Quantitative Aptitude Quiz – 67 (T’ S’ n D)

Hello Aspirants,
Welcome To D2G’s. This post contains some important questions on Time, Speed and Distance which can be asked in various competitive exams. Keep Learning.

1. A man drives 150 km from A to B in 3 hr and 20 mins, and returns to the place in 4 hr and 10 mins, If ‘S’ is the average speed of the entire trip then the average speed for the journey from A to B exceeds ‘S’ by
a) 5
b) 4.5
c) 5.5
d) 2.5

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Answer a) 5
Explanation:
A to B distance = 150
A to B time = 3 hr 20 min = 10/3
=>A to B speed = 150 / (10/3) = 45
B to A time = 4 hr 10 min = 25/6
B to A speed = 150 / (25/6) = 36
Average speed = 2*45*36 / (45 + 36) = 40
Then, It exceeds by 45 – 40 = 5.

2. A journalist travelled 1200 km by air, which formed 2/5 of his trip. He travelled one-third of the whole trip by car and rest by train. The distance travelled by train is:
a) 750
b) 800
c) 2600
d) None of these

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Answer b) 800
Explanation:
2/5 —-1200 =>1—-3000 = Total distance
1/3 of trip by car = 1000
Total distance till now = 2200
=>Remaining = 800 by train.

3. A train covers a distance of 880 km in 16 hr. What is the average speed of the car, if the speed fo the car is 30% less than the speed of the train?
a) 71.5
b) 38
c) 37.5
d) 38.5

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Answer d) 38.5
Explanation:
Train speed = 880 / 16 = 11/20
=>70% = 0.7 * 11/20 = 38.5.

4. If a student walks from his house to school at 8 kmph, he is late by 20 mins. However, if he walks at 10 kmph he is late by 5 mins only. The distance of the school from his house is:
a) 11 km
b) 10 km
c) 10.25 km
d) 10.5 km

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Answer b) 10 km
Explanation:
S1 / S2 = (t + 5) / (t + 20)
=> 8/10 = (t+5) / (t+20)
=>t = 55 min
=>Distance = 8*(55 + 20) / 60 = 10 km.

5. Two cars A and B started moving from the same point at the same time in opposite directions(A towards north and B towards south). If the speed of car A is 34.5 kmph and that of car B is 41.5 kmph, after how much time will they be 684 km apart? (in minutes)
a) 540
b) 30
c) 450
d) 480

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Answer a) 540
Explanation:
Relative speed = 34.5 + 41.5 = 76 (since opposite direction)
Time = D/T = 684 / 76 = 9hr = 540 min.

6. Rakesh rides his bike at an average speed of 30 kmph and reaches his destination in 6 hr. Suman covers the same distance in 4 hr. If Rakesh increases his average speed by 10 kmph and Suman increases his average speed 5 kmph. What would be the difference in their time taken to reach the destination?
a) 60
b) 40
c) 54
d) 58

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Answer c) 54
Explanation:
Distance travelled by Rakesh = Distance travelled by Suman = 30 * 6 = 180
Accor. to the Q’ Speed of Rakesh = 30 + 10 = 40
=> Time taken by Rakesh = 180 / 40 = 4.5
Now, Accor. to the Q’ Speed of Suman = 180 / (4+5) = 50
=>Time taken by Suman = 180 / 50 = 3.6
Time diff. = 0.9 hr or 0.9 * 60 = 54 min.

7. A driver of an ambulance sees a school bus 60 meters ahead of him. After 30 seconds the school bus is 90 meters behind. If the speed of the ambulance is 38 kmph, then what is the speed of the school bus?
a) 27
b) 25
c) 20
d) 19

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Answer c) 20
Explanation:
Speed = Total distance/Total Time
=>Relative speed = (90+60) / 30 = 5 mps
=>5 * 18/5 = 18 kmph
=>Relative Speed = Speed of ambulance – Speed of bus
=>Speed of school bus = 38 – 18 = 20 kmph.

8. 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
b) 10
c) 11
d) Can’t be determined

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Answer b) 10
Explanation:
Due to stoppages it covers 9 km less
Time taken to cover 9 km = (9/54) * 60 = 10 min.

9. By walking at 3/4 th of his usual speed, a man reaches office 20 minutes later than usual. What is his usual time?
a) 30 mins
b) 60 mins
c) 75 mins
d) 45 mins

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Answer b) 60
Explanation:
S1 / S2 = t2 / t1
=>S / (3/4 S) = (t + 20) / t
=>4/3 = (t + 20) / t
=>t = 60 mins.

10. Ram covers a part of the journey at 20 kmph and the balance at 70 kmph, taking total of 8 hr to cover the distance of 400 km. How many hours he been driving at 20 kmph?
a) 3 hr 12 m
b) 2 hr 20 m
c) 3 hr 15 m
d) 4 hr 12 m

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Answer a) 3 hr 12 m
Explanation:
Time taken @ 20 kmph = X
=> Time taken @ 70 kmph = 8 – X
Accor. to the Q’ 20X + (8 – X)*70 = 400
=>X = 160 / 50 = 16 / 5 = 3 hr 12 min.

Best Wishes,
D2G’s Team..