# Time Speed and Distance Aptitude Questions and Answers:

#### Overview:

 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.

1. 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?

1. $$62 \ seconds$$
2. $$65 \ seconds$$
3. $$60 \ seconds$$
4. $$67 \ seconds$$

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 $$Time \ taken \ (T) = \frac{l_1}{s_1 - s_2}$$ $$= \frac{650}{ 85 - 75}$$ $$= \frac{650}{10} = 65 \ seconds$$

1. 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?

1. $$700 \ meter$$
2. $$760 \ meter$$
3. $$750 \ meter$$
4. $$725 \ meter$$

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 $$Time \ taken \ (T) = \frac{l_1}{s_1 - s_2}$$ $$25 = \frac{l_1}{40 - 10}$$ $$l_1 = 750 \ meter$$

1. 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?

1. $$1850 \ meter$$
2. $$1963 \ meter$$
3. $$1950 \ meter$$
4. $$1850 \ meter$$

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 $$Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}$$ $$15 = \frac{l_1}{70 + 60}$$ $$l_1 = 1950 \ meter$$

1. 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. $$1.51 \ seconds$$
2. $$2.62 \ seconds$$
3. $$1.85 \ seconds$$
4. $$1.25 \ seconds$$

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 $$Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}$$ $$= \frac{500}{150 + 180}$$ $$= \frac{500}{330} = 1.51 \ seconds$$

1. 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?

1. $$12.25 \ km/hr$$
2. $$10.90 \ km/hr$$
3. $$9.50 \ km/hr$$
4. $$15.23 \ km/hr$$

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 $$Average \ distance = \frac{distance \ covered}{time \ taken}$$ $$= \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}}$$ $$= \frac{0.5 + 0.5}{\frac{0.5}{10} + \frac{0.5}{12}}$$ $$= \frac{1 \times 24 \times 20}{24 + 20}$$ $$= \frac{480}{44} = 10.90 \ km/hr$$

1. 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 $$x$$?

1. $$36.32 \ km/hr$$
2. $$30.50 \ km/hr$$
3. $$32.32 \ km/hr$$
4. $$33.33 \ km/hr$$

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 $$Average \ speed = \frac{distance \ covered}{time \ taken}$$ $$= \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}}$$ $$40 = \frac{0.5 + 0.5}{\frac{0.5}{x} + \frac{0.5}{50}}$$ $$40 = \frac{1 \times 2 \ x \times 100}{100 + 2 \ x}$$ $$4000 + 80 \ x = 200 \ x$$ $$x = \frac{4000}{120} = 33.33 \ km/hr$$

1. 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 $$x$$?

1. $$22.2 \ km/hr$$
2. $$33.2 \ km/hr$$
3. $$21.6 \ km/hr$$
4. $$22.5 \ km/hr$$

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 $$Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}$$ $$10 = \frac{100}{\frac{5}{18} \ (x + 30)}$$ $$10 = \frac{100 \times 18}{5 \ x + 150}$$ $$50 \ x + 1500 = 1800$$ $$x = \frac{300}{50} = 6 \ m/sec$$ $$x = 6 \times \frac{18}{5} = 21.6 \ km/hr$$

1. If a metro train covered $$300 \ km$$ distance in $$5 \ hours$$, then find the speed of the metro train?

1. $$65 \ km/hr$$
2. $$60 \ km/hr$$
3. $$62 \ km/hr$$
4. $$58 \ km/hr$$

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

Solution: Given, distance covered = $$300 \ km$$

time taken = $$5 \ hours$$

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

1. 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?

1. $$13.04 \ km/hr$$
2. $$12.58 \ km/hr$$
3. $$14.25 \ km/hr$$
4. $$13.98 \ km/hr$$

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 $$Average \ speed = \frac{distance \ covered}{time \ taken}$$ $$= \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}}$$ $$= \frac{60 + 40}{\frac{60}{12} + \frac{40}{15}}$$ $$= \frac{100 \times 60}{5 \times 60 + 4 \times 40}$$ $$= \frac{6000}{300 + 160}$$ $$= \frac{6000}{460} = 13.04 \ km/hr$$

1. 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.

1. $$20 \ km/hr$$
2. $$16 \ km/hr$$
3. $$18 \ km/hr$$
4. $$22 \ km/hr$$

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 $$Time \ taken \ (T) = \frac{l_m}{s_m}$$ $$2 = \frac{10}{s_m}$$ $$s_m = 5 \ m/sec$$ $$= 5 \times \frac{18}{5} = 18 \ km/hr$$