# Train and Platform Aptitude Questions & Formulas:

#### 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 (3): When a moving object is crossing another moving object and length of both moving objects are considered.

(a). When objects (may be trains) are moving in the same direction.$$\left[T = \frac{l_1 + l_2}{s_1 - s_2}\right]$$

Where,
$$T$$ = Time taken.
$$l_1$$ = Length of first moving object.
$$l_2$$ = Length of second moving object.
$$s_1$$ = Speed of first moving object.
$$s_2$$ = Speed of second moving object.

Example (1):Two trains, $$100 \ m$$ and $$150 \ m$$ long, moving in the same direction at the speed of $$600 \ m/sec$$ and $$500 \ m/sec$$ respectively. how much time the faster train will take to cross the slower train?

Solution: Given values,
length of first train $$(l_1) = 100 \ m$$
length of second train $$(l_2) = 150 \ m$$
speed of first train $$(s_1) = 600 \ m/sec$$
speed of second train $$(s_2) = 500 \ m/sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2}\right]$$ $$\left[Time \ taken \ (T) = \frac{100 + 150}{600 - 500}\right]$$ $$Time \ taken \ (T) = \frac{250}{100} = 2.5 \ sec$$

Example (2):Two trains, $$x \ m$$ and $$180 \ m$$ long, moving in the same direction at the speed of $$1200 \ m/sec$$ and $$1000 \ m/sec$$ respectively. If faster train takes $$8 \ sec$$ to cross the slower train, then find the value of $$x$$?

Solution: Given values,
length of first train $$(l_1) = x \ m$$
length of second train $$(l_2) = 180 \ m$$
speed of first train $$(s_1) = 1200 \ m/sec$$
speed of second train $$(s_2) = 1000 \ m/sec$$, time taken by the faster train to cross the slow train $$8 \ sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2}\right]$$ $$\left[8 = \frac{x + 180}{1200 - 1000}\right]$$ $$8 \times 200 = x + 180$$ $$x = 1420 \ m$$

(b). When objects (may be trains) are moving in opposite direction.$$\left[T = \frac{l_1 + l_2}{s_1 + s_2}\right]$$

Where,
$$T$$ = Time taken.
$$l_1$$ = Length of first moving object.
$$l_2$$ = Length of second moving object.
$$s_1$$ = Speed of first moving object.
$$s_2$$ = Speed of second moving object.

Example (1): A $$100 \ m$$ long train moving from Kanpur to Delhi at the speed of $$20 \ km/hr$$, and another $$50 \ m$$ long train moving fron Delhi to Kanpur at the speed of $$25 \ km/hr$$. How much time trains will take to pass each other?

Solution: Given values,
length of first train $$(l_1) = 100 \ m$$
length of second train $$(l_2) = 50 \ m$$
speed of first train $$(s_1) = 20 \ km/hr = 20 \times \frac{5}{18} \ m/sec$$
speed of second train $$(s_2) = 25 \ km/hr = 25 \times \frac{5}{18} \ m/sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2}\right]$$ $$\left[Time \ taken \ (T) = \frac{100 + 50}{(20 + 25) \times \frac{5}{18}}\right]$$ $$Time \ taken \ (T) = \frac{150}{45 \times \frac{5}{18}}$$ $$Time \ taken \ (T) = \frac{150 \times 18}{45 \times 5}$$ $$Time \ taken \ (T) = \frac{2700}{225} = 12 \ seconds$$

Example (2): A $$1000 \ m$$ long train moving from station $$A$$ to station $$B$$ at the speed of $$35 \ km/hr$$, and another $$1500 \ m$$ long train moving fron station $$B$$ to station $$A$$ at the speed of $$45 \ km/hr$$. How much time trains will take to pass each other?

Solution: Given values,
length of first train $$(l_1) = 1000 \ m$$
length of second train $$(l_2) = 1500 \ m$$
speed of first train $$(s_1) = 35 \ km/hr = 35 \times \frac{5}{18} \ m/sec$$
speed of second train $$(s_2) = 45 \ km/hr = 45 \times \frac{5}{18} \ m/sec$$, then $$\left[Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2}\right]$$ $$\left[Time \ taken \ (T) = \frac{1000 + 1500}{(35 + 45) \times \frac{5}{18}}\right]$$ $$Time \ taken \ (T) = \frac{2500}{80 \times \frac{5}{18}}$$ $$Time \ taken \ (T) = \frac{2500 \times 18}{80 \times 5}$$ $$Time \ taken \ (T) = \frac{45000}{400} = 112.5 \ sec$$