- Aptitude Test Preparation
- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude
- Aptitude Useful Resources
- Aptitude - Questions & Answers
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 ⇒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 understudy strolls from his home at 5/2 km/hr and achieves his school late by 6 min. Following day, he builds his pace by 1 km/hr and achieves a 6 min. before educational time. How far is the school from his home?
Answer : B
Explanation
Let the required distance be x km. then, x/ (5/2) - x/ (7/2) = 12/60 ( ∵difference between two times is 12 min.) ⇒ 2x/5 - 2x/7 = 1/5 ⇒ 14 x-10 x = 7 ⇒ 4x= 7 ⇒ x= 7/4 Required distance = 7/4 km
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?
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 - Sunil spreads a separation by strolling for 6 hours .While giving back his pace, diminishes by 1 km/hr and he takes 9 hr. to cover the same distance. What was his velocity consequently traveled?
Answer : A
Explanation
Let the speed in return journey be x km/hr. then 6(x+1) = 9x ⇒3x= 6 ⇒ x= 2 Hence, the speed in return journey is 2 km/hr
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 - A man performs 2/15 of the aggregate adventure by rail, 9/20 by transport and the remaining 10 km on cycle. His aggregate voyage is:
Answer : D
Explanation
Let the total journey be x km. then , 2x/15+ 9x/20 +10 = X ⇒ 8x+27 x+600 = 60 x ⇒ 25x = 600 ⇒x = 24 ∴ Total journey = 24 km
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?
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 - 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 - 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