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All the page numbers from book are added, beginning at page

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All the page numbers from book are added, beginning at page  [#permalink]

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New post Updated on: 05 May 2013, 22:53
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Question Stats:

54% (02:38) correct 46% (02:19) wrong based on 170 sessions

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All the page numbers from book are added, beginning at page 1. However one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?

A. 44
B. 45
C. 32
D. 12
E. 10

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Originally posted by GMATtracted on 05 May 2013, 20:15.
Last edited by Bunuel on 05 May 2013, 22:53, edited 1 time in total.
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Re: GMAT quant(problem solving)  [#permalink]

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New post 05 May 2013, 21:01
I did it using a long method. Trial and error!
1. If the number are 1, 2,...50 - Sum = ((1+50)/2)*50 = 1275.. too large
2. If the number are 1, 2,...45 - Sum = ((1+45)/2)*45 = 1035.. Almost there..

So, 1035-45 = 990. This means 10 was repeated twice!
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Re: GMAT quant(problem solving)  [#permalink]

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New post 05 May 2013, 21:31
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GMATtracted wrote:
All the page numbers from book are added, beginning at page 1. However one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?

A. 44
B. 45
C. 32
D. 12
E. 10


The sum of all the page numbers = \(\frac{n(n+1)}{2}\).We can quite easily see that for n=44, we have the sum as 44*45/2 = 22*45 = 990. Thus, the page number 10 must have been added twice.

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Re: GMAT quant(problem solving)  [#permalink]

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New post 05 May 2013, 21:38
the figure on the right, ABC is a right angled, equilateral triangle. if BC=x and DE=y, what is the area of the plot DBCE?
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Re: GMAT quant(problem solving)  [#permalink]

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New post 05 May 2013, 22:08
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Let 'n' be the total no. of pages in the book.
And, let the page with page no. 'p' be the one which was mistakenly added twice
Hence, the sum by including page 'p' twice would be:-
\(sum=\frac{n(n+1)}{2}+p=1000\)Therefore,\(\frac{n(n+1)}{2}=(1000-p)\)Now, examining the options carefully:-
If p=10 then \(\frac{n(n+1)}{2}=990\)or n(n+1)=2x990=2x9x2x5x11=(2x2x11).(9x5)=44x45
Hence, n=44
We see that by using p=10 we obtain a proper integral value of 'n'.
Therefore. page no. 10 was added twice
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All the page numbers from book are added, beginning at page  [#permalink]

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New post Updated on: 09 Nov 2017, 15:03
1
1
All the page numbers from book are added, beginning at page 1. However one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?

A. 44
B. 45
C. 32
D. 12
E. 10

let n=total pages in book
n(n+1)/2<1000
n(n+1)<2000
reference point: √2000~45
(45)(44)=1980
1980/2=990
1000-990=10
page 10 was added twice
E

Originally posted by gracie on 09 May 2016, 14:13.
Last edited by gracie on 09 Nov 2017, 15:03, edited 3 times in total.
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Re: All the page numbers from book are added, beginning at page  [#permalink]

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New post 16 Sep 2019, 16:56
The sum is given by: \(\frac{n(n+1)}{2} + repeated = 1000\)

And,
1. \(0 < repeated <= n\) (repeated page can't be outside of page numbers)
2. \(\frac{n(n+1)}{2}\) < 1000. (sum has to be smaller than before adding repeat number)

Take 2nd equation
\(n^2 + n < 2000\)

assuming \(n\) is very less than \(n^2\), we can also write: \(n^2 < 2000\)
\(n < 20 \sqrt{5}\)
\(n < 44.7\) (\sqrt{5} = 2.236)

now we can check for n = 44, 43, 42, etc.
when n=44, sum = 990 => repeated number = 10
when n=43, sum = 946 => repeated number = 54 (NOT possible since repeated number < 43)
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Re: All the page numbers from book are added, beginning at page   [#permalink] 16 Sep 2019, 16:56
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