Q16. Two persons are A and B are 350 metres apart. If A can run at the speed of 7 m/sec and B can run at the speed of 3 m/sec; find in how much time will they meet each other, if:
(i) They run in same direction?
(ii) They run in opposite direction?
Show Answer
(i) Given distance = 350 metres
$\Rightarrow$ A's speed = 7 m/sec
$\Rightarrow$ B's speed = 3 m/sec
$\Rightarrow$ If A and B run in the same direction, then their relative speed = $7 - 3 = 4$ m/sec
$\therefore$ time taken to meet eachother = $\dfrac{350}{4} = 87.5$ seconds
$\Rightarrow$ 87.5 seconds = 60 sec and 87.5 - 60 = 1 min. and 27.5 sec Ans
(ii) If A and B run in the same direction, then their relative speed = $7 + 3 = 10$ m/sec
$\Rightarrow$ time taken to meet eachother = $\dfrac{350}{10} = 35$ seconds Ans
$\Rightarrow$ A's speed = 7 m/sec
$\Rightarrow$ B's speed = 3 m/sec
$\Rightarrow$ If A and B run in the same direction, then their relative speed = $7 - 3 = 4$ m/sec
$\therefore$ time taken to meet eachother = $\dfrac{350}{4} = 87.5$ seconds
$\Rightarrow$ 87.5 seconds = 60 sec and 87.5 - 60 = 1 min. and 27.5 sec Ans
(ii) If A and B run in the same direction, then their relative speed = $7 + 3 = 10$ m/sec
$\Rightarrow$ time taken to meet eachother = $\dfrac{350}{10} = 35$ seconds Ans
Q17. Two boys start running together from the same place with speeds 6 km/hr and 4 km/hr. Find the distance between them after 6 minutes, if they run:
(i) In the same direction
(ii) In the opposite direction
Show Answer
(i) Given time = 6 min
$\Rightarrow$ Converting time into Hrs. = $\dfrac{6}{60} = \dfrac{1}{10}$
$\Rightarrow$ 1st boys' speed = 6 km/hr
$\Rightarrow$ 2nd boys' speed = 4 km/hr
$\Rightarrow$ If 2 boys run in the same direction, then their relative speed = $6 - 4 = 2$ km/hr
$\therefore$ Distance between them after 6 minutes = $2 \times \dfrac{1}{10} = \dfrac{2}{10} = 0.2$ km/hr Ans
(ii) If 2 boys run in opposite direction, then their relative speed = $6 + 4 = 10$ km/hr
$\therefore$ Distance between them after 6 minutes = $10 \times \dfrac{1}{10} = 1$ km/hr Ans
$\Rightarrow$ Converting time into Hrs. = $\dfrac{6}{60} = \dfrac{1}{10}$
$\Rightarrow$ 1st boys' speed = 6 km/hr
$\Rightarrow$ 2nd boys' speed = 4 km/hr
$\Rightarrow$ If 2 boys run in the same direction, then their relative speed = $6 - 4 = 2$ km/hr
$\therefore$ Distance between them after 6 minutes = $2 \times \dfrac{1}{10} = \dfrac{2}{10} = 0.2$ km/hr Ans
(ii) If 2 boys run in opposite direction, then their relative speed = $6 + 4 = 10$ km/hr
$\therefore$ Distance between them after 6 minutes = $10 \times \dfrac{1}{10} = 1$ km/hr Ans
Q18. P and Q are two railway stations 560 km apart. An express train leaves station P for station Q at the speed of 55 km/hr and a passenger train leaves station Q for station P at the same time but at the speed of 25 km/hr. Find, when and where will they meet.
Show Answer
$\Rightarrow$ Given distance = 560 km
$\Rightarrow$ Speed of express train = 55 km/hr
$\Rightarrow$ Speed of passenger train = 25 km/hr
$\because$ trains are travelling in opposite direction
$\therefore$ Relative speed = 55 + 25 = 80 km/hr
$\Rightarrow$ Time after which 2 trains will meet = $\dfrac{560}{80} = 7$ hrs Ans
$\Rightarrow$ distance they will meet from station P = $55 \times 7 = 385$ kms Ans
$\Rightarrow$ Speed of express train = 55 km/hr
$\Rightarrow$ Speed of passenger train = 25 km/hr
$\because$ trains are travelling in opposite direction
$\therefore$ Relative speed = 55 + 25 = 80 km/hr
$\Rightarrow$ Time after which 2 trains will meet = $\dfrac{560}{80} = 7$ hrs Ans
$\Rightarrow$ distance they will meet from station P = $55 \times 7 = 385$ kms Ans
Q19. A walks at the speed of 5 km/hr along a railway track. A train coming from behind him at 23 km/hr passes by him in 35 seconds. Find the lenght of the train.
