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 $$ \left[Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2}\right] $$ $$ \left[Time \ taken \ (T) = \frac{1000 + 1500}{(35 + 45) \times \frac{5}{18}}\right] $$ $$ Time \ taken \ (T) = \frac{2500}{80 \times \frac{5}{18}} $$ $$ Time \ taken \ (T) = \frac{2500 \times 18}{80 \times 5} $$ $$ Time \ taken \ (T) = \frac{45000}{400} = 112.5 \ sec $$

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