# 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 - 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 - A plane departed 30 min. later than the planned time and with a specific end goal to achieve the destination 1500 km away in time, it needed to build the pace by 250 km/hrs from the standard rate. Its standard rate is:

### Answer : D

### Explanation

Let the usual speed be x km/hr 1500/x - 1500/(x+250) =1/2 ⇒ 1/x - 1/(x+250) = 1/3000 ⇒ (x+250)- x /x(x+250) = 1/3000 ⇒x(x+250)= 75000 ⇒ x^{2}+250 x-75000=0 ⇒ x^{2}+ 1000x- 750 x- 75000= 0 ⇒x(x+1000)-750(x+1000) = 0 ⇒(x+1000)(x-750)=0 ⇒x = 750 ∴usual speed = 750 km/hr

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 strolls at a uniform rate of 4 km an hours and 4 hours after his begin, B cycles after him at the uniform rate of 10 km an hours. How a long way from the Beginning stage will B get A?

### Answer : D

### Explanation

Suppose B catches A after x hours. Then, Distance travelled by A in (x+4) hr. =distance travelled by B in x hours 4(x+4) =10x⇒6x=16⇒3x=8⇒x=8/3hrs. Distance travelled by B in 8/3 hrs= (8/3*10)km=80/3km=26.7km.

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 - A bullock truck needs to cover a separation of 80 km in 10 hours. On the off chance that it covers half of the excursion in 3/5 th of the time, what ought to be its velocity to cover the remaining separation in the time left?

### Answer : C

### Explanation

Distance left = (1/2 *80) km = 40 km Time left = {(1-3/5)*10} hrs = (2/5*10)= 4hrs. Speed required = 40/4 km/hr = 10 km/hr

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 - A man on visit ventures initial 160 km at 64 km/hr and the following 160 km at 80 km/hr. The normal rate for the entire excursion is:

### Answer : B

### Explanation

Average speed = 2xy/(x+y) km/hr = (2*64*80)/ (64+80) km/hr = (2*64*80)/144 km/hr = 640/9 km/hr = 71.11 km/hr

Q 10 - If the velocity of a railroad train is expanded by 5 km/hr from its ordinary pace, then it would have taken 2 hours less for a journey of 300 km. What is its ordinary velocity?

### Answer : B

### Explanation

Let the normal speed be x km/hr. Then 300/x - 300/(x+5) = 2 ⇒1/x - 1/(x+5) = 1/150 ⇒(x+5)-x/ x(x+5) = 1/150 ⇒ x^{2}+5x - 750 = 0 ⇒x^{2}+30x- 25x -750 = 0 ⇒x (x+30)-25(x+30) =0 ⇒(x+30) (x-25) = 0 ⇒ x= 25 ∴ Normal speed = 25 km/hr