Show Answer
$\Rightarrow$ Speed of walking man = 5 km/hr
$\Rightarrow$ Speed of train = 23 km/hr
$\because$ The man and train are moving in the same direction
$\therefore$ Relative speed = 23 - 5 = 18 km/hr
$\Rightarrow$ Converting time into m/sec = $18 \times \dfrac{5}{18} = 5$ m/sec
$\Rightarrow$ Given time = 35 sec
$\therefore$ Lenght of train = $35 \times 5 = 175$ Ans
$\Rightarrow$ Speed of train = 23 km/hr
$\because$ The man and train are moving in the same direction
$\therefore$ Relative speed = 23 - 5 = 18 km/hr
$\Rightarrow$ Converting time into m/sec = $18 \times \dfrac{5}{18} = 5$ m/sec
$\Rightarrow$ Given time = 35 sec
$\therefore$ Lenght of train = $35 \times 5 = 175$ Ans
Q20. A train, 125 m long, travelling at a uniform speed, passes a stationary man in 3.75 seconds. Find the speed of the train in km/hr.
Show Answer
$\Rightarrow$ Given, Distance = lenght of train = 125 metres
$\Rightarrow$ Given, time = 3.75 sec
$\therefore$ Speed = $\dfrac{125}{3.75} = \dfrac{125 \times 100}{375} = \dfrac{100}{3}$
$\Rightarrow$ Speed in km/hr = $\dfrac{100}{3} \times \dfrac{18}{5} = 20 \times 6 = 120$ km/hr Ans
$\Rightarrow$ Given, time = 3.75 sec
$\therefore$ Speed = $\dfrac{125}{3.75} = \dfrac{125 \times 100}{375} = \dfrac{100}{3}$
$\Rightarrow$ Speed in km/hr = $\dfrac{100}{3} \times \dfrac{18}{5} = 20 \times 6 = 120$ km/hr Ans
Q21. A train, 225 m long, travels at a uniform speed of 54 km/hr. How much time will it take to pass:
(i) A telegraph post?
(ii) A 150 m long platform?
(iii) A 105 m long bridge?
Show Answer
(i) Distance = 225 kms
$\Rightarrow$ Speed = 54 km/hr
$\Rightarrow$ Converting speed to m/sec = $54 \times \dfrac{5}{18} = 3 \times 5 = 15$ m/sec
$\therefore$ Time taken to pass a telegraph post = $\dfrac{225}{15} = 15$ seconds Ans
(ii) Distance = 225 + 150 = 375 kms
$\Rightarrow$ Speed = 15 km/hr
$\therefore$ Time = $\dfrac{375}{15} = 25$ seconds Ans
(iii) Distance = 225 + 105 = 330 kms
$\Rightarrow$ Speed = 15 km/hr
$\therefore$ Time = $\dfrac{330}{15} = 22$ seconds Ans
$\Rightarrow$ Speed = 54 km/hr
$\Rightarrow$ Converting speed to m/sec = $54 \times \dfrac{5}{18} = 3 \times 5 = 15$ m/sec
$\therefore$ Time taken to pass a telegraph post = $\dfrac{225}{15} = 15$ seconds Ans
(ii) Distance = 225 + 150 = 375 kms
$\Rightarrow$ Speed = 15 km/hr
$\therefore$ Time = $\dfrac{375}{15} = 25$ seconds Ans
(iii) Distance = 225 + 105 = 330 kms
$\Rightarrow$ Speed = 15 km/hr
$\therefore$ Time = $\dfrac{330}{15} = 22$ seconds Ans
Q22. A train, 140 m long, passes a telegrah in 14 seconds. Find:
(i) Its speed in km/hr.
(ii) Time taken by it to pass a 160 m long platform.
Show Answer
(i) Distance = 140 metres
$\Rightarrow$ Time = 14 sec
$\therefore$ Speed = $\dfrac{140}{14} = 10$ m/sec
$\Rightarrow$ Speed in km/hr = $10 \times \dfrac{18}{5} = 36$ km/hr Ans
(ii) Distance = 140 + 160 = 300 kms
$\Rightarrow$ Speed = 10 m/sec
$\therefore$ Time = $\dfrac{300}{10} = 30$ seconds Ans
$\Rightarrow$ Time = 14 sec
$\therefore$ Speed = $\dfrac{140}{14} = 10$ m/sec
$\Rightarrow$ Speed in km/hr = $10 \times \dfrac{18}{5} = 36$ km/hr Ans
(ii) Distance = 140 + 160 = 300 kms
$\Rightarrow$ Speed = 10 m/sec
$\therefore$ Time = $\dfrac{300}{10} = 30$ seconds Ans
Q23. A person is cycling, parallel to a railway track, at the speed of 10 km/hr. A train travelling in the same direction at 46 km/hr, passes him in 22 seconds. Find the lenght of the train?
