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If y is a positive integer, and |x| < 5 − y, then what is the least po

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If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post Updated on: 20 Mar 2018, 09:55
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Question Stats:

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If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?

A. 4
B. 1
C. 0
D. -1
E. -4

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Originally posted by Ayush1692 on 20 Mar 2018, 09:44.
Last edited by Bunuel on 20 Mar 2018, 09:55, edited 1 time in total.
Renamed the topic and edited the question.
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If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 20 Mar 2018, 10:25
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1
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?

A. 4
B. 1
C. 0
D. -1
E. -4

Don't remove brackets and solve. The question asks for the logic behind absolute value.

\(y\) is a positive integer

\(|x| < 5 − y\)

LHS is nonnegative - it is positive or 0.
Least value for x is ZERO

Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.

The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0

Answer C

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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 17 Jun 2019, 01:29
Since y is a positive integer so y=1,2,3,4,5,......
Now |x| must be >=0 since modulus of an expression cannot be zero. So, if we put the values of y starting with 5,4,3,2,1, we get
1) y=5 => |x|<0 (INVALID! as |x| must be non-negative)
2) y=4 => |x|<1 => this implies that magnitude of x must be less than. Option C as value for x satisfies this condition giving us the least possible integer for x.
3) y=3 => |x|<2 => This option gives us x such that it could be 0 or 1. Here also, 0 is the least possible value for x.

So correct option is C.
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 06 Jul 2019, 01:16
1
Why not -1.

Given y = 1 then
|x| < 5 - 1
|x| < 4
-4 < x < 4

Therefore, the best answer is -1 ?

Kindly advise
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 04 Aug 2019, 03:15
1
trailcb wrote:
Why not -1.

Given y = 1 then
|x| < 5 - 1
|x| < 4
-4 < x < 4

Therefore, the best answer is -1 ?

Kindly advise


Hey, I also used a similar method and got the answer wrong. Did you figure out why this approach is wrong
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 04 Aug 2019, 03:54
generis wrote:
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?

A. 4
B. 1
C. 0
D. -1
E. -4

Don't remove brackets and solve. The question asks for the logic behind absolute value.

\(y\) is a positive integer

\(|x| < 5 − y\)

LHS is nonnegative - it is positive or 0.
Least value for x is ZERO

Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.

The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0

Answer C

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This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Why can't x be a negative integer, in that case it should be -1 i.e. option D.
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 08 Aug 2019, 18:54
I don't understand why x can't be negative
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 08 Aug 2019, 19:59
2
Hi, I believe that the answer choices are wrong.

The least possible value of x is the smallest value that qualifies |x| < 5 − y (y is a positive number).

y=1 -> |x|<4 -> |x| can be 3 -> smallest x = -3
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If y is a positive integer, and |x| < 5 − y, then what is the least po  [#permalink]

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New post 21 Aug 2019, 06:28
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?

A. 4
B. 1
C. 0
D. -1
E. -4

--== 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 re-post 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.


Can someone confirm if the question did not originally mean to ask the least possible value of |x|? Bunuel, chetan2u
Because, otherwise, IMO, x can take -3.

The stem reads: |x| < 5 - y and y is positive. So, this implies |x| < 4. So, x > - 4. The least possible value that x can take in that case would be -3 (assuming X is also an integer).
Otherwise, the answer doesn't make much sense to me -> |-3| = 3 and it is clearly less than 5 - y, when y = 1.

So, perhaps, the stem should have read the least possible value of |x| instead of the least possible value of x.

Cheers
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If y is a positive integer, and |x| < 5 − y, then what is the least po   [#permalink] 21 Aug 2019, 06:28
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