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Grandpa is writing a book. Every morning he starts writing vigorously and fills a lot of pages. But post-lunch he goes through all that he's written that far (right from day one) and deletes one-fifth of it. He doesn't write anything more that day. At the end of the day the content of his book is still $20\%$ more than that at the end of the previous day.

How much percent new content does Grandpa write in the morning?

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Would the "homework" tag be appropriate on this Question? –  hardmath Nov 15 '11 at 13:51
    
Hello @hardmath, this is not technically my homework, we are supposed to do this problem in our next class (as classwork), I am just trying to work on it in advance. I don't mind any tag as long I am getting feedback form the community :) –  Quixotic Nov 15 '11 at 13:58
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4 Answers

up vote 3 down vote accepted

Say that after $n$ days he has $C(n)$ pages of content, and on day $n$ he writes $D(n)$ pages in the morning. Then $C(n+1)=\frac45(C(n)+D(n+1))=\frac65C(n)$. Just solve the second equation for $D(n+1)$ in terms of $C(n)$.

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And that is $\frac12=50\%$. –  Quixotic Nov 19 '11 at 3:28
    
@MaX: Looks good. –  Brian M. Scott Nov 19 '11 at 3:40
    
I have a doubt where do that 6/5 comes from ? I'm very sorry for my silly doubt sir :( –  vaidy_mit Jan 27 at 17:17
    
Yeah i got it ! :D C(n)+20% of C(n)=6/5 of C(n).Cool one sir ! :D –  vaidy_mit Jan 27 at 17:21
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Hint: you can only define it in terms of percentage increase. Grandpa will have to write more pages each morning to keep this up. Say he writes $x\%$ of the existing book in the morning. How much is left after he deletes $1/5$?

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Assume the content at wakeup is $C$, and let $x>0$ be the new content written during the morning, expressed as a fraction of $C$. According to your story you have

$$C\cdot (1+x)\cdot{4\over5}\ =\ {6\over 5}\ C\ .$$

Now solve for $x$ and multiply by 100 to express this quantitiy in $\%$.

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At the end of the day he deletes one-fifth of everything. This means that he deletes one-fifth of what he had written earlier, and one-fifth of what he wrote that day. So the four-fifths of the day's work that he added should make up for the one-fifth's of the original work he deleted, and still be one-fifth more beyond that.

So four-fifths of the day's work is two-fifths of the total earlier work, which means that the day's work is half of the total earlier work.

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