Show Answer
$\Rightarrow$ Speed of a cycling person = 10 km/hr
$\Rightarrow$ Speed of train = 46 km/hr
$\because$ The cycling person and train are travelling in same direction
$\therefore$ Relative speed = $46 - 10 = 36$ km/hr
$\Rightarrow$ Converting speed to m/sec = $36 \times \dfrac{5}{18} = 10$ m/sec
$\Rightarrow$ Given, time = 22 seconds
$\therefore$ Lenght of train = $22 \times 10 = 220$ metres Ans
$\Rightarrow$ Speed of train = 46 km/hr
$\because$ The cycling person and train are travelling in same direction
$\therefore$ Relative speed = $46 - 10 = 36$ km/hr
$\Rightarrow$ Converting speed to m/sec = $36 \times \dfrac{5}{18} = 10$ m/sec
$\Rightarrow$ Given, time = 22 seconds
$\therefore$ Lenght of train = $22 \times 10 = 220$ metres Ans
Q24. The speed of a boat in still water is 8 km/hr and the speed of the stream is 3 km/hr. Find:
(i) Time taken by boat to go 55 km downstream;
(ii) Time taken by boat to go 30 km upstream;
(iii) Distance covered by the boat in 12 hrs downstream.
Show Answer
$\Rightarrow$ Given, speed of boat in still water = 8 km/hr
$\Rightarrow$ Speed of stream = 3 km/hr
(i) Speed of boat downstream = $8 + 3 = 11$ km/hr
$\Rightarrow$ Distance to be covered = 55 kms
$\therefore$ Time taken = $\dfrac{55}{11} = 5$ hrs Ans
(ii) Speed of boat upstream = $8 - 3 = 5$ km/hr
$\Rightarrow$ Distance to be covered = 30 kms
$\therefore$ Time taken = $\dfrac{30}{5} = 6$ hrs Ans
(iii) Speed downstream = 11 km/hr
$\Rightarrow$ Time = 12 hrs
$\therefore$ Distance covered = $12 \times 11 = 132$ kms Ans
$\Rightarrow$ Speed of stream = 3 km/hr
(i) Speed of boat downstream = $8 + 3 = 11$ km/hr
$\Rightarrow$ Distance to be covered = 55 kms
$\therefore$ Time taken = $\dfrac{55}{11} = 5$ hrs Ans
(ii) Speed of boat upstream = $8 - 3 = 5$ km/hr
$\Rightarrow$ Distance to be covered = 30 kms
$\therefore$ Time taken = $\dfrac{30}{5} = 6$ hrs Ans
(iii) Speed downstream = 11 km/hr
$\Rightarrow$ Time = 12 hrs
$\therefore$ Distance covered = $12 \times 11 = 132$ kms Ans
Q25. A boat can travel with a speed of 22 km/hr in still water. If the speed of the sream is 8 km/hr; find:
(i) Time taken by it to go 135 kms downstream;
(ii) Time taken by it to go 105 kms downstream and return.
