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

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Intern
Joined: 06 Jan 2013
Posts: 24

Kudos [?]: 68 [0], given: 9

GPA: 3
WE: Engineering (Transportation)
All the page numbers from book are added, beginning at page [#permalink]

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05 May 2013, 20:15
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Difficulty:

75% (hard)

Question Stats:

63% (02:13) correct 38% (02:19) wrong based on 120 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
[Reveal] Spoiler: OA

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If you shut your door to all errors, truth will be shut out.

Last edited by Bunuel on 05 May 2013, 22:53, edited 1 time in total.
RENAMED THE TOPIC.

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Intern
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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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Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 627

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05 May 2013, 21:31
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.

E.
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Intern
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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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Intern
Joined: 06 Jan 2013
Posts: 24

Kudos [?]: 68 [0], given: 9

GPA: 3
WE: Engineering (Transportation)

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05 May 2013, 22:08
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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Kudos [?]: 68 [0], given: 9

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

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09 May 2016, 14:13
1
This post was
BOOKMARKED
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
E

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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08 Nov 2017, 10:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: All the page numbers from book are added, beginning at page   [#permalink] 08 Nov 2017, 10:19
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