Grandpa is writing a book. Every morning he starts writing vigorously

Question

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 does not 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?

A. 30%
B. 40%
C. 50%
D. 60%
E. 70%

A. 30%
B. 40%
C. 50%
D. 60%
E. 70%

Sk1003 · Answer

Ans: (C) 50%

Assume that on any given day, grandpa writes 'k%' new content
and that the size of content written thus far was 'N' (by the end of previous day)

Now after writing new content, the size of cumulative content = N * (1+k%)

After deleting 1/5th of the cumulative content, size = 0.8 * N * (1+k%)

As per the Q, this figure is 20% higher than N (size of content at the end of previous day)

Which gives us the equation:

0.8 * N * (1+k%) = 1.2 * N

N cancels out, so you're left with :

1+k% = 12/8

Which gives you k = 50

playthegame · Answer

Thanks for your solution, I was getting stuck as I was assuming numbers and I lost track and lost time so it was not my greatest attempt the first time, the second time I used variables and landed at a similar approach, most times algebra and arithmetic is king. Thanks.

PandCduo · Answer

Assume 1st day he writes 100 pages and then he deletes 1/5 so remaining 80 pages

Next day 20%more i.e 96 pages but this will be after he deletes 1/5th.

So he must have had written 120 pages in 2 days 120-20% = 96

Therefore in the morning he must have written 40 pages which is 50% higher

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playthegame · Answer

Yes right I started with this set of numbers when I first attempted this question then I lost my way. I think with variables I was able to better understand the equation. Appreciate your solution thanks.

luisdicampo · Answer

Deconstructing the Question

Let the book content at the end of the previous day be  P . 
In the morning Grandpa writes some new content. 
After lunch he deletes  one-fifth  of everything written so far, so  four-fifths remain . 
At the end of the day, the content is  20% more  than the previous day, so : 1.2P . 
We need the percent of  new content written in the morning  relative to  P .

Step-by-step

Let morning new content be  kP . 
Before deletion, total content is:

P + kP = P(1+k)

After deleting one-fifth, remaining content is:

\frac{4}{5} \cdot P(1+k)

This equals end-of-day content:

\frac{4}{5} \cdot P(1+k) = 1.2P

Cancel  P :

\frac{4}{5}(1+k) = 1.2

Solve:

1+k = 1.2 \cdot \frac{5}{4}

1+k = 1.5

k = 0.5

So Grandpa writes  50%  new content each morning.

Answer: 50%

Navdeep2000 · Answer

For the ease of understanding, let's do it for 2 days.

Let's say that grandpa writes x pages in the morning.
So, 1st morning, pages he writes =  x 
Post lunch, he deletes, (1/5) of it, meaning we have (4/5) of it remaining or  (4x/5)  in our case.

Next morning, he again writes x pages, totalling it to be (x)+(4x/5) =  (9x/5) 
Once again, post lunch, he deletes (1/5) of it, leaving us with (4/5)(9x/5) =  (36x/25)  pages

Now, as per the given, this  (36x/25)=1.2(4x/5) 
Simplifying it further, we get  (3x)=(2x) , meaning that we have 3x pages the next day compared to the 2x pages the previous day.
We just need to find the percent change between the two.

Therefore,  (3x-2x)/(2x)*100 = 50%

Option C

Grandpa is writing a book. Every morning he starts writing vigorously

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