May 19 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Sunday, May 19th at 7 AM PT May 19 07:00 PM EDT  08:00 PM EDT Some of what you'll gain: Strategies and techniques for approaching featured GMAT topics. Sunday May 19th at 7 PM ET May 20 10:00 PM PDT  11:00 PM PDT Practice the one most important Quant section  Integer Properties, and rapidly improve your skills. May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 26 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Sunday, May 26th at 7 am PT.
Author 
Message 
Manager
Status: Single
Joined: 05 Jun 2011
Posts: 98
Location: Shanghai China

A little difficult math problem, pls help me
[#permalink]
Show Tags
24 Jul 2011, 03:40
Question Stats:
88% (01:04) correct 12% (02:30) wrong based on 25 sessions
HideShow timer Statistics
The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x= a)4 b)5 c)6 d)7 e)8 I have no idea how to solve this ,pls help~~~ the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7 why x = 7 can be concluded from M=28? == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 31 May 2011
Posts: 78
Location: India
Concentration: Finance, International Business
GMAT Date: 12072011
GPA: 3.22
WE: Information Technology (Computer Software)

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
24 Jul 2011, 05:58
tracyyahoo wrote: The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=
a)4 b)5 c)6 d)7 e)8
I have no idea how to solve this ,pls help~~~
the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7
why x = 7 can be concluded from M=28? sum of first x terms where 1<x<10 is x(x+1)/2  Eq 1 Now we know that sum of terms in Arithmetic Progression is n/2[2a +d(n1)] where n = number of terms; here n= 10x. a = first number of the seq. here first term will be x+1 d = difference between 2 consecutive numbers; 1 in this case Hence the new expression is (10x)/2[2(x+1) + (10x+1)]  Eq 2 Now the question says Sum of X terms = 1 + sum of remaining terms of (10X) therefore putting eq1 and eq2. We get : x(x+1)/2 = 1+ (10x)/2[2(x+1) + (10x+1)] ... solving eq we will get: x^2 +x 56 = 0 or (x+8)(x7) = 0. there for X = 7 as 1< x < 10 Hence D



Intern
Joined: 17 May 2009
Posts: 30
Location: United States
GPA: 3.62
WE: Corporate Finance (Manufacturing)

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
24 Jul 2011, 17:10
We know that the sum of the integers from 1 to 10 inclusive is \((\frac{10+1}{2})*(101+1) = 5.5*10 = 55\)
\(55/2 = 27.5\), and the sum of the first x digits is one more than the sum of the rest, so the sum of the first x digits is 28 and the sum of the rest is 27.
List the numbers from 1 to 10:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
We can start adding the consecutive integers from 1 until we reach the integer x that brings the sum to 28:
\(1+2+3+4=10\)
\(10+5+6=21\)
\(21+7=28\)
\(x=7\)



Intern
Joined: 18 Jul 2011
Posts: 43

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
25 Jul 2011, 09:36
Both the replies given so far are correct, but I think there's a better way to approach this  use the answers. The answers are there to help you  you should always read them as part of the question.
One way to solve this problem would be to guess and check, then adjust your guess based on whether it was too high or too low. In these situation you should pick C as your first guess and check number. Because the GMAT always puts numerical answers in order, picking C will either get you the correct answer or eliminate 3 choices immediately. Then you only have to check one more answer
Guess C: 1 + 2 + 3 + 4 + 5 + 6 = 21 and 7 + 8 + 9 + 10 = 34. Too low
Now, obviously A and B can't be the answers, so we'll check D, and if that doesn't work E will be our answer.
Guess D: 21 + 7 = 28 and 34  7 = 27. Correct
Note that I used calculations I'd already done for my first guess in my second guess.
Most questions on the GMAT can be answered with a little common sense  students spending hours relearning algebra and geometry that they've forgotten since high school or college often overlook this approach.



