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15 Feb 2020, 19:35
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GMATBusters’ Quant Quiz Question -10
If y is an integer, and y = k+|k|, is y = 0?
A) √(-k)=3
B) \(\frac{(2-2y)}{\sqrt{(1-y)^2 }}\)=2
15 Feb 2020, 19:36
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OFFICIAL SOLUTION
If y is an integer, and y = k+|k|, is y = 0?
We know that
|k|= k, if k> or = 0
|k|= -k, if k< or = 0
hence y can be 2k , if k is positive or 0 , when k = 0 or negative
A) √(-k)=3
since (-k) is inside square root, it means - k is positive and hence k is negative.
So, y = 0
SUFFICIENT
B) \(\frac{(2-2y)}{\sqrt{(1-y)^2 }}\)=2
As\( \sqrt{(1-y)^2 }\) = |1-y|
so, the equation becomes
\(\frac{(2-2y)}{|1-y|}\)=2
hence , |1-y|= 1-y
or 1-y> or = 0
but since 1-y term is coming in denominator, it cannot be zero.
So, 1-y>0 or y<1
But we know that y can be 2k , if k is positive or 0 , when k = 0 or negative
so y = 0
SUFFICIENT
Answer D
shameekv1989
15 Feb 2020, 19:52
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If y is an integer, and y = k+|k|, is y = 0?
A) √(-k)=3
B) (2−2y)(1−y)2−−−−−−√(2−2y)(1−y)2=
A) Since we can not have -ve square root => k has to be negative by itself. For the square root to be equal to 3 k has to be -9
Therefore y = k+|k| = -9+|-9| = 0 => Sufficient
B) Solving the equation we get 1-y = |1-y|
if y = -2 => 1+2 = 1+2 = 3
if y = 0 => 1 =1
Hence y can have multiple values => Not Sufficient
Answer - A
A) √(-k)=3
B) (2−2y)(1−y)2−−−−−−√(2−2y)(1−y)2=
A) Since we can not have -ve square root => k has to be negative by itself. For the square root to be equal to 3 k has to be -9
Therefore y = k+|k| = -9+|-9| = 0 => Sufficient
B) Solving the equation we get 1-y = |1-y|
if y = -2 => 1+2 = 1+2 = 3
if y = 0 => 1 =1
Hence y can have multiple values => Not Sufficient
Answer - A
