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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-3 section-2

Each section contains 100 MCQs on Speed Time and Distance .
21
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{actual}}\,{\text{distance}}\,{\text{travelled}}\,{\text{be}}\,x\,km \cr & {\text{Then}},\,\frac{x}{{10}} = \frac{{x + 20}}{{14}} \cr & \Rightarrow 14x = 10x + 200 \cr & \Rightarrow 4x = 200 \cr & \Rightarrow x = 50\,km \cr} $$
22
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}}\,{\text{be}}\,x\,kmph \cr & {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \frac{{150}}{{100}}x = {\frac{3}{2}x} \,kmph \cr & \therefore \frac{{75}}{x} - \frac{{75}}{{\left( {3/2} \right)x}} = \frac{{125}}{{10 \times 60}} \cr & \Rightarrow \frac{{75}}{x} - \frac{{50}}{x} = \frac{5}{{24}} \cr & \Rightarrow x = {\frac{{25 \times 24}}{5}} \cr & \Rightarrow x = 120\,kmph \cr} $$
23
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?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Due}}\,{\text{to}}\,{\text{stoppages,}}\,{\text{it}}\,{\text{covers}}\,{\text{9}}\,{\text{km}}\,{\text{less}} \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{cover}}\,{\text{9}}\,{\text{km}} \cr & = \left( {\frac{9}{{54}} \times 60} \right)\min \cr & = 10\ \text{minutes} \cr} $$
24
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{duration}}\,{\text{of}}\,{\text{the}}\,{\text{flight}}\,{\text{be}}\,x\,{\text{hours}} \cr & {\text{Then}},\frac{{600}}{x} - \frac{{600}}{{x + \left( {1/2} \right)}} = 200 \cr & \Rightarrow \frac{{600}}{x} - \frac{{1200}}{{2x + 1}} = 200 \cr & \Rightarrow x\left( {2x + 1} \right) = 3 \cr & \Rightarrow 2{x^2} + x - 3 = 0 \cr & \Rightarrow (2x + 3)(x - 1) = 0 \cr & \Rightarrow x = 1\,\text{hour} \cr} $$
25
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \frac{{\left( {1/2} \right)x}}{{21}} + \frac{{\left( {1/2} \right)x}}{{24}} = 10 \cr & \Rightarrow \frac{x}{{21}} + \frac{x}{{24}} = 20 \cr & \Rightarrow 15x = 168 \times 20 \cr & \Rightarrow x = {\frac{{168 \times 20}}{{15}}} \cr & \Rightarrow x = 224\,km \cr} $$
26
The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,7x\,{\text{and}}\,8x\,{\text{km/hr}} \cr & {\text{Then}},\,8x = {\frac{{400}}{4}} = 100 \cr & \Rightarrow x = {\frac{{100}}{8}} = 12.5 \cr & \therefore {\text{Speed}}\,{\text{of}}\,{\text{first}}\,{\text{train}} \cr & = \left( {7 \times 12.5} \right){\text{km/hr}} \cr & = 87.5\,{\text{km/hr}} \cr} $$
27
A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Total}}\,{\text{time}}\,{\text{taken}} \cr & = \left( {\frac{{160}}{{64}} + \frac{{160}}{{80}}} \right)hrs \cr & = \frac{9}{2}\,hrs \cr & \therefore {\text{Average}}\,{\text{speed}} \cr & = \left( {320 \times \frac{2}{9}} \right)\,{\text{km/hr}} \cr & = 71.11\,{\text{km/hr}} \cr} $$
28
A car travelling with $$\frac{5}{7}$$ of its actual speed covers 42 km in 1 hour 40 minute 48 second. Find the actual speed of the car.
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Time}}\,{\text{taken = 1}}\,{\text{hr}}\,{\text{40}}\,{\text{min}}\,{\text{48}}\,{\text{sec}} \cr & = 1\,{\text{hr}}\,40\frac{4}{5}\,{\text{min}} = 1\frac{{51}}{{75}}\,hrs = \frac{{126}}{{75}}\,hrs \cr & {\text{Let}}\,{\text{the}}\,{\text{actual}}\,{\text{speed}}\,{\text{be}}\,x\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{5}{7}x \times \frac{{126}}{{75}} = 42 \cr & \Rightarrow x = {\frac{{42 \times 7 \times 75}}{{5 \times 126}}} \cr & \Rightarrow x = 35\,{\text{km/hr}} \cr} $$
29
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:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{Abhay's}}\,{\text{speed}}\,{\text{be}}\,{\text{x}}\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{{30}}{x} - \frac{{30}}{{2x}} = 3 \cr & \Rightarrow 6x = 30 \cr & \Rightarrow x = 5\,{\text{km/hr}} \cr} $$
30
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{by}}\,x\,{\text{km}} \cr & {\text{Then}},\,\frac{x}{{10}} - \frac{x}{{15}} = 2 \cr & \Rightarrow 3x - 2x = 60 \cr & \Rightarrow x = 60\,km \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,60\,km\,{\text{at}}\,10\,{\text{km/hr}} \cr & = {\frac{{60}}{{10}}} \,hrs = 6\,hrs \cr & {\text{So,}}\,{\text{Robert}}\,{\text{started}}\,{\text{6}}\,{\text{hours}}\,{\text{before}}\,2\,P.M.\,i.e.,\,at\,A.M. \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = {\frac{{60}}{5}} \,kmph = 12\,kmph \cr} $$