Speed Time And Distance(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

SPEED TIME AND DISTANCE MCQs

Time & Distance

Total Questions : 422 | Page 3 of 43 pages

Question 21. A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is

25 km/hr
24 km/hr
27 km/hr
26 km/hr

Discuss Question

Answer: Option B. -> 24 km/hr
Answer: (b)UsingRule 5,Required average speed= ${2 × 30 × 20}/{30 + 20}$[Since, Distance covered is same]= ${2 × 30 × 20}/50 = 24$ kmph

Question 22. A constant distance from Chennai to Bangalore is covered by Express train at 100 km/hr. If it returns to the same distance at 80 km/hr, then the average speed during the whole journey is

88.98 km/hr
88.89 km/hr
90.20 km/hr
88.78 km/hr

Discuss Question

Answer: Option B. -> 88.89 km/hr
Answer: (b)Using Rule 5,If same distance are covered at two different speed of x and y kmph,the average speed of journey = ${2xy}/{x + y}$= $({2 × 100 × 80}/{100 + 80})$ kmph= 88.89 kmph

Question 23. A man goes to a place on bicycle at speed of 16 km/hr and comes back at lower speed. If the average speed is 6.4 km/hr in total journey, then the return speed (in km/hr) is :

Discuss Question

Answer: Option B. -> 4
Answer: (b)Let the speed of cyclist while returning be x kmph.Average speed = ${2 × 16 × x}/{16 + x}$6.4 = ${32x}/{16 + x}$6.4 × 16 + 6.4x = 32x32x - 6.4x = 6.4 × 1625.6x = 6.4 × 16$x = {6.4 × 16}/{25.6}$ = 4 kmph.

Question 24. A man travels a distance of 24 km at 6 kmph. Another distance of 24 km at 8 kmph and a third distance of 24 km at 12 kmph. His average speed for the whole journey (in kmph) is

2$10/13$
9
8$2/3$
8

Discuss Question

Answer: Option D. -> 8
Answer: (d)Using Rule 2,Total distance= 24 + 24 + 24 = 72 km.Total time= $(24/6 + 24/8 + 24/12)$ hours= (4 + 3 + 2) hours = 9 hoursRequired average speed= $\text"Totaldistance"/ \text"Total time" = 72/9$ = 8 kmph.

Question 25. Walking at $3/4$ of his usual speed, a man reaches his office 20 minutes late. Then his usual time for walking to his office is :

30 minutes
40 minutes
45 minutes
1 hour

Discuss Question

Answer: Option D. -> 1 hour
Answer: (d)
Usual time = x minutes
New time = ${4x}/3$ minutes
$(Since, \text"Speed ∞ "\text"1"/\text"Time")$
According to the question,
${4x}/3 - x = 20$
$x/3 = 20$
x = 60 minutes i.e. 1 hour.

Question 26. Sarthak completed a marathon in 4 hours and 35 minutes. The marathon consisted of a 10 km run followed by 20 km cycle ride and the remaining distance again a run. He ran the first stage at 6 km/hr and then cycled at 16 km/ hr. How much distance did Sarthak cover in total, if his speed in the last run was just half that of his first run?

35 km.
45 km.
40 km.
5 km.

Discuss Question

Answer: Option A. -> 35 km.
Answer: (a)Let the total distance be x km.Time = $\text"Distance"/ \text"Speed"$According to the question,$10/6 + 20/16 + {x - 30}/3$= 4$35/60 = 4{7}/12$$5/3 + 5/4 + x/3 - 10 = 55/12$$x/3 + 5/3 + 5/4 - 10 = 55/12$$x/3 + ({20 + 15 - 120}/12) = 55/12$$x/3 - 85/12 = 55/12$$x/3 = 85/12 + 55/12 = 140/12$$x = 140/12 × 3$ = 35 km.

Question 27. A can give 40 metres start to B and 70 metres to C in a race of one kilometre How many metres start can B give to C in a race of one kilometre ?

31$1/4$ metre
32 metre
31$3/4$ metre
30 metre

Discuss Question

Answer: Option A. -> 31$1/4$ metre
Answer: (a)According to the question,When A covers 1000m, B covers= 1000 - 40 = 960 mand C covers =1000 - 70 = 930 mWhen B covers 960m, C covers 930 m.When B covers 1000m, C covers= $930/960 × 1000$ = 968.75 metreHence, B gives C a start of= 1000 - 968.75 = 31.25 metre

Question 28. Two persons ride towards each other from two places 55 km apart, one riding at 12km/hr and the other at 10 km/hr. In what time will they be 11 km apart?

1 hour and 30 minutes
2 hours and 45 minutes
2 hours
2 hours and 30 minutes

Discuss Question

Answer: Option C. -> 2 hours
Answer: (c)Relative speed= 12 + 10 = 22 kmphDistance covered= 55 - 11 = 44 km∴ Required time= $(44/22)$ hours = 2 hours

Question 29. A jeep is chasing a car which is 5km ahead. Their respective speed are 90 km/hr and 75 km/ hr. After how many minutes will the jeep catch the car ?

20 min.
25 min.
24 min.
18 min.

Discuss Question

Answer: Option A. -> 20 min.
Answer: (a)Relative speed= 95 - 75 = 15 kmphRequired time= $\text"Distance"/ \text"Relative speed"$= $5/15$ hours = $5/15 × 60$ minutes= 20 minutes

Question 30. A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then A's speed is equal to:

4 km/hr.
7 km/hr.
5 km/hr.
3 km/hr.

Discuss Question

Answer: Option A. -> 4 km/hr.
Answer: (a)Let, A's speed = x kmph.B's speed = (7 - x) kmphTime = $\text"Distance"/ \text"Speed"$According to the question,$24/x + 24/{7 - x} = 14$$24({7 - x + x}/{x(7 - x)})$ = 14${24 × 7}/{x(7 - x)}$ = 14x (7 - x) = 12 = 4 × 3 or 3 × 4x (7 - x) = 4 (7 - 4) or 3 (7 - 3)x = 4 or 3A's speed = 4 kmph.