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

The distance covered in two hour,
= 2 × 25 = 50 km
Time taken by first individual = (3h + 2h) = 5h
Then, the speed of predecessor,
= $$\frac{{50}}{5}$$ = 10 kmph.

Let the number of puffs abe,
$$3x \times \left( {3 \times \frac{1}{{150}}} \right) + \left( {13 \times 60 - 3x} \right) \times \frac{1}{{150}} = 6$$
On solving, We get, x = 20 puffs.

For every 15 minutes he takes a rest of 2 minutes.
Hence,
for 90 minutes of drive he would require 12 minutes of rest.
Thus, he will be traveling for 90 - 12 = 78 minutes.
In 60 minutes he covers 80 Km.
In 1 minute he would cover $$\frac{{80}}{{60}}$$ Km.
In 78 minutes he would cover $$\frac{{80}}{{60}} \times 78$$   = 104 Km.

Relative speed of the trains = 40 + 30 = 70 kmph.
Time taken by trains to collide,
= $$\frac{{560}}{{70}}$$
= 8h.
In 8h sparrow will cover,
= 8 × 80
= 640 km.

Time taken to collide the trains,
= one and half hour = $$\frac{3}{2}$$ h.
So, in $$\frac{3}{2}$$ h bird travels,
= $$\frac{{3 \times 60}}{2}$$
= 90 km.

In 20 minutes the difference between man and son,
= 20 × 20
= 400 m.
Distance traveled by dog when he goes towards the child,
= $$400 \times \frac{{60}}{{40}}$$
= 600 m and time required is 10 minutes.
In 10 min remaining distance between man and child,
= 400 - (20 × 10)
= 200 m.
Time taken by dog to meet the man,
= $$\frac{{200}}{{100}}$$
= 2 min

(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)

Remaining distance in 2 min,
= 200 - (2 × 20),
= 160 m.
Now, the time taken by dog to meet the child again,
= $$\frac{{160}}{{40}}$$
= 4 min.
In 4 min he covers = 4 × 60 = 240 m.
Now, remaining distance in 4 min = 160 - (4 × 20) = 80 m.
Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min.
Now remaining distance = 80 - (0.8 × 20) = 64 m.

Speed of tiger = 40 m/min
Speed of deer = 20 m/min.
Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m
Time take to catch = $$\frac{{400}}{{20}}$$ = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m

(6 - x) = (8 - 1.5x)
Or, x = 4 cm.
So, It will take 4 hours to burn it in such a way that they will remain equal in lengths.

Speed of man,
= 4 kmph = $$\frac{{4 \times 5}}{{18}}$$ m/sec
In 3 min (180 sec) man will go = $$\frac{{20 \times 180}}{8}$$ = 200 m.
That means the diagonal of the square field = 200 m.
Diagonal of square,
= Side of Square × $$\sqrt 2 $$  = 200 m.
→ Side of Square = $$\frac{{200}}{{\sqrt 2 }}$$
Area of Square = Side²
Area = $$\frac{{200 \times 200}}{2}$$   = 20000 sqm.

Train(T)__________ A_____5k____CAT__________B

T<-------x--------------><----12k-------------------------->

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
$$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & \Rightarrow 7x = 5\left( {x + 12k} \right) \cr & \Rightarrow \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$
  $${\text{or,}}\,\,6:1$$