# Time Speed and Distance Aptitude Questions with Solution:

#### 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. Raj travels from New Delhi to Jaipur with the speed of $$120 \ km/hr$$ and reached Jaipur in $$5 \ hours$$, then find the distance covered by the Raj?

1. $$350 \ km$$
2. $$400 \ km$$
3. $$550 \ km$$
4. $$600 \ km$$

Answer: (d) $$600 \ km$$

Solution: Given, speed of Raj = $$120 \ km/hr$$

time taken = $$5 \ hours$$, then $$Speed = \frac{distance \ covered}{time \ taken}$$ $$120 = \frac{distance \ covered}{5}$$ $$distance \ covered = 600 \ km$$

1. If a $$600 \ meter$$ long train crossed a $$120 \ meter$$ long platform in $$20 \ seconds$$, then find the speed of the train?

1. $$30 \ m/sec$$
2. $$32 \ m/sec$$
3. $$36 \ m/sec$$
4. $$38 \ m/sec$$

Answer: (c) $$36 \ m/sec$$

Solution: Given, length of the train $$(l_m) = 600 \ meter$$

length of the platform $$(l_s) = 120 \ meter$$

time taken by the train to cross the platform $$(T) = 20 \ seconds$$, then $$Time \ taken \ (T) = \frac{l_m + l_s}{s_m}$$ $$20 = \frac{600 + 120}{s_m}$$ $$s_m = \frac{720}{20} = 36 \ m/sec$$

1. A man goes from Bhopal to Nagpur at the speed of $$100 \ km/hr$$ and returns. If the average speed of the man during the whole journey is $$80 \ km/hr$$, then find the speed of the man when returns from Nagpur to Bhopal?

1. $$65.25 \ km/hr$$
2. $$66.67 \ km/hr$$
3. $$68.58 \ km/hr$$
4. $$68.21 \ km/hr$$

Answer: (b) $$66.67 \ km/hr$$

Solution: Given, speed of the man when travels from Bhopal to Nagpur $$(s_1) = 100 \ km/hr$$

Average speed of the man = $$80 \ km/hr$$

Let the distance between Bhopal and Nagpur = $$1 \ km$$, then $$Average \ speed = \frac{distance \ covered}{time \ taken}$$ $$Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}}$$ $$80 = \frac{1 + 1}{\frac{1}{100} + \frac{1}{s_2}}$$ $$80 = \frac{2 \times 100 \times s_2}{(s_2 + 100)}$$ $$80 \ s_2 + 8000 = 200 \ s_2$$ $$s_2 = \frac{8000}{120} = 66.67 \ km/hr$$

1. A women travels from city A to city B with the initial speed of $$30 \ km/hr$$ and increases the speed $$50 \ km/hr$$ with the rate of acceleration $$10 \ km/hr$$, then find the time taken by the women to achieve the final speed?

1. $$2 \ hours$$
2. $$3 \ hours$$
3. $$4 \ hours$$
4. $$5 \ hours$$

Answer: (a) $$2 \ hours$$

Solution: Given, initial speed of the women $$(V_i) = 30 \ km/hr$$

final speed of the women $$(V_f) = 50 \ km/hr$$

Acceleration $$(a) = 10 \ km/hr$$, then $$Acceleration \ (a) = \frac{V_f - V_i}{T}$$ $$10 = \frac{50 - 30}{T}$$ $$T = \frac{20}{10} = 2 \ hours$$

1. Two $$200 \ meter$$ and $$300 \ meter$$ long trains, moving in the same direction with the speed of $$150 \ m/sec$$ and $$x \ m/sec$$ respectively. If faster train will pass the slower train in $$10 \ seconds$$, then find the value of x?

1. $$105 \ m/sec$$
2. $$100 \ m/sec$$
3. $$110 \ m/sec$$
4. $$112 \ m/sec$$

Answer: (b) $$100 \ m/sec$$

Solution: Given, length of the first train $$(l_1) = 200 \ meter$$

length of the second train $$(l_2) = 300 \ meter$$

speed of the first train $$(s_1) = 150 \ m/sec$$

speed of the second train $$(s_2) = x \ m/sec$$

time taken by the faster train to pass the slower train $$(T) = 10 \ seconds$$, then $$Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2}$$ $$10 = \frac{200 + 300}{150 - x}$$ $$1500 - 10 \ x = 500$$ $$x = 100 \ m/sec$$

1. Two trains are moving towards each other with the speed of $$700 \ m/sec$$ and $$800 \ m/sec$$ respectively. If the length of the first train is $$2600 \ meter$$, then find time taken by the first train to cross the second train, where as the length of the second train is not considered?

