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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 is twice as quick as B and B is thrice as quick as C. The excursion secured by C in 42 min. will be secured by A in
Answer : A
Explanation
Let c speed be x meters/min. Then, B speed=3x meters /min and A speed =6x meters/ min. Ratio of speed of A and C =ratio of times taken by C and A 6x:x=42:ymin⇒6x/x=42/y⇒y=42/6min=7 min.
Q 3 - By strolling at 3/4 of his standard speed, a man achieves his office 20 min. later than Normal. His standard time is:
Answer : B
Explanation
At a speed of 3/4 of the usual speed , time taken = 4/3 of usual time ∴ (4/3 of usual time) - (usual time) = 20 min. Let the usual time be x min. then, (4x/3 - x) = 20 ⇒x = 60 min. ∴usual time is 60 min.
Q 4 - Bombay express left Delhi for Bombay at 14.30 hours, going at a rate of 60 kmph and Rajdhani express left Delhi for Bombay around the same time at 16.30 hours, going at a pace of 80 kmph. How far from Delhi will the two trains meet?
Answer : C
Explanation
Let the train meet at a distance of x km from Delhi. Then, x/60 - x/80 = 2 ⇒ 4x-3x = 480 ⇒x = 480 ∴ required distance = 480 km
Q 5 - A star is 8.1* 10ⁱ3km 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 * 10ⁱ3) km is covered in (1/3 *10⁵* 8.1*10ⁱ3) 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 man strolling at 3 km/hr crosses a square field corner to corner in 2 minutes. The zone of the field is:
Answer : C
Explanation
Speed =(3*5/18)m/sec. = 5/6 m/sec
Distance covered in 2 min. = (5/6* 2* 60) m = 100 m
Length of the diagonal of the square field = 100 m
Area = 1/2 * (diagonal) 2= (1/2 *100 *100 )m2= 5000 m2
= 5000/100 ares = 50 ares {1 are= 100 m2}
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 - 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 - 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 ⇒ x2+5x - 750 = 0 ⇒x2+30x- 25x -750 = 0 ⇒x (x+30)-25(x+30) =0 ⇒(x+30) (x-25) = 0 ⇒ x= 25 ∴ Normal speed = 25 km/hr