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October 26th, 2018, 04:56 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 491 Thanks: 1  How to solve this Late/Early/Usual Time type problem?
A person reaches office 15 mins late when travelling speed of 40 km/hr. A person reaches office 10 mins early when travelling speed of 50 km/hr. How much time he should travel if he reaches office by 1 hour late? Overall distance = 25*40*50/10*60 = 83.33m First person's time without considering late = 83.33/40 = 2.08hr after this how to solve it? Last edited by Ganesh Ujwal; October 26th, 2018 at 04:59 AM. 
October 26th, 2018, 10:00 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
v = required speed to get to office on time t = required time to get to office on time @v...............vt................>t @40.........40(t+1/4)..........>t+1/4 (15 min = 1/4 hr) @50.........50(t1/6)...........>t1/6 (10 min = 1/6 hr) 40(t+1/4) = 50(t1/6) Solve for t...then continue... 
October 26th, 2018, 07:34 PM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 491 Thanks: 1  
October 27th, 2018, 05:46 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
C'mon Ganesh. If it takes 1.83 hr to be on time, how long does it take to be 1 hour late? 
October 27th, 2018, 09:11 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,128 Thanks: 2336  The overall distance unit should be km, not m. The time you found, 2.08 hr (or exactly 2 hr 5 min) includes being 15 min late, so the journey time for arrival on time is exactly 1 hr 50 min. 
October 28th, 2018, 03:38 AM  #6 
Senior Member Joined: Aug 2014 From: India Posts: 491 Thanks: 1  
October 28th, 2018, 06:01 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,128 Thanks: 2336 
That would be exactly one hour more than the time found for arriving on time.

October 28th, 2018, 06:16 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  

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