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Aptitude Questions : Time and Distance Set 10

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Sheerin
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Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating question sample in Time and Distance with Explanation, which is common for all the IBPS,SBI,SSC and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

Questions Penned by Yogit

  1. The distance between two cities P and Q is 300km. A train starts from station P at 10 am with speed 80 km/hr towards Q. Another train starts from Q towards P with speed 40km/hr at 11 am . At what time do they meet.
    a) 12.20pm
    b) 12.40 pm
    c) 12.50 pm
    d) 1 pm
    Answer – c) 12.50 pm
    Solution:
    First train starts at 10am so in one hour it covers 80 km in one hour. Now distance b/w P and Q is 220. Suppose at some’ x’ km they meet. So,
    x/80 = (220-x)/40
    x = 440/3.
    The time after which they meet = (440/3)/80 = 11/6 i.e = 1hr 50 min.

  2. A car covers first 10 km with 40km/hr next 10km with 60 km and next 10 km in 20 km/hr. What is the average speed of the car.
    a) 320/11
    b) 330/11
    c) 350/11
    d) 360/11
    Answer – d) 360/11
    Solution:
    Average speed = (3*20*40*60)/800+2400+1200 = 360/11

  3. A train running with 72km/hr takes 20sec to cross a platform 200 m long. How much it take to cross a stationary train having same length
    a) 20 sec
    b) 30 sec
    c) 40 sec
    d) 50 sec
    Answer – a) 20 sec
    Solution:
    200+L =  72*(5/18)*20. L= 200m.
    400 = 72*(5/18)*t. So, t = 20sec.

  4. Two cities A and B are at a distance of 60 km from each other. Two persons P and Q start from First city at a speed of 10km/hr and 5km/hr respectively. P reached the second city B and returns back and meets Q at Y. Find the distance between A and Y. 
    a) 30 km
    b) 40 km
    c) 50 km
    d) 55 km
    Answer – b) 40 km
    Solution:
    Time taken by P to reach city B is 6hr. In 6 hr, distance covered by Q is 30km. Now at some x distance they will meet. So
    x/5 = (30-x)/10. X= 10.
    So distance b/w A and Y is 30+10 =40 km

  5. A man covers 2/3 distance at a speed of 30 km/hr and the remaining at 60 km/hr. If the total distance he covers is 300 km. Find the average speed of the man.
    a) 36 km/hr
    b) 38km/hr
    c) 40 km/hr
    d) 42km/hr
    Answer – a) 36 km/hr
    Solution:Average speed = 300/[(200/30)+(100/60)] = 36

  6. When priya travels from home to office with a speed of 40km/hr she reaches her office late by 20 minutes and when she travels with 60km/hr, she reach 10 minutes early. Find the distance between her office and home.
    a) 50 km
    b) 60 km
    c) 65 km
    d) 70 km
    Answer – b) 60 km
    Solution:
    D = 40(t+20/60) and D = 60(t-10/60)
    solve these two equation and get D.

  7.  A bus running at 3/5 of its usual speed reaches its destination in 15 hrs. If the bus runs at his usual speed, how much time would be saved.
    a) 6hr
    b) 7hr
    c) 8hr
    d) 10hr
    sol = > D = (3/5)*v*15
    Answer – a) 6hr
    Solution:
    Now running at usual speed, D = v*t = (3/5)*v*15, so time = 9hr.
    Hence saved 6hrs

  8. Excluding stoppages the speed of bus is 60 km/hr and including stoppages it average speed becomes 48km/hr. For how much time the bus stops.
    a) 10 min
    b) 12 min
    c) 14 min
    d) 16 min
    Answer – b) 12 min
    Solution:
    Due to stoppages bus covers 12 km less. To cover 12km with 60km/hr speed it takes 12min.

  9. Rahul takes 4 hr in walking at certain place and return back. While it takes 3 hrs in walking at certain place and riding back. Find the time rahul will take to ride both sides
    a) 2hr
    b) 3 hr
    c) 3.5 hr
    d) 1.5 hr
    Answer – a) 2hr
    Solution:
    W + W = 4. W =2
    W+ R = 3. R =1
    So to ride both direction, it will take 1+1 = 2 hrs.

  10. P walks with a speed of 6 km/hr and after 5 hr of his start, Q starts running towards P at a speed of 8 km/hr. At what distance from start will Q catch P.
    a) 100
    b) 110
    c) 120
    d) 140
    Answer – c) 120
    Solution:
    In 5hrs, P will cover 30 km. Now,  at some distance ‘x’. So P will cover X distance and Q will cover 30 + X.
    x/6 = (30+x)/8
    x = 90. So distance from start after which Q will catch P = 90+30 =120km.

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