# Train and Platform Aptitude Formulas & Examples:

#### Overview:

 Topic Included: Formulas, Definitions & Exmaples. Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Time Speed and Distance Aptitude Notes & Questions. Questions for practice: 10 Questions & Answers with Solutions.

We are discussing different cases of trains and platforms here to understand the scenario easily.

Case (4): When a moving object is crossing another moving object and length of second moving object is not considered.

(a). When objects (may be trains) are moving in same direction.$$\left[T = \frac{l_1}{s_1 - s_2}\right]$$

Where,
$$T$$ = Time taken.
$$l_1$$ = Length of first moving object.
$$s_1$$ = Speed of first moving object.
$$s_2$$ = Speed of second moving object.

Example (1): A $$100 \ m$$ long train and a $$10 \ m$$ car moving parallel in same direction, at the speed of $$30 \ km/hr$$ and $$20 \ km/hr$$ respectively. If the length of car is not considered then find how much time the train will take to pass the car?

Solution: Given values,
length of the train $$(l_1) = 100 \ m$$
Speed of the train $$(s_1) = 30 \ km/hr = 30 \times \frac{5}{18} \ m/sec$$
speed of the car $$(s_2) = 20 \ km/hr = 20 \times \frac{5}{18} \ m/sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1}{s_1 - s_2}\right]$$ $$\left[Time \ taken \ (T) = \frac{100}{(30 - 20) \ \times \frac{5}{18}}\right]$$ $$Time \ taken \ (T) = \frac{100 \times 18}{10 \times 5}$$ $$Time \ taken \ (T) = \frac{1800}{50} = 36 \ sec.$$

Example (2): Two $$x \ m$$, $$120 \ m$$ long trains moving parallel in same direction, at the speed of $$80 \ km/hr$$ and $$60 \ km/hr$$, respectively. If first train passed the second train in $$65 \ sec$$ while the length of second train is not considered then find the value of $$x$$?

Solution: Given values,
length of the train $$(l_1) = x \ m$$
Speed of the first train $$(s_1) = 80 \ km/hr = 80 \times \frac{5}{18} \ m/sec$$
speed of the second train $$(s_2) = 60 \ km/hr = 60 \times \frac{5}{18} \ m/sec$$, time taken by the first train to pass the second train $$(T) = 65 \ sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1}{s_1 - s_2}\right]$$ $$\left[65 = \frac{x}{(80 - 60) \times \frac{5}{18}}\right]$$ $$65 = \frac{x \times 18}{20 \times 5}$$ $$6500 = 18 \ x$$ $$x = 361.11 \ m$$

(b). When objects (may be trains) are moving in opposite direction.$$\left[T = \frac{l_1}{s_1 + s_2}\right]$$

Where,
$$T$$ = Time taken.
$$l_1$$ = Length of first moving object.
$$s_1$$ = Speed of first moving object.
$$s_2$$ = Speed of second moving object.

Example (1): A $$200 \ m$$ long train and a $$10 \ m$$ car moving parallel in opposite direction, at the speed of $$50 \ km/hr$$ and $$25 \ km/hr$$, respectively. If the length of car is not considered then at what time the car will pass the train?

Solution: Given values,
length of the first train $$(l_1) = 200 \ m$$
Speed of the train $$(s_1) = 50 \ km/hr = 50 \times \frac{5}{18} \ m/sec$$
speed of the car $$(s_2) = 25 \ km/hr = 25 \times \frac{5}{18} \ m/sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}\right]$$ $$\left[Time \ taken \ (T) = \frac{200}{(50 + 25) \ \times \frac{5}{18}}\right]$$ $$Time \ taken \ (T) = \frac{200 \times 18}{75 \times 5}$$ $$Time \ taken \ (T) = \frac{3600}{375} = 9.6 \ sec.$$

Example (2): Two $$x \ m$$, $$80 \ m$$ long trains moving parallel in opposite direction, at the speed of $$40 \ km/hr$$ and $$30 \ km/hr$$, respectively. If first train passed the second train in $$5 \ sec$$ while the length of second train is not considered then find the value of $$x$$?

Solution: Given values,
length of the first train $$(l_1) = x \ m$$
Speed of the first train $$(s_1) = 40 \ km/hr = 40 \times \frac{5}{18} \ m/sec$$
speed of the second train $$(s_2) = 30 \ km/hr = 30 \times \frac{5}{18} \ m/sec$$, time taken by the first train to pass the second train $$(T) = 5 \ sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}\right]$$ $$\left[5 = \frac{x}{(40 + 30) \times \frac{5}{18}}\right]$$ $$5 = \frac{x \times 18}{70 \times 5}$$ $$1750 = 18 \ x$$ $$x = 97.22 \ m$$