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November 29th, 2016, 11:59 AM   #1
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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?
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November 29th, 2016, 12:05 PM   #2
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$A t > C + B t$

$(A-B)t > C$

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

John has solved more codes than Neil after $t > \dfrac{C}{A-B}~days$
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November 29th, 2016, 12:37 PM   #3
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Quote:
Originally Posted by romsek View Post
$A t > C + B t$

$(A-B)t > C$

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
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November 29th, 2016, 03:12 PM   #4
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Originally Posted by swordf1sh View Post
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.
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