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# If y is an integer, and y

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Retired Moderator
Joined: 27 Oct 2017
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If y is an integer, and y  [#permalink]

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15 Feb 2020, 18:35
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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

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Re: If y is an integer, and y  [#permalink]

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

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Re: If y is an integer, and y  [#permalink]

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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 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

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Re: If y is an integer, and y  [#permalink]

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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 (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

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Re: If y is an integer, and y  [#permalink]

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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

(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
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Re: If y is an integer, and y  [#permalink]

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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 (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.

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Re: If y is an integer, and y  [#permalink]

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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) $$\frac{(2-2y)}{\sqrt{(1-y)^2}}$$ = 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.
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Re: If y is an integer, and y  [#permalink]

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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
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Re: If y is an integer, and y  [#permalink]

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

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.
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Re: If y is an integer, and y  [#permalink]

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16 Feb 2020, 05:16
This question is pretty straight forward.
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Re: If y is an integer, and y  [#permalink]

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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: (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