Time Speed and Distance Quantitative Aptitude Questions:

Answer: (d) \(40 \ km/hr\)

Solution: Given, distance covered = \(120 \ km\)

time taken = \(3 \ hours\)

then the speed of the women, $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{120}{3} = 40 \ km/hr $$

Answer: (b) \(18.4 \ seconds\)

Solution: Given, length of the train \((l_m) = 170 \ meter\)

length of the platform \((l_s) = 60 \ meter\)

speed of the train \((s_m) = 45 \ km/hr\) \(= 45 \times \frac{5}{18} \ m/sec\), then $$ Time \ taken = \frac{l_m + l_s}{s_m} $$ $$ = \frac{170 + 60}{45 \times \frac{5}{18}} $$ $$ = \frac{230 \times 18}{45 \times 5} $$ $$ = \frac{4140}{225} = 18.4 \ sec. $$

Answer: (d) \(160 \ km/hr\)

Solution: Given, speed of the train x \((s_1) = 80 \ km/hr\)

time taken by the train x to reach the point B \((T_1) = 16 \ hours\)

time taken by the train y to reach the point A \((T_2) = 4 \ hours\), then $$ \frac{s_1}{s_2} = \sqrt{\frac{T_2}{T_1}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{4}{16}} $$ $$ \frac{80}{s_2} = \sqrt{\frac{1}{4}} $$ $$ \frac{80}{s_2} = \frac{1}{2} $$ $$ s_2 = 160 \ km/hr $$

Answer: (b) \(97.2 \ seconds\)

Solution: Given, length of the first train \((l_1) = 120 \ meter\)

length of the second train \((l_2) = 150 \ meter\)

speed of the first train $$ (s_1) = 80 \ km/hr $$ $$ = 80 \times \frac{5}{18} \ m/sec $$ speed of the second train $$ (s_2) = 70 \ km/hr $$ $$ = 70 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1 + l_2}{s_1 - s_2} $$ $$ = \frac{120 + 150}{\frac{5}{18} \ (80 - 70)} $$ $$ = \frac{270 \times 18}{50} $$ $$ T = 97.2 \ seconds $$

Answer: (b) \(2 \ hours\)

Solution: Given, distance covered = \(80 \ km\)

speed = \(40 \ km/hr\), then $$ Speed = \frac{distance \ covered}{time \ taken} $$ $$ 40 = \frac{80}{T} $$ $$ T = 2 \ hours $$

Answer: (d) \(36 \ seconds\)

Solution: Given, length of the first train \((l_1) = 300 \ meter\)

speed of the train $$ (s_1) = 50 \ km/hr $$ $$ = 50 \times \frac{5}{18} \ m/sec $$ speed of the bicycle $$ (s_2) = 20 \ km/hr $$ $$ = 20 \times \frac{5}{18} \ m/sec $$ then, $$ time \ taken \ (T) = \frac{l_1}{s_1 - s_2} $$ $$ = \frac{300}{\frac{5}{18} \ (50 - 20)} $$ $$ = \frac{300 \times 18}{150} $$ $$ T = 36 \ seconds $$

Answer: (a) \(10 \ km/hr\)

Solution: Given, initial speed of the train \((V_i) = 30 \ km/hr\)

final speed of the train \((V_f) = 50 \ km/hr\)

time taken by the train to achieve the final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{V_f - V_i}{T} $$ $$ = \frac{50 - 30}{2} $$ $$ = \frac{20}{2} = 10 \ km/hr $$

Answer: (c) \(30 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = 40 \ km/hr\)

speed of stream \((S_s) = 10 \ km/hr\)

then speed of the girl in still water, $$ B_s = D_s - S_s $$ $$ B_s = 40 - 10 $$ $$ B_s = 30 \ km/hr $$

Answer: (d) \(7.5 \ km/hr\)

Solution: Given, speed of upstream \((U_s) = \frac{28}{4} = 7 \ km/hr\)

speed of downstream \((D_s) = \frac{32}{4} = 8 \ km/hr\)

then speed of the boat in still water, $$ B_s = \frac{D_s + U_s}{2} $$ $$ = \frac{8 + 7}{2} $$ $$ = \frac{15}{2} = 7.5 \ km/hr $$

Answer: (d) \(1.5 \ km/hr\)

Solution: Given, speed of downstream \((D_s) = \frac{14}{2} = 7 \ km/hr\)

speed of upstream \((U_s) = \frac{8}{2} = 4 \ km/hr\)

then speed of stream, $$ S_s = \frac{D_s - U_s}{2} $$ $$ = \frac{7 - 4}{2} $$ $$ = \frac{3}{2} = 1.5 \ km/hr $$