}{60} = \dfrac{54}{30} = \dfrac{27}{15} = \dfrac{9}{5}$
$\Rightarrow$ Now, speed = $\dfrac{72}{\dfrac{9}{5}}$
$\Rightarrow \dfrac{72 \times 5}{9} = 8 \times 5 = 40$ km/hr Ans
Q34. An electric train can travel at an average speed of 75 km per hour. Find:
(i) How far will it travel in 3 hours 40 minutes?
(ii) How long will it take to cover 600 km?
Show Answer
(i) Given, speed = 75 km/hr
$\Rightarrow$ and, Time = 3 hrs. 40 mins.
$\Rightarrow$ Converting Hours into minutes = $3 \times 60 + \dfrac{40}{60}$
$\Rightarrow 180 + \dfrac{40}{60} = \dfrac{220}{60} = \dfrac{11}{3}$
$\therefore$ Distance = $75 \times \dfrac{11}{3} = 25 \times 11 = 275$ kms Ans
(ii) Given, speed = 75 km/hr
$\Rightarrow$ Distance = 600 kms
$\therefore$ Time = $\dfrac{600}{75} = 80$ hrs Ans
$\Rightarrow$ and, Time = 3 hrs. 40 mins.
$\Rightarrow$ Converting Hours into minutes = $3 \times 60 + \dfrac{40}{60}$
$\Rightarrow 180 + \dfrac{40}{60} = \dfrac{220}{60} = \dfrac{11}{3}$
$\therefore$ Distance = $75 \times \dfrac{11}{3} = 25 \times 11 = 275$ kms Ans
(ii) Given, speed = 75 km/hr
$\Rightarrow$ Distance = 600 kms
$\therefore$ Time = $\dfrac{600}{75} = 80$ hrs Ans
Q35. A train 100 metres long is running at the speed fo 30 km an hour. In what time will it pass:
(i) A man standing near a railway track.
(ii) A bridge 160 metres long?
Show Answer
(i) $\Rightarrow$ Given, lenght of train = 100 m.
$\Rightarrow$ Speed of train = 30 km/hr
$\Rightarrow$ Distance travelled by train to pass a man standing near the railway track == Trains' own lenght
$\Rightarrow$ Distance = 100 m
$\Rightarrow$ time = $\dfrac{100}{30} = \dfrac{10}{3}$ km/hr
$\Rightarrow$ Converting km/hr into m/sec = $\dfrac{10}{3} \times \dfrac{18}{5} = 2 \times 6 = 12$ sec. Ans
(ii) Distance travelled by the train to cross a bridge == trains' own lenght $+$ bridges' lenght
$\Rightarrow$ Distance = $100 + 160 = 260$ kms
$\Rightarrow$ Time = $\dfrac{260}{30} = \dfrac{26}{3}$
$\Rightarrow$ Converting km/hr into m/sec = $\dfrac{26}{3} \times \dfrac{18}{5} = \dfrac{26 \times 6}{5} = \dfrac{156}{5} = 31.2$ sec. Ans
$\Rightarrow$ Speed of train = 30 km/hr
$\Rightarrow$ Distance travelled by train to pass a man standing near the railway track == Trains' own lenght
$\Rightarrow$ Distance = 100 m
$\Rightarrow$ time = $\dfrac{100}{30} = \dfrac{10}{3}$ km/hr
$\Rightarrow$ Converting km/hr into m/sec = $\dfrac{10}{3} \times \dfrac{18}{5} = 2 \times 6 = 12$ sec. Ans
(ii) Distance travelled by the train to cross a bridge == trains' own lenght $+$ bridges' lenght
$\Rightarrow$ Distance = $100 + 160 = 260$ kms
$\Rightarrow$ Time = $\dfrac{260}{30} = \dfrac{26}{3}$
$\Rightarrow$ Converting km/hr into m/sec = $\dfrac{26}{3} \times \dfrac{18}{5} = \dfrac{26 \times 6}{5} = \dfrac{156}{5} = 31.2$ sec. Ans
Q36. A man takes 120 steps a minute. If his stride is 75 cm long. Find his speed in kilometres per hour?
