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. |

- A \(650 \ meter\) long train and a motorbike, moving in the same direction. The speed of train and motorbike are \(85 \ m/sec\) and \(75 \ m/sec\) respectively. If length of the motorbike is not considered, then find how much time the train will take to pass the motorbike?
- \(62 \ seconds\)
- \(65 \ seconds\)
- \(60 \ seconds\)
- \(67 \ seconds\)

Answer: (b) \(65 \ seconds\)

Solution: Given, length of the train \((l_1) = 650 \ meter\)

speed of the train \((s_1) = 85 \ m/sec\)

speed of the motorbike \((s_2) = 75 \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{650}{ 85 - 75} $$ $$ = \frac{650}{10} = 65 \ seconds $$

Solution: Given, length of the train \((l_1) = 650 \ meter\)

speed of the train \((s_1) = 85 \ m/sec\)

speed of the motorbike \((s_2) = 75 \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{650}{ 85 - 75} $$ $$ = \frac{650}{10} = 65 \ seconds $$

- A bus and a bicycle, moving in the same direction. The speed of the bus and bicycle are \(40 \ m/sec\) and \(10 \ m/sec\) respectively. If length of the bicycle is not considered and the bus crosses the bicycle in \(25 \ seconds\), then find the length of the bus?
- \(700 \ meter\)
- \(760 \ meter\)
- \(750 \ meter\)
- \(725 \ meter\)

Answer: (c) \(750 \ meter\)

Solution: Given, speed of the bus \((s_1) = 40 \ m/sec\)

speed of the bicycle \((s_2) = 10 \ m/sec\)

time taken by the bus to pass the bicycle \((T) = 25 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ 25 = \frac{l_1}{40 - 10} $$ $$ l_1 = 750 \ meter $$

Solution: Given, speed of the bus \((s_1) = 40 \ m/sec\)

speed of the bicycle \((s_2) = 10 \ m/sec\)

time taken by the bus to pass the bicycle \((T) = 25 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ 25 = \frac{l_1}{40 - 10} $$ $$ l_1 = 750 \ meter $$

- Two trains A and B are moving in the opposite direction towards each other with the speed of \(70 \ m/sec\) and \(60 \ m/sec\) respectively. They will pass each otherin \(15 \ seconds\). If length of the train B is not considered, then find the length of the train A?
- \(1850 \ meter\)
- \(1963 \ meter\)
- \(1950 \ meter\)
- \(1850 \ meter\)

Answer: (c) \(1950 \ meter\)

Solution: Given, speed of the train A \((s_1) = 70 \ m/sec\)

speed of the train B \((s_2) = 60 \ m/sec\)

Train A will take time to pass the train B \((T) = 15 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 15 = \frac{l_1}{70 + 60} $$ $$ l_1 = 1950 \ meter $$

Solution: Given, speed of the train A \((s_1) = 70 \ m/sec\)

speed of the train B \((s_2) = 60 \ m/sec\)

Train A will take time to pass the train B \((T) = 15 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 15 = \frac{l_1}{70 + 60} $$ $$ l_1 = 1950 \ meter $$

- Two trains x and y are moving in the opposite direction towards each other with the speed of \(150 \ m/sec\) and \(180 \ m/sec\) respectively. If the length of the train x is \(500 \ meter\), then find how much time the train x will take to pass the train y. Where as the length of the train y is not considered?
- \(1.51 \ seconds\)
- \(2.62 \ seconds\)
- \(1.85 \ seconds\)
- \(1.25 \ seconds\)

Answer: (a) \(1.51 \ seconds\)

Solution: Given, speed of the train x \((s_1) = 150 \ m/sec\)

speed of the train y \((s_2) = 180 \ m/sec\)

length of the train x \((l_1) = 500 \ meter\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ = \frac{500}{150 + 180} $$ $$ = \frac{500}{330} = 1.51 \ seconds $$

Solution: Given, speed of the train x \((s_1) = 150 \ m/sec\)

speed of the train y \((s_2) = 180 \ m/sec\)

length of the train x \((l_1) = 500 \ meter\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ = \frac{500}{150 + 180} $$ $$ = \frac{500}{330} = 1.51 \ seconds $$

