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 train moves from the New Delhi railway station to Jaipur at the initial speed of \(45 \ km/hr\) and after \(0.5 \ hr\), speed of the train increases \(75 \ km/hr\), then find the rate of acceleration of the train?
- \(62 \ km/hr\)
- \(60 \ km/hr\)
- \(66 \ km/hr\)
- \(68 \ km/hr\)

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

Solution: Given, Initial speed of the train \((v_i) = 45 \ km/hr\)

final speed of the train \((v_f) = 75 \ km/hr\)

time taken by the train to achieve the final speed \((T) = 0.5 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{75 - 45}{0.5} $$ $$ = \frac{30}{0.5} = 60 \ km/hr $$

Solution: Given, Initial speed of the train \((v_i) = 45 \ km/hr\)

final speed of the train \((v_f) = 75 \ km/hr\)

time taken by the train to achieve the final speed \((T) = 0.5 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{75 - 45}{0.5} $$ $$ = \frac{30}{0.5} = 60 \ km/hr $$

- The initial speed of a car is \(25 \ km/hr\) and after \(2 \ hours\) the speed of car increases \(65 \ km/hr\), then find the rate of acceleration of car?
- \(20 \ km/hr\)
- \(22 \ km/hr\)
- \(24 \ km/hr\)
- \(25 \ km/hr\)

Answer: (a) \(20 \ km/hr\)

Solution: Given, initial speed of the car \((v_i) = 25 \ km/hr\)

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

time taken by the car to achieve final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{65 - 25}{2} $$ $$ = \frac{40}{2} = 20 \ km/hr $$

Solution: Given, initial speed of the car \((v_i) = 25 \ km/hr\)

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

time taken by the car to achieve final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{65 - 25}{2} $$ $$ = \frac{40}{2} = 20 \ km/hr $$

- A man travels from city A to city B with the initial speed of \(60 \ km/hr\) and achieved the speed \(120 \ km/hr\) with the acceleration rate \(30 \ km/hr\), then find the time taken by the man to achieve the final speed?
- \(1 \ hour\)
- \(2 \ hours\)
- \(3 \ hours\)
- \(4 \ hours\)

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

Solution: Given, Initial speed of the man \((v_i) = 60 \ km/hr\)

final speed of the man \((v_f) = 120 \ km/hr\)

Acceleration \((a) = 30 \ km/hr\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 30 = \frac{120 - 60}{T} $$ $$ T = \frac{60}{30} = 2 \ hours $$

Solution: Given, Initial speed of the man \((v_i) = 60 \ km/hr\)

final speed of the man \((v_f) = 120 \ km/hr\)

Acceleration \((a) = 30 \ km/hr\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 30 = \frac{120 - 60}{T} $$ $$ T = \frac{60}{30} = 2 \ hours $$

- An express train travels \(800 \ km\) in \(3 \ hours\) and another \(700 \ km\) in \(5 \ hours\), then find the average speed of the train?
- \(176.5 \ km/hr\)
- \(185.5 \ km/hr\)
- \(188.5 \ km/hr\)
- \(187.5 \ km/hr\)

Answer: (d) \(187.5 \ km/hr\)

Solution: Given, total distance covered by the train = \(800 \ km + 700 \ km\) = \(1500 \ km\)

total time taken by the train \(= 3 + 5 = 8 \ hours\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{1500}{8} = 187.5 \ km/hr $$

Solution: Given, total distance covered by the train = \(800 \ km + 700 \ km\) = \(1500 \ km\)

total time taken by the train \(= 3 + 5 = 8 \ hours\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{1500}{8} = 187.5 \ km/hr $$

- A man travels on bicycle with the initial speed of \(10 \ m/sec\) and accelerated with the rate of \(2 \ m/sec\), after \(60 \ seconds\) the man achieved the final speed. Find the final speed of the man?
- \(130 \ m/sec\)
- \(125 \ m/sec\)
- \(128 \ m/sec\)
- \(132 \ m/sec\)

Answer: (a) \(130 \ m/sec\)

Solution: Given, initial speed of the man \((v_i) = 10 \ m/sec\)

time taken by the man to achieve the final speed \((T) = 60 \ seconds\)

acceleration rate of the bicycle \((a) = 2 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 2 = \frac{v_f - 10}{60} $$ $$ 120 = v_f - 10 $$ $$ v_f = 130 \ m/sec $$

Solution: Given, initial speed of the man \((v_i) = 10 \ m/sec\)

time taken by the man to achieve the final speed \((T) = 60 \ seconds\)

acceleration rate of the bicycle \((a) = 2 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 2 = \frac{v_f - 10}{60} $$ $$ 120 = v_f - 10 $$ $$ v_f = 130 \ m/sec $$

