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51. The Ghaziabad - Hapur - Meerut EMU and the Meerut - Hapur - Ghaziabad EMU start at the same time from Ghaziabad and Meerut and proceed towards each other at 16 km/hr and 21 km/hr respectively. When they meet, it is found that one train has traveled 60 km more than the other . The distance between two stations is?

A. 440 km

B. 444 km

C. 445 km

D. 450 km

Discuss

Solution:

\begin{aligned} At the time of meeting , \\ let the distance travelled by the \\ first train be x km . \\ Then distance travelled by the \\ second train is (x + 60) km \\ ∴ \frac{x}{16} = \frac{x + 60}{21} \\ \Rightarrow 21 x = 16 x + 960 \\ \Rightarrow 5 x = 960 \Rightarrow x = 192 \\ Hence, \\ distance between two stations \\ =  (192 + 192 + 60) km \\ =  444 km . \end{aligned}

52. Two trains start simultaneously (with uniform speeds) from two stations 270 km apart, each to the opposite station; they reach their destinations in $6 \frac{1}{4}$ hours and 4 hours after they meet. The rate at which the slower train travels is :

A. 16 km/hr

B. 24 km/hr

C. 25 km/hr

D. 30 km/hr

Discuss

Solution:

\begin{aligned} Ratio of speeds \\ = \sqrt{4} : \sqrt{6 \frac{1}{4}} \\ = \sqrt{4} : \sqrt{\frac{25}{4}} \\ = 2 : \frac{5}{2} \\ = 4 : 5 \end{aligned}

Let the speeds of the two trains be 4x and 5x km/hr respectively
Then time taken by trains to meet each other

\begin{aligned} = (\frac{270}{4 x + 5 x}) hr \\ = (\frac{270}{9 x}) hr  = (\frac{30}{x}) hr \\ Time taken by slower train to travel \\ 270 km  = (\frac{270}{4 x}) hr \\ ∴ \frac{270}{4 x} = \frac{30}{x} + 6 \frac{1}{4} \\ \Rightarrow \frac{270}{4 x} - \frac{30}{x} = \frac{25}{4} \\ \Rightarrow \frac{150}{4 x} = \frac{25}{4} \\ \Rightarrow 100 x = 600 \\ \Rightarrow x = 6 \\ Hence speed of slower train \\ =  4 x \\ = 24 km/hr \end{aligned}

53. Two trains, A ans B start from stations X and Y towards each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach Y and X respectively after they meet. If train A is moving at 45 km/hr, then the speed of the train B is?

A. 60 km/hr

B. 64.80 km/hr

C. 54 km/hr

D. 37.5 km/hr

Discuss

Solution:

\begin{aligned} In these type of questions use the given \\ below formula to save your valuable time \\ \frac{S_{1}}{S_{2}} = \sqrt{\frac{T_{2}}{T_{1}}} \\ Where S_{1}, S_{2} and T_{1}, T_{2} are the respective \\ speeds and times of the objects \\ \Rightarrow \frac{45}{S_{2}} = \sqrt{3 \frac{1}{3} \div 4 \frac{4}{5}} \\ = S_{2} =  45 \times \frac{6}{5} =  54 km/hr \\ ∴ Required speed  =  54 km/hr \end{aligned}

54. A train passes by a lamp post at platform in 7 sec. and passes by the platform completely in 28 sec. If the length of the platform is 390m, then length of the train (in meters) is?

A. 120 m

B. 130 m

C. 140 m

D. 150 m

Discuss

Solution:

Length of train

= \frac{Length of the platform}{Difference in time}

× (Time taken to cross a lamp post)

\begin{aligned} = \frac{390}{28 - 7} \times 7 \\ = \frac{390}{21} \times 7 \\ = \frac{390}{3} \\ = 130 m \end{aligned}

55. A train moving at a rate of 36 km/hr crosses a standing man in 10 seconds. It will cross a platform 55 meters long in?

A. 6 second

B. 7 second

C. $15 \frac{1}{2}$ second

D. $5 \frac{1}{2}$ second

Discuss

Solution:

\begin{aligned} Length of the train \\ =  Speed \times time \\ =  36 km/hr \times 10 sec \\ =  36 \times \frac{5}{18} m/s \times 10 \sec \\ = 100 metres \\ Therefore, \\ Time taken by train to cross a plateform \\ of 55 metre long in time \\ = \frac{(100 + 55)}{36 \times \frac{5}{18}} \\ = \frac{155}{10} \\ Time = 15 \frac{1}{2} \sec \end{aligned}

56. Two trains start at the same time for two station A and B toward B and A respectively. If the distance between A and B is 220 km and their speeds are 50 km/hr and 60 km/hr respectively then after how much time will they meet each other?

A. 2 hr

B. $2 \frac{1}{2}$ hr

C. 3 hr

D. 1 hr

Discuss

Solution:

\begin{aligned} Relative speed \\ =   60  +  50 \\ =  110 km/h \\ Time taken \\ = \frac{220}{110} \\ =  2 hr \end{aligned}

57. A train 100 meter long meets a man going in opposite direction at 5 km/h and passes him in 7¹/₅ seconds. What is the speed of the train (in km/hr)?

A. 45 km/h

B. 60 km/h

C. 55 km/hr

D. 50 km/hr

Discuss

Solution:

\begin{aligned} Relative speed of man &  train \\ = \frac{100 \times 5}{36} \times \frac{18}{5} \\ =  50km/hr \\ ∴ speed of train \\ =  50 - 5 \\ =  45 km/hr \end{aligned}

58. A train takes 9 sec to cross a pole. If the speed of the train is 48 kmph, then length of the train is?

A. 150 m

B. 120 m

C. 90 m

D. 80 m

Discuss

Solution:

\begin{aligned} Time taken by train to cross a pole \\ =  9 sec \\ Distance covered in crossing a pole \\ =  length of train \\ Speed of the train \\ =  48 km/h \\ = (\frac{48 \times 5}{18}) m / \sec \\ = \frac{40}{3} m / \sec \\ ∴ Length of the train \\ =   Speed \times Time \\ = \frac{40}{3} \times 9 \\ =  120 m \end{aligned}

59. Two trains start at the same time from A and B and proceed toward each other at the sped of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have traveled 175 km more then the other. Find the distance between A and B?

A. 875 km

B. 785 km

C. 758 km

D. 857 km

Discuss

Solution:

\begin{aligned} Let the trains meet after t hours \\ Speed of train A \\ =  75 km/hr \\ Speed of train B \\ =  50 km/hr \\ Distance covered by train A \\ =  75 \times t  =  75t \\ Distance covered by train B \\ =  50 \times t  =  50t \\ Distance =  Speed \times Time \\ According to question \\ 75 t - 50 t = 175 \\ \Rightarrow 25 t = 175 \\ \Rightarrow t = \frac{175}{25} = 7 hour \\ ∴ Distance between A and B \\ =  75t + 50 t = 125 t \\ = 125 \times 7 = 875 km \end{aligned}

60. Two trains 180 meters and 120 meters in length are running towards each other on parallel tracks, one at the rate 65 km/hr and another at 55 km/hr. In how many seconds will they be cross each other from the moment they meet?

A. 6 seconds

B. 9 seconds

C. 12 seconds

D. 15 seconds

Discuss

Solution:

\begin{aligned} Time taken by trains to cross each \\ other in opposite direction \\ = \frac{l_{1} + l_{2}}{relative speed in opposite direction} \\ = \frac{(180 + 120)}{(65 + 55)} \\ = \frac{300}{120 \times \frac{5}{18}} \\ = 9 seconds \end{aligned}

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