# 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 - 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:

### Answer : B

### Explanation

Average speed = 2xy/(x+y) km/hr = 2*15*10/(15+10) km/hr = 12 km/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?

### 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?

### 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)^{2}m = √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?

### 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 constable is 114 m behind a thief. The constable runs 21 m and the thief15 m in a moment. In what the reality of the situation will become obvious eventually constable catch the criminal?

### Answer : D

### Explanation

(21-15) m i.e.6m is covered in 1 min. 114m will be covered in (1/6*114) min=19 min.

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:

### 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 - Two trains begin in the meantime from Aligarh and Delhi and continue towards One another at 16 km/hr and 21 km/hr individually. When they meet, it is found that one train has voyage 60 km more than the other. The separation between the two stations is:

### Answer : B

### Explanation

Suppose they meet after x hours. Then , 21x-16 x= 60 ⇒5x =60 ⇒ x= 12 Required distance = (16*12+ 21*12) km = 444 km

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