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

Quantitative Aptitude

SPEED TIME AND DISTANCE MCQs

Time And Distance, Time & Distance, Speed Time & Distance

Total Questions : 1223 | Page 5 of 123 pages

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. Abhays speed is:

5 kmph
6 kmph
6.25 kmph
7.5 kmph

Discuss Question

Answer: Option A. -> 5 kmph

Let Abhay's speed be x km/hr.

Then, \(\frac{30}{x}-\frac{30}{2x}=3\)

6x = 30

x = 5 km/hr.

Question 42.

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.?

8 kmph
11 kmph
12 kmph
14 kmph

Discuss Question

Answer: Option C. -> 12 kmph

Let the distance travelled by x km.

Then,\(\frac{x}{10}-\frac{x}{15}=2\)

3x - 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr = \(\left(\frac{60}{10}\right)hrs = 6hrs.\)

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Therefore Required speed = \(\left(\frac{60}{5}\right)kmph.=12kmph.\)

Question 43.

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

2 : 3
3 : 2
3 : 4
4 : 3

Discuss Question

Answer: Option C. -> 3 : 4

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then,\(\frac{120}{x}+\frac{480}{y}=8 \Rightarrow\frac{1}{x}+\frac{4}{y}=\frac{1}{15}.....(i)\)

And,\(\frac{200}{x}+\frac{400}{y}=\frac{25}{3} \Rightarrow\frac{1}{x}+\frac{2}{y}=\frac{1}{24}.....(ii)\)

Solving (i) and (ii), we get: x = 60 and y = 80.

Therefore Ratio of speeds = 60 : 80 = 3 : 4.

Question 44.

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:

14 km
15 km
16 km
17 km

Discuss Question

Answer: Option C. -> 16 km

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, \(\frac{x}{4}+\frac{(61-x)}{9}=9\)

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

Question 45.

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

35
\(36\frac{2}{3}\)
\(37\frac{1}{2}\)
40

Discuss Question

Answer: Option D. -> 40

Let distance = x km and usual rate = y kmph.

Then, \(\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60} \Rightarrow2y\left(y+3\right)=9x .......(i) \)

And, \(\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60} \Rightarrow y\left(y-2\right)=3x .......(ii)\)

On dividing (i) by (ii), we get: x = 40.

Question 46.

A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the speed of the train?

36 kmph
54 kmph
67 kmph
76 kmph
None of these

Discuss Question

Answer: Option B. -> 54 kmph
- Let the length of the train be x metres.
Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 seconds.
∴ x/8 =( x + 180) / 20 ⇔ 20x = 8(x + 180) ⇔ x = 120.
∴ Length of the train = 120 m.
Speed of the train = [120/8]m/sec =[15 x 18/5]kmph = 54 kmph.

Question 47.

Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?

27 7/9 m
28 m
29 m
30 2/7 m
None of these

Discuss Question

Answer: Option A. -> 27 7/9 m
- Relative speed = (40-20) km/hr = [20 x 5/18] m/sec = [50/9] m/sec.
Length of faster train = [50/9 x 5] m = 250/9 m = 27 7/9 m.

Question 48.

Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is:

120 m
140 m
160 m
180 m
None of these

Discuss Question

Answer: Option D. -> 180 m
- Relative speed = (36 + 45) km/hr
= [81 x 5/18] m/sec = [45/2] m/sec.
Length of train = [45/2 x 8] m = 180 m.

Question 49.

Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?

12 sec
14 sec
16 sec
20 sec
None of these

Discuss Question

Answer: Option A. -> 12 sec
- Speed of the first train = [120 / 10] m/sec = 12 m/sec.
Speed of the second train = [120 / 15] m/sec = 8 m/sec.
Relative speed = (12 + 8) = m/sec = 20 m/sec.
∴ Required time = (120 + 120) / 20 secc = 12 sec

Question 50.

Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:

12 kmph
24 kmph
36 kmph
48 kmph
None of these

Discuss Question

Answer: Option C. -> 36 kmph
- Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10.
∴ Speed of each train = 10 m/sec = [10 x 18/5] km/hr = 36 km/hr