Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Time Speed and Distance Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- If a \(250 \ m\) long train crossed a \(75 \ m\) long platform with the speed of \(60 \ km/hr\), then find the time taken by the train to cross the platform?
- \(22.5 \ seconds\)
- \(19.5 \ seconds\)
- \(20.5 \ seconds\)
- \(25.5 \ seconds\)

Answer: (b) \(19.5 \ seconds\)

Solution: Given, length of the train \((l_m) = 250 \ m\)

length of the platform \((l_s) = 75 \ m\)

speed of the train \((s_m) = 60 \ km/hr = 60 \times \frac{5}{18} = \frac{50}{3} \ m/sec\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ = \frac{250 + 75}{\frac{50}{3}} $$ $$ = \frac{325 \times 3}{50} = 19.5 \ sec. $$

Solution: Given, length of the train \((l_m) = 250 \ m\)

length of the platform \((l_s) = 75 \ m\)

speed of the train \((s_m) = 60 \ km/hr = 60 \times \frac{5}{18} = \frac{50}{3} \ m/sec\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ = \frac{250 + 75}{\frac{50}{3}} $$ $$ = \frac{325 \times 3}{50} = 19.5 \ sec. $$

- If a \(380 \ m\) long train crossed a \(125 \ m\) long platform in \(12 \ sec\), then find the speed of the train?
- \(42.083 \ m/sec\)
- \(40.205 \ m/sec\)
- \(45.525 \ m/sec\)
- \(38.325 \ m/sec\)

Answer: (a) \(42.083 \ m/sec\)

Solution: Given, length of the train \((l_m) = 380 \ m\)

length of the platform \((l_s) = 125 \ m\)

time taken by the train to cross the platform \((T) = 12 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 12 = \frac{380 + 125}{s_m} $$ $$ s_m = \frac{505}{12} = 42.083 \ m/sec $$

Solution: Given, length of the train \((l_m) = 380 \ m\)

length of the platform \((l_s) = 125 \ m\)

time taken by the train to cross the platform \((T) = 12 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 12 = \frac{380 + 125}{s_m} $$ $$ s_m = \frac{505}{12} = 42.083 \ m/sec $$

- If a car covered \(280 \ km\) in \(4 \ hours\) then find the speed of the car?
- \(68 \ km/hr\)
- \(75 \ km/hr\)
- \(70 \ km/hr\)
- \(72 \ km/hr\)

Answer: (c) \(70 \ km/hr\)

Solution: Given, distance covered by the car = \(280 \ km\), time taken = \(4 \ hours\), then the speed of the car $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ speed = \frac{280}{4} = 70 \ km/hr $$

Solution: Given, distance covered by the car = \(280 \ km\), time taken = \(4 \ hours\), then the speed of the car $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ speed = \frac{280}{4} = 70 \ km/hr $$

- If a train crossed a \(160 \ m\) long platform in \(2.5 \ seconds\), with the speed of \(130 \ m/sec\), then find the length of the train?
- \(168 \ meters\)
- \(175 \ meters\)
- \(155 \ meters\)
- \(165 \ meters\)

Answer: (d) \(165 \ meters\)

Solution: Given, length of the platform \((l_s) = 160 \ meters\)

speed of the train \((s_m) = 130 \ m/sec\)

time taken \((T) = 2.5 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 2.5 = \frac{l_m + 160}{130} $$ $$ l_m = 325 - 160 = 165 \ meters $$

Solution: Given, length of the platform \((l_s) = 160 \ meters\)

speed of the train \((s_m) = 130 \ m/sec\)

time taken \((T) = 2.5 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 2.5 = \frac{l_m + 160}{130} $$ $$ l_m = 325 - 160 = 165 \ meters $$

- If a \(425 \ meters\) long train crossed a platform in \(3.6 \ seconds\) with the speed of \(160 \ m/sec\), then find the length of the platform?
- \(157 \ meters\)
- \(151 \ meters\)
- \(152 \ meters\)
- \(156 \ meters\)

Answer: (b) \(151 \ meters\)

Solution: Given, length of the train \((l_m) = 425 \ meters\)

speed of the train \((T) = 3.6 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 3.6 = \frac{425 + l_s}{160} $$ $$ l_s = 576 - 425 = 151 \ meters $$

Solution: Given, length of the train \((l_m) = 425 \ meters\)

speed of the train \((T) = 3.6 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 3.6 = \frac{425 + l_s}{160} $$ $$ l_s = 576 - 425 = 151 \ meters $$

