Train

Concept

The math of train is the integral part of the concept of Time, Distance and Speed related questions. This type of questions is a little bit different from the basic speed, distance and time questions and requires somewhat different approach to solve.

The concept of the math of trains is based on the principles of mechanics, which describe how objects move and interact with each other. The math of trains involves the study of forces, motion, energy, and momentum. It is used to understand how trains move along tracks, how they accelerate and decelerate, and how they can be controlled. Basically this types of maths is required to calculate the average speed that is required by the train to cover a particular distance, time at a certain speed to cover a distance and distance covered at a certain speed within a time period.

Tips to solve problems on Train:

The only basic formula: Total distance covered by the train = Speed of the Train x Time taken
If the lengths of two trains are D1 and D2 and speed of them S1 and S2 are given, then the time taken by trains to cross each other = $\displaystyle \left(\frac{D1+D2}{S1+S2}\right)$

if they goes in opposite direction.
If the lengths of two trains are D1 and D2 and speed of them S1 and S2 are given, then the time taken by trains to cross each other = $\displaystyle \left(\frac{D1+D2}{S1-S2}\right)$
if they goes in same direction.

Q. A train of length 120 m passes a pole in 6 seconds. What is the speed of the train in km/h?

A) 72 km/h

The speed of the train is $\displaystyle \frac{120}{6} = 20 \text{ m/s}.$ Converting to km/h, we get $\displaystyle \left(20 \times \frac{18}{5}\right) = 72 \text{ km/h}.$

B) 80 km/h

C) 88 km/h

D) 96 km/h

Discuss

Q. A train of length 150 m crosses a platform of length 200 m in 15 seconds. Another train of length 180 m crosses the same platform in 18 seconds. What is the ratio of the speeds of the two trains?

A) 4:5 `

B) 5:6

The speed of the first train is $\displaystyle \frac{150+200}{15} = 23.33 \text{ m/s}$ . The speed of the second train is $\displaystyle \frac{180+200}{18} = 21.11 \text{ m/s}$ The ratio of the speeds is 23.33:21.11, which can be simplified to 5:6.

C) 6:7

D) 7:9

Discuss

Q. Two trains, A and B, start from stations X and Y, respectively, and travel towards each other at constant speeds of 60 km/h and 40 km/h, respectively. When they meet, train A has traveled 120 km more than train B. What is the distance between stations X and Y?

A) 360 km

B) 400 km

C) 440 km

Let t be the time taken by the trains to meet. Then, we have: • (60t) + (40t) = d, where d is the distance between X and Y. • Also, we have: • (60t) - (40t) = 120, since train A has traveled 120 km more than train B. • Solving these two equations simultaneously, we get: • t = 3 hours and d = 440 km.

D) 480 km

Discuss

Q. A train travels at a constant speed of 60 km/h from station A to station B, which are 300 km apart. The train stops for 10 minutes at station B and then returns to station A at a constant speed of 80 km/h. What is the average speed of the train for the entire trip?

A) 66 km/h

B) 68 km/h

The time taken by the train to go from A to B is $\displaystyle \frac{300}{60} = 5$ The time taken by the train to return from B to A is $\displaystyle \frac{300}{80} = 3.75$ hours. The total time for the trip is $\displaystyle 5 + \frac{10}{60} + 3.75 = 9$ hours. The total distance covered by the train is 300 + 300 = 600 km. The average speed of the train is $\displaystyle \frac{600}{9} = 66.67 \text{ km/h}$ , which is closest to option B.

C) 70 km/h

D) 72 km/h

Discuss

Q. Two trains, X and Y, start from opposite ends of a 120 km long track and move towards each other at constant speeds of 40 km/h and 50 km/h, respectively. At the same time, a bird starts flying from the front of train X towards train Y at a constant speed of 60 km/h. When the bird reaches train Y, it turns around and flies back to train X, and so on, until the trains collide. How far does the bird fly in total?

