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. |

- Two trains x and y, start moving at the same time from stations A and B respectively towards each other. After passing each other, trains take \(20 \ hours\) and \(5 \ hours\) to reach stations A and B respectively. If the train x is moving at the speed of \(65 \ km/hr\), then find the speed of the train y?
- \(125 \ km/hr\)
- \(135 \ km/hr\)
- \(130 \ km/hr\)
- \(120 \ km/hr\)

Answer: (c) \(130 \ km/hr\)

Solution: Given, speed of the train x \((s_1) = 65 \ km/hr\)

time taken by the train x to reach the station B \((T_1) = 20 \ hours\)

time taken by the train y to reach the station A \((T_2) = 5 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{5}{20}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{65}{s_2} = \frac{1}{2} $$ $$ s_2 = 130 \ km/hr $$

Solution: Given, speed of the train x \((s_1) = 65 \ km/hr\)

time taken by the train x to reach the station B \((T_1) = 20 \ hours\)

time taken by the train y to reach the station A \((T_2) = 5 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{5}{20}} $$ $$ \frac{65}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{65}{s_2} = \frac{1}{2} $$ $$ s_2 = 130 \ km/hr $$

- A train crossed a pole in \(10 \ seconds\), if the length of the train is \(260 \ meter\), then find the speed of the train?
- \(28 \ m/sec\)
- \(26 \ m/sec\)
- \(25 \ m/sec\)
- \(24 \ m/sec\)

Answer: (b) \(26 \ m/sec\)

Solution: Given, length of the train \((l_m) = 260 \ meter\)

time taken by the train to cross the train \((T) = 10 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 10 = \frac{260}{s_m} $$ $$ s_m = 26 \ m/sec $$

Solution: Given, length of the train \((l_m) = 260 \ meter\)

time taken by the train to cross the train \((T) = 10 \ seconds\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 10 = \frac{260}{s_m} $$ $$ s_m = 26 \ m/sec $$

- A man goes from station A to station B at the speed of \(40 \ km/hr\) and returns from station B to station A at the speed of \(60 \ km/hr\). If the man takes \(20 \ hours\) on traveling then find the distance between station A and B?
- \(435 \ km\)
- \(475 \ km\)
- \(465 \ km\)
- \(480 \ km\)

Answer: (d) \(480 \ km\)

Solution: Given, speed of the man from station A to B = \(40 \ km/hr\)

speed of the man from station B to A = \(60 \ km/hr\)

time taken by the man on traveling = \(20 \ hours\)

Let the distance between between the two stations = \(d \ km\), then $$ \frac{d}{40} + \frac{d}{60} = 20 $$ $$ d \ \left(\frac{1}{40} + \frac{1}{60}\right) = 20 $$ $$ d \ \left(\frac{3 + 2}{120}\right) = 20 $$ $$ d = \frac{120 \times 20}{5} = 480 \ km $$

Solution: Given, speed of the man from station A to B = \(40 \ km/hr\)

speed of the man from station B to A = \(60 \ km/hr\)

time taken by the man on traveling = \(20 \ hours\)

Let the distance between between the two stations = \(d \ km\), then $$ \frac{d}{40} + \frac{d}{60} = 20 $$ $$ d \ \left(\frac{1}{40} + \frac{1}{60}\right) = 20 $$ $$ d \ \left(\frac{3 + 2}{120}\right) = 20 $$ $$ d = \frac{120 \times 20}{5} = 480 \ km $$

- A train crossed a pole at the speed of \(45 \ km/hr\). If the length of the train is \(80 \ meter\), then find the time taken by the train tp cross the pole? where as the length of the pole is not considered.
- \(6.4 \ seconds\)
- \(5.6 \ seconds\)
- \(5.8 \ seconds\)
- \(6.9 \ seconds\)

Answer: (a) \(6.4 \ seconds\)

