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 November 29th, 2016, 11:59 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, 12:05 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 $A t > C + B t$ $(A-B)t > C$ if $A \leq B$ then John will never catch up. Otherwise John has solved more codes than Neil after $t > \dfrac{C}{A-B}~days$ Thanks from swordf1sh
November 29th, 2016, 12:37 PM   #3
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Quote:
 Originally Posted by romsek $A t > C + B t$ $(A-B)t > C$ if $A \leq B$ then John will never catch up. Otherwise John has solved more codes than Neil after $t > \dfrac{C}{A-B}~days$

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, 03:12 PM   #4
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Quote:
 Originally Posted by swordf1sh A confusion for me now, 'The last statement should be Neil has solved more codes than John, isn't? Sorry I might be wrong
oh.. jeeze the problem is worded poorly.

so my previous answer is all wrong.

it should be this

$A t + C < B t$

$C < (B-A)t$

if $B \leq A$ then Neil will never catch up. Otherwise

$t > \dfrac{C}{B-A}$ is the minimum # of days for Neil to have solved more codes than John.

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

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