GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 09 Dec 2019, 07:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 06 Jan 2013
Posts: 25
GPA: 4
WE: Engineering (Transportation)
All the page numbers from book are added, beginning at page  [#permalink]

### Show Tags

Updated on: 05 May 2013, 22:53
16
00:00

Difficulty:

85% (hard)

Question Stats:

54% (02:41) correct 46% (02:21) wrong based on 173 sessions

### HideShow timer Statistics

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

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

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.
RENAMED THE TOPIC.
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 583

### Show Tags

05 May 2013, 21:31
1
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.
_________________
Current Student
Joined: 06 Jan 2013
Posts: 25
GPA: 4
WE: Engineering (Transportation)

### Show Tags

05 May 2013, 22:08
1
1
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
_________________
If you shut your door to all errors, truth will be shut out.
VP
Joined: 07 Dec 2014
Posts: 1217
All the page numbers from book are added, beginning at page  [#permalink]

### Show Tags

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
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.
Intern
Joined: 04 Sep 2012
Posts: 14
Location: India
WE: Marketing (Consumer Products)

### Show Tags

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!
Intern
Joined: 05 May 2013
Posts: 1

### Show Tags

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?
Intern
Joined: 03 Nov 2017
Posts: 4
GMAT 1: 710 Q50 V35
GPA: 3.3
Re: All the page numbers from book are added, beginning at page  [#permalink]

### Show Tags

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)
Re: All the page numbers from book are added, beginning at page   [#permalink] 16 Sep 2019, 16:56
Display posts from previous: Sort by