Problems on Trains: Concept
Problems on Trains, like concepts of speed, distance, and time, focus on evaluating a train’s speed, distance traveled, and time spent under various conditions.
The weightage of questions asked from this topic in the quantitative aptitude component of various government exams frequently ranges between 1 or 2 marks, and the most common exams that contain this topic in its syllabus include Bank, Insurance, SSC, RRB, and other important government exams.
Important formulas
Problems on Trains formulas |
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| The train’s speed is calculated as | $$ \frac{the \ total \ distance \ reached}{divided \ by \ the \ time \ taken} $$ |
| If the length of two trains is specified, say a and b, and the trains are moving in opposite directions with speeds of x and y, then the time taken for trains to cross each other | $$\frac{(a+b)}{(x+y)}$$ |
| If the length of two trains is specified, say a and b, and they are moving in the same direction, with speeds x and y consequently, the time taken to cross each other is | $$\frac{(a+b)}{(x-y)}$$ |
| When the starting time of two trains is the same from x and y towards each other, and after crossing each other, they needed t1 and t2 time to reach y and x respectively, then the ratio between the speed of two trains | $$ √t2: √t1 $$ |
| If two trains leave x and y stations at times t1 and t2, respectively, and travel at speeds L and M, then the distance from x where two trains meet is | $$ = (t2 – t1) × \frac {product \ of \ speed}{difference\ in \ speed} $$ |
| If a train’s average speed without a stoppage is x and it covers the same distance with a stoppage at an average speed of y, then the rest time per hour | $$ = \frac {difference\ in \ average\ speed}{speed \ without \ stoppage}$$ |
| If two trains of equal lengths and different speeds take t1 and t2 time to pass a pole, then the time taken for them to cross each other if they are running in opposing directions | $$ = \frac{2×t1×t2}{t2+t1}$$ |
| If two trains of equal lengths and different speeds take t1 and t2 time to cross a pole, then the time taken for them to cross each other if they are heading in the same direction | $$ =\frac{2×t1×t2}{t2-t1} $$ |
Tips & Tricks for Solving Train Problems
To help hopefuls prepare for and conquer the quantitative aptitude section, here are a few pointers that will help you solve train problems faster and more efficiently:
- Always read the question carefully and do not hasten to answer it, as train-based questions are frequently given in a complex manner.
- After reading the questions, try to apply a formula to them; this may result in a direct solution and save you time.
- Don’t guess if you’re not sure. Since there is negative marking in competitive exams, make sure you don’t make assumptions and answer the questions.
- If you are unable to answer, do not overcomplicate the question or devote too much effort to solving it.
- In case of confusion, you can also refer to the options supplied in the objective type papers and try to discover the answer using the options given.
Practice Quizzes
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Problems on Trains Quiz 1 – Coming Soon |
Problems on Trains Quiz 2 – Coming Soon | Problems on Trains Quiz 3 – Coming Soon |
Sample Questions: Problems on Trains
Q1. A train takes 90 seconds to cross a poll with speed of 60 km/h. Find the length of the train?
a) 1500 m
b) 1250 m
c) 1350 m
d) 1400 m
Answer: (A)
Speed of train = 60 km/h × 5/18 × 90 =1500m
Q2. A train crosses a pole in 15 sec and a 100 m long platform in 25 sec, what is the length of the train?
a) 90 m
b) 120 m
c) 150 m
d) 180 m
Answer: C
Let the speed of train be x m/s
Length of the train = 15×x = 15x
Train crosses 100 m long platform in 25 sec
Total length which the train is covering (15x + 100) m
Time taken = 25 sec
15x + 100 = 25 × x
100 = 10x
10 m/s = x
Length of the train = 10 × 15 = 150 m
Q3. How many poles a train crosses when travelling with a speed of 45 kmph in 4 hour. If the distance between two poles is 50 m and poles are counting from starting.
a) 1001
b) 3601
c) 3000
d) 1801
Answer: B
Distance = Speed × Time = 180 × 1000 = 180000
No. of poles = 180000/50 = 3600
Q4. A train travel with a speed of 7 kmph and another train travel with a speed of 14 kmph. If they cross each other then find the average speed.
a) 17/3 kmph
b) 28/3 kmph
c) 29/3 kmph
d) 19/3 kmph
Answer: B
Average Speed = (2 × 7 × 14)/ (7 + 14)
= 196/21
= 28/3 kmph
Q5. A train starts from station A at the speed of 50 km/hr and another train starts from station B after 30 min at the speed of 150 km/hr towards each other. Find the distance of their meeting point from Station A. The distance between A and B is 725 km.
a) 200 km
b) 250 km
c) 225 km
d) None of these
Answer: A
Distance travelled by first train in ½ hour = 50 × ½ = 25 km
Remaining distance = 725 – 25 = 700
Time = 700/(150+50) = 3.5 hour
Relative speed = 150 + 50 = 200 kmph
Distance travelled by first train in (3.5+0.5) hour = 50 × 4 = 200 km
Q6. Train A of length 250 m crosses a pole in 25 seconds. Train B travelling at the same speed crosses a 400 m long platform in 1 minute. Find the time taken by train B to cross train A, if the train A is stationary.
a) 45 seconds
b) 42 seconds
c) 40 seconds
d) 48 seconds
Answer: A
Speed of train A = 250/25 = 10 m/sec
So, speed of train B = 10 m/sec
Let the le o B ‘x’ .
Then, (x + 400)/10 = 60
x = 200 m
Total length of both trains = (250 + 200) = 450 m
So, time taken by train B to cross train A = 450/10 = 45 seconds.
Q7. At starting 1/3rd of passenger left and 96 passengers boarding the train, again 1/2 of passenger left and 12 passengers boarding the train now there are total 248 passengers are in the train. Find how many passengers are there at starting?
a) 865
b) 664
c) 564
d) 789
Answer: C
According to the question,
248 – 12 = 236
1/2 of passenger left the train it means passenger = 2 ×236 = 472
If 96 passengers already board the train then = 472 – 96 =376
1/3 rd of passenger left the train it means 2/3x = 376
x = 188 × 3 = 564
So, initially there are total 564 passengers.
Q8. The Length of train A is 200 m and length of train B is 400 m. They travel in same direction they will take 30 sec and they travel in opposite direction they will take 6 sec to reach the destination. Find the total speed of the train.
a) 6000 km/hr
b) 3000 km/hr
c) 7200 km/hr
d) 3600 km/hr
Answer: D
Speed = Distance/time
(s1 + s2)/ (s1 – s2) = 5/1
s1/ s2 = 3/2
When train travel in opposite direction,
600 = 5x × 6
600 = 30x
x = 600/30
x = 20
Total speed = 5x = 5 × 20 = 100 m/sec = 1000 × 18/5 = 3600 km/hr
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Frequently Asked Questions About Problems on Trains
Q1. What is the definition of a train problem?
Problems on Trains: Concept, Tips, Tricks, and Examples
Train Problems – A Basic Concept. Train problems, like the notion of speed, distance, and time, are centered on evaluating a train’s speed, distance traveled, and time spent under various conditions.
Q2. What is the formula for speed and time train problems?
The formula t = d/s indicates that time equals distance divided by speed, which ultimately yields the time measured in train problems. Ans. To find the speed in train problems, use the formula speed = distance divided by time.
Q3. What is the most important formula for a train?
Remember some important formulas for train problems to get quick solutions.
x km/hr = x*(5/18) m/s, but x m/s equals x*(18/5) km/hr.
The time it takes for a train of length/meters to pass a pole, a single post, or a standing man is equivalent to the time it takes to cover/meters.