Show Answer
$\Rightarrow$ Speed per minute = $120 \times 75$
$\Rightarrow$ Speed per hour = $120 \times 75 \times 60$
$\Rightarrow$ Speed per hour = $\dfrac{120 \times 75 \times 60}{1000 \times 100}$
$\Rightarrow \dfrac{12 \times 75 \times 6}{10 \times 100} = \dfrac{6 \times 75 \times 6}{5 \times 100} = \dfrac{6 \times 3 \times 6}{5 \times 4}
\Rightarrow \dfrac{3 \times 3 \times 6}{5 \times 2} = \dfrac{3 \times 3 \times 3}{5} = \dfrac{27}{5} = 5.4$ km/hr Ans
$\Rightarrow$ Speed per hour = $120 \times 75 \times 60$
$\Rightarrow$ Speed per hour = $\dfrac{120 \times 75 \times 60}{1000 \times 100}$
$\Rightarrow \dfrac{12 \times 75 \times 6}{10 \times 100} = \dfrac{6 \times 75 \times 6}{5 \times 100} = \dfrac{6 \times 3 \times 6}{5 \times 4}
\Rightarrow \dfrac{3 \times 3 \times 6}{5 \times 2} = \dfrac{3 \times 3 \times 3}{5} = \dfrac{27}{5} = 5.4$ km/hr Ans
Q37. A man takes two steps in walking 120 metres. If he takes 3 steps in 1 second, find his speed in:
(i) m/sec
(ii) km/hr
Show Answer
$\Rightarrow$ Given, a man takes two steps to walk = 120 metres
$\therefore$ Metres covered per step = $\dfrac{120}{150} = \dfrac{4}{5} = 0.8$ m/step
(i) If the man takes 3 steps per minute, then his speed in m/sec = $3 \times 0.8 = 2.4$ m/sec Ans
(ii) Speed in km/hr = $2.4 \times \dfrac{18}{5}$
$\Rightarrow = \dfrac{24}{10} \times \dfrac{18}{5} = \dfrac{24}{5} \times \dfrac{9}{5} = \dfrac{216}{25} = 8.64$ km/hr Ans
$\therefore$ Metres covered per step = $\dfrac{120}{150} = \dfrac{4}{5} = 0.8$ m/step
(i) If the man takes 3 steps per minute, then his speed in m/sec = $3 \times 0.8 = 2.4$ m/sec Ans
(ii) Speed in km/hr = $2.4 \times \dfrac{18}{5}$
$\Rightarrow = \dfrac{24}{10} \times \dfrac{18}{5} = \dfrac{24}{5} \times \dfrac{9}{5} = \dfrac{216}{25} = 8.64$ km/hr Ans
Q38. How long will a man take to walk 600 metres if he walks at the rate of 4 km/hr?
Show Answer
$\Rightarrow$ Given, distance = 600 m
$\Rightarrow$ Given, Time = 4 km/hr
$\Rightarrow$ Converting hours into minutes = $\dfrac{4 \times 1000}{60} = \dfrac{ 1000}{15} = \dfrac{200}{3}$
$\Rightarrow$ Now, Time = $\dfrac{600}{\dfrac{200}{3}} = \dfrac{600 \times 3}{200} = 3 \times 3 = 9$ minutes Ans
$\Rightarrow$ Given, Time = 4 km/hr
$\Rightarrow$ Converting hours into minutes = $\dfrac{4 \times 1000}{60} = \dfrac{ 1000}{15} = \dfrac{200}{3}$
$\Rightarrow$ Now, Time = $\dfrac{600}{\dfrac{200}{3}} = \dfrac{600 \times 3}{200} = 3 \times 3 = 9$ minutes Ans
Q39. A train passes a platform 60 metres long in 20 seconds and a man standing on the platform in 12 seconds. Find the speed of the train in km/hr.
Show Answer
$\Rightarrow$ Let lenght of train be = $x$ metres
$\Rightarrow$ Given, lenght of platform = 60 metres
$\therefore$ Distance = $x + 60$ metres
$\Rightarrow$ Time taken to cross the platform = 20 sec.
$\therefore$ Speed of train = $\dfrac{x + 60}{20}$
$\Rightarrow$ Given, Time taken to cross the man = $\dfrac{x}{12}$
So, we have the equation:
$\Rightarrow \dfrac{x + 60}{20} = \dfrac{x}{12}$
$\Rightarrow 3x + 180 = 5x$ $\text{(L.C.M of 12 and 20 = 60)}$
$\Rightarrow 2x = 180$
$\Rightarrow x = \dfrac{180}{2} = 90$
$\Rightarrow$ Now, total distance = $90 + 60 = 150$
$\Rightarrow$ Speed of train to cross the platform = $\dfrac{150}{20} = \dfrac{15}{2}$
$\Rightarrow$ Converting m/sec to km/hr = $\dfrac{15}{2} \times \dfrac{18}{5} = 3 \times 9 = 27$ km/hr Ans
$\Rightarrow$ Given, lenght of platform = 60 metres
$\therefore$ Distance = $x + 60$ metres
$\Rightarrow$ Time taken to cross the platform = 20 sec.
