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 \(450 \ meter\) and \(550 \ meter\) long trains moving in the same direction at the speed of \(85 \ km/hr\) and \(75 \ km/hr\) respectively. Find the time taken by the faster train to cross the slower train?
- \(250 \ seconds\)
- \(360 \ seconds\)
- \(380 \ seconds\)
- \(320 \ seconds\)

Answer: (b) \(360 \ seconds\)

Solution: Given, length of the first train \((l_1) = 450 \ meter\)

length of the second train \((l_2) = 550 \ meter\)

speed of the first train \((s_1) = 85 \ km/hr = 85 \times \frac{5}{18} \ m/sec\)

speed of the second train \((s_2) = 75 \ km/hr = 75 \times \frac{5}{18} \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{450 + 550}{\frac{5}{18} \ (85 - 75)} $$ $$ = \frac{1000 \times 18}{10 \times 5} $$ $$ = \frac{18000}{50} = 360 \ sec $$

Solution: Given, length of the first train \((l_1) = 450 \ meter\)

length of the second train \((l_2) = 550 \ meter\)

speed of the first train \((s_1) = 85 \ km/hr = 85 \times \frac{5}{18} \ m/sec\)

speed of the second train \((s_2) = 75 \ km/hr = 75 \times \frac{5}{18} \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{450 + 550}{\frac{5}{18} \ (85 - 75)} $$ $$ = \frac{1000 \times 18}{10 \times 5} $$ $$ = \frac{18000}{50} = 360 \ sec $$

- Two \(700 \ meter\) and \(800 \ meter\) long trains, moving in the same direction. Faster train takes \(200 \ seconds\) to cross the slower train. If speed of the first train is \(70 \ km/hr\), then find the speed of the second train?
- \(42.98 \ km/hr\)
- \(40.36 \ km/hr\)
- \(45.62 \ km/hr\)
- \(46.25 \ km/hr\)

Answer: (a) \(42.98 \ km/hr\)

Solution: Given, length of the first train \((l_1) = 700 \ meter\)

length of the second train \((l_2) = 800 \ meter\)

speed of the first train \((s_1) = 70 \ km/hr = 70 \times \frac{5}{18} \ m/sec\)

time taken by the faster train to cross the slower one = \(200 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 200 = \frac{700 + 800}{70 \times \frac{5}{18} - s_2} $$ $$ 200 = \frac{1500 \times 18}{70 \times 5 - 18 \ s_2} $$ $$ 200 = \frac{27000}{350 - 18 \ s_2} $$ $$ 70000 - 3600 \ s_2 = 27000 $$ $$ 3600 \ s_2 = 43000 $$ $$ s_2 = \frac{43000}{3600} = 11.94 \ m/sec $$ $$ = 11.94 \times \frac{18}{5} \ km/hr $$ $$ = 42.98 \ km/hr $$

Solution: Given, length of the first train \((l_1) = 700 \ meter\)

length of the second train \((l_2) = 800 \ meter\)

speed of the first train \((s_1) = 70 \ km/hr = 70 \times \frac{5}{18} \ m/sec\)

time taken by the faster train to cross the slower one = \(200 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 200 = \frac{700 + 800}{70 \times \frac{5}{18} - s_2} $$ $$ 200 = \frac{1500 \times 18}{70 \times 5 - 18 \ s_2} $$ $$ 200 = \frac{27000}{350 - 18 \ s_2} $$ $$ 70000 - 3600 \ s_2 = 27000 $$ $$ 3600 \ s_2 = 43000 $$ $$ s_2 = \frac{43000}{3600} = 11.94 \ m/sec $$ $$ = 11.94 \times \frac{18}{5} \ km/hr $$ $$ = 42.98 \ km/hr $$

- The \(250 \ meter\) long train x is moving from station A to station B, at the speed of \(40 \ km/hr\) and the \(300 \ meter\) long train y is moving from station B to station A. If both the trains taken \(25 \ seconds\) to pass each other, then find the speed of the train y moving from station B to A?
- \(9.4 \ m/sec\)
- \(11.6 \ m/sec\)
- \(10.89 \ m/sec\)
- \(12.59 \ m/sec\)

Answer: (c) \(10.89 \ m/sec\)

