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

Let distance = x km and usual rate = y kmph. Then,
$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr & {\text{or,}}\,y\left( {y + 10} \right) = 10x\,.\,.\,.\,.\,.\,.\,.\,.\,\,\left( 1 \right) \cr} $$
Now, in the 2^nd scenario with a further increase in speed the driver could have saved another 45 min = $$\frac{3}{4}$$ hrs.
Therefore, total time saved
$$ = 1 + \frac{3}{4} = \frac{7}{4}\,{\text{hrs}}{\text{.}}$$
Putting it in equation, we get
$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr & {\text{or,}}\,y\left( {y + 20} \right) = \frac{{80x}}{7}\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
On dividing (1) by (2), we get y = 60
Substituting y = 60 in (1), we get :
x = 420 km.

L.C.M. of 40, 50, 60 and 30
= 600 minutes
= 10 hours
So, they meet again 10 hours after they start
i.e., 7.00 pm

Let the original speed of Aeroplane be x km/hr
$$\eqalign{ & \frac{{1200}}{{\left( {x - 300} \right)}} - \frac{{1200}}{x} = 2 \cr & \frac{{x - x + 300}}{{\left( {x - 300} \right)x}} = \frac{2}{{1200}} \cr} $$

Original time :
$$\eqalign{ & t = \frac{{1200}}{{600}} \cr & t = 2{\text{ hours}} \cr} $$

Speed of the first train :
= $$\frac{150}{15}$$ = 10 m/s
Time taken by trains to cross each other = 12 sec
And, relative speed of two trains :
= $$\frac{150 + 150}{12}$$
= 25 m/s
∴ Speed of the second train :
= (25 - 10) × $$\frac{18}{5}$$
= 54 km/hr

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Actual}}\,\,\,\,\,\,\,\,\,\,\,{\text{Reduced}} \cr & {\text{Ratio of speed}} = {\text{15}}\,\,\,\,\,\,\,\,\,{\text{:}}\,\,\,\,\,\,\,\,\,\,{\text{14}} \cr & {\text{Ratio of time}}\,\,\, = {\text{14}}\,\,\,\,\,\,\,\,\,{\text{:}}\,\,\,\,\,\,\,\,\,{\text{15}} \cr & {\text{14}} \to {\text{28 hrs}} \cr & {\text{15}} \to {\text{30 hrs}} \cr & {\text{So, in 2 hrs it travels 10 kms}} \cr & {\text{Speed = }}\frac{{10}}{2} = 5{\text{ km/hr}} \cr} $$

The man covers $$\frac{9}{20}$$ of the journey by bus
∴ Remaining journey :
= 1 - $$\frac{9}{20}$$
= $$\frac{11}{20}$$
⇒ According to the question,
$$\frac{11}{20}$$ of the journey :
= $$\frac{20}{11}$$ × 10
= 18.18 km

V_rel. = 63 - 3 = 60 km/hr
T = $$\frac{500 × 18}{60 × 5}$$  sec = 30 sec
∴ Required time = 30 seconds

Speed of the train is = 180 km/hr
$$\because $$ 1 km/hr = $$\frac{5}{18}$$ m/s
Since,
1 km = 1000 metres
1 hr = 60 × 60 seconds
∴ 1 km/hr = $$\frac{1000}{60 × 60}$$
                = $$\frac{5}{18}$$ m/s
∴ Speed in m/s :
= 180 × $$\frac{5}{18}$$
= 50 m/s

$$\eqalign{ & {\text{Crossing time }} = \frac{{{l_1} + {l_2}}}{{{\text{Speed}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{270 + 180}}{{36 \times \frac{5}{{18}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{450}}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45{\text{ seconds}} \cr} $$

Relative speed of policeman with respect to thief = (11 - 10) = 1 Km/hr
Now the relative distance covered by policeman in 6 min
$$\eqalign{ & = {\text{Speed}}\,\, \times \,\,{\text{Time}} \cr & = 1 \times \frac{6}{{60}} \cr & = \frac{1}{{10}}\,{\text{km}} \cr & = 100\,{\text{m}} \cr} $$
The distance between the policeman and thief after 6 min
= 200 - 100
= 100 m