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

Let the total distance be 3x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{3} + \frac{x}{4} + \frac{x}{5} = \frac{{47}}{{60}} \cr & \Leftrightarrow \frac{{47x}}{{60}} = \frac{{47}}{{60}} \cr & \Leftrightarrow x = 1 \cr} $$
∴ Total distance = (3 × 1) km = 3 km

Let speed of jogging be x km/hr
Total time taken :
$$\eqalign{ & = \left( {\frac{9}{6}{\text{hrs}} + 1.5{\text{ hrs}}} \right) \cr & = 3{\text{ hrs}} \cr} $$
Total distance covered = $$\left( {9 + 1.5x} \right){\text{ km}}$$
$$\eqalign{ & \therefore \frac{{9 + 1.5x}}{3} = 9 \cr & \Leftrightarrow 9 + 1.5x = 27 \cr & \Leftrightarrow \frac{3}{2}x = 18 \cr & \Leftrightarrow x = \left( {18 \times \frac{2}{3}} \right) \cr & \Leftrightarrow x = 12{\text{ kmph}} \cr} $$

Let the actual distance travelled be x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{{10}} = \frac{{x + 20}}{{14}} \cr & \Leftrightarrow 14x = 10x + 200 \cr & \Leftrightarrow 4x = 200 \cr & \Leftrightarrow x = 50{\text{ km}} \cr} $$

Distance covered by Aryan in 5 min :
= (40 × 5) m = 200 m
Relative speed of Rahul w.r.t. Aryan :
= (50 - 40) m/min = 10 m/min
Time taken to cover 200 m at relative speed :
$$\eqalign{ & = \left( {\frac{{200}}{{10}}} \right)\min \cr & = 20{\text{ min}} \cr} $$
Distance covered by the dog in 20 min :
= (60 × 20) m
= 1200 m

In the same time, they cover 110 km and 90 km respectively.
∴ Ratio of their speed
= 110 : 90
= 11 : 9

$$\eqalign{ & {\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & = \left( {\frac{{650 + 850}}{{12 + 18}}} \right)\,{\text{kmph}} \cr & = \left( {\frac{{1500}}{{30}}} \right)\,{\text{kmph}} \cr & \therefore {\text{Average speed}} = 50{\text{ kmph}} \cr} $$

Number of gaps between 21 telephone posts = 20
Distance travelled in 1 minute
= (50 × 20) m
= 1000 m
= 1 km
∴ Speed = 60 km/hr

Let the speed of Kim be a and that of Om be b
Distance between point A and B = 400 km
Then,
$$\frac{{400}}{a} - \frac{{400}}{b} = 1$$
Let,
$$\eqalign{ & \frac{1}{a} = x{\text{ and }}\frac{1}{b} = y \cr & 400x - 400y = 1.....(i) \cr} $$
Speed of km doubles and she will take lets time i.e., 1 hr 30 min than 0 metre
Again,
$$\eqalign{ & \frac{{400}}{b} - \frac{{400}}{{2b}} = \frac{3}{2} \cr & \therefore 400y - 200x = \frac{3}{2} \cr & 800y - 400x = 3 \cr} $$
Solving (i) and (ii), we get
$$\eqalign{ & 400x - 400y = 1 \cr & \frac{{ - 400x + 800y = 3}}{{400y = 4}} \cr & \therefore y = \frac{4}{{100}} = \frac{1}{{100}}{\text{ km}} \cr & \therefore b = 100{\text{ km}} \cr & {\text{Now,}} \cr & \frac{{400}}{a} - \frac{{400}}{{100}} = 4 \cr & {\text{or, }}\frac{{400}}{a} \cr & \therefore a = 80{\text{ kmph}} \cr} $$

$$\eqalign{ & 25{\text{ m/}}\sec : \cr & = \left( {25 \times \frac{{18}}{5}} \right){\text{km/hr }} \cr & = 90{\text{ km/hr}} \cr & {\text{And, 25 m/}}\sec : \cr & = \left( {25 \times 60} \right){\text{ m/}}\sec \cr & = 1500{\text{ m/}}\sec \cr} $$
So, all the three speeds are equal.

Distance covered in first 2 hours :
= (70 × 2) km
= 140 km
Distance covered in next 2 hours :
= (80 × 2) km
= 160 km
Remaining distance :
= 345 - (140 + 160)
= 45 km
Speed in the fifth hour = 90 km/hr
Time taken to cover 45 km :
= $$\frac{45}{90}$$ hr
= $$\frac{1}{2}$$ hr
∴ Total time taken :
= 2 + 2 + $$\frac{1}{2}$$
= $$4\frac{1}{2}$$ hrs