Time Speed and Distance Aptitude Questions with Solution:

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

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

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

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

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

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

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

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

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

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

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

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