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

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

$$\eqalign{ & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & = \frac{{162}}{{15 \times \frac{{18}}{5}}}{\text{hr}} \cr & = \frac{{162 \times 5}}{{15 \times 18}}{\text{hr}} \cr & = 3{\text{ hrs}} \cr} $$

Distance = Speed × Time
Distance = 45 min × 80
After incrassation the speed $$ = \frac{{80 \times 225}}{{100}} = 180{\text{ km/hr}}$$
$$\eqalign{ & 45 \times 80 \times \frac{8}{5} = 180 \times t\left\{ {{s_1} \times {t_1} = {s_2} \times {t_2}} \right\} \cr & t = 32{\text{ minutes}} \cr} $$

Let the speed of the train A be x km/hr
Ratio of speed of the A before to after = x : 2x
As we know,
Time is inversely proportional to speed.
Ratio of time of A before to after to cover distance = 2x : x
Time difference = 2x - x = $$2\frac{2}{3} + 1\frac{1}{3}$$
⇒ x = $$\frac{8}{3} + \frac{4}{3}$$
⇒ x = $$\frac{{12}}{3}$$
⇒ x = 4 hr
Time taken by A to cover distance = 2x = 2 × 4 = 8 hr
∴ Speed of train A = $$\frac{{416}}{8}$$ = 52 km/hr

$$\eqalign{ & {\text{Total distance}} = \frac{{{S_1} \times {S_2}}}{{{S_1} + {S_2}}} \times {\text{Total time}} \cr & = \frac{{5\frac{1}{2} \times 4}}{{5\frac{1}{2} + 4}} \times 4\frac{3}{4} \cr & = \frac{{\frac{{11}}{2} \times 4}}{{\frac{{11}}{2} + 4}} \times \frac{{19}}{4} \cr & = \frac{{22}}{{\frac{{19}}{2}}} \times \frac{{19}}{4} \cr & = \frac{{44}}{{19}} \times \frac{{19}}{4} \cr & = 11{\text{ km}} \cr & {\text{One side distance}} = 11{\text{ km}} \cr & {\text{Both side distance}} = 2 \times 11 = 22{\text{ km}} \cr} $$

$$\eqalign{ & x + y = 70 \cr & \frac{{30}}{x} + \frac{{30}}{y} = \frac{{21}}{{10}} \cr & 30\left[ {\frac{{x + y}}{{xy}}} \right] = \frac{{21}}{{10}} \cr & \frac{{30 \times 70}}{{xy}} = \frac{{21}}{{10}} \cr & xy = 1000 \cr & {\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy \cr & {\left( {x - y} \right)^2} = 4900 - 4000 \cr & {\left( {x - y} \right)^2} = 900 \cr & \left( {x - y} \right) = 30\,{\text{km/hr}} \cr} $$

Distance = 140 km
y > x
Same direction:
y - x = $$\frac{{140}}{7}$$ = 20 . . . . . .(i)
Opposite direction:
y + x = $$\frac{{140}}{1}$$ = 140 . . . . . . (ii)
Add equation (i) and (ii)
2y = 160
y = 80 km/hr

$$\eqalign{ & \underleftrightarrow {\,\,\,\,\,\,\,\,\,\,150{\text{ km}}\,\,\,\,\,\,\,\,\,\,} \cr & {\text{Initial speed}} = \frac{{150}}{5} = 30{\text{ km/h}} \cr & {\text{New speed}} = \frac{1}{3} \times 30 = 10{\text{ km/h}} \cr & {\text{Time taken to cover 150 km}} = \frac{{150}}{{10}} = 15{\text{ h}} \cr & {\text{Time difference}} = 15 - 5 = 10{\text{ hour}} \cr} $$

$$60\% = \frac{3}{5}$$
Let the speed of the man be 5x
60% of the speed = 5x × $$\frac{3}{5}$$ = 3x
Ratio of speed of man before and after = 5x : 3x
As we know, speed is inversely proportional to time.
Time ratio of man before and after = 3x : 5x
According to the question
5x - 3x = 1 hr 40 min
5x - 3x = $$\left( {1 + \frac{{40}}{{60}}} \right){\text{hr}}$$
2x = $$\frac{5}{3}$$
x = $$\frac{5}{{3 \times 2}}$$
x = $$\frac{5}{6}$$ hr
Required time = 3x = 3 × $$\frac{5}{6}$$ = $$2\frac{1}{2}$$ hr

The distance of 13 km covered by the speed of 65 km/hr in time
$$\eqalign{ & \to \frac{{13}}{{65}} = \frac{1}{5} \times 60 = 12\,{\text{min/hr}} \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr & \left( {\frac{{{\text{Faster speed}} - {\text{Slower speed}}}}{{{\text{Faster speed}}}} \times 60} \right){\text{min/hr}} \cr & = \frac{{65 - 52}}{{65}} \times 60 \cr & = 12\,{\text{min/hr}} \cr} $$