Time Speed and Distance Aptitude Questions and Answers:

Answer: (b) \(65 \ seconds\)

Solution: Given, length of the train \((l_1) = 650 \ meter\)

speed of the train \((s_1) = 85 \ m/sec\)

speed of the motorbike \((s_2) = 75 \ m/sec\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{650}{ 85 - 75} $$ $$ = \frac{650}{10} = 65 \ seconds $$

Answer: (c) \(750 \ meter\)

Solution: Given, speed of the bus \((s_1) = 40 \ m/sec\)

speed of the bicycle \((s_2) = 10 \ m/sec\)

time taken by the bus to pass the bicycle \((T) = 25 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ 25 = \frac{l_1}{40 - 10} $$ $$ l_1 = 750 \ meter $$

Answer: (c) \(1950 \ meter\)

Solution: Given, speed of the train A \((s_1) = 70 \ m/sec\)

speed of the train B \((s_2) = 60 \ m/sec\)

Train A will take time to pass the train B \((T) = 15 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 15 = \frac{l_1}{70 + 60} $$ $$ l_1 = 1950 \ meter $$

Answer: (a) \(1.51 \ seconds\)

Solution: Given, speed of the train x \((s_1) = 150 \ m/sec\)

speed of the train y \((s_2) = 180 \ m/sec\)

length of the train x \((l_1) = 500 \ meter\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ = \frac{500}{150 + 180} $$ $$ = \frac{500}{330} = 1.51 \ seconds $$

Answer: (b) \(10.90 \ km/hr\)

Solution: Given, speed of the girl during first half \((s_1) = 10 \ km/hr\)

speed of the girl during second half \((s_2) = 12 \ km/hr\)

Let total distance of the journey = \(1 \ km\), then $$ Average \ distance = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{0.5 + 0.5}{\frac{0.5}{10} + \frac{0.5}{12}} $$ $$ = \frac{1 \times 24 \times 20}{24 + 20} $$ $$ = \frac{480}{44} = 10.90 \ km/hr $$

Answer: (d) \(33.33 \ km/hr\)

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

speed of the car during the second half \((s_2) = 50 \ km/hr\)

average speed of the car = \(40 \ km/hr\)

Let total distance of the whole journey = \(1 \ km\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ 40 = \frac{0.5 + 0.5}{\frac{0.5}{x} + \frac{0.5}{50}} $$ $$ 40 = \frac{1 \times 2 \ x \times 100}{100 + 2 \ x} $$ $$ 4000 + 80 \ x = 200 \ x $$ $$ x = \frac{4000}{120} = 33.33 \ km/hr $$

Answer: (c) \(21.6 \ km/hr\)

Solution: Given, speed of the bus A \((s_1) = x \ km/hr = x \times \frac{5}{18} \ m/sec\)

speed of the bus B \((s_2) = 30 \ km/hr = 30 \times \frac{5}{18} \ m/sec\)

length of the bus \((l_1) = 100 \ meter\)

time taken by the buses to pass each other \((T) = 10 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_1}{s_1 + s_2} $$ $$ 10 = \frac{100}{\frac{5}{18} \ (x + 30)} $$ $$ 10 = \frac{100 \times 18}{5 \ x + 150} $$ $$ 50 \ x + 1500 = 1800 $$ $$ x = \frac{300}{50} = 6 \ m/sec $$ $$ x = 6 \times \frac{18}{5} = 21.6 \ km/hr $$

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

Solution: Given, distance covered = \(300 \ km\)

time taken = \(5 \ hours\)

then the speed of the metro train, $$ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{300}{5} = 60 \ km/hr $$

Answer: (a) \(13.04 \ km/hr\)

Solution: Given, first distance covered by the man \((d_1) = 60 \ km\)

second remaining distance covered by the man \((d_2) = 40 \ km\)

speed of the man to cover first distance \((s_1) = 12 \ km/hr\)

speed of the man to cover second remaining distance \((s_2) = 15 \ km/hr\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{60 + 40}{\frac{60}{12} + \frac{40}{15}} $$ $$ = \frac{100 \times 60}{5 \times 60 + 4 \times 40} $$ $$ = \frac{6000}{300 + 160} $$ $$ = \frac{6000}{460} = 13.04 \ km/hr $$

Answer: (c) \(18 \ km/hr\)

Solution: Given, length of the car \((l_m) = 10 \ meter\)

time taken by the car to cross the pole \((T) = 2 \ seconds\), then $$ Time \ taken \ (T) = \frac{l_m}{s_m} $$ $$ 2 = \frac{10}{s_m} $$ $$ s_m = 5 \ m/sec $$ $$ = 5 \times \frac{18}{5} = 18 \ km/hr $$