# Shortcut Methods to Solve Problems on Train- Part 2

## Effective for IBPS PO - SBI PO Exam

**Dear Reader,**

Here we will start a series of Quantitative Aptitude Shortcut Tricks for your upcoming SBI - IBPS - SSC and Other Government Competitive Exams. We will try to cover up all topics of the quantitative Aptitude Sections from which question was generally asked.

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**Trick - 1**

- Two trains of the same length but with different speeds pass a static pole in ${{t}_{1}}$ seconds and ${{t}_{2}}$ seconds respectively. They are moving in the same direction. The time they will take to cross each other is given by

$\left(
\frac{2{{t}_{1}}{{t}_{2}}}{{{t}_{2}}-{{t}_{1}}} \right)\sec onds$

- Two trains of the length ${{l}_{1}}$ m and ${{l}_{2}}$ m respectively with different speeds pass a static pole in ${{t}_{1}}$ seconds and ${{t}_{2}}$ seconds respectively. When are moving in the opposite direction, they will cross each other in

$\left[
\frac{\left( {{l}_{1}}+{{l}_{2}}
\right){{t}_{1}}{{t}_{2}}}{{{t}_{2}}{{l}_{1}}+{{t}_{1}}{{l}_{2}}} \right]\sec
onds$

**Trick - 3**

- If a train l metres long moving at a speed of x km/hr crosses another train in t seconds, then the time taken by it to cover its own length is given by

$\frac{18}{5}\left(
\frac{l}{x} \right)$

**Case 1 :**If $\frac{18}{5}\left( \frac{l}{x} \right)$ <t, trains are moving in the same case.

**Case 2 :**If $\frac{18}{5}\left( \frac{l}{x} \right)$ >t, trains are moving in opposite directions.

**Trick - 4**

- A train running at x km/hr takes ${{t}_{1}}$ seconds to pass a platform. Next it takes ${{t}_{2}}$ seconds to pass a man walking at y km/hr in the same direction, then the length of the train is $\left[ \frac{5}{18}\left( x-y \right){{t}_{2}} \right]$ metres and that of platform is $\frac{5}{18}\left[ x\left( {{t}_{1}}-{{t}_{2}} \right)+y{{t}_{2}} \right]$ metres.

**Trick -5**

- L meters long train crosses a bridge of length ${{L}_{1}}$ metres in T seconds. Time taken by train to cross a platform of ${{L}_{2}}$ metres is given by

$\left[
\frac{L+{{L}_{2}}}{L+{{L}_{1}}} \right]T$ seconds

**Trick -6**

- Two trains are moving in opposite directions at x km/hr and y km/hr (where x>y) if the faster train crosses a man in the slower train in t seconds, then the length of the faster train is given by

$\left[
\frac{5}{18}\left( x+y \right)t \right]metres$

**Questions for Practice**

**Q1. Two trains of the same length but with different speeds pass a static pole in 4 seconds and 5 seconds respectively. In what time will they cross each other when they are moving in the same direction ?**

Q2. Two trains of the length 100 m and 125 m respectively with different speeds pass a static pole in 4 seconds and 7 seconds respectively. In what time will they cross each other when they are moving in the opposite direction

Q2. Two trains of the length 100 m and 125 m respectively with different speeds pass a static pole in 4 seconds and 7 seconds respectively. In what time will they cross each other when they are moving in the opposite direction

**Q3.**

**A train 105 m long moving at a speed of 54 km per/hr crosses another train in 6 seconds. Then which of the following is true ?**

**(a) Trains are moving in the same direction.**

**(b)**

**Trains are moving in the opposite direction.**

**(c) The other train is not moving.**

**(d) Data inadequate**

**Q4. A train running at 25 km/hr takes 18 seconds to pass a platform. Next it takes 9 seconds to pass a man walking at 5 km/hr in the same direction. Find the length of the train and that of platform.**

**Q5.**

**150 meters long train crosses a platform of length 250 metres in 30 seconds. Find the time for train to cross a bridge of 130 metres.**

**Q6. Two trains are moving in opposite directions at 50 km/hr and 40 km/hr. The faster train crosses a man in the slower train in 3 seconds. Find the length of the faster train**.

__Answers__**Answer 1. 40 seconds**

**Answer 2. 5.25 seconds**

**Answer 3. B is the correct answer**

**Answer 4. 75 m**

**Answer 5. 21 sec**

**Answer 6. 75 m**

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