- A girl coveres half distance of her journey with the speed of \(10 \ km/hr\) and remaining half distance with the speed \(12 \ km/hr\), then find the average speed of the girl?
- \(12.25 \ km/hr\)
- \(10.90 \ km/hr\)
- \(9.50 \ km/hr\)
- \(15.23 \ km/hr\)

Answer: (b) \(10.90 \ km/hr\)

Solution: Given, speed of the girl during first half \((s_1) = 10 \ km/hr\)

speed of the girl during second half \((s_2) = 12 \ km/hr\)

Let total distance of the journey = \(1 \ km\), then $$ Average \ distance = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.5 + 0.5}{\frac{0.5}{10} + \frac{0.5}{12}} $$ $$ = \frac{1 \times 24 \times 20}{24 + 20} $$ $$ = \frac{480}{44} = 10.90 \ km/hr $$

Solution: Given, speed of the girl during first half \((s_1) = 10 \ km/hr\)

speed of the girl during second half \((s_2) = 12 \ km/hr\)

Let total distance of the journey = \(1 \ km\), then $$ Average \ distance = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.5 + 0.5}{\frac{0.5}{10} + \frac{0.5}{12}} $$ $$ = \frac{1 \times 24 \times 20}{24 + 20} $$ $$ = \frac{480}{44} = 10.90 \ km/hr $$

- A car coveres first half distance of the journey with the speed of \(x \ km/hr\) and second half with the speed of \(50 \ km/hr\). If the average speed of the car during the whole jouney is \(40 \ km/hr\), then find the value of \(x\)?
- \(36.32 \ km/hr\)
- \(30.50 \ km/hr\)
- \(32.32 \ km/hr\)
- \(33.33 \ km/hr\)

Answer: (d) \(33.33 \ km/hr\)

Solution: Given, speed of the car during the first half \((s_1) = x \ km/hr\)

speed of the car during the second half \((s_2) = 50 \ km/hr\)

average speed of the car = \(40 \ km/hr\)

Let total distance of the whole journey = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 40 = \frac{0.5 + 0.5}{\frac{0.5}{x} + \frac{0.5}{50}} $$ $$ 40 = \frac{1 \times 2 \ x \times 100}{100 + 2 \ x} $$ $$ 4000 + 80 \ x = 200 \ x $$ $$ x = \frac{4000}{120} = 33.33 \ km/hr $$

Solution: Given, speed of the car during the first half \((s_1) = x \ km/hr\)

speed of the car during the second half \((s_2) = 50 \ km/hr\)

average speed of the car = \(40 \ km/hr\)

Let total distance of the whole journey = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 40 = \frac{0.5 + 0.5}{\frac{0.5}{x} + \frac{0.5}{50}} $$ $$ 40 = \frac{1 \times 2 \ x \times 100}{100 + 2 \ x} $$ $$ 4000 + 80 \ x = 200 \ x $$ $$ x = \frac{4000}{120} = 33.33 \ km/hr $$

- Two buses A and B are moving in the opposite direction towards each other with speed of \(x \ km/hr\) and \(30 \ km/hr\) respectively. The length of the bus A is \(100 \ meter\), where as the length of the bus B is not considered. If they passes each other in \(10 \ seconds\), then find the value of \(x\)?
- \(22.2 \ km/hr\)
- \(33.2 \ km/hr\)
- \(21.6 \ km/hr\)
- \(22.5 \ km/hr\)

Answer: (c) \(21.6 \ km/hr\)

Solution: Given, speed of the bus A \((s_1) = x \ km/hr = x \times \frac{5}{18} \ m/sec\)

speed of the bus B \((s_2) = 30 \ km/hr = 30 \times \frac{5}{18} \ m/sec\)

length of the bus \((l_1) = 100 \ meter\)

time taken by the buses to pass each other \((T) = 10 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 10 = \frac{100}{\frac{5}{18} \ (x + 30)} $$ $$ 10 = \frac{100 \times 18}{5 \ x + 150} $$ $$ 50 \ x + 1500 = 1800 $$ $$ x = \frac{300}{50} = 6 \ m/sec $$ $$ x = 6 \times \frac{18}{5} = 21.6 \ km/hr $$

