# Shortcut Methods to Solve Problems on Train- Part 2

## Effective for IBPS PO - SBI PO Exam

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

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

Trick - 2
• 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

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.

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.