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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 ⇒x^{2}+10x-3575=0⇒x^{2}+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 - Two train approach one another at 30 km/hr and 27 km/hr from two spot 342 km separated. After how long will they meet?

### Answer : B

### Explanation

Suppose the two trains meet after x hours. Then, 30x+27 x= 342 ⇒ 57 x = 342 ⇒ x = 342/57 = 6. So the two trains will meet after 6 hours.

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 - 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 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 - A auto ventures a separation of 840 km at a uniform pace. On the off chance that the velocity of the auto is 10 km/hr more, it takes 2 hours less to cover the same separation. The first speed of the auto was

### Answer : C

### Explanation

Let the original speed be x km/hr then, 840/x-840/(x+10) = 2 ⇒ 840(x+10)-840 x = 2x(x+10) ⇒2x^{2}+20 x-8400 = 0 ⇒x^{2}+10x- 4200= 0 ⇒(x+70) (x-60) = 0 ⇒x = 60 ∴ Original speed = 60 km/hr

Q 8 - If a train keeps running at 40 km/hr, it achieves its destination late by 11 minutes. In any case, in the event that it keeps running at 50 km/hr, it is late by 5 minute just. The right time for the train to cover its trip, is:

### Answer : C

### Explanation

Let the required time be x minutes. Distance covered in (x+11) min at 40 km/hr Distance covered in (x+5) min at 50km/hr ∴(x+11)/60*(x+5)/60*50⇒4(x+11) =5(x+5) ⇒x= (44-25) =19. Hence, the required time is 19 minutes.

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 - Sweta runs 9 km at a pace of 6 km/hr. At what pace would she have to go amid the following 1.5 hours to have a normal velocity of 9 km/hr for the whole running session?

### Answer : C

### Explanation

Time taken to cover 9 km = 9/6 hrs. = 1.5 hrs Total time taken = (1.5 +1.5) hrs = 3 hrs. Total distance covered in 3 hrs = (9*3) = 27 km In 1.5 hrs, distance covered = (27-9) km = 18 km Required speed = 18/ (3/2) km/hr = (18*2/3) km/hr = 12 km/hr