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# M20-15

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Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135171 [0], given: 12664

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16 Sep 2014, 01:08
Expert's post
4
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Difficulty:

15% (low)

Question Stats:

78% (00:54) correct 22% (01:02) wrong based on 111 sessions

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Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135171 [0], given: 12664

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135171 [0], given: 12664

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16 Sep 2014, 01:08
Official Solution:

Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10

Backsolve: $$@(@(@8)) = 8 - 1 - 1 - 1 = 5$$.

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Kudos [?]: 135171 [0], given: 12664

Retired Moderator
Joined: 26 Nov 2012
Posts: 594

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30 Jun 2016, 08:52
Bunuel wrote:
Official Solution:

Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10

Backsolve: $$@(@(@8)) = 8 - 1 - 1 - 1 = 5$$.

Hi Bunuel,

@x given and given that if $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0.

and given @a , but here how we can consider a as positive.

Since there is no negative sign then you considerd a as +ve ? Please let me know.

Kudos [?]: 179 [0], given: 45

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135171 [0], given: 12664

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30 Jun 2016, 08:58
msk0657 wrote:
Bunuel wrote:
Official Solution:

Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10

Backsolve: $$@(@(@8)) = 8 - 1 - 1 - 1 = 5$$.

Hi Bunuel,

@x given and given that if $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0.

and given @a , but here how we can consider a as positive.

Since there is no negative sign then you considerd a as +ve ? Please let me know.

The question is solved by backsolving, by substituting the options.
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Kudos [?]: 135171 [0], given: 12664

Senior Manager
Joined: 15 Sep 2011
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23 Jul 2016, 18:53
Top Contributor
The symbol represents one step in the number line, whether the numbe is negative or positive. Therefore, since the symbol appears three times, the $$a$$ must be either $$a+3$$ or $$a-3$$. Furthermore, since both hypothetical results are positive (5+3, 5-3), only the former -- 8-- works since all positive numbers are deducted one.

Kudos [?]: 427 [0], given: 45

Intern
Joined: 19 Nov 2012
Posts: 2

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

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26 Jan 2017, 08:22
hello,

i understood Bunuel's solution. However, i'd a question: how does one decide if one should use back substitution or equations to solve? in this case, i tried solving with equations. since we don't know sign of a, i solved for both and got 2 and 8 as answers then back substituted to check ( since both 2 and 8 are in answer choices) and then found only 9 works. any advise?

thanks.

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Manager
Joined: 23 Nov 2016
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Location: United States (NY)
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06 Feb 2017, 22:04
sanjay1810 wrote:
hello,

i understood Bunuel's solution. However, i'd a question: how does one decide if one should use back substitution or equations to solve? in this case, i tried solving with equations. since we don't know sign of a, i solved for both and got 2 and 8 as answers then back substituted to check ( since both 2 and 8 are in answer choices) and then found only 9 works. any advise?

thanks.

If you take a step back and look at the operation it's pretty simple. If x > 0, subtract 1. If x < 0, add one. The initial number then definitely can't be negative or lower than 5 since you'd then have no way to get up to 5. Hence, if you want to get to 5 you have to start higher, since eventually you'll just keep getting alternating values of 0, -1, 0, -1, 0, -1.

If the operation isn't that simple it's a game time decision I'd think. I typically just backsolve. The only time I'd consider not backsolving is if the answer options are nasty or if the operation is "normal" and just a function - f(k) or f(x) or whatever.

Kudos [?]: 13 [0], given: 20

Intern
Joined: 15 Jan 2014
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Location: India
Concentration: Technology, Strategy
Schools: Haas '19
GMAT 1: 650 Q49 V30
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WE: Information Technology (Consulting)

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10 Feb 2017, 09:34
Without backsolving :

@(@(@a))=5

Consider full underline part as 1 unit say X
then above equation becomes: @(X)=5
Now X is positive so we can substitute @X=X−1
X−1 = 5 ==> X is 6

Now again following above step :
@(@a)=6 ==> X-1 = 6 ==> 7
Now last iteration :
@a= 7 ==> X-1 = 7 ==> X=8 .

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Senior Manager
Joined: 08 Jun 2015
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03 Oct 2017, 05:58
+1 for D.
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Location: India
Concentration: Marketing, General Management
GPA: 1
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12 Oct 2017, 23:09
Bunuel wrote:
Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10

Kudos [?]: 13 [0], given: 75

Math Expert
Joined: 02 Sep 2009
Posts: 42529

Kudos [?]: 135171 [1], given: 12664

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12 Oct 2017, 23:12
1
KUDOS
Expert's post
aashishagarwal2 wrote:
Bunuel wrote:
Operation $$@$$ is defined as $$@x = x + 1$$ if $$x$$ is negative and $$@x = x - 1$$ if $$x$$ is positive or 0. If $$@(@(@a)) = 5$$, what is the value of $$a$$ ?

A. -7
B. -5
C. 2
D. 8
E. 10

Check the topics below. There you can find all the articles and questions we have.
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Kudos [?]: 135171 [1], given: 12664

Re: M20-15   [#permalink] 12 Oct 2017, 23:12
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# M20-15

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