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

According to the question,
Let the speed of the faster train = x km/hr
The speed of the slower train = y km/hr
$$\eqalign{ & {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & x + y = \frac{{100}}{1}{\text{ km/hr and }} \cr & x - y = \frac{{100}}{5} = 20{\text{ km/hr}} \cr & x + y = 100.....(i) \cr & x - y = 20.....(ii) \cr} $$
Solve equation (i) and (ii), we get
x = 60 km/hr
y = 40 km/hr
Speed of faster car = 60 km/hr

According to the question,
$$\eqalign{ & \therefore 50{\text{ m}} = {\text{20 m}} \cr & \therefore {\text{1m}} = \frac{{20}}{{50}}{\text{m}} \cr & \therefore {\text{1000 m}} = \left( {\frac{{20}}{{50}} \times 1000} \right){\text{ m}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 400{\text{ metres}} \cr} $$

1 yard = 0.9144 m
100 yards = (100 × 0.9144) m
                = 91.44 m
∴ Required difference :
= (100 - 91.44) m
= 8.56 m

Time taken to cover 600 km
$$\eqalign{ & = \left( {\frac{{600}}{{100}}} \right)hr \cr & = 6hrs \cr} $$
Number of stoppages
$$\eqalign{ & = \frac{{600}}{{75}} - 1 \cr & = 7 \cr} $$
Total time of stoppages
= (3 × 7) min
= 21 min
Hence, total time taken = 6 hrs 21 min

Let the length of the journey be x km.
Then,
$$\eqalign{ & \Rightarrow \frac{{\frac{1}{3}x}}{{60}} + \frac{{\frac{2}{3}x}}{{80}} = 5 \cr & \Rightarrow \frac{x}{{180}} + \frac{x}{{120}} = 5 \cr & \Rightarrow \frac{{5x}}{{360}} = 5 \cr & \Rightarrow x = 360\,\text{ km} \cr} $$

Let the whole distance travelled be x km and the average speed of the car for the whole journey be y km/hr
Then,
$$\eqalign{ & \Leftrightarrow \frac{{\left( {\frac{x}{3}} \right)}}{{10}} + \frac{{\left( {\frac{x}{3}} \right)}}{{20}} + \frac{{\left( {\frac{x}{3}} \right)}}{{60}} = \frac{x}{y} \cr & \Leftrightarrow \frac{x}{{30}} + \frac{x}{{60}} + \frac{x}{{180}} = \frac{x}{y} \cr & \Leftrightarrow \frac{1}{{18}}y = 1 \cr & \Leftrightarrow y = 18{\text{ km/hr}} \cr} $$

Let the distance be x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{3} - \frac{x}{{3.75}} = \frac{1}{2} \cr & \Leftrightarrow 2.5x - 2x = 3.75 \cr & \Leftrightarrow x = \frac{{3.75}}{{0.50}} \cr & \Leftrightarrow x = \frac{{15}}{2} \cr & \Leftrightarrow x = 7.5{\text{ km}} \cr} $$

Suppose the thief is overtaken x hrs after 2.30 pm
The, distance covered by the thief in x hrs = Distance covered by owner in $$\left( {x - \frac{1}{2}} \right)$$  hrs
$$\eqalign{ & \therefore 60x = 75\left( {x - \frac{1}{2}} \right) \cr & \Leftrightarrow 15x = \frac{{75}}{2} \cr & \Leftrightarrow x = \frac{5}{2}{\text{ hrs}} \cr} $$
So, the theif is overtaken at 5 pm

Let the speed of the faster and slower cyclists be x km/hr and y km/hr respectively
Then,
$$\eqalign{ & \frac{k}{{x - y}} = r \cr & \left( {x - y} \right)r = k.....(i) \cr & {\text{And,}} \cr & \frac{k}{{x + y}} = t \cr & \left( {x + y} \right)t = k.....(ii) \cr} $$
From (i) and (ii), we have :
$$\eqalign{ & \Rightarrow \left( {x - y} \right)r = \left( {x + y} \right)t \cr & \Rightarrow xr - yr = xt + yt \cr & \Rightarrow xr - xt = yr + yt \cr & \Rightarrow x\left( {r - t} \right) = y\left( {r + t} \right) \cr & \Rightarrow \frac{x}{y} = \frac{{r + t}}{{r - t}} \cr} $$

Let the distance covered by the cyclist be x and the time taken be y
Then, required ratio :
$$\eqalign{ & = \frac{{\frac{1}{2}x}}{{2y}}:\frac{x}{y} \cr & = \frac{1}{4}:1 \cr & = 1:4 \cr} $$