Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 27 Oct 2017
Posts: 1787
WE: General Management (Education)

If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 18:35
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) \(\frac{(22y)}{\sqrt{(1y)^2 }}\)=2
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Retired Moderator
Joined: 27 Oct 2017
Posts: 1787
WE: General Management (Education)

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 18:36
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) \(\frac{(22y)}{\sqrt{(1y)^2 }}\)=2 As\( \sqrt{(1y)^2 }\) = 1y so, the equation becomes \(\frac{(22y)}{1y}\)=2 hence , 1y= 1y or 1y> or = 0 but since 1y term is coming in denominator, it cannot be zero. So, 1y>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
_________________



Director
Joined: 14 Dec 2019
Posts: 659
Location: Poland
WE: Engineering (Consumer Electronics)

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 18:52
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 1y = 1y
if y = 2 => 1+2 = 1+2 = 3 if y = 0 => 1 =1
Hence y can have multiple values => Not Sufficient
Answer  A



Manager
Joined: 16 Sep 2011
Posts: 128

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 19:07
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 (1y)^2 = 2 (1 y) /sqroot (1y)^2 = 1 (1y) = sqroot (1y)^2 here y can be positive as well as negative
A is the answer



Manager
Joined: 04 Jun 2019
Posts: 79
Location: United States

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 19:08
(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
(22y)=2(√(1−y)^2) divide by 2 => 1y=√(1−y)^2 => (1y)^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
Joined: 22 Feb 2018
Posts: 741

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 19:19
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 (1y)/√(1−y)^2 =2, (1y)/√(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.



Senior Manager
Joined: 31 May 2018
Posts: 432
Location: United States
Concentration: Finance, Marketing

Re: If y is an integer, and y
[#permalink]
Show Tags
Updated on: 16 Feb 2020, 02:00
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) \(\frac{(22y)}{\sqrt{(1y)^2}}\) = 2
\(\frac{(1y)}{(1y)}\) = 1 from here y not equal to 1 (since euation will be undefined)
since (1y) is positive (1y) will have to be > 0 this implies y<1y 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
D is the answer
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
Joined: 28 Jul 2016
Posts: 1019
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)

Re: If y is an integer, and y
[#permalink]
Show Tags
15 Feb 2020, 21:17
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
Joined: 25 Mar 2013
Posts: 11

Re: If y is an integer, and y
[#permalink]
Show Tags
16 Feb 2020, 04:11
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 = \(\frac{(2−2y)}{\sqrt{(1−y)^2}}\) = \(\frac{2(1y)}{\sqrt{(1−y)^2}}\) Since RHS = 2, \((1y)\) 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
Joined: 13 Nov 2019
Posts: 33
Location: India
Concentration: International Business, Marketing
GPA: 4

Re: If y is an integer, and y
[#permalink]
Show Tags
16 Feb 2020, 05:16
This question is pretty straight forward.
Attachments
New Doc 20200216 08.09.13_8.jpg [ 151.4 KiB  Viewed 339 times ]



Manager
Joined: 05 May 2016
Posts: 142
Location: India

Re: If y is an integer, and y
[#permalink]
Show Tags
16 Feb 2020, 06:12
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: (1y)/((1y)^2) = 1 => 1y =1, => y=0. Sufficient.
Answer D




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




