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November 29th, 2016, 10:59 AM  #1 
Newbie Joined: Nov 2016 From: United States Posts: 2 Thanks: 0  Day when second person overtakes first person
John and Neil like coding but John does more than Neil. Neil decides to go ahead of John. Lets say John solves A codes per day and Neil solves B codes per day. John is already C problems ahead of Neil. What is the minimum number of days it will take for Neil to have solved more codes than John?

November 29th, 2016, 11:05 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 
$A t > C + B t$ $(AB)t > C$ if $A \leq B$ then John will never catch up. Otherwise John has solved more codes than Neil after $t > \dfrac{C}{AB}~days$ 
November 29th, 2016, 11:37 AM  #3  
Newbie Joined: Nov 2016 From: United States Posts: 2 Thanks: 0  Quote:
A confusion for me now, 'The last statement should be Neil has solved more codes than John, isn't? Sorry I might be wrong  
November 29th, 2016, 02:12 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  Quote:
so my previous answer is all wrong. it should be this $A t + C < B t$ $C < (BA)t $ if $B \leq A$ then Neil will never catch up. Otherwise $t > \dfrac{C}{BA}$ is the minimum # of days for Neil to have solved more codes than John.  

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day, go ahead, minimum days, overtakes, per day, person, problem 
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