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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.

Q 1 - A competitor runs 200 meters race in 24 seconds. His rate is:

Answer : D

Explanation

Speed = 200/24 m/sec = (200/24* 18/5) km/hr = 30 km/hr

Q 2 - The velocities of A and B are in the proportion 3:4. A takes 20 min. more than B to achieve a destination. In what time does A achieve the destination?

Answer : A

Explanation

Let the time taken by A be x hrs.
Then, time taken by B = (x-20/60) hrs = (x-1/3) hrs
Ratio of speeds = inverse ratio of time taken
∴3:4 =(x- 1/3): x ⇒3x-1/3x = 3/4
⇒12x- 4 = 9x
⇒3x= 4 ⇒x= 4/3 hrs
Required time = 4/3 hrs.

Q 3 - Two train approach one another at 30 km/hr and 27 km/hr from two spot 342 km separated. After how long will they meet?

Answer : B

Explanation

Suppose the two trains meet after x hours. Then,
30x+27 x= 342 ⇒ 57 x = 342 ⇒ x = 342/57 = 6.
So the two trains will meet after 6 hours.

Q 4 - R and S begin strolling towards one another at 10 am at pace of 3 km/hr and 4 km/hr individually. They were at first 17.5 km separated. At what time do they meet?

Answer : B

Explanation

Suppose they meet after x hours. then,
3x+4x = 17.5 ⇒ 7x = 17.5 ⇒x = 2.5 hours
So they meet at 12.30 pm

Q 5 - Sunil spreads a separation by strolling for 6 hours .While giving back his pace, diminishes by 1 km/hr and he takes 9 hr. to cover the same distance. What was his velocity consequently traveled?

A - 2km/hr

B - 3km/hr

C - 5km/hr

D - cannot be resolved

Answer : A

Explanation

Let the speed in return journey be x km/hr. then
6(x+1) = 9x
⇒3x= 6 ⇒ x= 2
Hence, the speed in return journey is 2 km/hr

Q 6 - A quick prepare takes 3 hours not exactly the moderate train for a voyage of 600 km. On the off chance that the velocity of the moderate train is 10 km/hr not exactly the quick prepare, the pace of the moderate train is:

Answer : C

Explanation

Let the speed of the train be x km/hr and (x+10) km/hr. Then,
600/x-600/(x+10) =3⇒1/x-1/(x+10) =1/200
⇒(x+10)-x/x(x+10) =1/200⇒x²+10x-2000=0
⇒(x+50) (x-40) =0⇒x=40.
∴speed of slow train=40km/hr.

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 - Laxman needs to cover 6 km in 45 minutes. On the off chance that he covers one portion of the separation in 2/3 rd time, what ought to be his rate to cover the remaining separation in the remaining time?

Answer : A

Explanation

Time left = (1/3*45/60)hr = 1/4 hr, distance left = 3 km
Speed required = 3/ (1/4) km/hr = 12 km/hr

Q 9 - The proportion between the rates of going of An and B is 2:3 and in this manner A takes 10 minute more than the time taken by B to achieve a destination. In the event that A had strolled at twofold speed, he would have secured the separation in

Answer : D

Explanation

Ratio of time taken by A and B = 1/2: 1/3
Suppose B takes x min. Then, A takes (x+10) min.
∴(x+10): x= 1/2: 1/3 = 3:2 ⇒ x+10/x = 3/2 ⇒ 2x+20 = 3x ⇒x = 20
Thus B takes 20 min. and A takes 30 min.
AT double speed A would covers it in 15 min.

Q 10 - A certain separation is secured by cyclist at a sure speed. On the off chance that a jogger covers a large portion of the separation in twofold the time, the proportion of the rate of the jogger to that of the cyclist is:

Answer : C

Explanation

Let distance = d meters and time taken by cyclist = t sec.
Speed of the cyclist = d/t m/sec.
Again, distance = d/2 meters, time taken by jogger = 2t sec.
Speed of the jogger = (d/2)/2t m/sec. = d/4t m/sec.
Ratio of speeds of jogger and cyclist = d/4t: d/t =   1/4:1 = 1:4

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