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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 - The proportion between the paces of two trains is 7:8. On the off chance that the second prepare keeps running in 5 hours 400 km, the pace of the first prepare is :
Answer : A
Explanation
Let the speed of first train be 7x km/hr. Then the speed of the second train is 8x km/hr. But speed of the second train=400/5km/hr=80 km/hr ∴8x=80⇒x=10. Hence the speed of first train is (7*10) km/hr=70 km/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 - A train secured a separation at a uniform velocity. On the off chance that the train had been 6 km/hr speedier, it would have taken 4 hours not exactly the booked time, and if the train were slower by 6 km/hr, the train would have taken 6 hours more than the planned time. The length of the trip is?
Answer : B
Explanation
Let the required distance be x km and uniform speed by y km/hr x/y - x/(y+6) = 4 ...(a) x/(y-6) - x/y = 6 ...(b) ⇒xy+6x-xy = 4y (y+6) and xy - xy +6x = 6y (y-6) ⇒4y2+24 y - 6x = 0 and 6y2- 36y - 6x = 0 ⇒2y2- 60 y = 0 ⇒ 2y (y-30) = 0 ⇒y = 30 ∴x/30 - x/36 = 4 ⇒6x- 5x = 720 ⇒x = 720 km
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
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?
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 - Renu began cycling along the limits of a square field ABCD from corner point A. after thirty minutes, he came to the corner point C, slantingly inverse to A. In the event that his rate was 8 km/hr, the zone of the field is:
Answer : B
Explanation
Length of diagonal = (8*1/2 )km = 4 km Area of the field = (1/2 *4*4) sq. km = 8 sq. km
Q 8 - A train leaves Meerut at 6 am and achieves Delhi at 10 am. Another train leaves Delhi at 8am and ranges Meerut at 11.30 am. At what time do the two trains cross one another?
Answer : D
Explanation
Let the distance between Meerut and Delhi be x km. Average speed of train from Meerut = x/4 km/hr Suppose they meet y hrs. After 6 am. Then, (X/4*y)+2x/7 * (y-2) = x ⇒ y/4+ 2(y-2)/7 = 1 ⇒7y+8(y-2) = 28 ⇒15 y= 44 ⇒ y = 44/15 hrs = 2 hrs. 56 min. So, the trains meet at 8.56 am
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 - 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