# 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 proportion between the paces of two trains is 7:8. On the off chance that the second prepare keeps running in 5 hours 400 km, the pace of the first prepare is :

### Answer : A

### Explanation

Let the speed of first train be 7x km/hr. Then the speed of the second train is 8x km/hr. But speed of the second train=400/5km/hr=80 km/hr ∴8x=80⇒x=10. Hence the speed of first train is (7*10) km/hr=70 km/hr.

Q 3 - A auto going with 5/7 of its typical rate covers 42 km in 1 hr. 40 min. 48sec.What is the typical pace of the auto?

### Answer : D

### Explanation

Let the usual speed be x km/hr. 42/(5x/7)= 126/75 ⇒ 42*7/5x = 42/25 ⇒5x = (25*7) ⇒x = (5*7) = 35

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* 10ⁱ^{3}km 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 * 10ⁱ^{3}) km is covered in (1/3 *10⁵* 8.1*10ⁱ^{3}) 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 - 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⇒x^{2}+10x-2000=0 ⇒(x+50) (x-40) =0⇒x=40. ∴speed of slow train=40km/hr.

Q 7 - The separation between two stations A and B is 220 km. A train leaves a towards B at 80 km/hr. After thirty minutes, another train leaves B towards An at 100 km/hr. The separation of the point where the two trains meet, from A is:

### Answer : A

### Explanation

Let the required distance be x km. then, x/80 - (220-x)/100 = 1/2 ⇒ 5x-4 (220-x) = 200 ⇒9x = 1800 ⇒x = 120 Hence the required distance is 120 km.

Q 8 - X and y are two stations 500 km separated. A train begins from X and moves towards Y at 20 km/hr. Another train begins from Y in the meantime and moves towards X at 30 km/hr .How a long way from X will they cross one another?

### Answer : B

### Explanation

Suppose they meet x km from X. then, X/20 = (500-x)/30 = 30 x = 10000- 20 x ⇒50x = 10000 ⇒x = 200 So, they meet 200 km from X.

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