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Bunuel
Is K an even number ?


(1) |K - |K|| = K + |K|

(2) \(\sqrt K=K\)


from 1 : given relation stands true only at K = 0 so sufficient

from 2 sqrtk= sqrt k* sqrtk
or 1=sqrtk
we get k =+/-1 so in sufficient IMO A is correct.
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The answer is A.
1) This is true only for 0.
(any positive number will give 0 on the left side and a positive number on the right. and negative number will give a positive number on the left and 0 on the right).
Sufficient!
2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient!

@DavidTutorexamPAL

for stament 2:
sqrtk = k is given
and we can say
sqrtk= sqrtk * sqrtk

OR
1=sqrtk
or k= +/-1 .. which is in sufficient..
I am not sure why have you said that "2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient![/quote]"
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DavidTutorexamPAL
The answer is A.
1) This is true only for 0.
(any positive number will give 0 on the left side and a positive number on the right. and negative number will give a positive number on the left and 0 on the right).
Sufficient!
2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient!

@DavidTutorexamPAL

for stament 2:
sqrtk = k is given
and we can say
sqrtk= sqrtk * sqrtk

OR
1=sqrtk
or k= +/-1 .. which is in sufficient..
I am not sure why have you said that "2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient!
"[/quote]

Not sure I understand the question...
The equation in 2) is indeed true for 1 and -1, as you mention, but it is also true for 0 (root0=0). Thus, k can be either even (0) or odd (1, -1), and this statement is insufficient.
Is this clearer?
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The answer is A.
1) This is true only for 0.
(any positive number will give 0 on the left side and a positive number on the right. and negative number will give a positive number on the left and 0 on the right).
Sufficient!
2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient!

@DavidTutorexamPAL

for stament 2:
sqrtk = k is given
and we can say
sqrtk= sqrtk * sqrtk

OR
1=sqrtk
or k= +/-1 .. which is in sufficient..
I am not sure why have you said that "2) this is true for both x=0 (an even number) and x=1 (an odd number). insufficient!
"

Not sure I understand the question...
The equation in 2) is indeed true for 1 and -1, as you mention, but it is also true for 0 (root0=0). Thus, k can be either even (0) or odd (1, -1), and this statement is insufficient.
Is this clearer?[/quote]

DavidTutorexamPAL:
Yes i did get that it would be true for both 0, +/-1 .. what I meant was that if you simplify stmt 2 you would get k = +/-1 only .. not 0 ..

In DS my approach is usually to simplify the given relation than plug in values..

anyways statement 2 is in sufficient ... thanks for replying..
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[quote=Yes i did get that it would be true for both 0, +/-1 .. what I meant was that if you simplify stmt 2 you would get k = +/-1 only .. not 0 ..

In DS my approach is usually to simplify the given relation than plug in values..

anyways statement 2 is in sufficient ... thanks for replying..

Ok, now I understand.
This is not true - and in fact, would cause a mistake!
Remember, when simplifying, we can never divide or multiply by a variable which may be 0. in this case, doing so makes us miss that x=0 is a possible solution as well. (which in turn should actually make us select D, since 1 and -1 are both odd.).
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Archit3110
[quote=Yes i did get that it would be true for both 0, +/-1 .. what I meant was that if you simplify stmt 2 you would get k = +/-1 only .. not 0 ..

In DS my approach is usually to simplify the given relation than plug in values..

anyways statement 2 is in sufficient ... thanks for replying..

Ok, now I understand.
This is not true - and in fact, would cause a mistake!
Remember, when simplifying, we can never divide or multiply by a variable which may be 0. in this case, doing so makes us miss that x=0 is a possible solution as well. (which in turn should actually make us select D, since 1 and -1 are both odd.).


DavidTutorexamPAL : OK I understood your point .. thanks
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Bunuel
Is K an even number ?


(1) |K - |K|| = K + |K|

(2) \(\sqrt K=K\)

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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(1) |K - |K|| = K + |K|

(2) √K‾‾=K

Simplifying (2), squaring both sides
k = k*k => k(k-1)=0 so k=1 or k =0 not sufficient

Simplifying(1)
for k>0
|k-k| = k+k => 0 = 2k => k=0

For k< 0
|k+k| = 0 => k=0 sufficient

hence (A)
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For B
K = 0 or 1 NS
For A ( i used 2 approaches )
|K - |K|| = K + |K|
if I remove this modulus i will get K -|K| = K+|K|
2|K|= 0 then K=0
or K -|K| = -(K+|K|)
2K=0
both equations are giving K= 0 then A is sufficient
I am not sure whether my approach for A is right because number plugging was quick approach
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Bunuel
Is K an even number ?


(1) |K - |K|| = K + |K|

(2) \(\sqrt K=K\)

Asked: Is K an even number ?


(1) |K - |K|| = K + |K|
K = 0
SUFFICIENT

(2) \(\sqrt K=K\)[
K = 0 or 1
NOT SUFFICIENT

IMO A
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