$\therefore$ Speed of train = $\dfrac{x + 60}{20}$
$\Rightarrow$ Given, Time taken to cross the man = $\dfrac{x}{12}$
So, we have the equation:
$\Rightarrow \dfrac{x + 60}{20} = \dfrac{x}{12}$
$\Rightarrow 3x + 180 = 5x$ $\text{(L.C.M of 12 and 20 = 60)}$
$\Rightarrow 2x = 180$
$\Rightarrow x = \dfrac{180}{2} = 90$
$\Rightarrow$ Now, total distance = $90 + 60 = 150$
$\Rightarrow$ Speed of train to cross the platform = $\dfrac{150}{20} = \dfrac{15}{2}$
$\Rightarrow$ Converting m/sec to km/hr = $\dfrac{15}{2} \times \dfrac{18}{5} = 3 \times 9 = 27$ km/hr Ans
Q40. A can complete the journey in 10 hours, the first half at the rate of 21 km per hour and second half at the rate of 24 km per hour. Find the total journey in kilometres.
Show Answer
$\Rightarrow$ Let total lenght of journey be = $x$ kms
$\Rightarrow$ Given, speed in 1st half of the journey = 21 kms
$\therefore$ Time taken to complete 1st half of the journey = $\dfrac{\dfrac{x}{2}}{21} = \dfrac{x}{2 \times 21} = \dfrac{x}{42}$
Similarly:
$\Rightarrow$ Given, speed in 1st half of the journey = 24 kms
$\therefore$ Time taken to complete 1st half of the journey = $\dfrac{\dfrac{x}{2}}{24} = \dfrac{x}{2 \times 24} = \dfrac{x}{48}$
$\Rightarrow$ Given, total time to complete the journey = 10 hrs
So, we have the equation:
$\Rightarrow \dfrac{x}{42} + \dfrac{x}{48} = 10$
$\Rightarrow 8x + 7x = 3360$ $\text{(L.C.M of 42 and 48 = 336)}$
$\Rightarrow 15x = 3360$
$\Rightarrow x = \dfrac{3360}{15} = 224$ kms Ans
$\Rightarrow$ Given, speed in 1st half of the journey = 21 kms
$\therefore$ Time taken to complete 1st half of the journey = $\dfrac{\dfrac{x}{2}}{21} = \dfrac{x}{2 \times 21} = \dfrac{x}{42}$
Similarly:
$\Rightarrow$ Given, speed in 1st half of the journey = 24 kms
$\therefore$ Time taken to complete 1st half of the journey = $\dfrac{\dfrac{x}{2}}{24} = \dfrac{x}{2 \times 24} = \dfrac{x}{48}$
$\Rightarrow$ Given, total time to complete the journey = 10 hrs
So, we have the equation:
$\Rightarrow \dfrac{x}{42} + \dfrac{x}{48} = 10$
$\Rightarrow 8x + 7x = 3360$ $\text{(L.C.M of 42 and 48 = 336)}$
$\Rightarrow 15x = 3360$
$\Rightarrow x = \dfrac{3360}{15} = 224$ kms Ans
Q41. A man cycles from A to B, a distance of 21 km, in 1 hour 40 minutes. The road from A is level for 13 km and then is up-hill to B. The man's average speed on level is 15 km/hr. Find his average speed up-hill.
Show Answer
$\Rightarrow$ Given, total distance = 21 kms
$\Rightarrow$ Given, total time = 1 hr. 40 min.
$\Rightarrow$ Converting time to minutes = $\dfrac{60 + 40}{60}$
$\Rightarrow \dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$ mins.
$\Rightarrow$ Given, distance of level road = 13 kms
$\Rightarrow$ Given, speed on level road = 15 km/hr
$\therefore$ Time taken to cover level road distance = $\dfrac{13}{15}$
$\Rightarrow$ Distance of uphill road = $21 - 13 = 8$ kms
$\therefore$ Time taken to cover uphill road distance = $\dfrac{5}{3} - \dfrac{13}{15} = \dfrac{25 - 13}{15} = \dfrac{12}{15}$
$\Rightarrow$ Speed to cover uphill road = $\dfrac{8}{\dfrac{12}{15}}$
$\Rightarrow \dfrac{8 \times 15}{12} = \dfrac{120}{12} = 10$ km/hr Ans
$\Rightarrow$ Given, total time = 1 hr. 40 min.
