Time Speed and Distance Quantitative Aptitude Questions:

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{450 + 550}{\frac{5}{18} \ (85 - 75)} $$ $$ = \frac{1000 \times 18}{10 \times 5} $$ $$ = \frac{18000}{50} = 360 \ sec $$

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 200 = \frac{700 + 800}{70 \times \frac{5}{18} - s_2} $$ $$ 200 = \frac{1500 \times 18}{70 \times 5 - 18 \ s_2} $$ $$ 200 = \frac{27000}{350 - 18 \ s_2} $$ $$ 70000 - 3600 \ s_2 = 27000 $$ $$ 3600 \ s_2 = 43000 $$ $$ s_2 = \frac{43000}{3600} = 11.94 \ m/sec $$ $$ = 11.94 \times \frac{18}{5} \ km/hr $$ $$ = 42.98 \ km/hr $$

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ 25 = \frac{250 + 300}{40 \times \frac{5}{18} + s_2} $$ $$ 25 = \frac{550 \times 18}{200 + 18s_2} $$ $$ 5000 + 450 \ s_2 = 9900 $$ $$ 450 \ s_2 = 4900 $$ $$ s_2 = 10.89 \ m/sec $$

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 + s_2} $$ $$ = \frac{150 + 160}{45 + 50} $$ $$ = \frac{310}{95} = 3.26 \ sec $$

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 $$ Time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ 5 = \frac{60 + 80}{x - 200} $$ $$ 5 \ x - 1000 = 140 $$ $$ x = \frac{1140}{5} = 228 \ m/sec $$

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

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 $$ Relative \ speed = s_1 - s_2 $$ $$ = 180 - 160 = 20 \ m/sec $$ $$ = 20 \times \frac{18}{5} = 72 \ km/hr $$

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

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 $$ Relative \ speed = s_1 + s_2 $$ $$ = 25 + 35 = 60 \ km/hr $$

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 $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{600 + 300}{200} = 4.5 \ sec $$