Train and Platform Aptitude Formulas, Definitions, & 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 (1): When a moving object (may be train) is crossing a stationary object (may be platform) and length of stationary object is considered then $$\left[T = \frac{l_m + l_s}{s_m}\right]$$

Where,
$$T$$ = Time taken.
$$l_m$$ = Length of moving object.
$$l_s$$ = Length of stationary object.
$$s_m$$ = Speed of moving object.

Example (1): A $$150$$ meter long train is crossing a 50 meter platform, at the speed of $$100 \ km/hr$$. How much time the train will take to cross the platform?

Solution: Given values, Length of the train $$(l_m) = 150 \ m$$, length of the platform $$(l_s) = 50 \ m$$, and Speed of the train $$(s_m) = 100 \ km/hr$$ = $$100 \times \frac{5}{18}$$ \ meter/sec, then $$\left[Time \ taken \ (T) = \frac{l_m + l_s}{s_m}\right]$$ $$\left[Time \ taken \ (T) = \frac{150 + 50}{100 \times \frac{5}{18}}\right]$$ $$Time \ taken \ (T) = \frac{200 \times 18}{500} = 7.2 \ sec$$

Example (2): A $$500$$ meter long train crossed a platform in $$30 \ sec$$, at the speed of $$80 \ km/hr$$. find out the length of the platform?

Solution: Given values, Length of the train $$(l_m) = 500 \ m$$, time taken by the train to cross the platform $$(T) = 30 \ sec$$, and Speed of the train $$(s_m) = 80 \ km/hr$$ = $$80 \times \frac{5}{18}$$ \ meter/sec, then $$\left[Time \ taken \ (T) = \frac{l_m + l_s}{s_m}\right]$$ $$\left[30 = \frac{500 + l_s}{80 \times \frac{5}{18}}\right]$$ $$\frac{30 \times 80 \times 5}{18} = 500 + l_s$$ $$l_s = 166.67 \ meter$$

Example (3): A $$650$$ meter long train crossed a $$120 \ m$$ platform in $$12 \ sec$$, then find out the speed of the train?

Solution: Given values, Length of the train $$(l_m) = 650 \ m$$, time taken by the train to cross the platform $$(T) = 12 \ sec$$ length of the platform $$(l_s) = 120 \ m$$, then Speed of the train, $$\left[Time \ taken \ (T) = \frac{l_m + l_s}{s_m}\right]$$ $$\left[12 = \frac{650 + 120}{s_m}\right]$$ $$s_m = \frac{770}{12} = 64.167 \ m/sec$$

Example (4): A train crossed a $$250 \ m$$ platform in $$45 \ sec$$, at the speed of $$180 \ km/hr$$. find out the length of the train?

Solution: Given values, Length of the platform $$(l_s) = 250 \ m$$, time taken by the train to cross the platform $$(T) = 45 \ sec$$, and Speed of the train $$(s_m) = 180 \ km/hr$$ = $$180 \times \frac{5}{18}$$ \ meter/sec, then $$\left[Time \ taken \ (T) = \frac{l_m + l_s}{s_m}\right]$$ $$\left[45 = \frac{l_m + 250}{180 \times \frac{5}{18}}\right]$$ $$\frac{45 \times 180 \times 5}{18} = l_m + 250$$ $$l_m = 2,000 \ meter$$