A) 72 km

B) 80 km

C) 90 km

[The trains will collide after $\displaystyle \frac{120}{40+50} = 1.5$ hours. In this time, the bird will fly at a constant speed of 60 km/h, so it will cover a distance of $\displaystyle 60 \times 1.5 = 90 \text{ km}$

D) 100 km

Discuss

Q. A train leaves station P at noon and travels at a constant speed of x km/h towards station Q, which is 180 km away. Another train leaves station Q at 1 pm and travels at a constant speed of y km/h towards station P. The two trains meet at station R, which is between P and Q. If x > y and the distance from P to R is three times the distance from Q to R, what is the value of x/y?

A) 2

B) 3

C) 4

[Let t be the time taken by the first train to reach R from P. Then, the time taken by the second train to reach R from Q is (t-1). Since the distance from P to R is three times the distance from Q to R, we have: xt = 3y*(t-1). Simplifying, we get: $\displaystyle \frac{x}{y} = \frac{3t-3}{t}$ , Since x > y, we have: $\displaystyle \frac{3t-3}{t} > 1$ Now, we get t > $\displaystyle t > \frac{3+\sqrt{5}}{2}$ or t is less than $\displaystyle \frac{3-\sqrt{5}}{2}$ . Since t must be positive, we reject the second option. Substituting $\displaystyle t = \frac{3+\sqrt{5}}{2} \text{ in } \frac{x}{y} = \frac{3t-3}{t}$ , we get: $\displaystyle \frac{x}{y} = \frac{9+3\sqrt{5}-6}{3+\sqrt{5}}$ . Simplifying, we get: $\displaystyle \frac{x}{y} = \frac{3+\sqrt{5}}{1+\sqrt{5}}$ . Rationalizing, we get: $\displaystyle \frac{x}{y} = \frac{4+2\sqrt{5}}{2}$ Dividing by two, we get: $\displaystyle \frac{x}{y} = 2+\sqrt{5}$ This is closest to option C.]

D) 5

Discuss

Q. A train has a length of 150 m and crosses a bridge of length 250 m in t seconds. If the speed of the train is increased by 20%, how long will it take to cross a tunnel of length 350 m?

A) $\displaystyle \frac{t \times 7}{6}$ seconds

B) $\displaystyle \frac{t \times 8}{7}$ seconds

C) $\displaystyle \frac{t \times 9}{8}$ seconds

D) $\displaystyle \frac{t \times 10}{9}$ seconds

The original speed of the train is $\displaystyle \frac{150+250}{t} = \frac{400}{t} \text{ m/s}$ . The increased speed of the train is $\displaystyle \frac{400}{t} \times 1.2 = \frac{480}{t} \text{ m/s}$ . The time taken by the train to cross the tunnel is $\displaystyle \frac{150+350}{\frac{480}{t}} = \frac{500t}{480} \text{ seconds}$ . Simplifying, we get: $\displaystyle \frac{500t}{480} = \frac{t \times 25}{24} \text{ seconds}$ . This is equivalent to: $\displaystyle \frac{t \times 10}{9} \text{ seconds}$

Discuss

Q. A train travels at a constant speed of 72 km/h on a straight track. A car starts from a point 60 km behind the train and accelerates uniformly from 0 to 108 km/h in 30 seconds. How long will it take for the car to overtake the train?

