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 women covered \(120 \ km\) in \(3 \ hours\), then find the speed of the women?
- \(25 \ km/hr\)
- \(30 \ km/hr\)
- \(35 \ km/hr\)
- \(40 \ km/hr\)

Answer: (d) \(40 \ km/hr\)

Solution: Given, distance covered = \(120 \ km\)

time taken = \(3 \ hours\)

then the speed of the women, $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{120}{3} = 40 \ km/hr $$

Solution: Given, distance covered = \(120 \ km\)

time taken = \(3 \ hours\)

then the speed of the women, $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{120}{3} = 40 \ km/hr $$

- If a \(170 \ meter\) long train crossed a \(70 \ meter\) long platform with the speed of \(45 \ km/hr\), then find the time taken by train to cross the platform?
- \(15.6 \ seconds\)
- \(18.4 \ seconds\)
- \(12.6 \ seconds\)
- \(11.8 \ seconds\)

Answer: (b) \(18.4 \ seconds\)

Solution: Given, length of the train \((l_m) = 170 \ meter\)

length of the platform \((l_s) = 60 \ meter\)

speed of the train \((s_m) = 45 \ km/hr\) \(= 45 \times \frac{5}{18} \ m/sec\), then $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{170 + 60}{45 \times \frac{5}{18}} $$ $$ = \frac{230 \times 18}{45 \times 5} $$ $$ = \frac{4140}{225} = 18.4 \ sec. $$

Solution: Given, length of the train \((l_m) = 170 \ meter\)

length of the platform \((l_s) = 60 \ meter\)

speed of the train \((s_m) = 45 \ km/hr\) \(= 45 \times \frac{5}{18} \ m/sec\), then $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{170 + 60}{45 \times \frac{5}{18}} $$ $$ = \frac{230 \times 18}{45 \times 5} $$ $$ = \frac{4140}{225} = 18.4 \ sec. $$

- Two trains x and y, start moving at the same time from points A and B respectively towards each other. After passing each other, trains take \(16 \ hours\) and \(4 \ hours\) to reach points A and B respectively. If the train x is moving at the speed of \(80 \ km/hr\), then find the speed of train y?
- \(120 \ km/hr\)
- \(135 \ km/hr\)
- \(145 \ km/hr\)
- \(160 \ km/hr\)

Answer: (d) \(160 \ km/hr\)

Solution: Given, speed of the train x \((s_1) = 80 \ km/hr\)

time taken by the train x to reach the point B \((T_1) = 16 \ hours\)

time taken by the train y to reach the point A \((T_2) = 4 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{4}{16}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{80}{s_2} = \frac{1}{2} $$ $$ s_2 = 160 \ km/hr $$

Solution: Given, speed of the train x \((s_1) = 80 \ km/hr\)

time taken by the train x to reach the point B \((T_1) = 16 \ hours\)

time taken by the train y to reach the point A \((T_2) = 4 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{4}{16}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{80}{s_2} = \frac{1}{2} $$ $$ s_2 = 160 \ km/hr $$

- Two \(120 \ meter\) and \(150 \ meter\) long trains moving in the same direction at the speed of \(80 \ km/hr\) and \(70 \ km/hr\) respectively. Find the time taken by the faster train to cross the slower train?
- \(95.5 \ seconds\)
- \(97.2 \ seconds\)
- \(96.6 \ seconds\)
- \(92.3 \ seconds\)

Answer: (b) \(97.2 \ seconds\)

Solution: Given, length of the first train \((l_1) = 120 \ meter\)

length of the second train \((l_2) = 150 \ meter\)

speed of the first train $$ (s_1) = 80 \ km/hr $$ $$ = 80 \times \frac{5}{18} \ m/sec $$ speed of the second train $$ (s_2) = 70 \ km/hr $$ $$ = 70 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{120 + 150}{\frac{5}{18} \ (80 - 70)} $$ $$ = \frac{270 \times 18}{50} $$ $$ T = 97.2 \ seconds $$

Solution: Given, length of the first train \((l_1) = 120 \ meter\)

length of the second train \((l_2) = 150 \ meter\)

speed of the first train $$ (s_1) = 80 \ km/hr $$ $$ = 80 \times \frac{5}{18} \ m/sec $$ speed of the second train $$ (s_2) = 70 \ km/hr $$ $$ = 70 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{120 + 150}{\frac{5}{18} \ (80 - 70)} $$ $$ = \frac{270 \times 18}{50} $$ $$ T = 97.2 \ seconds $$

- If a man covered \(80 \ km\) distance with the speed of \(40 \ km/hr\), then find the time taken by the man to cover the distance?
- \(1 \ hour\)
- \(2 \ hours\)
- \(3 \ hours\)
- \(4 \ hours\)

Answer: (b) \(2 \ hours\)

Solution: Given, distance covered = \(80 \ km\)

speed = \(40 \ km/hr\), then $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ 40 = \frac{80}{T} $$ $$ T = 2 \ hours $$