Solution: Given, speed of the train \((s_m) = 45 \ km/hr = 45 \times \frac{5}{18} \ m/sec\)

length of the train \((l_m) = 80 \ meter\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ = \frac{80}{45 \times \frac{5}{18}} $$ $$ = \frac{80 \times 18}{45 \times 5} $$ $$ = \frac{1440}{225} = 6.4 \ sec $$

Solution: Given, speed of the train \((s_m) = 45 \ km/hr = 45 \times \frac{5}{18} \ m/sec\)

length of the train \((l_m) = 80 \ meter\), then $$ time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ = \frac{80}{45 \times \frac{5}{18}} $$ $$ = \frac{80 \times 18}{45 \times 5} $$ $$ = \frac{1440}{225} = 6.4 \ sec $$

- Two trains x and y, start moving at the same time from station A and B respectively towards each other. The train x is moving at the speed of \(45 \ km/hr\) and the train y is moving at the speed of \(55 \ km/hr\). If the distance between two stations A and B is \(500 \ km\), then find how far from station A both the trains will cross each other?
- \(250 \ km\)
- \(225 \ km\)
- \(265 \ km\)
- \(230 \ km\)

Answer: (b) \(225 \ km\)

Solution: Given, distance between two stations = \(500 \ km\)

speed of the train x = \(45 \ km/hr\)

speed of the train y = \(55 \ km/hr\)

time taken by the trains to meet each other = \(\frac{500}{45 + 55}\) = \(5 \ hours\)

distance from station A when both the trains will cross each other $$ = 5 \times 45 = 225 \ km $$

Solution: Given, distance between two stations = \(500 \ km\)

speed of the train x = \(45 \ km/hr\)

speed of the train y = \(55 \ km/hr\)

time taken by the trains to meet each other = \(\frac{500}{45 + 55}\) = \(5 \ hours\)

distance from station A when both the trains will cross each other $$ = 5 \times 45 = 225 \ km $$

- A man crosses a \(250 \ meter\) long bridge in \(3 \ minutes\), then find the speed of the man?
- \(1.25 \ m/sec\)
- \(1.05 \ m/sec\)
- \(1.38 \ m/sec\)
- \(1.63 \ m/sec\)

Answer: (c) \(1.38 \ m/sec\)

Solution: Given, length of the bridge = \(250 \ meter\)

time taken by the man to cross the bridge \(= 3 \times 60 = 180 \ seconds\), then $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{250}{180} = 1.38 \ m/sec $$

Solution: Given, length of the bridge = \(250 \ meter\)

time taken by the man to cross the bridge \(= 3 \times 60 = 180 \ seconds\), then $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{250}{180} = 1.38 \ m/sec $$

- A car goes from city A to city B at the speed of \(75 \ km/hr\) and returns from city B to A. If the average speed of the car during the whole journey is \(65 \ km/hr\), then find the speed of the car returns from city B to city A?
- \(57.35 \ km/hr\)
- \(56.25 \ km/hr\)
- \(46.25 \ km/hr\)
- \(58.62 \ km/hr\)

Answer: (a) \(57.35 \ km/hr\)

Solution: Given, speed of the car from city A to B \((s_1) = 75 \ km/hr\)

Average speed of the car = \(65 \ km/hr\)

Let the distance between city A and city B = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 65 = \frac{1 + 1}{\frac{1}{75} + \frac{1}{s_2}} $$ $$ 65 = \frac{2 \times 75 \times s_2}{(s_2 + 75)} $$ $$ 65 \ s_2 + 4875 = 150 \ s_2 $$ $$ s_2 = \frac{4875}{85} = 57.35 \ km/hr $$

Solution: Given, speed of the car from city A to B \((s_1) = 75 \ km/hr\)

Average speed of the car = \(65 \ km/hr\)

