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:
Q 2 - A man beginning from his home covers a separation at 15 km/hr and come back to the beginning spot at 10 km/hr. His normal velocity amid entire excursion is:
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?
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 - 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?
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 - 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:
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 ventures 35 km halfway at 4 km/hr and at 5 km/hr. in the event that he covers previous separation at 5 km/hr and later separation at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the entire separation at unique rate is:
Suppose the man covers first distance in x hrs and the second distance in y hrs. then,
4x+5 y = 35 ... (a) And 5x+4 y = 37 ...(b)
On solving (a) and (b), we get x= 5, y = 3
Total time taken = (5+3) = 8 hrs.
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?
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
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:
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