Solution: Given, distance covered = \(80 \ km\)

speed = \(40 \ km/hr\), then $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ 40 = \frac{80}{T} $$ $$ T = 2 \ hours $$

- A \(300 \ meter\) long train and a bicycle, moving in the same direction, the speed of the train and bicycle are \(50 \ km/hr\) and \(20 \ km/hr\) respectively. If length of the bicycle is not considered, then find the time taken by the train to cross the bicycle?
- \(30 \ seconds\)
- \(32 \ seconds\)
- \(34 \ seconds\)
- \(36 \ seconds\)

Answer: (d) \(36 \ seconds\)

Solution: Given, length of the first train \((l_1) = 300 \ meter\)

speed of the train $$ (s_1) = 50 \ km/hr $$ $$ = 50 \times \frac{5}{18} \ m/sec $$ speed of the bicycle $$ (s_2) = 20 \ km/hr $$ $$ = 20 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{300}{\frac{5}{18} \ (50 - 20)} $$ $$ = \frac{300 \times 18}{150} $$ $$ T = 36 \ seconds $$

Solution: Given, length of the first train \((l_1) = 300 \ meter\)

speed of the train $$ (s_1) = 50 \ km/hr $$ $$ = 50 \times \frac{5}{18} \ m/sec $$ speed of the bicycle $$ (s_2) = 20 \ km/hr $$ $$ = 20 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{300}{\frac{5}{18} \ (50 - 20)} $$ $$ = \frac{300 \times 18}{150} $$ $$ T = 36 \ seconds $$

- A train is moving from point A to point B with the initial speed of \(30 \ km/hr\) and after \(2 \ hours\), speed of the train increases \(50 \ km/hr\), then find the rate of the acceleration of the train?
- \(10 \ km/hr\)
- \(12 \ km/hr\)
- \(13 \ km/hr\)
- \(15 \ km/hr\)

Answer: (a) \(10 \ km/hr\)

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

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

time taken by the train to achieve the final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{V_f - V_i}{T} $$ $$ = \frac{50 - 30}{2} $$ $$ = \frac{20}{2} = 10 \ km/hr $$

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

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

time taken by the train to achieve the final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{V_f - V_i}{T} $$ $$ = \frac{50 - 30}{2} $$ $$ = \frac{20}{2} = 10 \ km/hr $$

- A girl swims downstream with the speed of \(40 \ km/hr\). If the speed of stream is \(10 \ km/hr\), then find the speed of the girl in still water?
- \(25 \ km/hr\)
- \(28 \ km/hr\)
- \(30 \ km/hr\)
- \(32 \ km/hr\)

Answer: (c) \(30 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = 40 \ km/hr\)

speed of stream \((S_s) = 10 \ km/hr\)

then speed of the girl in still water, $$ B_s = D_s - S_s $$ $$ B_s = 40 - 10 $$ $$ B_s = 30 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = 40 \ km/hr\)

speed of stream \((S_s) = 10 \ km/hr\)

then speed of the girl in still water, $$ B_s = D_s - S_s $$ $$ B_s = 40 - 10 $$ $$ B_s = 30 \ km/hr $$

- If a boat goes \(28 \ km\) upstream and \(32 \ km\) downstream, taking \(4 \ hours\) each time, then find the speed of the boat in still water?
- \(10.5 \ km/hr\)
- \(8.5 \ km/hr\)
- \(5.6 \ km/hr\)
- \(7.5 \ km/hr\)

Answer: (d) \(7.5 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{28}{4} = 7 \ km/hr\)

speed of downstream \((D_s) = \frac{32}{4} = 8 \ km/hr\)

then speed of the boat in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{8 + 7}{2} $$ $$ = \frac{15}{2} = 7.5 \ km/hr $$

Solution: Given, speed of upstream \((U_s) = \frac{28}{4} = 7 \ km/hr\)

speed of downstream \((D_s) = \frac{32}{4} = 8 \ km/hr\)

then speed of the boat in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{8 + 7}{2} $$ $$ = \frac{15}{2} = 7.5 \ km/hr $$

- If a man swims \(14 \ km\) downstream and \(8 \ km\) upstream, taking \(2 \ hours\) each time, then find the speed of stream?
- \(1.0 \ km/hr\)
- \(1.3 \ km/hr\)
- \(1.4 \ km/hr\)
- \(1.5 \ km/hr\)

Answer: (d) \(1.5 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = \frac{14}{2} = 7 \ km/hr\)

speed of upstream \((U_s) = \frac{8}{2} = 4 \ km/hr\)

then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{7 - 4}{2} $$ $$ = \frac{3}{2} = 1.5 \ km/hr $$

Solution: Given, speed of downstream \((D_s) = \frac{14}{2} = 7 \ km/hr\)

speed of upstream \((U_s) = \frac{8}{2} = 4 \ km/hr\)

then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{7 - 4}{2} $$ $$ = \frac{3}{2} = 1.5 \ km/hr $$

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