Let the distance between city A and city B = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 65 = \frac{1 + 1}{\frac{1}{75} + \frac{1}{s_2}} $$ $$ 65 = \frac{2 \times 75 \times s_2}{(s_2 + 75)} $$ $$ 65 \ s_2 + 4875 = 150 \ s_2 $$ $$ s_2 = \frac{4875}{85} = 57.35 \ km/hr $$

- A \(125 \ meter\) long train passes a bridge at the speed of \(65 \ km/hr\) in \(10 \ seconds\), then find the length of the bridge?
- \(56.5 \ meter\)
- \(55.5 \ meter\)
- \(58.5 \ meter\)
- \(54.5 \ meter\)

Answer: (b) \(55.5 \ meter\)

Solution: Given, length of the train = \(125 \ meter\)

speed of the train = \(65 \ km/hr\) = \(65 \times \frac{5}{18} = 18.05 \ m/sec\)

time taken by the train to cross the bridge = \(10 \ seconds\)

distance covered by the train in \(10 \ seconds\) with the speed of \(18.05 \ m/sec\) = \(18.05 \times 10\) = \(180.5 \ meter\), then length of the bridge $$ = 180.5 - length \ of \ the \ train $$ $$ = 180.5 - 125 = 55.5 \ meter $$

Solution: Given, length of the train = \(125 \ meter\)

speed of the train = \(65 \ km/hr\) = \(65 \times \frac{5}{18} = 18.05 \ m/sec\)

time taken by the train to cross the bridge = \(10 \ seconds\)

distance covered by the train in \(10 \ seconds\) with the speed of \(18.05 \ m/sec\) = \(18.05 \times 10\) = \(180.5 \ meter\), then length of the bridge $$ = 180.5 - length \ of \ the \ train $$ $$ = 180.5 - 125 = 55.5 \ meter $$

- A train crosses a pole in \(8 \ seconds\) with the speed of \(150 \ km/hr\), if the length of the pole is not considered, then find the length of the train?
- \(333.36 \ meter\)
- \(320.25 \ meter\)
- \(350.36 \ meter\)
- \(325.45 \ meter\)

Answer: (a) \(333.36 \ meter\)

Solution: Given, speed of the train \((s_m) = 150 \ km/hr\) = \(150 \times \frac{5}{18}\) = \(41.67 \ m/sec\)

time taken by the train to cross the pole = \(8 \ seconds\), then $$ time \ taken = \frac{l_m}{s_m} $$ $$ 8 = \frac{l_m}{41.67} $$ $$ l_m = 333.36 \ meter $$

Solution: Given, speed of the train \((s_m) = 150 \ km/hr\) = \(150 \times \frac{5}{18}\) = \(41.67 \ m/sec\)

time taken by the train to cross the pole = \(8 \ seconds\), then $$ time \ taken = \frac{l_m}{s_m} $$ $$ 8 = \frac{l_m}{41.67} $$ $$ l_m = 333.36 \ meter $$

- A \(1000 \ meter\) long train crossed a \(250 \ meter\) long bridge with the speed of \(200 \ km/hr\). If length of the bridge is considered then find the time taken by the train to cross the bridge?
- \(20.50 \ seconds\)
- \(24.25 \ seconds\)
- \(22.49 \ seconds\)
- \(25.62 \ seconds\)

Answer: (c) \(22.49 \ seconds\)

Solution: Given, length of the train \((l_m) = 1000 / meter\)

length of the bridge \((l_s) = 250 \ meter\)

speed of the train \((s_m) = 200 \ km/hr\) = \(200 \times \frac{5}{18} = 55.56 \ m/sec\), then $$ time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{1000 + 250}{55.56} = 22.49 \ seconds $$

Solution: Given, length of the train \((l_m) = 1000 / meter\)

length of the bridge \((l_s) = 250 \ meter\)

speed of the train \((s_m) = 200 \ km/hr\) = \(200 \times \frac{5}{18} = 55.56 \ m/sec\), then $$ time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{1000 + 250}{55.56} = 22.49 \ seconds $$

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