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# M20-15

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Math Expert
Joined: 02 Sep 2009
Posts: 50016

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16 Sep 2014, 01:08
2
2
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Difficulty:

25% (medium)

Question Stats:

75% (01:03) correct 25% (01:03) wrong based on 163 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

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Math Expert
Joined: 02 Sep 2009
Posts: 50016

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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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Joined: 26 Nov 2012
Posts: 593

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 50016

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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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Senior Manager
Joined: 15 Sep 2011
Posts: 335
Location: United States
WE: Corporate Finance (Manufacturing)

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23 Jul 2016, 18:53
Top Contributor
1
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.
Intern
Joined: 19 Nov 2012
Posts: 26

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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.
Manager
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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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.
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Joined: 15 Jan 2014
Posts: 34
Location: India
Concentration: Technology, Strategy
Schools: Haas '19
GMAT 1: 650 Q49 V30
GPA: 2.5
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 .
Senior Manager
Joined: 08 Jun 2015
Posts: 451
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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03 Oct 2017, 05:58
+1 for D.
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Joined: 04 Jul 2017
Posts: 61
Location: India
Concentration: Marketing, General Management
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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

Math Expert
Joined: 02 Sep 2009
Posts: 50016

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12 Oct 2017, 23:12
1
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

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

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