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

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?

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 - If an understudy strolls from his home to class at 5km/hrs, he is late by 30 min. However, on the off chance that he strolls at 6 km/hr. he is late by 5 min. just. The separation of his school from his home is:

Answer : D

Explanation

Let the required distance be x km. then,
x/5 - x/6 = 25/7   (difference between two times is 25 min.)
⇒ 12x- 10 x = 25 ⇒2x = 25   ⇒ x= 25/2 km = 12.5 km

Q 4 - A and B begin all the while from a sure point in North and South bearings on engine cycles. The velocity of A is 80 km/hr and that Of B is 65 km/hr. What is the separation in the middle of A and B following 12 minutes?

Answer : B

Explanation

Required distance = sum of distance covered by A and B
= {(80*12/60) + (65*12/60)} = (16+13) = 29 km

Q 5 - A star is 8.1* 10ⁱ³km far from the earth. Assume light goes at the pace of 3.0* 10⁵ km for every second. To what extent will it take light from star to achieve the earth?

Answer : B

Explanation

(3*10⁵) km is covered in 1 sec.
(8.1 * 10ⁱ³) km is covered in (1/3 *10⁵* 8.1*10ⁱ³) sec = (2.7 *10⁸*1/60*1/60) hrs = (2.7*10⁶)/36 hrs= (2.7 *100*10⁴)/36 hrs = (7.5 *10⁴) hrs.

Q 6 - By what amount of percent must a driver expand his pace so as to lessen the time by 20%, taken to cover a sure separation?

Answer : B

Explanation

Distance= (time*speed) =t*x.
Let the required increase in speed be p%. Then,
(80%of t)*(100+p)/100=x=t*x
⇒80/100*t*(100+p)/100*x=t*x⇒4 (100+p/500=1⇒4p=100⇒p=25.
∴Required increase in speed=25%.

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?

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 - 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 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

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

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