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Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K

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Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 04:36
1
1
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

55% (01:31) correct 45% (01:25) wrong based on 123 sessions

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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:01
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!
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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:22
Bunuel wrote:
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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:24
DavidTutorexamPAL wrote:
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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:30
Archit3110 wrote:
DavidTutorexamPAL wrote:
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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:33
DavidTutorexamPAL wrote:
Archit3110 wrote:
DavidTutorexamPAL wrote:
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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:42
Archit3110 wrote:
[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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 06 Dec 2018, 05:46
DavidTutorexamPAL wrote:
Archit3110 wrote:
[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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 24 Dec 2018, 00:18
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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K  [#permalink]

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New post 24 Dec 2018, 16:56
(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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Re: Is K an even number ? (1) |K - |K|| = K + |K| (2) K^(1/2) = K &nbs [#permalink] 24 Dec 2018, 16:56
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