Show Answer
$\Rightarrow$ Given, speed of boat in still water = 22 km/hr
$\Rightarrow$ and, speed of stream = 8 km/hr
(i) Speed of boat downstream = $22 + 8 = 30$ km/hr
$\Rightarrow$ Distance to be covered = 135 kms
$\therefore$ Time taken = $\dfrac{135}{30} = \dfrac{9}{2} = 4\dfrac{1}{2}$ hrs Ans
(ii) Speed of boat downstream = 30 km/hr
$\Rightarrow$ Speed of boat upstream = $22 - 8 = 14$ km/hr
$\Rightarrow$ Time taken to cover the distance downstream = $\dfrac{105}{30} = \dfrac{21}{6} = \dfrac{7}{2} = 3\dfrac{1}{2}$ hrs
$\Rightarrow$ Time taken to cover the distance upstream = $\dfrac{105}{14} = \dfrac{15}{2} = 7\dfrac{1}{2}$ hrs
$\therefore$ Total time taken:
$\Rightarrow 3\dfrac{1}{2} + 7\dfrac{1}{2} = \dfrac{7}{2} + \dfrac{15}{2} = \dfrac{7 + 15}{2} = \dfrac{22}{2} = 11$ hrs Ans
$\Rightarrow$ and, speed of stream = 8 km/hr
(i) Speed of boat downstream = $22 + 8 = 30$ km/hr
$\Rightarrow$ Distance to be covered = 135 kms
$\therefore$ Time taken = $\dfrac{135}{30} = \dfrac{9}{2} = 4\dfrac{1}{2}$ hrs Ans
(ii) Speed of boat downstream = 30 km/hr
$\Rightarrow$ Speed of boat upstream = $22 - 8 = 14$ km/hr
$\Rightarrow$ Time taken to cover the distance downstream = $\dfrac{105}{30} = \dfrac{21}{6} = \dfrac{7}{2} = 3\dfrac{1}{2}$ hrs
$\Rightarrow$ Time taken to cover the distance upstream = $\dfrac{105}{14} = \dfrac{15}{2} = 7\dfrac{1}{2}$ hrs
$\therefore$ Total time taken:
$\Rightarrow 3\dfrac{1}{2} + 7\dfrac{1}{2} = \dfrac{7}{2} + \dfrac{15}{2} = \dfrac{7 + 15}{2} = \dfrac{22}{2} = 11$ hrs Ans
Q26. A train is running at a uniform speed passes a bridge 275 m long in 15 seconds and another bridge 42 m long in 21 seconds. Find:
(i) Lenght of train.
(ii) Speed of train in km/hr.
Show Answer
(i) $\because$ It is said that the train travels at a uniform speed
$\therefore$ The train will pass the 2 bridges at the same speed
$\Rightarrow$ Let trains' lenght be = $x$ metres
$\Rightarrow$ We now the equation:
$\Rightarrow \dfrac{x + 275}{15} = \dfrac{x + 425}{21}$
$\Rightarrow 21(x + 275) = 15(x + 425)$
$\Rightarrow 21x + 5775 = 15x + 6375$
$\Rightarrow 21x - 15x = 6375 - 5775$
$\Rightarrow 6x = 600$
$\Rightarrow x = \dfrac{600}{6} = 100$ metres Ans
(ii) Speed of train = $\dfrac{100 + 275}{15} = \dfrac{375}{15} = \dfrac{75}{3} = 25$ m/sec
$\Rightarrow$ Converting speed into km/hr = $25 \times \dfrac{18}{5} = 5 \times 18 = 90$ km/hr Ans
$\therefore$ The train will pass the 2 bridges at the same speed
$\Rightarrow$ Let trains' lenght be = $x$ metres
$\Rightarrow$ We now the equation:
$\Rightarrow \dfrac{x + 275}{15} = \dfrac{x + 425}{21}$
$\Rightarrow 21(x + 275) = 15(x + 425)$
$\Rightarrow 21x + 5775 = 15x + 6375$
$\Rightarrow 21x - 15x = 6375 - 5775$
$\Rightarrow 6x = 600$
$\Rightarrow x = \dfrac{600}{6} = 100$ metres Ans
(ii) Speed of train = $\dfrac{100 + 275}{15} = \dfrac{375}{15} = \dfrac{75}{3} = 25$ m/sec
$\Rightarrow$ Converting speed into km/hr = $25 \times \dfrac{18}{5} = 5 \times 18 = 90$ km/hr Ans
Q27. The speed of a boat in still water is 10 km/hr. If the boat goes 37.5 km upstream in 5 hours; find the speed of the stream.
Show Answer
$\Rightarrow$ Let the speed of stream be = $x$ km/hr
$\Rightarrow$ Given, speed of boat in still water = 10km/hr
$\Rightarrow$ Distance covered upstream = 37.5 kms
$\Rightarrow$ Time taken = 5 hrs
$\Rightarrow$ We have the equation:
$\Rightarrow 10 - x = \dfrac{37.5}{5} = \dfrac{375}{5 \times 10} = \dfrac{75}{10} = 7.5$
$\Rightarrow x = 10 - 7.5 = 2.5$ km/hr Ans
$\Rightarrow$ Given, speed of boat in still water = 10km/hr
$\Rightarrow$ Distance covered upstream = 37.5 kms
$\Rightarrow$ Time taken = 5 hrs
$\Rightarrow$ We have the equation:
$\Rightarrow 10 - x = \dfrac{37.5}{5} = \dfrac{375}{5 \times 10} = \dfrac{75}{10} = 7.5$
$\Rightarrow x = 10 - 7.5 = 2.5$ km/hr Ans
Q28. The speed of a boat in still water is 9 km/hr. If the boat goes 54 km downstream in 5 hours; find the speed of the stream.