Solution: Given, speed of the bus A \((s_1) = x \ km/hr = x \times \frac{5}{18} \ m/sec\)

speed of the bus B \((s_2) = 30 \ km/hr = 30 \times \frac{5}{18} \ m/sec\)

length of the bus \((l_1) = 100 \ meter\)

time taken by the buses to pass each other \((T) = 10 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 10 = \frac{100}{\frac{5}{18} \ (x + 30)} $$ $$ 10 = \frac{100 \times 18}{5 \ x + 150} $$ $$ 50 \ x + 1500 = 1800 $$ $$ x = \frac{300}{50} = 6 \ m/sec $$ $$ x = 6 \times \frac{18}{5} = 21.6 \ km/hr $$

- If a metro train covered \(300 \ km\) distance in \(5 \ hours\), then find the speed of the metro train?
- \(65 \ km/hr\)
- \(60 \ km/hr\)
- \(62 \ km/hr\)
- \(58 \ km/hr\)

Answer: (b) \(60 \ km/hr\)

Solution: Given, distance covered = \(300 \ km\)

time taken = \(5 \ hours\)

then the speed of the metro train, $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{300}{5} = 60 \ km/hr $$

Solution: Given, distance covered = \(300 \ km\)

time taken = \(5 \ hours\)

then the speed of the metro train, $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{300}{5} = 60 \ km/hr $$

- A man covered \(60 \ km\) with the speed of \(12 \ km/hr\) and remaining \(40 \ km\) with the speed of \(15 \ km/hr\), then find the average speed of the man during the whole journey?
- \(13.04 \ km/hr\)
- \(12.58 \ km/hr\)
- \(14.25 \ km/hr\)
- \(13.98 \ km/hr\)

Answer: (a) \(13.04 \ km/hr\)

Solution: Given, first distance covered by the man \((d_1) = 60 \ km\)

second remaining distance covered by the man \((d_2) = 40 \ km\)

speed of the man to cover first distance \((s_1) = 12 \ km/hr\)

speed of the man to cover second remaining distance \((s_2) = 15 \ km/hr\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{60 + 40}{\frac{60}{12} + \frac{40}{15}} $$ $$ = \frac{100 \times 60}{5 \times 60 + 4 \times 40} $$ $$ = \frac{6000}{300 + 160} $$ $$ = \frac{6000}{460} = 13.04 \ km/hr $$

Solution: Given, first distance covered by the man \((d_1) = 60 \ km\)

second remaining distance covered by the man \((d_2) = 40 \ km\)

speed of the man to cover first distance \((s_1) = 12 \ km/hr\)

speed of the man to cover second remaining distance \((s_2) = 15 \ km/hr\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{60 + 40}{\frac{60}{12} + \frac{40}{15}} $$ $$ = \frac{100 \times 60}{5 \times 60 + 4 \times 40} $$ $$ = \frac{6000}{300 + 160} $$ $$ = \frac{6000}{460} = 13.04 \ km/hr $$

- A car crosses a pole in \(2 \ seconds\), if the length of the car is \(10 \ meter\), then find the speed of the car? where as the length of pole is not considered.
- \(20 \ km/hr\)
- \(16 \ km/hr\)
- \(18 \ km/hr\)
- \(22 \ km/hr\)

Answer: (c) \(18 \ km/hr\)

Solution: Given, length of the car \((l_m) = 10 \ meter\)

time taken by the car to cross the pole \((T) = 2 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 2 = \frac{10}{s_m} $$ $$ s_m = 5 \ m/sec $$ $$ = 5 \times \frac{18}{5} = 18 \ km/hr $$

Solution: Given, length of the car \((l_m) = 10 \ meter\)

time taken by the car to cross the pole \((T) = 2 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 2 = \frac{10}{s_m} $$ $$ s_m = 5 \ m/sec $$ $$ = 5 \times \frac{18}{5} = 18 \ km/hr $$

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