$\Rightarrow$ Converting time to minutes = $\dfrac{60 + 40}{60}$
$\Rightarrow \dfrac{100}{60} = \dfrac{10}{6} = \dfrac{5}{3}$ mins.
$\Rightarrow$ Given, distance of level road = 13 kms
$\Rightarrow$ Given, speed on level road = 15 km/hr
$\therefore$ Time taken to cover level road distance = $\dfrac{13}{15}$
$\Rightarrow$ Distance of uphill road = $21 - 13 = 8$ kms
$\therefore$ Time taken to cover uphill road distance = $\dfrac{5}{3} - \dfrac{13}{15} = \dfrac{25 - 13}{15} = \dfrac{12}{15}$
$\Rightarrow$ Speed to cover uphill road = $\dfrac{8}{\dfrac{12}{15}}$
$\Rightarrow \dfrac{8 \times 15}{12} = \dfrac{120}{12} = 10$ km/hr Ans
Q42. A boy walking at the rate of 6 km/hr takes 2 hours 40 minutes to reach a place. How long will another boy walking at 5 km/hr take to walk double the above distance?
Show Answer
$\Rightarrow$ Given, speed of 1st boy = 6 km/hr
$\Rightarrow$ Time taken by 1st boy to reach destination = 2 hrs. 40 mins.
$\Rightarrow$ Converting time to minutes = $(60 + 60) + \dfrac{60 + 40}{60}$
$\Rightarrow \dfrac{160}{60} = \dfrac{16}{6} = \dfrac{8}{3}$
$\therefore$ Distance covered by 1st boy = $6 \times \dfrac{8}{3} = 2 \times 8 = 16$ kms
$\Rightarrow$ Double the distance = $16 \times 2 = 32$ kms
$\Rightarrow$ Speed of 2nd boy = 5 km/hr
$\Rightarrow$ Time taken by 2nd boy to cover double the distance = $\dfrac{32}{5} = 6.4$
$\therefore 6.4$ converts to 6 hrs and 40 minutes Ans
$\Rightarrow$ Time taken by 1st boy to reach destination = 2 hrs. 40 mins.
$\Rightarrow$ Converting time to minutes = $(60 + 60) + \dfrac{60 + 40}{60}$
$\Rightarrow \dfrac{160}{60} = \dfrac{16}{6} = \dfrac{8}{3}$
$\therefore$ Distance covered by 1st boy = $6 \times \dfrac{8}{3} = 2 \times 8 = 16$ kms
$\Rightarrow$ Double the distance = $16 \times 2 = 32$ kms
$\Rightarrow$ Speed of 2nd boy = 5 km/hr
$\Rightarrow$ Time taken by 2nd boy to cover double the distance = $\dfrac{32}{5} = 6.4$
$\therefore 6.4$ converts to 6 hrs and 40 minutes Ans
Q43. I have to be at a certain place at a certain time and find that I shall be 15 minutes too late, if I walk at 4 km/hr and 10 minutes too soon, if I walk at 6 km/hr. How far I have to walk?
Show Answer
$\Rightarrow$ Let distance be = $x$ kms
$\therefore$ Time taken at speed 4 km/hr = $\dfrac{x}{4}$
$\Rightarrow$ Also, time taken at speed 6 km/hr = $\dfrac{x}{6}$
$\because$ Walking time difference = $15 + 10 = 25$
$\Rightarrow$ Converting min. into hrs. = $\dfrac{25}{60} = \dfrac{5}{12}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{4} - \dfrac{x}{6} = \dfrac{5}{12}$
$\Rightarrow 3x - 2x = 5x \text{(L.C.M of 4, 6 and 12 = 12)}$
$\Rightarrow x = 5$
Hence, required total distance = $5$ kms Ans
$\therefore$ Time taken at speed 4 km/hr = $\dfrac{x}{4}$
$\Rightarrow$ Also, time taken at speed 6 km/hr = $\dfrac{x}{6}$
$\because$ Walking time difference = $15 + 10 = 25$
$\Rightarrow$ Converting min. into hrs. = $\dfrac{25}{60} = \dfrac{5}{12}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{4} - \dfrac{x}{6} = \dfrac{5}{12}$
$\Rightarrow 3x - 2x = 5x \text{(L.C.M of 4, 6 and 12 = 12)}$
$\Rightarrow x = 5$
Hence, required total distance = $5$ kms Ans
Q44. One-third of a certain distance is covered at 10 km/hr and the reamining at 20 km/hr. Find the average speed for the whole distance covered.
