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

Let the time taken by A to cover 1 km = x sec
Time taken by B and C to cover the same distance = (x + 25) sec and (x + 55) sec

$$\eqalign{ & \frac{{\text{A}}}{{\text{C}}} = \frac{{29}}{{40}} = \frac{x}{{x + 55}} \cr & \Rightarrow 29x + 1595 = 40x \cr & \Rightarrow x = \frac{{1595}}{{11}} \cr & \Rightarrow x = 145 \cr} $$
∴ Time taken by A
= 145 sec
= 2 minutes 25 seconds

Let length of train be x and speed be S
S = $$\frac{x + 50}{14}$$ , also S = $$\frac{x}{10}$$
Then,
$$\frac{x + 50}{14}$$ = $$\frac{x}{10}$$
5x + 250 = 7x
2x = 250
x = 125
∴ Speed :
= $$\frac{125}{10}$$ × $$\frac{18}{5}$$
= 45 km/hr

Speed of the train :
= $$\frac{10 × 60}{12}$$
= 50 km/hr
Now, new speed :
= 50 - 5
= 45 km/hr
And, required time :
= $$\frac{10}{45}$$
= $$\frac{2}{9}$$ × 60
= 13 minutes 20 seconds

Circumference of the circle = 2πr
= 2πr
= 2 × $$\frac{22}{7}$$ × 42
= 264 cm
Distance cover in 1 sec
= 264 × 5
= 1320 cm/sec

Relative speed = 24 - 18 = 6 km/hr
Time required by faster train to overtake slower train :
= $$\frac{27}{6}$$
= $$4\frac{1}{2}$$ hr
∴ Distance between Q and R :
= 18 × $$4\frac{1}{2}$$
= 81 km

Average speed = $$\frac{2xy}{x + y}$$
x = 10 km/hr
y = 12 km/hr
Average speed = $$\frac{{2 \times 10 \times 12}}{{\left( {10 + 12} \right)}}$$
∴ Average speed = 10.9 km/hr

$$\eqalign{ & \,\,{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}} \cr & 190\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,200 \cr & 180\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,200 \cr & \,\,{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}} \cr & 171\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,180\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,200 \cr} $$

⇒ C can give a start of 200 - 171 = 29 metres to A

$$\eqalign{ & {\text{Usual speed}}\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,{\text{New speed}} \cr & \,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{35{\text{ km/h}}}^7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,\,\,\,\,\, \downarrow \times 5}\limits_{{\text{25 km/h}}}^5 \cr} $$
$$\because $$ Train covers 42 kms in 1 hr, 40 min, 48 sec with the speed of $$\frac{5}{7}$$ of its speed.
Then its new speed :
$$ = \frac{{{\text{Distance }}}}{{{\text{Time}}}} = \frac{{42{\text{ km}}}}{{\frac{{504}}{{300}}{\text{hr}}}}$$

\[\left\{ \begin{gathered} {\text{1 hr 40 min 48 sec}} \hfill \\ {\text{1hr + 40 min + }}\frac{{48}}{{60}}{\text{min}} \hfill \\ {\text{1 hr + }}\left( {40 + \frac{4}{5}} \right)\min \hfill \\ 1{\text{ hr}} + \frac{{204}}{5}\min \hfill \\ \left( {1 + \frac{{204}}{{5 \times 60}}} \right){\text{ hr}} = \frac{{504}}{{300}}{\text{ hr }} \hfill \\ \end{gathered} \right\}\]

$$\eqalign{ & = \frac{{42}}{{504}} \times 300{\text{ km/hr}} \cr & {\text{ = 25 km/hr}} \cr & \because 5{\text{ units = 25 km/hr}} \cr & \,\,\,\,{\text{1 unit = 5 km/hr}} \cr & \because {\text{Usual speed = 7 units}} \cr & \because {\text{Usual speed = 7}} \times {\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,= 35{\text{ km/hr}} \cr} $$

$$\eqalign{ & {L_t} = {L_P} = l \cr & S = {\text{90 km /hr }} \cr & \,\,\,\,{\text{ = }}\frac{{90 \times 1000}}{{60}}{\text{ m/min}} \cr & \,\,\,\,\, = 1500\text{ m/min} \cr & \Rightarrow l = {L_t} = {L_P} = \frac{{1500}}{2} = 750{\text{ meters}} \cr} $$

1 m/sec = $$\frac{18}{5}$$ km/hr
$$\frac{10}{3}$$ m/sec = $$\frac{10}{3}$$ × $$\frac{18}{5}$$
                = 12 km/hr