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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
A. 4
B. 1
C. 0
D. -1
E. -4
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20 Mar 2018, 10:25
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Ayush1692
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.A. 4
B. 1
C. 0
D. -1
E. -4
\(y\) 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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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.
17 Jun 2019, 01:29
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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.
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.
