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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 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 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
⇒x²+10x-3575=0⇒x²+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)²m = √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 - By what amount of percent must a driver expand his pace so as to lessen the time by 20%, taken to cover a sure separation?

Answer : B

Explanation

Distance= (time*speed) =t*x.
Let the required increase in speed be p%. Then,
(80%of t)*(100+p)/100=x=t*x
⇒80/100*t*(100+p)/100*x=t*x⇒4 (100+p/500=1⇒4p=100⇒p=25.
∴Required increase in speed=25%.

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 - The proportion between the rates of strolling of A and B is 2:3. In the event that the time taken by B to cover a sure separation is 36 minutes, the time taken by A to cover that much separation is

Answer : B

Explanation

Ratio of time taken = 1/2:1/3 = 3:2
Time taken by B = 36 min. let the time taken by A be x min.
∴x/36 = 3/2 ⇒x = (3*36/2) min. = 54 min

Q 9 - Excluding stoppages, the velocity of a transport is 54 km/hr and including stoppages, it is 45 km/hr. For how long does the transport stop every hour?

Answer : B

Explanation

Due to stoppage, it covers 9 km less per hour.
Time taken to cover 9 km = (9/54*60) min. = 10 min.

Q 10 - A agriculturist voyaged a separation of 61 km in 9 hours. He voyaged halfway by walking at 4 km/hr and incompletely on bike at 9 km/hr. The separation went by walking is:

Answer : C

Explanation

Let the distance travelled on foot be x km
Then, distance covered on bicycle = (61-x) km
∴x/4 + 61-x/9 =9 ⇒ 9x+4(61-x)= 324 ⇒ 5x = (324-244)=  80 ⇒x = 16
Distance covered on foot = 16 km

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