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

Speed Time And Distance MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Speed Time And Distance with easy and logical explanations. Speed Time And Distance MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-8 section-4

Each section contains 100 MCQs on Speed Time and Distance .
71
To reach point B from point A at 4 pm, Sara will have to travel at an average speed of 18 kmph. She will reach point B at 3 pm if she travels at an average speed of 24 kmph. At what average speed should Sara travels to reach point B at 2 pm ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Difference between time = 1 hour
Distance between point AB = x km
According to the question,
$$ \Rightarrow \frac{x}{{18}} - \frac{x}{{24}} = 1$$
L.C.M. of 18 and 24 = 72
$$\eqalign{ & \Rightarrow \frac{{4x - 3x}}{{72}} = 1 \cr & \Rightarrow x = 72{\text{ km}} \cr} $$
Time taken at 18 km/hr to cover 72 km
$$ = \frac{{72}}{{18}} = 4{\text{ hours}}$$
∴ Speed to cover 72 km in 2 hours :
$$ = \frac{{72}}{2} = 36{\text{ km/hr}}$$
72
A person travels three equal distance at a speed of x km/hr, y km/hr and z km/hr respectively. What is the average speed for the whole journey ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let each distance be equal to d
Then, total distance travelled = 3d
Total time taken :
$$\eqalign{ & = \left( {\frac{d}{x} + \frac{d}{y} + \frac{d}{z}} \right){\text{hr}} \cr & = \frac{{d\left( {xy + yz + zx} \right)}}{{xyz}}{\text{ hr}} \cr} $$
∴ Average speed :
$$\eqalign{ & = \left[ {3d \times \frac{{xyz}}{{d\left( {xy + yz + zx} \right)}}} \right]{\text{km/hr}} \cr & = \frac{{3xyz}}{{\left( {xy + yz + zx} \right)}}{\text{ km/hr}} \cr} $$
73
A car takes 15 minutes less to cover a distance of 75 km, if it increases its speed by 10 km/hr from its usual speed. How much time would it take to cover a distance of 300 km using this speed ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the usual speed be x km/hr
Then,
$$\eqalign{ & \Leftrightarrow \frac{{75}}{x} - \frac{{75}}{{x + 10}} = \frac{{15}}{{60}} \cr & \Leftrightarrow x\left( {x + 10} \right) = 3000 \cr & \Leftrightarrow {x^2} + 10x - 3000 = 0 \cr & \Leftrightarrow \left( {x + 60} \right)\left( {x - 50} \right) = 0 \cr & \Leftrightarrow x = 50 \cr} $$
∴ Required time :
$$\eqalign{ & = \left( {\frac{{300}}{{60}}} \right){\text{hrs}} \cr & = 5{\text{ hrs}} \cr} $$
74
A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 metres ahead of the policeman, then the time required for the policeman to catch the thief will be :
Discuss
Answer & Solution
Answer: Option B
Solution:
Relative speed = (10 - 8) km/hr = 2 km/hr
Required time = Time taken to cover 100 m at relative speed :
$$\eqalign{ & = \left( {\frac{{100}}{{2000}}} \right){\text{hr}} \cr & = \left( {\frac{1}{{20}}} \right){\text{hr}} \cr & = \left( {\frac{1}{{20}} \times 60} \right){\text{min}} \cr & = 3\text{ minutes} \cr} $$
75
Two planes move along a circle of circumference 1.2 kms with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction one plane overtakes the other every 60 seconds. The speed of the slower plane is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Let their speeds be x m/sec and y m/sec respectively.
Then,
$$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$
And,
$$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$
Adding (i) and (ii), we get :
2x = 100 or x = 50
Putting x = 50 in (i), we get : y = 30
Hence, speed of slower plane :
= 30 m/sec
= 0.03 km/sec
76
A student goes to school at the rate of $$2\frac{1}{2}$$ km/hr and reaches 6 min late. If he travels at the speed of 3 km/hr he is 10 min early. What is the distance to the school ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the distance between school and home be = D
According to the given information :
$$\frac{D}{{2\frac{1}{2}}} - \frac{D}{3} = 16{\text{ minutes}}$$
$$\eqalign{ & \therefore \frac{{2D}}{5} - \frac{D}{3} = \frac{{16}}{{60}} \cr & \Rightarrow \frac{{6D - 5D}}{{15}} = \frac{{16}}{{60}} \cr & \Rightarrow D = \frac{{16}}{{60}} \times 15 \cr & \Rightarrow D = 4{\text{ km}} \cr} $$
77
The distance between two railway stations is 1176 km. To cover this distance, an express train takes 5 hours less than a passenger train while the average speed of the passenger train is 70 km/h less than that of the express train. The time taken by the passenger train to complete the travel is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the speed of express train = x
Passenger train = x - 70
$$\frac{{1176}}{{x - 70}} - \frac{{1176}}{x} = 5$$
At x = 168 is satisfy
$$\eqalign{ & \frac{{1176}}{{98}} - \frac{{1176}}{{168}} = 5 \cr & 12 - 7 = 5 \cr & 5 = 5 \cr} $$
The time of passenger train $$ = \frac{{1176}}{{98}} = 12{\text{ hours}}$$