1. $$1.73 \ seconds$$
2. $$1.25 \ seconds$$
3. $$2.35 \ seconds$$
4. $$2.37 \ seconds$$

Answer: (a) $$1.73 \ seconds$$

Solution: Given, speed of the first train $$(s_1) = 700 \ m/sec$$

speed of the second train $$(s_2) = 800 \ m/sec$$

length of the first train $$(l_1) = 2600 \ meter$$, then $$Time \ taken \ (T) = \frac{l_1}{s_1 + s_2}$$ $$= \frac{2600}{700 + 800}$$ $$T = 1.73 \ seconds$$

1. A girl travels on car with the initial speed of $$100 \ m/sec$$ and accelerated with $$10 \ m/sec$$ after $$15 \ seconds$$ the girl achieved final speed. Find the final speed of the car?

1. $$220 \ m/sec$$
2. $$260 \ m/sec$$
3. $$250 \ m/sec$$
4. $$270 \ m/sec$$

Answer: (c) $$250 \ m/sec$$

Solution: Given, initial speed of the car $$(V_i) = 100 \ m/sec$$

time taken by the car to achieve the final speed $$(T) = 15 \ seconds$$

acceleration rate of the car $$(a) = 10 \ m/sec$$, then $$Acceleration \ (a) = \frac{V_f - V_i}{T}$$ $$10 = \frac{V_f - 100}{15}$$ $$V_f = 250 \ m/sec$$

1. A boat goes downstream with the speed of $$55 \ km/hr$$. If the speed of the boat in still water is $$35 \ km/hr$$, then find the speed of stream?

1. $$15 \ km/hr$$
2. $$18 \ km/hr$$
3. $$20 \ km/hr$$
4. $$25 \ km/hr$$

Answer: (c) $$20 \ km/hr$$

Solution: Given, speed of downstream $$(D_s) = 55 \ km/hr$$

speed of the boat in still water $$(B_s) = 35 \ km/hr$$, then $$B_s = D_s - S_s$$ $$35 = 55 - S_s$$ $$S_s = 20 \ km/hr$$

1. If the boat goes $$30 \ km$$ upstream and $$40 \ km$$ downstream, taking $$2 \ hours$$ each time, then find the speed of the boat in still water?

1. $$17.5 \ km/hr$$
2. $$15.5 \ km/hr$$
3. $$20.5 \ km/hr$$
4. $$12.5 \ km/hr$$

Answer: (a) $$17.5 \ km/hr$$

Solution: Given, speed of upstream $$(U_s) = \frac{30}{2} = 15 \ km/hr$$

speed of downstream $$(D_s) = \frac{40}{2} = 20 \ km/hr$$

then the speed of the boat in still water, $$B_s = \frac{D_s + U_s}{2}$$ $$= \frac{15 + 20}{2}$$ $$= \frac{35}{2} = 17.5 \ km/hr$$

1. A boat goes upstream from point A to point B, taking $$4 \ hours$$ and returns downstream from point B to point A. If the speed of stream is $$20 \ km/hr$$ and speed of the boat in still water is $$40 \ km/hr$$, then find the distance between point A and B?

1. $$80 \ km$$
2. $$70 \ km$$
3. $$60 \ km$$
4. $$50 \ km$$

Answer: (c) $$60 \ km$$

Solution: Given, speed of the stream $$(S_s) = 20 \ km/hr$$

speed of the boat in still water $$(B_s) = 40 \ km/hr$$, then

speed of upstream $$(U_s) = B_s - S_s$$ $$= 40 - 20 = 20 \ km/hr$$

speed of downstream $$(D_s) = B_s + S_s$$ $$= 40 + 20 = 60 \ km/hr$$

Let the distance between point A and B = $$x \ km$$, then $$\frac{x}{60} + \frac{x}{20} = 4$$ $$\frac{x + 3 \ x}{60} = 4$$ $$x + 3 \ x = 240$$ $$x = 60 \ km$$