Practice MCQ Questions and Answer on Speed Time and Distance
1.
There are 8 equidistant points A, B, C, D, E, F, G and H in the clockwise direction on the periphery of a circle. In a time interval t, a person reaches from A to C with uniform motion while another person reaches the point E from the point B during the same time interval with uniform motion. Both the persons move in the same direction along the circumference of the circle and start at the same instant. How much time after the start, will the two persons meet each other ?
(A) 4t
(B) 7t
(C) 9t
(D) Never
Solution:
Distance covered by first person in time t = $$\frac{2}{8}$$ round = $$\frac{1}{4}$$ round Distance covered by second person in time t = $$\frac{3}{8}$$ round Speed of the first person = $$\frac{1}{4t}$$ Speed of the second person = $$\frac{3}{8t}$$ Since the two persons start from A and B respectively, so they shall meet each other when there is a difference of $$\frac{7}{8}$$ round between the two. Relative speed of A and B : $$\eqalign{ & = \left( {\frac{3}{{8t}} - \frac{1}{{4t}}} \right) \cr & = \frac{1}{{8t}} \cr} $$ Time taken to cover $$\frac{7}{8}$$ round at this speed : $$\eqalign{ & = \left( {\frac{7}{8} \times 8t} \right) \cr & = 7t \cr} $$
2.
A man travels the distance of his journey $$\frac{3}{4}$$ by bus, $$\frac{1}{6}$$ by rickshaw and remaining 2 km on foot. The total distance travelled by the man is :
(A) 12 km
(B) 18 km
(C) 20 km
(D) 24 km
Solution:
Let the man travels 1 unit distance So, remaining distance $$\eqalign{ & = 1 - \left( {\frac{1}{6} + \frac{3}{4}} \right) \cr & = 1 - \frac{{22}}{{24}} \cr & = \frac{1}{{12}} \cr & \because \frac{1}{{12}}{\text{ unit}} = {\text{2 km}} \cr & {\text{So, 1 unit}} = 24{\text{ km}} \cr} $$
3.
A square playground measure 1127.6164 sq.m. If a man walks $$2\frac{9}{{20}}$$ m a minute, the time taken by him to walk one round around it is approximately :
(A) 50.82 minutes
(B) 54.82 minutes
(C) 54.62 minutes
(D) 50.62 minutes
Solution:
Let side of the square playground is $$x$$ Ans Area : $$\eqalign{ & {x^2} = 1127.6164 \cr & x = \sqrt {1127.6164} \cr & x = 33.58 \cr} $$ Perimeter of playground : $$\eqalign{ & = 4x \cr & = 4 \times 33.58 \cr & = 134.32 \cr} $$ Time taken to complete 1 round : $$\eqalign{ & = \frac{{134.32}}{{2\frac{9}{{20}}}} \cr & = \frac{{134.32}}{{49}} \times 20 \cr & = 54.82\,\,\text{minutes} \cr} $$
4.
A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speed is :
(A) 1 : 16
(B) 4 : 1
(C) 2 : 1
(D) 1 : 4
Solution:
Let the original speed be S1 and time t1 and distance be D. Now, $$\eqalign{ & \frac{{ {\frac{D}{2}} }}{{2{t_1}}} = {S_2} \cr & {S_2} = \frac{D}{{4{t_1}}}\,{\text{and}}\,{S_1} = \frac{D}{{{t_1}}} \cr & {\text{Thus}}, \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{4}{1} = 4:1 \cr} $$
5.
A railway engine passes two bridges of lengths 400 m and 235 m in 100 seconds and 60 seconds, respectively. Twice the length of the railway engine (in m) is:
(A) 24
(B) 25
(C) 12.5
(D) 12
Solution:
Let length of engine = $$x$$ m According to question $$\eqalign{ & \frac{{x + 400}}{{100}} = \frac{{x + 235}}{{60}} \cr & \Rightarrow 6x + 2400 = 10x + 2350 \cr & \Rightarrow 4x = 50 \cr & \Rightarrow 2x = 25{\text{ m}} \cr} $$
6.
A cyclist moving on a circular track of radius 100 meters completes one revolution in 2 minutes. What is the average speed of cyclist (approx.)?
A thief is noticed by a policeman from a distance of 200 metres the thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km/hr and 11 km/hr respectively. What is the distance between them after 6 minutes ?
(A) 100 m
(B) 190 m
(C) 200 m
(D) 150 m
Solution:
Relative speed of policeman with respect to thief = (11 - 10) = 1 Km/hr Now the relative distance covered by policeman in 6 min $$\eqalign{ & = {\text{Speed}}\,\, \times \,\,{\text{Time}} \cr & = 1 \times \frac{6}{{60}} \cr & = \frac{1}{{10}}\,{\text{km}} \cr & = 100\,{\text{m}} \cr} $$ The distance between the policeman and thief after 6 min = 200 - 100 = 100 m
8.
A man can row 15 km/h in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in the river. The speed of current (in km/h) :
(A) 6 km/h
(B) 6.5 km/h
(C) 4.5 km/h
(D) 5 km/h
Solution:
Let the speed of the current be x km/h. Now, according to the question, $$\frac{{15 + x}}{{15 - x}} = \frac{2}{1}$$ 15 + x = 30 - 2x 3x = 30 - 15 3x = 15 x = 5 km/h
9.
A train travelling at a speed of 30 m/sec crosses a platform, 600 metres long in 30 seconds. The length (in metres) of train is :
(A) 120 metres
(B) 150 metres
(C) 200 metres
(D) 300 metres
Solution:
Total distance covered by the train in 30 seconds with the speed of 30 m/s is = 30 × 30 m/s = 900 metres Total distance - train's length + platform's length 900 = train's length + 600 (when train crosses platform it covers length equal to length of train + length of platform) Train's length = 900 - 600 Train's length = 300 metres
10.
A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $$\frac{5}{{12}}$$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?
(A) 1 : 6
(B) 3 : 5
(C) 6 : 1
(D) 5 : 4
Solution:
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 $$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & \Rightarrow 7x = 5\left( {x + 12k} \right) \cr & \Rightarrow \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$ $${\text{or,}}\,\,6:1$$