Solution: Given, length of the train x \((l_1) = 250 \ meter\)

length of the train y \((l_2) = 300 \ meter\)

speed of the train x = \((s_1) = 40 \ km/hr = 40 \times \frac{5}{18} \ m/sec\)

time taken by the trains to pass each other \((T) = 25 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ 25 = \frac{250 + 300}{40 \times \frac{5}{18} + s_2} $$ $$ 25 = \frac{550 \times 18}{200 + 18s_2} $$ $$ 5000 + 450 \ s_2 = 9900 $$ $$ 450 \ s_2 = 4900 $$ $$ s_2 = 10.89 \ m/sec $$

Solution: Given, length of the train x \((l_1) = 250 \ meter\)

length of the train y \((l_2) = 300 \ meter\)

speed of the train x = \((s_1) = 40 \ km/hr = 40 \times \frac{5}{18} \ m/sec\)

time taken by the trains to pass each other \((T) = 25 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ 25 = \frac{250 + 300}{40 \times \frac{5}{18} + s_2} $$ $$ 25 = \frac{550 \times 18}{200 + 18s_2} $$ $$ 5000 + 450 \ s_2 = 9900 $$ $$ 450 \ s_2 = 4900 $$ $$ s_2 = 10.89 \ m/sec $$

- Two metro trains \(150 \ meter\) and \(160 \ meter\) long, moving in the opposite direction with the speed of \(45 \ m/sec\) and \(50 \ m/sec\) respectively. Find how much time the metro trains will take to pass each other?
- \(3.42 \ seconds\)
- \(5.25 \ seconds\)
- \(4.36 \ seconds\)
- \(3.26 \ seconds\)

Answer: (d) \(3.26 \ seconds\)

Solution: Given, length of the first metro train \((l_1) = 150 \ meter\)

length of the second metro train \((l_2) = 160 \ meter\)

speed of the first metro train \((s_1) = 45 \ m/sec\)

speed of the second metro train \((s_2) = 50 \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ = \frac{150 + 160}{45 + 50} $$ $$ = \frac{310}{95} = 3.26 \ sec $$

Solution: Given, length of the first metro train \((l_1) = 150 \ meter\)

length of the second metro train \((l_2) = 160 \ meter\)

speed of the first metro train \((s_1) = 45 \ m/sec\)

speed of the second metro train \((s_2) = 50 \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ = \frac{150 + 160}{45 + 50} $$ $$ = \frac{310}{95} = 3.26 \ sec $$

- Two buses \(60 \ meter\) and \(80 \ meter\) long, moving in the same direction with the speed of \(x \ m/sec\) and \(200 \ m/sec\) respectively. If faster bus will pass the slower one in \(5 \ seconds\), then find the value of \(x\)?
- \(225 \ m/sec\)
- \(228 \ m/sec\)
- \(232 \ m/sec\)
- \(230 \ m/sec\)

Answer: (b) \(228 \ m/sec\)

Solution: Given, length of the first bus \((l_1) = 60 \ meter\)

length of the second bus \((l_2) = 80 \ meter\)

speed of the first bus \((s_1) = x \ m/sec\)

speed of the second bus \((s_2) = 200 \ m/sec\)

time taken by the faster bus to cross the slower one \((T) = 5 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 5 = \frac{60 + 80}{x - 200} $$ $$ 5 \ x - 1000 = 140 $$ $$ x = \frac{1140}{5} = 228 \ m/sec $$

Solution: Given, length of the first bus \((l_1) = 60 \ meter\)

length of the second bus \((l_2) = 80 \ meter\)

speed of the first bus \((s_1) = x \ m/sec\)

speed of the second bus \((s_2) = 200 \ m/sec\)

time taken by the faster bus to cross the slower one \((T) = 5 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 5 = \frac{60 + 80}{x - 200} $$ $$ 5 \ x - 1000 = 140 $$ $$ x = \frac{1140}{5} = 228 \ m/sec $$

- A man covered half of his distance with the speed of \(120 \ m/sec\) and second half covered with the speed of \(140 \ m/sec\). Find the average speed of the man during the whole journey?
- \(126.25 \ m/sec\)
- \(128.05 \ m/sec\)
- \(129.23 \ m/sec\)
- \(127.63 \ m/sec\)

Answer: (c) \(129.23 \ m/sec\)

Solution: Given, speed of the man during one half = \(120 \ m/sec\)

speed of the man during second half = \(140 \ m/sec\)

Let total distance = \(1 \ meter\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.5 + 0.5}{\frac{0.5}{120} + \frac{0.5}{140}} $$ $$ = \frac{1 \times 240 \times 280}{280 + 240} $$ $$ = \frac{67200}{520} = 129.23 \ m/sec $$