Show Answer
$\Rightarrow$ Let distance be = $x$ kms
$\Rightarrow$ Let average speed for whole distance be = $y$ kms
$\Rightarrow$ Time taken to complete $\dfrac{1}{3}$ of journey = $\dfrac{x}{3 \times 10} = \dfrac{x}{30}$
$\because$ Remaining journey = $1 - \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\therefore$ Time taken complete $\dfrac{2}{3}$ of journey = $\dfrac{2x}{3 \times 20} = \dfrac{2x}{60}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{30} + \dfrac{2x}{60} = \dfrac{x}{y}$
$\Rightarrow \dfrac{2x + 2x}{60} = \dfrac{x}{y}$
$\Rightarrow \dfrac{4x}{60} = \dfrac{x}{y}$
$\Rightarrow y(4x) = 60x$
$\Rightarrow y = \dfrac{60x}{4x} = 15$
Hence, required average speed = $15$ km/hr Ans
$\Rightarrow$ Let average speed for whole distance be = $y$ kms
$\Rightarrow$ Time taken to complete $\dfrac{1}{3}$ of journey = $\dfrac{x}{3 \times 10} = \dfrac{x}{30}$
$\because$ Remaining journey = $1 - \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\therefore$ Time taken complete $\dfrac{2}{3}$ of journey = $\dfrac{2x}{3 \times 20} = \dfrac{2x}{60}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{30} + \dfrac{2x}{60} = \dfrac{x}{y}$
$\Rightarrow \dfrac{2x + 2x}{60} = \dfrac{x}{y}$
$\Rightarrow \dfrac{4x}{60} = \dfrac{x}{y}$
$\Rightarrow y(4x) = 60x$
$\Rightarrow y = \dfrac{60x}{4x} = 15$
Hence, required average speed = $15$ km/hr Ans
Q45. A bus travels the first $\dfrac{1}{3}$ of a certain distance with a speed of 20 km/hr, the next $\dfrac{1}{3}$ distance with a speed of 40 km/hr and the last $\dfrac{1}{3}$ distance with a speed of 120 km/hr. Find average speed of bus for the whole journey?
Show Answer
$\Rightarrow$ Let distance for whole journey be = $x$ kms
$\Rightarrow$ Let average speed for whole distance be = $y$ kms
$\Rightarrow$ Time taken to complete first $\dfrac{1}{3}$ of journey at 20 km/hr = $\dfrac{x}{3 \times 20} = \dfrac{x}{60}$
$\Rightarrow$ Time taken to complete second $\dfrac{1}{3}$ of journey at 40 km/hr = $\dfrac{x}{3 \times 40} = \dfrac{x}{120}$
$\Rightarrow$ Time taken to complete third $\dfrac{1}{3}$ of journey at 120 km/hr = $\dfrac{x}{3 \times 120} = \dfrac{x}{360}$
$\Rightarrow$ Given, total time taken = $\dfrac{x}{y}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{60} + \dfrac{x}{120} + \dfrac{x}{360} = \dfrac{x}{y}$
$\Rightarrow \dfrac{6x + 3x + x}{360} = \dfrac{x}{y}$
$\Rightarrow \dfrac{10x}{360} = \dfrac{x}{y}$
$\Rightarrow y(10x) = 360x$
$\Rightarrow y = \dfrac{360x}{10x} = 36$
Hence, required average speed = $36$ km/hr Ans
$\Rightarrow$ Let average speed for whole distance be = $y$ kms
$\Rightarrow$ Time taken to complete first $\dfrac{1}{3}$ of journey at 20 km/hr = $\dfrac{x}{3 \times 20} = \dfrac{x}{60}$
$\Rightarrow$ Time taken to complete second $\dfrac{1}{3}$ of journey at 40 km/hr = $\dfrac{x}{3 \times 40} = \dfrac{x}{120}$
$\Rightarrow$ Time taken to complete third $\dfrac{1}{3}$ of journey at 120 km/hr = $\dfrac{x}{3 \times 120} = \dfrac{x}{360}$
$\Rightarrow$ Given, total time taken = $\dfrac{x}{y}$
As such, we have the equation:
$\Rightarrow \dfrac{x}{60} + \dfrac{x}{120} + \dfrac{x}{360} = \dfrac{x}{y}$
$\Rightarrow \dfrac{6x + 3x + x}{360} = \dfrac{x}{y}$
$\Rightarrow \dfrac{10x}{360} = \dfrac{x}{y}$
$\Rightarrow y(10x) = 360x$
$\Rightarrow y = \dfrac{360x}{10x} = 36$
Hence, required average speed = $36$ km/hr Ans