Speed & Distance - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Speed & Distance. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - A man finishes 30 km of a voyage at 6km/hr and the staying 40km of the venture in 5 hr.His normal pace for the entire voyage is:

A - 70/11 km/hr

B - 7 km/hr

C - 15/2 km

D - 8 km/hr

Answer : B

Explanation

Total distance = (30+40)km= 70 km
Total time taken = (30/6 +5) hrs =10 hrs
Average speed = 70/10 km/hr = 7 km/hr

Q 2 - The proportion between the paces of two trains is 7:8. On the off chance that the second prepare keeps running in 5 hours 400 km, the pace of the first prepare is :

A - 70 km/hr

B - 200 km/hr

C - 250 km/hr

D - 350 km/hr

Answer : A

Explanation

Let the speed of first train be 7x km/hr.
Then the speed of the second train is 8x km/hr.
But speed of the second train=400/5km/hr=80 km/hr
∴8x=80⇒x=10.
Hence the speed of first train is (7*10) km/hr=70 km/hr.

Q 3 - By strolling at 3/4 of his standard speed, a man achieves his office 20 min. later than Normal. His standard time is:

A - 30 min

B - 60 min

C - 75 min

D - 1 hr.30 min

Answer : B

Explanation

At a speed of 3/4 of the usual speed , time taken = 4/3 of usual time
∴ (4/3 of usual time) - (usual time) = 20 min.
Let the usual time be x min. then, (4x/3 - x) = 20 ⇒x = 60 min.
∴usual time is 60 min.

Q 4 - Two auto begins in the meantime from one point and moves along two streets at right edges to one another. Their rates are 36 km/hr and 48 km/hr individually. After 15 sec, the contrast between them will be?

A - 150 m

B - 250 m

C - 300 m

D - 400 m

Answer : B

Explanation

36 km/hr = (36*5/18)m/sec= 10 m/sec.
Distance covered in 15 sec. A= (10*15) m = 150 m
48 km/hr = (48*5/18) m/sec = 40/3 m/sec.
Distance covered in 15 sec. = B = (40/3 *15) m = 200 m
Distance between A and B = AB= √ (150)2+ (200)2m = √62500 m = 250 m

Q 5 - A taken 2 hr. more than B to walk d km. If A twofold his rate then he can make it in 1 hours less than B. How much time does B require for strolling d km?

A - d/2 hrs

B - 3 hrs

C - 4 hrs

D - 2d/3 hrs

Answer : C

Explanation

Suppose B takes x hours to walk d km.
Then, A takes (x+2) hours to walk d km.
With double of the speed, A will take 1/2 (x+2) hours.
∴ x- 1/2 (x+2) = 1 ⇒2x- (x+2) = 2 ⇒ x = 4
Hence B takes 4 hours to walk d km.

Q 6 - A thief takes an auto at 1.30 pm and drives it at 45 km/hr. the burglary is found at 2 p.m and the proprietor sets off in another auto at 50 km/hr. HE will surpass the criminal at:

A - 3.30 pm

B - 4 p.m

C - 4.30 pm

D - 6 pm

Answer : B

Explanation

Distance covered by thief in 1/2 hour =20 km.
Clearly, 20 km is compensated by the owner at a relative speed of 10km/hr in 2 hours.
So, he overtakes the thief at 4p.m.

Q 7 - Renu began cycling along the limits of a square field ABCD from corner point A. after thirty minutes, he came to the corner point C, slantingly inverse to A. In the event that his rate was 8 km/hr, the zone of the field is:

A - 64 sq. km

B - 8 sq. km

C - 4 sq. km

D - cannot be resolved

Answer : B

Explanation

Length of diagonal = (8*1/2 )km = 4 km
Area of the field = (1/2 *4*4) sq. km = 8 sq. km

Q 8 - A train leaves Meerut at 6 am and achieves Delhi at 10 am. Another train leaves Delhi at 8am and ranges Meerut at 11.30 am. At what time do the two trains cross one another?

A - 9.26 am

B - 9.am

C - 8.36 am

D - 8.56 am

Answer : D

Explanation

Let the distance between Meerut and Delhi be x km.
Average speed of train from Meerut = x/4 km/hr
Suppose they meet y hrs. After 6 am. Then,
(X/4*y)+2x/7 * (y-2) = x ⇒ y/4+ 2(y-2)/7 = 1 ⇒7y+8(y-2) = 28
⇒15 y= 44 ⇒ y = 44/15 hrs = 2 hrs. 56 min.
So, the trains meet at 8.56 am

Q 9 - A auto covers four progressive 3 km extends at 10 km/hr, 20 km/hr, 30 km/hr and 60 km/hr separately. Its normal rate over this separation is:

A - 10 km/hr

B - 20 km/hr

C - 25 km/hr

D - 30 km/hr

Answer : B

Explanation

Total distance = (3*4) km = 12 kms
Total time taken = (3/10+3/20+3/30+ 3/60) = (36+18+12+6)/120 hrs.
= 72/120 hrs = 3/5 hrs.
Average speed = 12/ (3/5) km/hr = (12*5)/3 = 20 km/hr

Q 10 - Sweta runs 9 km at a pace of 6 km/hr. At what pace would she have to go amid the following 1.5 hours to have a normal velocity of 9 km/hr for the whole running session?

A - 9 km/hr

B - 10 km/hr

C - 12 km/hr

D - 14 km/hr

Answer : C

Explanation

Time taken to cover 9 km = 9/6 hrs. = 1.5 hrs
Total time taken = (1.5 +1.5) hrs = 3 hrs.
Total distance covered in 3 hrs = (9*3) = 27 km
In 1.5 hrs, distance covered = (27-9) km = 18 km
Required speed = 18/ (3/2) km/hr = (18*2/3) km/hr = 12 km/hr

aptitude_speed_distance.htm
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