Solution: Given, speed of the man during one half = \(120 \ m/sec\)

speed of the man during second half = \(140 \ m/sec\)

Let total distance = \(1 \ meter\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.5 + 0.5}{\frac{0.5}{120} + \frac{0.5}{140}} $$ $$ = \frac{1 \times 240 \times 280}{280 + 240} $$ $$ = \frac{67200}{520} = 129.23 \ m/sec $$

- Two cars moving in the same direction with the speed of \(180 \ m/sec\) and \(160 \ m/sec\) respectively. Find the relative speed of the cars?
- \(75 \ km/hr\)
- \(70 \ km/hr\)
- \(76 \ km/hr\)
- \(72 \ km/hr\)

Answer: (d) \(72 \ km/hr\)

Solution: Given, speed of the first car \((s_1) = 180 \ m/sec\)

speed of the second car \((s_2) = 160 \ m/sec\), then $$ Relative \ speed = s_1 - s_2 $$ $$ = 180 - 160 = 20 \ m/sec $$ $$ = 20 \times \frac{18}{5} = 72 \ km/hr $$

Solution: Given, speed of the first car \((s_1) = 180 \ m/sec\)

speed of the second car \((s_2) = 160 \ m/sec\), then $$ Relative \ speed = s_1 - s_2 $$ $$ = 180 - 160 = 20 \ m/sec $$ $$ = 20 \times \frac{18}{5} = 72 \ km/hr $$

- A car covered \(0.8 \ \%\) distance of the journey at the speed of \(60 \ m/sec\) and covered \(0.2 \ \%\) of the journey with the speed of \(80 \ m/sec\), then find the average speed of the car?
- \(66.25 \ m/sec\)
- \(65.35 \ m/sec\)
- \(63.15 \ m/sec\)
- \(64.25 \ m/sec\)

Answer: (c) \(63.15 \ m/sec\)

Solution: Given, speed of the car during \(0.8 \ \%\) of distance = \(60 \ m/sec\)

speed of the car remaining \(0.2 \ \%\) of distance = \(80 \ m/sec\)

Let total distance = \(1 \ meter\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.8 + 0.2}{\frac{0.8}{60} + \frac{0.2}{80}} $$ $$ = \frac{1}{\frac{1}{75} + \frac{1}{400}} $$ $$ = \frac{1 \times 30000}{400 + 75} $$ $$ = \frac{30000}{475} $$ $$ Average \ speed = 63.15 \ m/sec $$

Solution: Given, speed of the car during \(0.8 \ \%\) of distance = \(60 \ m/sec\)

speed of the car remaining \(0.2 \ \%\) of distance = \(80 \ m/sec\)

Let total distance = \(1 \ meter\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.8 + 0.2}{\frac{0.8}{60} + \frac{0.2}{80}} $$ $$ = \frac{1}{\frac{1}{75} + \frac{1}{400}} $$ $$ = \frac{1 \times 30000}{400 + 75} $$ $$ = \frac{30000}{475} $$ $$ Average \ speed = 63.15 \ m/sec $$

- Two cars, moving in the opposite direction with the speed of \(25 \ km/hr\) and \(35 \ km/hr\) respectively. Find the relative speed of the cars?
- \(65 \ km/hr\)
- \(60 \ km/hr\)
- \(62 \ km/hr\)
- \(58 \ km/hr\)

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

Solution: Given, speed of the first car \((s_1) = 25 \ km/hr\)

speed of the second car \((s_2) = 35 \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 25 + 35 = 60 \ km/hr $$

Solution: Given, speed of the first car \((s_1) = 25 \ km/hr\)

speed of the second car \((s_2) = 35 \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 25 + 35 = 60 \ km/hr $$

- If a \(600 \ meter\) long train crossed a \(300 \ meter\) platform with the speed of \(200 \ m/sec\). If the length of the platform is considered then find the time taken by the train to cross the platform?
- \(4.5 \ seconds\)
- \(4.2 \ seconds\)
- \(5.5 \ seconds\)
- \(5.2 \ seconds\)

Answer: (a) \(4.5 \ seconds\)

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

length of the platform \((l_s) = 300 \ meter\)

speed of the train \((s_m) = 200 \ m/sec\), then $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{600 + 300}{200} = 4.5 \ sec $$

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

length of the platform \((l_s) = 300 \ meter\)

speed of the train \((s_m) = 200 \ m/sec\), then $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{600 + 300}{200} = 4.5 \ sec $$

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