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

Total distance travelled :
$$\eqalign{ & = \left( {30 \times \frac{{12}}{{60}} + 45 \times \frac{8}{{60{\text{ }}}}} \right){\text{ km}} \cr & = {\text{12 km}} \cr} $$
Total time taken :
= (12 + 8) min
= 20 min
= $$\frac{1}{3}$$ hr
∴ Average speed :
= (12 × 3) km/hr
= 36 km/hr

Let distance = x km and usual rate y kmph
Then,
$$\eqalign{ & \Rightarrow \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr & \Rightarrow y\left( {y + 10} \right) = 10x.....(i) \cr} $$
And,
$$\eqalign{ & \Rightarrow \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr & \Rightarrow y\left( {y + 20} \right) = \frac{{80x}}{7}.....(ii) \cr} $$
On dividing (i) by (ii) we get :
y = 60
Substituting y = 60 in (i), we get: x = 420 km

Quantity of petrol consumed in 1 hour:
$$\eqalign{ & = \left( {\frac{{40}}{{10}} + \frac{1}{2}} \right){\text{ litres}} \cr & = 4\frac{1}{2}{\text{ litres}} \cr} $$
Time for which the fuel lasted :
$$\eqalign{ & = \left[ {\frac{{18}}{{\left( {4\frac{1}{2}} \right)}}} \right]{\text{ hrs}} \cr & = \left( {18 \times \frac{2}{9}} \right){\text{ hrs}} \cr & = 4{\text{ hrs}} \cr} $$
∴ Required distance = (40 × 4) km = 160 km

Relative speed = (200 + 200) km/hr = 400 km/hr
Distance covered in 1 hr 30 min
$$\eqalign{ & = \left( {400 \times \frac{1}{2} + 200 \times \frac{1}{2} + 100 \times \frac{1}{2}} \right){\text{km}} \cr & = \left( {200 + 100 + 50} \right){\text{km}} \cr & = 350{\text{ km}} \cr} $$
[$$\because $$ Speed reduces by half every half an hour]
Hence, distance between the trains :
= (425 - 350) km
= 75 km

Speed of Ashok = 39 km/ph
Speed of Rahul = 47 km/ph
Distance between place A and place B = 637 km

Distance covered by Ashok (from 8 am to 10 am) in 2 hours = 2 × 39 = 78 km
∴ Remaining distance = 637 - 78 = 559
Relative speed = 39 + 47 = 86 km/ph
∴ Time taken to travel 559 km = $$\frac{559}{86}$$ = 6.5 hours
So, they meet at = (10 am + 6.5 hrs) = 4.30 pm

$${\text{Speed }} = \frac{{{\text{Distance }}}}{{{\text{Time}}}}$$    or
$$\frac{{{\text{Distance }}}}{{{\text{Time}}}} = {\text{Speed}}$$
⇒ $$\frac{1}{r} = p - q$$

$$\eqalign{ & \text{Distance}=\left(240\times5\right) \text{ km} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ = 1200 km} \cr & \therefore \text{ Required Speed:} \cr & = \frac{\text{Distance}}{\text{Time}} \cr & = \frac{1200}{\frac{5}{3}} \cr & = \left( {1200 \times \frac{3}{5}} \right){\text{ km/hr}} \cr & = 720{\text{ kmph}} \cr} $$

Total time taken :
$$\eqalign{ & = \left( {\frac{{160}}{{64}} + \frac{{160}}{{80}}} \right){\text{ hrs}} \cr & = \frac{9}{2}{\text{ hrs}} \cr} $$
∴ Average speed :
$$\eqalign{ & = \left( {320 \times \frac{2}{9}} \right){\text{ km/hr}} \cr & = 71.11{\text{ km/hr}} \cr} $$

Let speed of the car be x kmph
Then, speed of the train :
$$ = \frac{{150}}{{100}}x = \left( {\frac{3}{2}x} \right){\text{ kmph}}$$
$$\eqalign{ & \therefore \frac{{75}}{x} - \frac{{75}}{{\frac{3}{2}x}} = \frac{{125}}{{10 \times 60}} \cr & \Leftrightarrow \frac{{75}}{x} - \frac{{50}}{x} = \frac{5}{{24}} \cr & \Leftrightarrow x = \left( {\frac{{25 \times 24}}{5}} \right) \cr & \Leftrightarrow x = 120{\text{ kmph}} \cr} $$

Perimeter of the field
= 2(400 + 300) m
= 1400 m
= 1.4 km
Since A and B move in the same direction, so they will first meet each other when there is a difference of one round i.e., 1.4 km between the two.
Relative speed of A and B = (3 - 2.5) km = 0.5 km/hr
Time take to cover 1.4 km at this speed :
$$\eqalign{ & = \left( {\frac{{1.4}}{{0.5}}} \right){\text{ hr}} \cr & = 2\frac{4}{5}{\text{ hr}} \cr & = 2{\text{ hr 48 min}} \cr} $$
So, they shall first cross each other at 9.48 am
And again 2 hr 48 min after 9.48 am i,e., 12.36 pm
Thus, till 12.30 pm they will cross each other once.