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 train goes at 82.6km/hr. What number of meters will it go in 15 minutes?

A - 20.65 m

B - 206.5 m

C - 2065 m

D - 20650 m

Answer : D

Explanation

82.6 km/hr = (82.6*5/18)m/sec = 413/18 m/sec
Distance covered in 15 min = (413/18*15 *60) m =20650 m

Q 2 - A auto covers a separation of 715 km at a steady speed. On the off chance that the pace of the auto would have been 10 km/hr all the more, then it would have taken 2 hours less to cover the same separation. What is the first speed of the auto?

A - 45km/hr

B - 50 km/hr

C - 55 km/hr

D - 65 km/hr

Answer : C

Explanation

Let the constant speed be x km/hr. Then,
715/x-715/(x+10) =2⇒1/x-1/(x+10) =2/715
⇒(x+10)-x/x(x+10) =2/715⇒x(x+10) =3575
⇒x2+10x-3575=0⇒x2+65x-55x-3575=0
⇒x(x+65)-55(x+65)=0
⇒(x+65)(x-55)=0
⇒x=55.
∴Original speed of the car is 55km/hr.

Q 3 - A understudy strolls from his home at 5/2 km/hr and achieves his school late by 6 min. Following day, he builds his pace by 1 km/hr and achieves a 6 min. before educational time. How far is the school from his home?

A - 5/4 km

B - 7/4 km

C - 9/4 km

D - 11/4 km

Answer : B

Explanation

Let the required distance be x km. then,
x/ (5/2) - x/ (7/2) = 12/60 (  ∵difference between two times is 12 min.)
⇒ 2x/5 - 2x/7 = 1/5   ⇒ 14 x-10 x = 7 ⇒ 4x= 7 ⇒ x= 7/4
Required distance = 7/4 km

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 - The velocities of three autos in the proportion 2:3:4. The proportion of the times taken by these autos to venture to every part of the same separation is:

A - 2:3:4

B - 4:3:2

C - 4:3:6

D - 6:4:3

Answer : D

Explanation

Ratio of time taken = 1/2: 1/3: 1/4 = 6:4:3

Q 6 - Two trains begin from stations A and B and travel towards one another at 50 km/hr and 60 km/hr separately. At the season of their meeting, the second prepare has voyage 120 km more than the first. The separation in the middle of A and B is:

A - 990 kms

B - 1200 kms

C - 1320 kms

D - 1440 kms

Answer : C

Explanation

Let the two train meet after x hours . Then,
60x-50x-=120⇒ 10x =120⇒x=12hrs.
Distance AB = (distance covered by slower train) + (distance covered by fast train)
= [(150*12)+(60*12)]km=(600+720)km=1320km.

Q 7 - A man performs 2/15 of the aggregate adventure by rail, 9/20 by transport and the remaining 10 km on cycle. His aggregate voyage is:

A - 31.2 kms

B - 38.4 kms

C - 32.8 kms

D - 24 kms

Answer : D

Explanation

Let the total journey be x  km. then ,
2x/15+ 9x/20 +10 = X ⇒ 8x+27 x+600 = 60 x ⇒ 25x = 600 ⇒x = 24
∴ Total journey = 24 km

Q 8 - The proportion between the rates of strolling of A and B is 2:3. In the event that the time taken by B to cover a sure separation is 36 minutes, the time taken by A to cover that much separation is

A - 24 min

B - 54 min

C - 48 min

D - none of these

Answer : B

Explanation

Ratio of time taken = 1/2:1/3 = 3:2
Time taken by B = 36 min. let the time taken by A be x min.
∴x/36 = 3/2 ⇒x = (3*36/2) min. = 54 min

Q 9 - A man on visit ventures initial 160 km at 64 km/hr and the following 160 km at 80 km/hr. The normal rate for the entire excursion is:

A - 35.55 km/hr

B - 71.11 km/hr

C - 36 km/hr

D - 72 km/hr

Answer : B

Explanation

Average speed = 2xy/(x+y) km/hr = (2*64*80)/ (64+80) km/hr
=   (2*64*80)/144 km/hr = 640/9 km/hr = 71.11 km/hr

Q 10 - A train voyaged separations of 10 km, 20 km and 30 km at rates of 50 km/hr, 60 km/hr and 90 km/hr separately. The normal rate of the train was

A - 60 km/hr

B - 66.67 km/hr

C - 69.23 km/hr

D - 65 km/hr

Answer : C

Explanation

Total distance covered = (10+20+30) km = 60 km
Total time taken = (10/50 + 20/60+ 30/90) hr = (1/5+1/3+1/3) hr = 13/15 hr.
Average speed = (60* 15/13) km/hr = (900/13) km/hr = 69.23 km/hr
aptitude_speed_distance.htm
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