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. |

- 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?
- \(350 \ km\)
- \(400 \ km\)
- \(550 \ km\)
- \(600 \ km\)

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

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

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

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

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

- If a \(600 \ meter\) long train crossed a \(120 \ meter\) long platform in \(20 \ seconds\), then find the speed of the train?
- \(30 \ m/sec\)
- \(32 \ m/sec\)
- \(36 \ m/sec\)
- \(38 \ m/sec\)

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 $$ Time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 20 = \frac{600 + 120}{s_m} $$ $$ s_m = \frac{720}{20} = 36 \ m/sec $$

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 $$ Time \ taken \ (T) = \frac{l_m + l_s}{s_m} $$ $$ 20 = \frac{600 + 120}{s_m} $$ $$ s_m = \frac{720}{20} = 36 \ m/sec $$

- 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?
- \(65.25 \ km/hr\)
- \(66.67 \ km/hr\)
- \(68.58 \ km/hr\)
- \(68.21 \ km/hr\)

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 $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 80 = \frac{1 + 1}{\frac{1}{100} + \frac{1}{s_2}} $$ $$ 80 = \frac{2 \times 100 \times s_2}{(s_2 + 100)} $$ $$ 80 \ s_2 + 8000 = 200 \ s_2 $$ $$ s_2 = \frac{8000}{120} = 66.67 \ km/hr $$

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 $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ Average \ speed = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 80 = \frac{1 + 1}{\frac{1}{100} + \frac{1}{s_2}} $$ $$ 80 = \frac{2 \times 100 \times s_2}{(s_2 + 100)} $$ $$ 80 \ s_2 + 8000 = 200 \ s_2 $$ $$ s_2 = \frac{8000}{120} = 66.67 \ km/hr $$

- 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?
- \(2 \ hours\)
- \(3 \ hours\)
- \(4 \ hours\)
- \(5 \ hours\)

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

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

- 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?
- \(105 \ m/sec\)
- \(100 \ m/sec\)
- \(110 \ m/sec\)
- \(112 \ m/sec\)

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 10 = \frac{200 + 300}{150 - x} $$ $$ 1500 - 10 \ x = 500 $$ $$ x = 100 \ m/sec $$

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 10 = \frac{200 + 300}{150 - x} $$ $$ 1500 - 10 \ x = 500 $$ $$ x = 100 \ m/sec $$

- 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?
- \(1.73 \ seconds\)
- \(1.25 \ seconds\)
- \(2.35 \ seconds\)
- \(2.37 \ seconds\)

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 $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ = \frac{2600}{700 + 800} $$ $$ T = 1.73 \ seconds $$

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 $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ = \frac{2600}{700 + 800} $$ $$ T = 1.73 \ seconds $$

- 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?
- \(220 \ m/sec\)
- \(260 \ m/sec\)
- \(250 \ m/sec\)
- \(270 \ m/sec\)

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 $$ Acceleration \ (a) = \frac{V_f - V_i}{T} $$ $$ 10 = \frac{V_f - 100}{15} $$ $$ V_f = 250 \ m/sec $$

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 $$ Acceleration \ (a) = \frac{V_f - V_i}{T} $$ $$ 10 = \frac{V_f - 100}{15} $$ $$ V_f = 250 \ m/sec $$

- 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?
- \(15 \ km/hr\)
- \(18 \ km/hr\)
- \(20 \ km/hr\)
- \(25 \ km/hr\)

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 $$ B_s = D_s - S_s $$ $$ 35 = 55 - S_s $$ $$ S_s = 20 \ km/hr $$

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

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

- 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?
- \(17.5 \ km/hr\)
- \(15.5 \ km/hr\)
- \(20.5 \ km/hr\)
- \(12.5 \ km/hr\)

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, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{15 + 20}{2} $$ $$ = \frac{35}{2} = 17.5 \ km/hr $$

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, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{15 + 20}{2} $$ $$ = \frac{35}{2} = 17.5 \ km/hr $$

- 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?
- \(80 \ km\)
- \(70 \ km\)
- \(60 \ km\)
- \(50 \ km\)

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 $$ \frac{x}{60} + \frac{x}{20} = 4 $$ $$ \frac{x + 3 \ x}{60} = 4 $$ $$ x + 3 \ x = 240 $$ $$ x = 60 \ km $$

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 $$ \frac{x}{60} + \frac{x}{20} = 4 $$ $$ \frac{x + 3 \ x}{60} = 4 $$ $$ x + 3 \ x = 240 $$ $$ x = 60 \ km $$

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