Speed & Distance - Online Quiz


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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.

Questions and Answers

Q 1 - A train goes at 82.6km/hr. What number of meters will it go in 15 minutes?

A - 20.65 m

B - 206.5 m

C - 2065 m

D - 20650 m

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?

A - 45km/hr

B - 50 km/hr

C - 55 km/hr

D - 65 km/hr

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 - 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:

A - 720 km/hr

B - 730 km/hr

C - 740 km/hr

D - 750 km/hr

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
⇒ x2+250 x-75000=0
⇒ x2+ 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 - R and S begin strolling towards one another at 10 am at pace of 3 km/hr and 4 km/hr individually. They were at first 17.5 km separated. At what time do they meet?

A - 11.30 am

B - 12.30 pm

C - 1.30 pm

D - 2.30 pm

Answer : B

Explanation

Suppose they meet after x hours. then,
3x+4x = 17.5 ⇒ 7x = 17.5 ⇒x = 2.5 hours
So they meet at 12.30 pm

Q 5 - A kid is running at a velocity of p km/hr to cover a separation of 1 km. in any case, , because of dangerous ground , his velocity is lessened by q km/hr (p>q). On the off chance that he takes r hours to cover the separation, them

A - 1/r = (p-q)

B - r = (p-q)

C - 1/r = (p+q)

D - r = (p+q)

Answer : A

Explanation

Actual speed = (p-q) km/hr, time taken = r hrs.
Distance = (speed*time)
∴ 1 = (p-q) r ⇒1/r = (p-q)

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?

A - 20%

B - 25%

C - 30%

D - none of these

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 - 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?

A - 8 km/hr

B - 6.4 km/hr

C - 10 km/hr

D - 20 km/hr

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?

A - 300 kms

B - 200 kms

C - 120 kms

D - 40 kms

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 - 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?

A - 9 min

B - 10 min

C - 12 min

D - 20 min

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 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

A - 60 km/hr

B - 66.67 km/hr

C - 69.23 km/hr

D - 65 km/hr

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


aptitude_speed_distance.htm

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