Intern
Joined: 16 Jul 2011
Posts: 14
Location: Canada

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
25 Jul 2011, 10:29
I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently, Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l) We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10 The sum of the first x terms = x(1 + x)/2 Now we have "10  x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10  x )(x + 1 + 10)/2 = (10  x)(x + 11)/2 But x(x + 1)/2  1= (10  x)(x + 11)/2 Now solving above equation we get a quadratic equation x^2 + x  56 = 0 and we can factor it as (x + 8) (x  7) = 0 Hence x = 8, 7 Reject 8 as it is not in page numbers and our answer is x = 7 which is given by choice d.
_________________



Manager
Status: Single
Joined: 05 Jun 2011
Posts: 98
Location: Shanghai China

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
25 Jul 2011, 18:26
Thank u . ur post is most comprehensable.
uote="2ndgrademath"]I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently, Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)
We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10
The sum of the first x terms = x(1 + x)/2
Now we have "10  x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10  x )(x + 1 + 10)/2 = (10  x)(x + 11)/2
But x(x + 1)/2  1= (10  x)(x + 11)/2
Now solving above equation we get a quadratic equation x^2 + x  56 = 0 and we can factor it as
(x + 8) (x  7) = 0 Hence x = 8, 7 Reject 8 as it is not in page numbers and our answer is x = 7 which is given by choice d.[/quote]
Posted from my mobile device



Manager
Status: Single
Joined: 05 Jun 2011
Posts: 98
Location: Shanghai China

Re: A little difficult math problem, pls help mew
[#permalink]
Show Tags
25 Jul 2011, 22:09
why x(x1)/2  1? I don"t follow this. Pls help me to explaim this 2ndgrademath wrote: I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently, Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)
We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10
The sum of the first x terms = x(1 + x)/2
Now we have "10  x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10  x )(x + 1 + 10)/2 = (10  x)(x + 11)/2
But x(x + 1)/2  1= (10  x)(x + 11)/2
Now solving above equation we get a quadratic equation x^2 + x  56 = 0 and we can factor it as
(x + 8) (x  7) = 0 Hence x = 8, 7 Reject 8 as it is not in page numbers and our answer is x = 7 which is given by choice d. Posted from my mobile device



Intern
Joined: 07 Jun 2011
Posts: 45

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
26 Jul 2011, 00:34
I went directly into the answers and started eliminating.
the only answer choice that satisfies the given condition is 7



Intern
Joined: 16 Jul 2011
Posts: 14
Location: Canada

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
27 Jul 2011, 09:23
Hi tracyyahoo Thank you very much for your appreciation. And we take away one from the left side because of the following reason: Because the sum of first "x" terms, [which is x(1 + x)/2], is 1 more than the sum of 10x terms. So, to make them equal we need to subtract 1 from the greater. Or you can add one to the smaller sum. Hope it will help.
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9218
Location: Pune, India

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
27 Jul 2011, 22:26
tracyyahoo wrote: The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=
a)4 b)5 c)6 d)7 e)8
I have no idea how to solve this ,pls help~~~
the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7
why x = 7 can be concluded from M=28? The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them. Case 1: Using Options If I add 1 to 6, I get 6*7/2 = 21 Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34) Nope. We need to go to D and do the same there. We see that it works. Case 2: Using Algebra Sum of first x pages = x(x+1)/2 Sum of all pages = 10*11/2 = 55 Sum of all pages except first x pages = 55  x(x+1)/2 Given: x(x+1)/2 = 1 + 55  x(x+1)/2 x(x+1) = 56 = 7*8 x must be 7.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 05 Jul 2010
Posts: 329

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
27 Jul 2011, 22:45
This is how I would solve:
S(n) = n*(n+1)/2 ... S = Sum S(10) = 10*11/2 = 55
We need to find "largest" n: S(n) > S(10)  S(n) => S(n) > 55  S(n) => 2*S(n) > 55 => n*(n+1) > 55 => n*(n+1) >= 56 = 7*8 => n=7
Note: Here the solution came easily as 56=7*8. But you might have something like n*(n+1) < 45 normally in a question ... then you need to just do some mental math to find largest n, such that n*(n+1) < 45... This is how I would start thinking... 5*6 = 30, 6*7 = 42, so n = 6. So be prepared to do some easy mental math in the Test. From all the tests I have given, I have seen atleast one problem that has Sum of an arithmetic progression (S(n) is the simplest form of Arithmetic pregression) combined with inequality.
Hope this helps.