Alternate Solution:-
We can go through option
1176 → 1 + 1 + 7 + 6 = 15 → It is divisible by 3.
So, we have to check option, which one is divisible by 3, option A and D
→ Now we can check 18 and 12.
78
One third of a certain journey is covered at the speed of 80 km/hr one fourth of the journey at the speed of 50 km/hr and the rest at the speed of 100 km/hr what will be the average speed (in km/hr) for the whole journey?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total Journey}}}}{{{\text{Time Taken}}}} \cr & \Rightarrow {\text{Remaining distance}} \cr & = 1 - \left( {\frac{1}{3} + \frac{1}{4}} \right) = \frac{5}{{12}}{\text{ km}} \cr & \therefore {\text{Average speed}} \cr & = \frac{1}{{\frac{1}{{3 \times 80}} + \frac{1}{{50 \times 4}} + \frac{{1 \times 5}}{{100 \times 12}}}} \cr & = \frac{1}{{\frac{{5 + 6 + 5}}{{1200}}}} \cr & = \frac{{1200}}{{16}} \cr & = 75{\text{ km/hr}}{\text{.}} \cr} $$
79
A car covers 15 km, 20 km, 30 km and 12 km at speeds of 20 km/h, 30 km/h, 40 km/h and 30 km/h, respectively. The average speed of the car for the total journey is:
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{15 + 20 + 30 + 12}}{{\frac{3}{4} + \frac{2}{3} + \frac{3}{4} + \frac{2}{5}}} \cr & = \frac{{27}}{{\frac{3}{2} + \frac{2}{3} + \frac{2}{5}}} \cr & = \frac{{77 \times 30}}{{45 + 20 + 12}} \cr & = \frac{{77 \times 30}}{{77}} \cr & {\text{Average speed}} = 30{\text{ km/hr}}{\text{.}} \cr} $$
80
A covered a distance of 240 km at a certain speed. Had his speed been 8 km/h less, then the time taken would have been one hour more for covering the same distance. How much time (in hours) will he takes to cover a distance of 480 km at his original speed?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Distance}} = \frac{{{S_1} \times {S_2}}}{{{S_1} - {S_2}}} \times {\text{difference of time}} \cr & {\text{240}} = \frac{{S\left( {S - 8} \right)}}{{S - S + 8}} \times 1 \cr & 240 \times 8 = S \times \left( {S - 8} \right) \cr & 48 \times 40 = S \times \left( {S - 8} \right) \cr & S = 48{\text{ km/hr}}{\text{.}} \cr & {\text{ = }}\frac{{480}}{{48}} = 10{\text{ hrs}} \cr} $$