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### Question 1

A train running at 4km/h leaves a railway station 21 hours later than another train, and meets it in 7 hours. What is the speed of the other train?

### Question 2

Two trains coming from opposite direction and running respectively at 24km/h and 48km/h cross each other in 11sec. What is the length of each train if the two trains are equal?

**A**

110 m.

**B**

111 m.

**C**

109 m.

**D**

112 m.

**Soln.**

**Ans: a**

The trains cover a distance equal to the sum of their lengths at a relative speed 24 + 48 = 72km/h × (5/18), or 20m/s. We can use the speed distance formula: sum of lengths = 20 × 11 = 220m. Halving this we get the length of one train = 110m.

### Question 3

A train is running at a speed of 56m/s. If it takes 3sec to move past a telegraph pole, then what is its length?

### Question 4

Two trains running in opposite directions cross each other in 31 seconds. They, respectively, take 14 and 42 seconds to cross a man standing on the platform. What is the ratio of their speeds?

**A**

$1{6/11}$.

**B**

$2{4/5}$.

**C**

${6/13}$.

**D**

$3{11/13}$.

**Soln.**

**Ans: a**

Let the ratio of their speeds by r. If the speed of one train is v, then the speed of the other is rv. By the speed and distance formula, the sum of their lengths is $(v × 14) + (rv × 42)$ which should equal the value obtained from the time they take to cross each other,i.e., $(v + rv) × 31)$. So $v × (14 + r × 42$ = $v × (1 + r) × 31).$ Cancelling v and solving for r we get ${17/11}$.

### Question 5

A train is running at a speed of 82km/h. If it takes 36sec to move past a telegraph pole, then what is its length?

**A**

820 meters.

**B**

821 meters.

**C**

819 meters.

**D**

822 meters.

**Soln.**

**Ans: a**

The total distance to be covered is equal to the length of the train. By the time and distance formula, we get length = time × speed × (5/18), which gives $36 × 82 × (5/18)$ = 820m. Please note that 5/18 is the conversion from km/h to m/s.

This Blog Post/Article "Problems on Trains Quiz Set 003" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Updated on 2020-02-07. Published on: 2016-05-13