- A bus starts moving and after \(2 \ seconds\) achieves the speed \(175 \ km/hr\). If the bus acclerating with the rate of \(10 \ m/sec\), then find the initial speed of the bus?
- \(111.255 \ km/hr\)
- \(115.936 \ km/hr\)
- \(102.996 \ km/hr\)
- \(108.672 \ km/hr\)

Answer: (c) \(102.996 \ km/hr\)

Solution: Given, final speed of the bus \((v_f) = 175 \ km/hr\) = \(175 \times \frac{5}{18}\)

time taken by the bus to achieve the final speed \((T) = 2 \ seconds\)

rate of accleration of the bus \((a) = 100 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 10 = \frac{175 \times \frac{5}{18} - v_i}{2} $$ $$ 20 \times 18 = 175 \times 5 - 18 \ v_i $$ $$ 18 \ v_i = 515 $$ $$ v_i = 28.61 \ m/sec $$ $$ v_i = 28.61 \times \frac{18}{5} \ km/hr $$ $$ v_i = 102.996 \ km/hr $$

Solution: Given, final speed of the bus \((v_f) = 175 \ km/hr\) = \(175 \times \frac{5}{18}\)

time taken by the bus to achieve the final speed \((T) = 2 \ seconds\)

rate of accleration of the bus \((a) = 100 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 10 = \frac{175 \times \frac{5}{18} - v_i}{2} $$ $$ 20 \times 18 = 175 \times 5 - 18 \ v_i $$ $$ 18 \ v_i = 515 $$ $$ v_i = 28.61 \ m/sec $$ $$ v_i = 28.61 \times \frac{18}{5} \ km/hr $$ $$ v_i = 102.996 \ km/hr $$

- A man covered a certain distance in \(25 \ seconds\) with the average speed of \(500 \ m/sec\), then find distance covered by the man?
- \(10.5 \ km\)
- \(15.5 \ km\)
- \(13.5 \ km\)
- \(12.5 \ km\)

Answer: (d) \(12.5 \ km\)

Solution: Given, average speed of the man = \(500 \ m/sec\)

time taken by the man to cover the distance = \(25 \ seconds\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ 500 = \frac{distance \ covered}{25} $$ $$ distance \ covered = 12,500 \ meter $$ $$ = 12,500 \times \frac{1}{1000} = 12.5 \ km $$

Solution: Given, average speed of the man = \(500 \ m/sec\)

time taken by the man to cover the distance = \(25 \ seconds\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ 500 = \frac{distance \ covered}{25} $$ $$ distance \ covered = 12,500 \ meter $$ $$ = 12,500 \times \frac{1}{1000} = 12.5 \ km $$

- Two cars are moving in the opposite direction towards each other with the speed of \(150 \ km/hr\) and \(110 \ km/hr\) respectively, then find the relative speed of the cars?
- \(263 \ km/hr\)
- \(260 \ km/hr\)
- \(262 \ km/hr\)
- \(258 \ km/hr\)

Answer: (b) \(260 \ km/hr\)

Solution: Given, speed of the first car \((s_1) = 150 \ km/hr\)

speed of the second car \((s_2) = 110 \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 150 + 110 = 260 \ km/hr $$

Solution: Given, speed of the first car \((s_1) = 150 \ km/hr\)

speed of the second car \((s_2) = 110 \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 150 + 110 = 260 \ km/hr $$

- A man travels from station A to station B. He covered first \(200 \ km\) with the speed of \(75 \ km/hr\) and covered second \(200 \ km\) with the speed of \(100 \ km/hr\). Find the average speed of the man during the whole journey?
- \(85.7 \ km/hr\)
- \(86.5 \ km/hr\)
- \(82.6 \ km/hr\)
- \(87.6 \ km/hr\)

Answer: (a) \(85.7 \ km/hr\)

Solution: $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{200 + 200}{\frac{200}{75} + \frac{200}{100}} $$ $$ = \frac{400 \times 300}{800 + 600} $$ $$ = \frac{120000}{1400} = 85.7 \ km/hr $$

Solution: $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{200 + 200}{\frac{200}{75} + \frac{200}{100}} $$ $$ = \frac{400 \times 300}{800 + 600} $$ $$ = \frac{120000}{1400} = 85.7 \ km/hr $$

- A girl covered a certain distance with the speed of \(10 \ m/sec\). Find the speed of the girl in \(km/hr\)?
- \(30 \ km/hr\)
- \(36 \ km/hr\)
- \(38 \ km/hr\)
- \(32 \ km/hr\)

Answer: (b) \(36 \ km/hr\)

Solution: speed of the girl = \(10 \ m/sec\), then $$ = 10 \times \frac{18}{5} \ km/hr $$ $$ = 36 \ km/hr $$

Solution: speed of the girl = \(10 \ m/sec\), then $$ = 10 \times \frac{18}{5} \ km/hr $$ $$ = 36 \ km/hr $$

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