Speed, Distance and Time Practice Test Sample Paper - 1 with Answer Key and Solution [Quantitative Aptitude]- Download in PDF

Here is solved questions of speed, distance and time [quantitative aptitude] for SBI PO and IBPS exams. Questions regarding finding speed, distance and time are asked in almost all entrance examinations and candidates need to practice this section very hard to score good marks and hence good position in merit list. You can also download this question paper in PDF to save it for future references or take print out.

 

Speed, Distance and Time [Quantitative Aptitude] Model Test Paper - 1

 

Maximum Marks: 10     /      Duration: 10 Mins

Q.1. How many minutes does Aditya take to cover a distance of 400m, if he runs at a speed of 20 km/hr?

A) 20 min
B) 2 min
C) 1 1/5 min
D) 1 2/5 min
E) None of these

Explanation (Show/Hide)

Distance = 400m = 0.4 Km, Speed = 20 km/hr
Time = Distance / Speed = 1/50 Hr
1 hr = 60 min => 1/50 hr = 60/50 min = 1 1/5 min.

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Q.2. A man walking at the rate of 5 km/ hr crosses a bridge in 15 minutes. The length of the bridge (in meter) is:

A) 600
B) 750
C) 100
D) 1250
E) None of these

Explanation (Show/Hide)

Given Speed = 5 km/hr, time = 15 min = 1/4 hr.
Distance = Speed * Time = 5 * (1/4) km = 1.25 km = 1250 meter.

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Q.3. An express train traveled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?

A) 6 hrs 21 mins
B) 6 hrs 24 mins
C) 6 hrs 27 mins
D) 6 hrs 30 mins
E) None of these

Explanation (Show/Hide)

Time taken to cover 600 km = 6 hrs.
Number of stoppages = 600/75 - 1 = 7
Total time of stoppage = 7 * 3 - 21 mins.
Hence total time taken by train to reach its destination is 6hrs 21 mins.

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Q.4. A train when moves at an average speed of 40 kmph, reaches its destination on time. When its average speed becomes 35 kmph, then it reaches its destination 15 minutes late. Find the length of journey.

A) 30 km
B) 40 km
C) 70 km
D) 80 km
E) None of these

Explanation (Show/Hide)

Let distance be D, Time = T
T = D/40, T+(15/60) = D/35
=> D/40 = D/35 - 1/4
D = 70 km.

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Q.5. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:

A) 50 km
B) 56 km
C) 70 km
D) 80 km
E) None of these

Explanation (Show/Hide)

Let the original distance be D D/10 = (D+20)/14
D = 50

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Q.6. A person travels equal distances with speeds of 3 km/hr, 4 km/hr and 5 km/hr and takes a total time of 47 minutes. The total distance (in km) is :

A) 2
B) 3
C) 4
D) 5
E) None of these

Explanation (Show/Hide)

Let the distance traveled be 3D kms = D + D + D.
T1 = D/3, T2 = D/4, T3 = D/5
T1 + T2 + T3 = 47 mins = 47/60 hrs
=> D = 1 km. and total distance = 3D = 3 km

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Q.7. A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be:

A) 10 min
B) 11 min 20 sec
C) 13 min
D) 13 min 20 sec
E) None of these

Explanation (Show/Hide)

Speed of the train = 10 / (12/60) = 50 km/hr
After decreasing speed by 5 km/hr
time = 10 / 45 hrs = (10/45) * 3600 sec = 800 sec = 13 min 20 sec

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Q.8. In covering a certain distance, the speeds of A and B are in the ratio of 3 : 4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is:

A) 1 hr
B) 1 1/2 hr
C) 2 hr
D) 2 1/2 hr
E) None of these

Explanation (Show/Hide)

Let time taken by A = T mins, time taken by B = (T - 30) mins
As ratio of their speed = 3:4 so ratio of their time = 4:3
=> 4/3 = T/(T-30)
=> T = 120 mins = 2 hour.

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Q.9. A man traveled from the village to the post office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village

A) 20 km
B) 27 km
C) 42 km
D) 37 km
E) None of these

Explanation (Show/Hide)

Average speed = 2xy/(x+y) = 200/29 km/hr
5 hour 48 minute = 5 4/5 hour
Distance traveled in 5 4/5 hour = (200/29) * (29/5) = 40 km.
Distance from village to post office = 40/2 = 20 km.

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Q.10. A farmer traveled a distance of 61 km in 9 hours. He traveled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance traveled on foot is:

A) 14 km
B) 15 km
C) 16 km
D) 17 km
E) None of these

Explanation (Show/Hide)

Avg speed of the farmer = 61/9 km/hr
Let distance traveled on foot = D, distance traveled on bicycle= 61 - D
Avg speed = {(D1 + D2)*S1*S2}/(S1*D2 + S2*D1) = 61*4*9/ (4*(61-D) + 9D)
=> D = 16 km.

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