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 $$ \left[Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}\right] $$ $$ \left[5 = \frac{x}{(40 + 30) \times \frac{5}{18}}\right] $$ $$ 5 = \frac{x \times 18}{70 \times 5} $$ $$ 1750 = 18 \ x $$ $$ x = 97.22 \ m $$

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