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 $$ \left[Time \ taken \ (T) = \frac{l_m + l_s}{s_m}\right] $$ $$ \left[45 = \frac{l_m + 250}{180 \times \frac{5}{18}}\right] $$ $$ \frac{45 \times 180 \times 5}{18} = l_m + 250 $$ $$ l_m = 2,000 \ meter $$

Lec 1: Introduction to Time, Speed and Distance
Exercise-1
Lec 2: Concept of Train and Platform Case (1)
Exercise-2
Lec 3: Concept of Train and Platform Case (2)
Exercise-3
Lec 4: Concept of Train and Platform Case (3)
Exercise-4
Lec 5: Concept of Train and Platform Case (4)
Exercise-5
Lec 6: Concept of Acceleration
Exercise-6
Lec 7: Concept of Boat and Stream Case (1) and Case (2)
Exercise-7
Lec 8: Concept of Boat and Stream Case (3)
Exercise-8
Exercise-9
Exercise-10