Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Distance}} = 5a \times \frac{1}{2} = 27{\text{ km}} \cr & {\text{Time}} = 20{\text{ min}} = \frac{1}{3}{\text{ hr}} \cr & {\text{Speed}} = \frac{{27}}{{\frac{1}{3}}} = 81{\text{ km/hr}} \cr} $$

$$\eqalign{ & {\text{Let distance}} = 1200{\text{ km}} \cr & {\text{Average speed}} = \frac{{1200}}{{\frac{{800}}{{45}} + \frac{{300}}{{60}} + \frac{{100}}{{75}}}} \cr & = \frac{{1200}}{{\frac{{160}}{9} + 5 + \frac{4}{3}}} \cr & = \frac{{1200 \times 9}}{{217}} \cr & = 49.77{\text{ km/h}} \cr} $$

$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{210 + 198}}{{\frac{{210}}{{60}} + \frac{{198}}{{66}}}} \cr & = 62.769 \cr & = 62.8\,{\text{km/hr}} \cr} $$

Speed = 40 km/hr
Distance = 30 km
Time = $$\frac{{30}}{{40}}$$ × 60 = 45 min
Stoppage time = 5 × 5 = 25 min
Total time = 70 min
$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{30}}{{70}} \times 60 \cr & = \frac{{180}}{7} \cr & = 25.7 \cr} $$

Relative speed = 23 - 21 = 2 m/s
Time taken by the policeman = $$\frac{{97}}{2}$$ = 48.5 sec to catch the thief.

Distance same
Speed = 15 : 3

The distance between A and B = 15 × 2 = 30 km

$$\eqalign{ & {\text{Diameter}} = 210{\text{ cm}} \cr & {\text{Radius}} = 105{\text{ cm}} \cr & {\text{Circumference}} = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 105 \cr & = 44 \times 15 \cr & = 660{\text{ cm}} \cr & {\text{1 hr}} \to {\text{120 km}} = 120000{\text{ m}} \cr & {\text{1 min}} \to \frac{{120000}}{{60}} = 2000{\text{ m}} \cr & {\text{Number of revolution}} = \frac{{2000 \times 100}}{{660}} = 303.03 \cr} $$

Let right time for the train to cover its journey (in minutes) be $$t$$ hours, then
As we know,
Distance = Speed × Time
According to the question,
$$\eqalign{ & 60 \times \left( {t + \frac{{20}}{{60}}} \right) = 75 \times \left( {t - \frac{{44}}{{60}}} \right) \cr & \Rightarrow 4\left( {t + \frac{1}{3}} \right) = 5\left( {t - \frac{{11}}{{15}}} \right) \cr & \Rightarrow 4t + \frac{4}{3} = 5t - \frac{{11}}{3} \cr & \Rightarrow 5t - 4t = \frac{{11}}{3} + \frac{4}{3} \cr & \Rightarrow t = \frac{{11 + 4}}{3} \cr & \Rightarrow t = 5{\text{ hours}} \cr & {\text{Distance }}\left( {\text{D}} \right) = 60 \times \left( {5 + \frac{{20}}{{60}}} \right) \cr & \Rightarrow {\text{D}} = 60 \times \left( {5 + \frac{1}{3}} \right) \cr & \Rightarrow {\text{D}} = 60 \times \frac{{16}}{3} \cr & \Rightarrow {\text{D}} = 320{\text{ km}} \cr} $$
Distance between H and O is 320 km
Average speed (S) = $$\frac{{320}}{5}$$ = 64 km/hr
∴ Sum of numerical value of D and S = 320 + 64 = 384

$$\eqalign{ & {\text{Let the length of train}} \Rightarrow x \cr & 36 \times \frac{5}{{18}} = \frac{x}{{25}} \cr & x = 250 \cr & {\text{Now,}} \cr & 36 \times \frac{5}{{18}} = \frac{{350 + 250}}{t} \cr & t = 60{\text{ second}} \cr} $$

$$\eqalign{ & \frac{{5x}}{5} + \frac{x}{5} + \frac{{4x}}{6} = \frac{{112}}{{60}} \cr & \Rightarrow \frac{{30x + 6x + 20x}}{{30}} = \frac{{112}}{{60}} \cr & \Rightarrow 56x = 56 \cr & \Rightarrow x = 1 \cr & \therefore {\text{Total distance}} = 5x + x + 4x \cr & = 10x \cr & = 10{\text{ km}} \cr} $$