GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 May 2020, 15:26 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If y is an integer, and y

Author Message
TAGS:

### Hide Tags

Retired Moderator V
Joined: 27 Oct 2017
Posts: 1787
WE: General Management (Education)
If y is an integer, and y  [#permalink]

### Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 40% (01:39) correct 60% (01:46) wrong based on 40 sessions

### HideShow timer Statistics

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

_________________
Retired Moderator V
Joined: 27 Oct 2017
Posts: 1787
WE: General Management (Education)
Re: If y is an integer, and y  [#permalink]

### Show Tags

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

_________________
Director  P
Joined: 14 Dec 2019
Posts: 659
Location: Poland
GMAT 1: 570 Q41 V27
WE: Engineering (Consumer Electronics)
Re: If y is an integer, and y  [#permalink]

### Show Tags

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

Manager  S
Joined: 16 Sep 2011
Posts: 128
Re: If y is an integer, and y  [#permalink]

### Show Tags

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

Manager  G
Joined: 04 Jun 2019
Posts: 79
Location: United States
Re: If y is an integer, and y  [#permalink]

### Show Tags

(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
Director  V
Joined: 22 Feb 2018
Posts: 741
Re: If y is an integer, and y  [#permalink]

### Show Tags

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.

Senior Manager  P
Joined: 31 May 2018
Posts: 432
Location: United States
Concentration: Finance, Marketing
Re: If y is an integer, and y  [#permalink]

### Show Tags

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

A) √(-k)=3
from this statement, we get K = -9
y = k+|k| = -9 +|-9| = -9+9 = 0
is y = 0?---YES
so, statement A is SUFFICIENT

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

$$\frac{(1-y)}{|(1-y)|}$$ = 1
from here y not equal to 1 (since euation will be undefined)

since |(1-y)| is positive
(1-y) will have to be > 0
this implies y<1---y can be = 0,-1,-2,-3....... (y integer)

since y is of form = k+|k|--(y is integer)
if k>0 y = 2k (but +ve value of y is not possible)
if k<0 y = 0
if k = 0 y = 0

from here we know only possible value of y = 0
so, SUFFICIENT

Originally posted by shridhar786 on 15 Feb 2020, 20:43.
Last edited by shridhar786 on 16 Feb 2020, 02:00, edited 1 time in total.
VP  V
Joined: 28 Jul 2016
Posts: 1019
Location: India
Concentration: Finance, Human Resources
Schools: ISB '18 (D)
GPA: 3.97
WE: Project Management (Investment Banking)
Re: If y is an integer, and y  [#permalink]

### Show Tags

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
_________________

Keep it simple. Keep it blank
Intern  B
Joined: 25 Mar 2013
Posts: 11
Re: If y is an integer, and y  [#permalink]

### Show Tags

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.
Intern  S
Joined: 13 Nov 2019
Posts: 33
Location: India
GPA: 4
Re: If y is an integer, and y  [#permalink]

### Show Tags

This question is pretty straight forward.
Attachments New Doc 2020-02-16 08.09.13_8.jpg [ 151.4 KiB | Viewed 339 times ]

Manager  S
Joined: 05 May 2016
Posts: 142
Location: India
Re: If y is an integer, and y  [#permalink]

### Show Tags

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. Re: If y is an integer, and y   [#permalink] 16 Feb 2020, 06:12

# If y is an integer, and y  