Practice MCQ Questions and Answer on Speed Time and Distance
11.
A and B start moving from places X to Y and Y to X, respectively, at the same into on the same day. After crossing each other, A and B take $$5\frac{4}{9}$$ÃÂÃÂ hours and 9 hours, respectively, to each their respective destinations. If the speed of A is 33 km/h, then the speed (in km/h) of B is:
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour ?
(A) 9 min
(B) 10 min
(C) 12 min
(D) 20 min
Solution:
Due to stoppages, it covers 9 k less. Time taken to cover 9 km : $$\eqalign{ & = \left( {\frac{9}{{54}} \times 60} \right){\text{ min}} \cr & = 10\min \cr} $$
13.
A runs $$\frac{7}{4}$$ times as fast as B. If A gives B a start of 300 m, how far must the winning post be if both A and B have to end the race at same time?
(A) 1400 m
(B) 700 m
(C) 350 m
(D) 210 m
Solution:
.ABReason (ST=D) Speed74GivenTime 47Since, Speed ∝ $$\frac{1}{{{\text{Time}}}}$$Distance4 7Distance ∝ time Now, 7x - 4x = 300 (A runs 7x m and B runs 4x) x = 100 7x = 7 × 100 = 700 m Winning post is 700 m away, As A runs 700 m to complete the race.
14.
A person walks a distance from point A to B at 15 km/h, and from point B to A at 30 km/h. If he takes 3 hours to complete the journey, then what is the distance from point A to B?
(A) 25 km
(B) 10 km
(C) 15 km
(D) 30 km
Solution:
Distance same Speed = 15 : 3 The distance between A and B = 15 × 2 = 30 km
15.
How much time does a train 50 m long, moving at 68 km/hr takes to pass another train 75 m long moving at 50 km/hr in the same direction ?
(A) 5 sec
(B) 10 sec
(C) 20 sec
(D) 25 sec
Solution:
Total distance covered by the length of both trains : = 50 m + 75 m And, their relative speed in same direction : = 68 - 50 = 18 km/hr $$\because $$ (Speed subtracted in same direction) Then, the time to cross each other will be : $$\eqalign{ & = \frac{{125{\text{ m/s}}}}{{18{\text{ kms}}}} \cr & = \frac{{125{\text{ }} \times {\text{18}}}}{{18 \times 5{\text{ }}}}{\text{m/s}} \cr & \left\{ {\because 1{\text{ km/hr = }}\frac{5}{{18}}{\text{ m/s}}} \right\} \cr & = 25\sec \cr} $$
16.
A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour ?
In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
The speed of electric train is 25% more than that of steam engine train. What is the time taken by an electric train to cover a distance which a steam engine takes 4 hours 25 minutes to cover ?
(A) $$3\frac{1}{10}$$ hrs
(B) $$3\frac{11}{15}$$ hrs
(C) $$3\frac{11}{12}$$ hrs
(D) $$3\frac{8}{15}$$ hrs
Solution:
Let the speed of steam engine train be x Then, speed of electric train = 125% of x = $$\frac{5x}{4}$$ Time taken by steam engine : = 4 hrs 25 min = $$4\frac{25}{60}$$ hrs = $$4\frac{5}{12}$$ hrs Let the time taken by electric train be $$t$$ hours Then, $$\eqalign{ & \Rightarrow x:\frac{{5x}}{4}::t:4\frac{5}{{12}} \cr & \Rightarrow 1:\frac{5}{4}::t:\frac{{53}}{{12}} \cr & \Rightarrow \frac{5}{4}t = \frac{{53}}{{12}} \cr & \Rightarrow t = \left( {\frac{{53}}{{12}} \times \frac{4}{5}} \right) \cr & \Rightarrow t = \frac{{53}}{{15}} \cr & \Rightarrow t = 3\frac{8}{{15}}{\text{ hrs}} \cr} $$
19.
An individual is cycling at a speed of 25 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier ?
(A) 15 kmph
(B) 12 kmph
(C) 10 kmph
(D) 8 kmph
Solution:
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.
20.
A car after traveling 18 km from a point A developed some problem in the engine and the speed became $$\frac{4}{5}$$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
(A) 25 kmph
(B) 30 kmph
(C) 20 kmph
(D) 35 kmph
Solution:
He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time. Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min Thus, $$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours. Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.