Affairscloud Math-Time & Distance, Part-20

Time and Distance Set 20

1. The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?
   A. 28 kmph
   B. 25 kmph
   C. 20 kmph
   D. 24 kmph
   
   

   Answer – C. 20 kmph
   
   Explanation :
   80/x – 80/(x+4) = 1
   x(x+20) – 16(x+20) = 0
   x = 16kmph
   Increased speed = 20 kmph
   
2. Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school?
   A. 16 km
   B. 12 km
   C. 15 km
   D. 10 km
   
   

   Answer – D. 10 km
   
   20*25/(25-20)*6/60=10.
   
3. Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the ratio of time taken by A and B to meet at M?
   A. 1:3
   B. 1:2
   C. 1:4
   D. 3:2
   E. None of these
   
   

   Answer – D. 3:2 
   
   Explanation :
   Time taken by B to reach at M = 4h
   Time taken by A to reach at M = 6h
   Ratio = 6:4 = 3:2
   
4. Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the extra time taken by A to reach at S?
   A. 6hrs 20 minutes
   B. 6hrs 40 minutes
   C. 6hrs 30 minutes
   D. 6hrs 10 minutes
   
   

   Answer – B. 6hrs 40 minutes
   
   Explanation :
   Time taken by A to reach at Q = 800/40 = 20 hours
   Time taken by B to reach at Q = 800/60 = 13 hours and 20 min
   A takes 6hr 40 minutes extra time to reach at Q.
   
5. Ajay covers certain distance with his own speed but when he reduces his speed by 10kmph his time duration for the journey increases by 40 hours while if he increases his speed by 5 kmph from his original speed he takes 10 hours less than the original time taken. Find the distance covered by him. 
   A. 1000 km
   B. 1200 km
   C. 1500 km
   D. 1800 km
   
   
   Answer – C. 1500 km 
   
   Explanation :
   x/(y – 10)  – x/y = 40
   x = 4y(y-10) —(i)
   x/y  – x/(y + 5) = 10
   x = 2y(y + 5) — (ii) From (i) and (ii) => y = 25; x = 1500
   
6. The driver of an ambulance sees a college bus 40 m ahead of him after 20 seconds, the college bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the college bus?
   A. 10 kmph
   B. 12 kmph
   C. 15 kmph
   D. 22 kmph
   
   

   Answer – B. 12 kmph
   
   Explanation :
   Relative Speed = (Total distance)/total time
   = (60+40) /20 = 5 m/s = (5*18)/5 = 18 kmph
   Relative Speed = (speed of ambulance – speed of College bus)
   Speed of College bus = speed of ambulance – relative speed.
   = 30-18 = 12 kmph.
   
7. Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the distance from R, where B overtakes A?
   A. 260 km
   B. 235 km
   C. 240 km
   D. 300 km
   
   

   Answer – C. 240 km
   
   Explanation :
   Distance between R and M = 4 * 60 = 240
   
8. Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, 5/6 hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit?
   A. 20 kmph
   B. 10 kmph
   C. 15 kmph
   D. 25 kmph
   
    

   Answer – A. 20 kmph
   
   Explanation :
   Let second rabbit takes x hr with speed s2
   First rabbit takes x-5/6 hr with speed s1
   Total distance = 50km
   S1 = 50/(x-(5/6))
   S2= 50/x
   As they cross each other in 1hr…
   Total speed = s1 + s2
   Now, T = D / S
   50/(s1+s2) = 1
   x = 5/2, 1/3
   Put x= 5/2 in s2 –> 20km/hr
   
9. Pranav walked at 5 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 25 kmph. If he took 10 hours for the entire journey, what part of journey did he travelled by auto if the average speed of the entire journey be 17 kmph
   A. 750 km
   B. 100 km
   C. 150 km
   D. 200 km
   
   

   Answer – C. 150 km
   
   Explanation :
   

   Total distance = 17*10=170

   Let Journey travelled by auto in x hr

   25 * x + (10-x ) 5 = 170

   25 x + 50 – 5x = 170

   x = 6

   Required Distance = 6 * 25 = 150 km
   
10. Aravind started for the station half a km from his home walking at 1 km/h to catch the train in time. After 3 minutes he realised that he had forgotten a document at home and returned with increased, but constant speed to get it succeded in catching the train. Find his latter speed in kmph?
    A. 1.25
    B. 1.1
    C. 11/9
    D. 2
    
    

    Answer – C. 11/9
    
    Explanation :
    

    Distance covered in 3 minutes = 3*(1000/60) = 50

    Now he has to cover (500+50)m in (30-3) minutes

    Required speed = (550/1000)/(27/60) = 11/9 km/h