A) 50 Seconds

B) 60 Seconds

The acceleration of the car is $latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B108 0%7D%7B30%7D+%3D+3$ . The relative speed of the car with respect to the train is $\displaystyle 108-72 = 36 \text{ km/h}$ . The distance between the car and the train when they start moving is 60 km. Using the formula s = ut + (1/2)at^2 , where s is the distance, u is the initial speed, a is the acceleration and t is the time, we get: • $\displaystyle 60 = 0 \times t + \frac{1}{2} \times 3.6t^2$ • Simplifying, we get: • $\displaystyle t^2 = 33.33$ • Taking the positive root, we get: • $\displaystyle t^2 = 33.33$ • This is approximately equal to: • $\displaystyle t = 5.77 \times 3600 \text{ seconds}$ • Converting to seconds, we get: • $\displaystyle t = 5.77 \times 3600$ seconds • This is approximately equal to: • $\displaystyle t = 20772$ seconds • This is the time taken by the car to cover the initial distance of 60 km. After that, the car will overtake the train in $\displaystyle \frac{60}{36} = 1.67 \text{ hours}$ , which is equivalent to $\displaystyle 1.67 \times 3600 = 6006$ seconds. The total time taken by the car to overtake the train is 20772 + 6006 = 26778 seconds, which is closest to option B.

C) 70 Seconds

D) 80 Seconds

Discuss

Q. A train travels at a constant speed of x km/h for the first half of its journey and at a constant speed of y km/h for the second half of its journey. The total time taken by the train is t hours and the total distance covered by the train is d km. What is the value of x/y in terms of t and d?

A) $\displaystyle \frac{t^2}{4d}$

B) $\displaystyle \frac{4d}{t^2}$

C) $\displaystyle \frac{t^2}{2d}$

D) $\displaystyle \frac{2d}{t^2}$

Let ( h ) be the time taken by the train for the first half of its journey. Then, the time taken by the train for the second half of its journey is ( t-h ). Since the distance covered by the train in each half is equal to $\displaystyle \frac{d}{2}$ , we have: The time ( x ) multiplied by ( h ) equals half the distance, $\displaystyle xh = \frac{d}{2}$ . The time ( y ) multiplied by ( t-h ) also equals half the distance, $\displaystyle y(t-h) = \frac{d}{2}$ . Solving for the ratio of ( x ) to ( y ), we get: The ratio $\displaystyle \frac{x}{y}$ is equal to the half distance divided by the product of ( t-h ) and ( h ), $\displaystyle \frac{x}{y} = \frac{\frac{d}{2}}{(t-h)h}$ . Simplifying, we find that $\displaystyle \frac{x}{y}$ is equal to the distance divided by twice the product of ( t ), ( h ), minus twice the square of ( h ), $\displaystyle \frac{x}{y} = \frac{d}{2th-2h^2}$ . Substituting ( h ) with $\displaystyle \frac{t}{2}$ , we get $\displaystyle \frac{x}{y} = \frac{d}{\frac{t^2}{2}}$ , $\displaystyle \frac{x}{y} = \frac{d}{\frac{t^2}{2}}$ . Multiplying by two, we conclude that $\displaystyle \frac{x}{y}$ is equal to twice the distance divided by the square of ( t ), $\displaystyle \frac{x}{y} = \frac{2d}{t^2}$ .

Discuss

Q. A train leaves station A at noon and travels at a constant speed of a km/h towards station B, which is b km away. Another train leaves station B at 1 pm and travels at a constant speed of c km/h towards station A. If a > c and b > c, at what time will the two trains meet?

A) $\displaystyle \left(\frac{b-c}{a+c}\right) \times 60$ minutes past noon

B) $\displaystyle \left(\frac{b-c}{a-c}\right) \times 60$ minutes past noon

Let t be the time taken by the first train to meet the second train from noon. Then, the time taken by the second train to meet the first train from 1 pm is (t-1). Since the distance covered by the trains is equal when they meet, we have: • at = c(t-1) + b • Simplifying, we get: • (a-c)*t = b-c • Solving for t, we get: • $\displaystyle t = \frac{b-c}{a-c}$ • Converting to minutes, we get: • $\displaystyle t = \left(\frac{b-c}{a-c}\right) \times 60$ minutes •Adding to noon, we get the time of meeting as: • $\displaystyle \left(\frac{b-c}{a-c}\right) \times 60$ minutes past noon

C) $\displaystyle \left(\frac{b-a}{a+c}\right) \times 60$ minutes past noon

D) $\displaystyle \left(\frac{b-a}{a-c}\right) \times 60$ minutes past noon

Discuss

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