Intern
Status: If I play my cards right, I can work this to my advantage
Joined: 18 Jul 2011
Posts: 14
Location: India
Concentration: Operations, Social Entrepreneurship
GMAT Date: 11122011
WE: Information Technology (Telecommunications)

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
28 Jul 2011, 05:52
VeritasPrepKarishma wrote: tracyyahoo wrote: The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=
a)4 b)5 c)6 d)7 e)8
I have no idea how to solve this ,pls help~~~
the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7
why x = 7 can be concluded from M=28? The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them. Case 1: Using Options If I add 1 to 6, I get 6*7/2 = 21 Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34) Nope. We need to go to D and do the same there. We see that it works. Case 2: Using Algebra Sum of first x pages = x(x+1)/2 Sum of all pages = 10*11/2 = 55 Sum of all pages except first x pages = 55  x(x+1)/2 Given: x(x+1)/2 = 1 + 55  x(x+1)/2 x(x+1) = 56 = 7*8 x must be 7. yep picking numbers is what helped me ... good that this question was for lesser number of pages Else Probly we might have to resort to formulae



Intern
Joined: 24 Jun 2011
Posts: 38

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
28 Jul 2011, 11:00
Its in your explanation itself. Quote: the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7 Sum of x = m which is x(x+1)/2 = m and M = 28 => x (x+1)/2 = 28 => x(x+1) = 56. since x is positive we get x = 7.



Intern
Joined: 27 Feb 2011
Posts: 40

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
28 Jul 2011, 12:56
I felt substitution was easy ..
sum of consecutive 10 numbers = 55
try substituting answers to satisfy the condition ..



Intern
Joined: 02 Aug 2011
Posts: 43
Location: Canada
Concentration: Finance, Healthcare
GPA: 3.4
WE: Analyst (Consulting)

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
09 Aug 2011, 13:07
Can't we just guess and check? I guessed 7 and got it right by chance, but even if we guessed 6 first it wouldn't be hard.
Guess: 6 1+6+2+5+3+4 = 7*3 = 21 7+8+9+10 = 19 + 16 = 35 Wrong, too small
Guess 7: 1+7+2+6+3+5+4 = 3*8+4= 28 8+9+10 = 27 Correct



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9218
Location: Pune, India

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
09 Aug 2011, 19:36
ssarkar wrote: How you getting 7 if you use case1  using options
If we add 1 to 7, then we get 7*8/2=28 Adding 8+9+10, we get 27. ( average = 27/3=9) Then how we get 7. Is the fact that 10*11/2=55 and 28+27=55. The condition that the answer should satisfy is this: If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x If x = 7, the sum of page number up to and including x is 1+2+3..+6+7 = 7*8/2 = 28 (The sum of n consecutive numbers starting from 1 is n(n+1)/2) The sum of the page numbers following page number x = 8+9+10 = 27 Since 28 is one more than 27, our condition is satisfied. So answer is 7.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Director
Joined: 03 May 2007
Posts: 773
Schools: University of Chicago, Wharton School

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
09 Aug 2011, 21:33
tracyyahoo wrote: The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=
a)4 b)5 c)6 d)7 e)8
I have no idea how to solve this ,pls help~~~
the explanation says that suppose the sum of x is m, then 2m1=1+2+3+.......+10. M=28, so x = 7
why x = 7 can be concluded from M=28? Another approach: Sum of the page numbers from 1 to 10 = 55 The sum of the page numbers up to and including page number x of the report = a The sum of the page numbers following page number x = b a + b = 55 ..........i a  b = 1.............ii Add i and ii: a = 28 and b = 27. what makes of b = 27? 10+9? no. 10+9+8? yes. so x has to be 7.



Director
Joined: 01 Feb 2011
Posts: 646

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
21 Aug 2011, 19:09
1+2...+x = 1+(x+1)...10
1+2+..+x = 1 + (1+2...10) (1+2+...x)
2(1+2...x) = 1+55
x(x+1) = 56
x=7



Manager
Joined: 20 Aug 2011
Posts: 128

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
21 Aug 2011, 21:21
Acc. to the question, the page numbers are between 1 and 10 Let x be the point where we divide the page numbers in two parts so no. of terms between 1 & x inclusive is x and between 10 & x is 10x first term of series from 1 to x is 1 & last term is x using sum of series formula s1=x(x+1)/2 similarly for the next set first term is x+1 and last term is 10 s2=(10x)(10+x+1)/2=(10x)*(11+x)/2 acc. to given conditions (10x)*(11+x)+2=x(x+1) 112xx2=x2+x 2(x2+x56)=0 x=7 or 8 x can't be negative as it's a page no. so x7 Hope it helps
_________________
Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.



Manager
Joined: 25 Dec 2010
Posts: 65

Re: A little difficult math problem, pls help me
[#permalink]
Show Tags
23 Aug 2011, 09:06
Substituting the given values of X would be easy in this case.




Re: A little difficult math problem, pls help me
[#permalink]
23 Aug 2011, 09:06



Go to page
1 2
Next
[ 22 posts ]



