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

Let the distance between A and B to the distance between B and C be 3a and 5a respectively
According to the question
Time taken by man to travel from A and B at a speed of x km/hr = $$\frac{{3a}}{x}$$
Time taken by man to travel from B and C at a speed of 50 km/hr = $$\frac{{5a}}{{50}} \Rightarrow \frac{a}{{10}}$$
Average speed of the entire journey = $$\frac{{{\text{Total distance}}}}{{{\text{Total time taken}}}}$$
$$\eqalign{ & \Rightarrow \frac{{3a + 5a}}{{\frac{{3a}}{x} + \frac{a}{{10}}}} = 40 \cr & \Rightarrow \frac{{8a}}{{\frac{{3a}}{x} + \frac{a}{{10}}}} = 40 \cr & \Rightarrow 8a = 40a\left( {\frac{3}{x} + \frac{1}{{10}}} \right) \cr & \Rightarrow \frac{1}{5} = \frac{{30 + x}}{{10x}} \cr & \Rightarrow 2x = 30 + x \cr & \Rightarrow 2x - x = 30 \cr & \Rightarrow x = 30\,{\text{km/hr}} \cr} $$
Now,
The value of (x - 10) : (x + 1) = (30 - 10) : (30 + 1)
⇒ 20 : 31
∴ The required value is 20 : 31

$$\eqalign{ & \frac{x}{{30}} + \frac{x}{{20}} = 5 \cr & \frac{{2x + 3x}}{{60}} = 5 \cr & x = 60 \cr & {\text{Total distance}} = 2x \cr & = 2 \times 60 \cr & = 120{\text{ km}} \cr} $$

$$\eqalign{ & {\text{Time}} = \frac{6}{8} = \frac{3}{4}{\text{ hours}} = 45{\text{ minutes}} \cr & {\text{Including rest}} = 45 + 5 \times 4 = 65{\text{ minutes}} \cr} $$

\[\begin{array}{*{20}{c}} {}&{{\mathbf{Car}}}&{}&{{\mathbf{Train}}} \\ {}&{100}&{}&{220} \\ {{\text{Speed}} \to }&5&:&{11} \\ {{\text{Time}} \to }&{11}&:&5 \end{array}\]
$$\eqalign{ & {\text{Actual speed of car}} = \frac{{950}}{{19}} = 50{\text{ km/m}} \cr & {\text{5 unit}} \to {\text{50 km/h}} \cr & {\text{1 unit}} \to 1{\text{0 km/h}} \cr & {\text{11 unit}} \to 11{\text{0 km/h}} \cr & {\text{Distance cover by train in }}3\frac{1}{2}{\text{ hour}} \cr & = 110 \times \frac{7}{2} \cr & = \frac{{770}}{2} \cr & = 385{\text{ km}} \cr} $$

Let the total time = 5 hours
The distance covered in 5 hours = 5 × 60 = 300 km

Speed = $$\frac{{120}}{3}$$ = 40 km/hr

Y's speed greater than X's speed.
Y > X
X + Y = 100 km/hr . . . . . . (i)
$$\eqalign{ & {\text{Total time}} = 3{\text{ hr }}45{\text{ min}} = 3\frac{{45}}{{60}} = \frac{{15}}{4}{\text{ hr}} \cr & {\text{Distance}} = \frac{{{S_1} \times {S_2}}}{{{S_1} + {S_2}}} \times {\text{Total time}} \cr & 90 = \frac{{X \times Y}}{{X + Y}} \times \frac{{15}}{4} \cr & X + Y = \frac{{90 \times 100 \times 4}}{{15}} \cr & X + Y = 2400{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr} $$
Solve equation (i) and (ii)
X = 40; Y = 60     [Y > X]
Ratio of X's speed and Y's speed = 40 : 60 = 2 : 3

Let the speed of truck is = $$x$$ km/h
Their relative speed in same direction = (45 - $$x$$) km/h (Here (45 - $$x$$) has been written because bus crosses the truck which is running 150 meters ahead of it. i.e. Truck speed will be lower than that of bus)
According to the question
$$\eqalign{ & {\text{Time}} = \frac{{{\text{Total distance}}}}{{{\text{Total speed}}}} \cr & \frac{{150}}{{\left( {45 - x} \right) \times \frac{5}{{18}}}} = 30 \cr & \frac{{150 \times 18}}{{\left( {45 - x} \right) \times 5}} = 30 \cr & x = 27{\text{ km/h}} \cr} $$
So, speed of the truck is 27 km/h

$${\frac{4}{5}^{{\text{th}}}}$$ part of distance $$ = 160 \times \frac{4}{5} = 128$$
Remaining distance = 32 km
$${\frac{2}{3}^{{\text{th}}}}$$ of time $$ = 15 \times \frac{2}{3} = 10{\text{ hr}}$$
Remaining time = 5 hr
Speed $$ = \frac{{32}}{5} = 6.4{\text{ km/hr}}$$

$$\eqalign{ & {\text{Time}} = \frac{{250}}{{\left( {36 + 54} \right) \times \frac{5}{{18}}}} \cr & = \frac{{250 \times 18}}{{90 \times 5}} \cr & = 10{\text{ second}} \cr} $$

Time : Speed = 2x : 3x
Distance = 2x × 3x
600 = 6x²
x = 10
Time = 2x = 2 × 10 = 20 hours