- A train is moving at the speed of \(280 \ km/hr\) and one another train also moving parallel to the first train in the same direction with the speed of \(230 \ km/hr\), then find the relative speed of the trains?
- \(50 \ km/hr\)
- \(56 \ km/hr\)
- \(52 \ km/hr\)
- \(58 \ km/hr\)

Answer: (a) \(50 \ km/hr\)

Solution: Given, speed of the first train \((s_1) = 280 \ km/hr\)

speed of the second train \((s_2) = 230 \ km/hr\), then $$ Relative \ speed = s_1 - s_2 $$ $$ = 280 - 230 = 50 \ km/hr $$

Solution: Given, speed of the first train \((s_1) = 280 \ km/hr\)

speed of the second train \((s_2) = 230 \ km/hr\), then $$ Relative \ speed = s_1 - s_2 $$ $$ = 280 - 230 = 50 \ km/hr $$

- If the speed of a man is \(45 \ m/sec\), then find the speed of the man in \(km/hr\)?
- \(158 \ km/hr\)
- \(162 \ km/hr\)
- \(165 \ km/hr\)
- \(156 \ km/hr\)

Answer: (b) \(162 \ km/hr\)

Solution: speed of the man = \(45 \ m/sec\), then $$ = 45 \times \frac{18}{5} = 162 \ km/hr $$

Solution: speed of the man = \(45 \ m/sec\), then $$ = 45 \times \frac{18}{5} = 162 \ km/hr $$

- If a train covered first \(10 \ km\) at the speed of \(80 \ km/hr\) and second \(20 \ km\) at the speed of \(100 \ km/hr\), then find the average speed of the train?
- \(98.5 \ km/hr\)
- \(93.2 \ km/hr\)
- \(92.3 \ km/hr\)
- \(94.6 \ km/hr\)

Answer: (c) \(92.3 \ km/hr\)

Solution: $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{10 + 20}{\frac{10}{80} + \frac{20}{100}} $$ $$ = \frac{30 \times 400}{50 + 80} $$ $$ = \frac{12000}{130} = 92.3 \ km/hr $$

Solution: $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{10 + 20}{\frac{10}{80} + \frac{20}{100}} $$ $$ = \frac{30 \times 400}{50 + 80} $$ $$ = \frac{12000}{130} = 92.3 \ km/hr $$

- If a train travels from A to B at the speed of \(x_1 \ km/hr\) and travels from B to A at the speed of \(x_2 \ km/hr\), then find the average speed of the train?
- \(\frac{x_1 + x_2}{2 \ x_1 \ x_2} \ km/hr\)
- \(\frac{2 \ x_1 \ x_2}{x_1 + x_2} \ km/hr\)
- \(\frac{x_1 - x_2}{2 \ x_1 \ x_2} \ km/hr\)
- \(\frac{2 \ x_1 \ x_2}{x_1 - x_2} \ km/hr\)

Answer: (b) \(\frac{2 \ x_1 \ x_2}{x_1 + x_2} \ km/hr\)

Solution: Let the distance between A to B = \(d \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d + d}{\frac{d}{x_1} + \frac{d}{x_2}} $$ $$ \frac{2d \times x_1 \ x_2}{d \ (x_2 + x_1)} $$ $$ = \frac{2 \ x_1 \ x_2}{x_1 + x_2} \ km/hr$$

Solution: Let the distance between A to B = \(d \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d + d}{\frac{d}{x_1} + \frac{d}{x_2}} $$ $$ \frac{2d \times x_1 \ x_2}{d \ (x_2 + x_1)} $$ $$ = \frac{2 \ x_1 \ x_2}{x_1 + x_2} \ km/hr$$

- If two trains moving parallel in the opposite direction at the speed of \(2P \ km/hr\) and \(3P \ km/hr\) respectively. Find the relative speed of the train?
- \(6P \ km/hr\)
- \(5P \ km/hr\)
- \(8P \ km/hr\)
- \(7P \ km/hr\)

Answer: (b) \(5P \ km/hr\)

Solution: Given, speed of the first train \((s_1) = 2P \ km/hr\)

speed of the second train \((s_2) = 3P \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 2P + 3P = 5P \ km/hr $$

Solution: Given, speed of the first train \((s_1) = 2P \ km/hr\)

speed of the second train \((s_2) = 3P \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 2P + 3P = 5P \ km/hr $$

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