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General GMAT Forum Moderator V
Joined: 15 Jan 2018
Posts: 772
Concentration: General Management, Finance
GMAT 1: 720 Q50 V37 Adam tears off a page from his textbook. The sum of the remaining page  [#permalink]

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3 00:00

Difficulty:   65% (hard)

Question Stats: 33% (02:29) correct 67% (02:14) wrong based on 18 sessions

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Adam tears off a page from his textbook. The sum of the remaining page numbers is 10000. What are the two page-numbers on the torn-off pages of his textbook?

A. 8 and 9

B. 33 and 34

C. 77 and 78

D. 97 and 98

E. 76 and 77

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Math Expert V
Joined: 02 Aug 2009
Posts: 8250
Re: Adam tears off a page from his textbook. The sum of the remaining page  [#permalink]

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1
DisciplinedPrep wrote:
Adam tears off a page from his textbook. The sum of the remaining page numbers is 10000. What are the two page-numbers on the torn-off pages of his textbook?

A. 8 and 9 or 76 and 77

B. 33 and 34 or 5 and 6

C. 77 and 78 or 8 and 9

D. 97 and 98 or 77 and 78

E. 76 and 77 or 5 and 6

Let the numbers be a and b, and total pages be n, so sum of n pages = $$\frac{n(n+1)}{2}=10000+a+b....n(n+1)=20000+2(a+b)$$
Thus 20000+2(a+b) should be the product of two consecutive numbers n and n+1..

You can check from choices or look for the squareroot of 20000, which is closer to 140..
So if 140 pages are there..14*141=19740<20000, so take next value of n as 141
141*142=20022, so look for 2(a+b) = 22...a+b=11...From choices 5 and 6 fit in
Take next value of n as 142, so 142*143=20306...look for 2(a+b)=306...a+b=153...Only 76+77=153 fits in

E
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Intern  B
Joined: 30 Apr 2019
Posts: 3
Re: Adam tears off a page from his textbook. The sum of the remaining page  [#permalink]

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chetan2u wrote:
DisciplinedPrep wrote:
Adam tears off a page from his textbook. The sum of the remaining page numbers is 10000. What are the two page-numbers on the torn-off pages of his textbook?

A. 8 and 9 or 76 and 77

B. 33 and 34 or 5 and 6

C. 77 and 78 or 8 and 9

D. 97 and 98 or 77 and 78

E. 76 and 77 or 5 and 6

Let the numbers be a and b, and total pages be n, so sum of n pages = $$\frac{n(n+1)}{2}=10000+a+b....n(n+1)=20000+2(a+b)$$
Thus 20000+2(a+b) should be the product of two consecutive numbers n and n+1..

You can check from choices or look for the squareroot of 20000, which is closer to 140..
So if 140 pages are there..14*141=19740<20000, so take next value of n as 141
141*142=20022, so look for 2(a+b) = 22...a+b=11...From choices 5 and 6 fit in
Take next value of n as 142, so 142*143=20306...look for 2(a+b)=306...a+b=153...Only 76+77=153 fits in

E

Hi, I did not get your explanation like how you solved it and what approach you used. Can you explain me a different approach?
Senior Manager  G
Joined: 16 Feb 2015
Posts: 355
Location: United States
Re: Adam tears off a page from his textbook. The sum of the remaining page  [#permalink]

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akanksha10899 wrote:
chetan2u wrote:
DisciplinedPrep wrote:
Adam tears off a page from his textbook. The sum of the remaining page numbers is 10000. What are the two page-numbers on the torn-off pages of his textbook?

A. 8 and 9 or 76 and 77

B. 33 and 34 or 5 and 6

C. 77 and 78 or 8 and 9

D. 97 and 98 or 77 and 78

E. 76 and 77 or 5 and 6

Let the numbers be a and b, and total pages be n, so sum of n pages = $$\frac{n(n+1)}{2}=10000+a+b....n(n+1)=20000+2(a+b)$$
Thus 20000+2(a+b) should be the product of two consecutive numbers n and n+1..

You can check from choices or look for the squareroot of 20000, which is closer to 140..
So if 140 pages are there..14*141=19740<20000, so take next value of n as 141
141*142=20022, so look for 2(a+b) = 22...a+b=11...From choices 5 and 6 fit in
Take next value of n as 142, so 142*143=20306...look for 2(a+b)=306...a+b=153...Only 76+77=153 fits in

E

Hi, I did not get your explanation like how you solved it and what approach you used. Can you explain me a different approach?

Dear akanksha10899 ,

This Approach is called Average.
Below URL contains the concept & related questions like this.
https://mba.hitbullseye.com/cat/Average-Questions.php

Regards,
Rajat Chopra
VP  V
Joined: 19 Oct 2018
Posts: 1301
Location: India
Re: Adam tears off a page from his textbook. The sum of the remaining page  [#permalink]

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1
n(n+1)/2= 10000+x+x+1

n^2+n= 20000+4x+2

n^2+n- (20002+4x)=0

As n is a real number and an integer; hence, b^2-4ac must be a perfect square

b^2-4ac=1+4(20002+4x)= 16x+80009

We can use the options to get to the answer.

16*76+80009= 285^2

E

DisciplinedPrep wrote:
Adam tears off a page from his textbook. The sum of the remaining page numbers is 10000. What are the two page-numbers on the torn-off pages of his textbook?

A. 8 and 9

B. 33 and 34

C. 77 and 78

D. 97 and 98

E. 76 and 77 Re: Adam tears off a page from his textbook. The sum of the remaining page   [#permalink] 06 Jan 2020, 01:21
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