If y is an integer, and y

Question

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

GMATBusters · Answer

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 · Answer

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

pk123 · Answer

If y is an integer, and y = k+|k|, is y = 0?
y = 0 can only happen as k <0 as mod of k (|k|)is always positive
so we need to prove whether k is negative
option 1)sqrt (-k) = 3 which can happen only when k =-9...so k is negative
option 2) (2- 2y) /sqroot (1-y)^2 = 2
(1- y) /sqroot (1-y)^2 = 1
(1-y) = sqroot (1-y)^2
here y can be positive as well as negative

A is the answer

KhulanE · Answer

(A) √(-k)=3 => (√(-k))^2=3^2 => (-k)=9 => k=-9
y = k+|k|=-9+|-9|=-9+9=0 y=0 OK
(B) (2−2y)/(√(1−y)^2) =2 y cannot be 1

(2-2y)=2(√(1−y)^2) divide by 2 => 1-y=√(1−y)^2 => (1-y)^2=(√(1−y)^2) ^2=(1−y)^2, for any y except 1, the equation is true, so y cannot be 0
A is right

Raxit85 · Answer

If y is an integer, and y = k+|k|, is y = 0?
Y can be 0, if k=0 or k=-ve.

A) √(-k)=3 means k = -ve because, square root is positive and it can only be when k is negative to become ultimately positive. So, k=-ve. sufficient
B) (2−2y)/√(1−y)^2 = 2, 
2 (1-y)/√(1−y)^2 =2,
(1-y)/√(1−y)^2 = 1
If y=0, then above equation can be proven.
If y = 1, then equation can not be proven as it may lead -ve number on RHS.
If y= greater than 1, then also equation can not be proven as it may lead -ve number on RHS.
If y = -ve (less than -1), then also equation can not be proven as it may lead -ve number on RHS.
So, y = 0 only. So, sufficient.

D is the answer.

shridhar786 · Answer

If y is an integer, and y = k+|k|, is y = 0?
A) √(-k)=3
B) \(\frac{(2−2y)}{√(1−y)^2}=2\)

1)
√-k has to be positive
thus k is negative

example
√(- (-9)) = 3
any negative value when substitutes in y = k+|k| = 0 ex
-3+3 =0
sufficient

2)
\(\frac{(2−2y)}{√(1−y)^2}=2\)
\(y \neq{1}\)
now RHS = 2
thus LHS also has to be

\(\frac{2(1−y)}{√(1−y)^2}\)
thus \(\frac{(1−y)}{|(1−y)|}\) has to be positive
or y < 1 since y is an integer
y <=0
for all such value y = k+|k| any negative values satisfying this value will result in y = 0
thus sufficient

D

globaldesi · Answer

If y is an integer, and y = k+|k|, is y = 0?
A) √(-k)=3
B)  \frac{(2−2y)}{√(1−y)^2}=2

1)
√-k has to be positive
thus k is negative

example
√(- (-9)) = 3
any negative value when substitutes in y = k+|k| = 0 ex
-3+3 =0
sufficient

2)
 \frac{(2−2y)}{√(1−y)^2}=2 
 y 
eq{1} 
now RHS = 2
thus LHS also has to be

\frac{2(1−y)}{√(1−y)^2} 
thus  \frac{(1−y)}{|(1−y)|}  has to be positive
or y < 1 since y is an integer 
y <=0
for all such value y = k+|k| any negative values satisfying this value will result in y = 0
thus sufficient

D

aditibarjatya · Answer

Solution :

Question 10 :  If y is an integer, and y = k+|k|, is  y = 0 ?

A)  \sqrt{(-k)}=3

B)  \frac{(2−2y)}{\sqrt{(1−y)^2}}  = 2

In this question, it is given that y is an integer and y = k+|k|. We need to find out whether y = 0.

First statement :  \sqrt{(-k)}=3  
 This statement implies that k < 0 as -k is a perfect square and square root of -k is equal to 3. Thus, k must be equal to -9 and -(-9) = 9  
 From first statement, we know that  k = - 9  .  Substituting the value of equation k in equation y = k+|k|, we get,  y = -9 + 9 = 0   
 Therefore  y = 0 . Therefore, first statement is sufficient.

Second statement :  \frac{(2−2y)}{\sqrt{(1−y)^2}}  = 2

LHS =  \frac{(2−2y)}{\sqrt{(1−y)^2}}  =  \frac{2(1-y)}{\sqrt{(1−y)^2}}  
 
 Since RHS = 2,  (1-y)  in the numerator would cancel out the  \sqrt{(1−y)^2}  in the denominator.  
 Clearly,  \sqrt{(1−y)^2}  =  1 - y . Then,  1 - y > 0  or  y < 1 . 
 It is given that y = k+|k| and we know that y < 1 and y is an integer. Thus, either y is 0 or y is a negative integer.  
 But, y or k+|k| < 1 only when  k < 0 . Thus,  k+|k| = 0  and  y = 0 . 
 Second statement also is sufficient.

Option D  is the correct answer as either of the statements alone are sufficient to answer the question.

ajaymahadev · Answer

This question is pretty straight forward.

bM22 · Answer

Solution:

Statement A. √(-k)=3

-k =9 => k=-9, Since, y = k+ |k| => y = -9+9 = 0.  Sufficient .

B) (2−2y)/\sqrt{(1−y)^2}=2

implies: (1-y)/((1-y)^2) = 1 => 1-y =1, => y=0.  Sufficient .

Answer D

If y is an integer, and y

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