Show Answer
$\Rightarrow$ Let speed of steam be = $x$ km/hr
$\Rightarrow$ Given, speed of still water = 9 km/hr
$\Rightarrow$ Distance covered downstream = 54 kms
$\Rightarrow$ Time taken = 4 hrs
$\Rightarrow$ We have following the equation:
$\Rightarrow 9 + x = \dfrac{54}{4}$
$\Rightarrow 4(9 + x) = 54$
$\Rightarrow 36 + 4x = 54$
$\Rightarrow 4x = 54 - 36 = 18$
$\Rightarrow x = \dfrac{18}{4} = 4.5$ km/hr Ans
$\Rightarrow$ Given, speed of still water = 9 km/hr
$\Rightarrow$ Distance covered downstream = 54 kms
$\Rightarrow$ Time taken = 4 hrs
$\Rightarrow$ We have following the equation:
$\Rightarrow 9 + x = \dfrac{54}{4}$
$\Rightarrow 4(9 + x) = 54$
$\Rightarrow 36 + 4x = 54$
$\Rightarrow 4x = 54 - 36 = 18$
$\Rightarrow x = \dfrac{18}{4} = 4.5$ km/hr Ans
Q29. Two trains start at the same time from two stations and proceed towards each other with speeds 40 km/hr and 45 km/hr. When they meet, it is found that one train has travelled 40 km more than the other. Find the distance between the two stations.
Show Answer
$\Rightarrow$ Let time to reach the meeting point by both trains be = $x$
$\Rightarrow$ So, we have the equation:
$\Rightarrow 40 \times x = 45 \times x + 40$
$\Rightarrow 40x = 45x + 40$
$\Rightarrow 45x - 40x = 40$
$\Rightarrow 5x = 40$
$\Rightarrow x = \dfrac{40}{5} = 8$ hrs
$\because$ Time taken by 2 trains to reach meeting point is 8 hrs
$\therefore$ Distance travelled by the 1st train = $40 \times 8 = 320$ kms
$\Rightarrow$ Distance travelled by the 2nd train = $45 \times 8 = 360$ kms
$\therefore$ Distance between 2 stations = $320 + 360 = 680$ kms Ans
$\Rightarrow$ So, we have the equation:
$\Rightarrow 40 \times x = 45 \times x + 40$
$\Rightarrow 40x = 45x + 40$
$\Rightarrow 45x - 40x = 40$
$\Rightarrow 5x = 40$
$\Rightarrow x = \dfrac{40}{5} = 8$ hrs
$\because$ Time taken by 2 trains to reach meeting point is 8 hrs
$\therefore$ Distance travelled by the 1st train = $40 \times 8 = 320$ kms
$\Rightarrow$ Distance travelled by the 2nd train = $45 \times 8 = 360$ kms
$\therefore$ Distance between 2 stations = $320 + 360 = 680$ kms Ans
Q30. A train travelling at 60 km/hr leaves station A at 3.00 PM and another train leaves the same station in the same direction at 5.00 PM at the speed of 75 km/hr. How many kms from station A will the two trains be together?
Show Answer
$\Rightarrow$ Let distance where both trains will be together be = $x$
$\therefore$ Time taken by 1st train to cover $x$ kms = $\dfrac{x}{60}$
$\Rightarrow$ Time taken by 2nd train to cover $x$ kms = $\dfrac{x}{75}$
$\Rightarrow$ Time gap between 2 trains = $5$PM $- 3$PM = $2$ hrs
$\Rightarrow$ As such, 1st train will take 2 hrs more to cover $x$ kms
$\Rightarrow$ So, we have the equation:
$\Rightarrow \dfrac{x}{60} - \dfrac{x}{75} = 2$
$\Rightarrow \dfrac{5x - 4x}{300} = 2$
$\Rightarrow \dfrac{x}{300} = 2$
$\Rightarrow x = 2 \times 300 = 600$ kms Ans
$\therefore$ Time taken by 1st train to cover $x$ kms = $\dfrac{x}{60}$
$\Rightarrow$ Time taken by 2nd train to cover $x$ kms = $\dfrac{x}{75}$
$\Rightarrow$ Time gap between 2 trains = $5$PM $- 3$PM = $2$ hrs
$\Rightarrow$ As such, 1st train will take 2 hrs more to cover $x$ kms
$\Rightarrow$ So, we have the equation:
$\Rightarrow \dfrac{x}{60} - \dfrac{x}{75} = 2$
$\Rightarrow \dfrac{5x - 4x}{300} = 2$
$\Rightarrow \dfrac{x}{300} = 2$
$\Rightarrow x = 2 \times 300 = 600$ kms Ans