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14 Jun 2018, 04:49
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
41% (02:09) correct
59%
(01:48)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
Is x an integer?
(1) \(x^2=2y\), where y is an integer.
(2) x*y is an even number.
New question
(1) \(x^2=2y\), where y is an integer.
(2) x*y is an even number.
New question
14 Jun 2018, 09:17
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statement 1
x^2=2y
x=sqrt(2*integer) is not definitely an integer not suff.
statement 2
even number means that it is an integer, only integers have +-ve nature but anyway it's not suff
1 case: x=1 y=2
2 case x=1.25 y=4
statements 1&2
We can't get more information as above
P.S I'm very cautious about Chetan's questions, traps everywhere
x^2=2y
x=sqrt(2*integer) is not definitely an integer not suff.
statement 2
even number means that it is an integer, only integers have +-ve nature but anyway it's not suff
1 case: x=1 y=2
2 case x=1.25 y=4
statements 1&2
We can't get more information as above
P.S I'm very cautious about Chetan's questions, traps everywhere
14 Jun 2018, 09:27
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Bookmarks
To find: Is x an integer?
Statement 1:
\(X^2 = 2y\) and y is an integer.
y = 2 then x = +2 or -2. but yes an integer.
y = 4 then x = + or -\(2\sqrt{2}\). x is not an integer.
Statement 1 is insufficient.
Statement 2:
x*y is an even number.
x = 1/2 and y = 24. It satisfies even number condition but x is not an integer.
x = 2 and y = 4. It satisfies even number condition but x is an integer.
Statement 2 is insufficient.
From Statement 1 and Statement 2:
From \(X^2 = 2y\) we can say, \(X = \sqrt{2} * \sqrt{y}\)
And as, x*y is an even number. This implies x * y is even integer ( as even number cannot be a fraction ).
and we already know that y is an integer ( From statemnt 1 ).
We can rewrite x * y as \(\sqrt{2} * \sqrt{y} * y\)= even integer.
\(\sqrt{2y} * y\) = even integer.
\(\sqrt{2y}\) this should result in integer then only as a whole x * y will result as evn number.
This implies x should be an integer.
The correct choice is C.
Statement 1:
\(X^2 = 2y\) and y is an integer.
y = 2 then x = +2 or -2. but yes an integer.
y = 4 then x = + or -\(2\sqrt{2}\). x is not an integer.
Statement 1 is insufficient.
Statement 2:
x*y is an even number.
x = 1/2 and y = 24. It satisfies even number condition but x is not an integer.
x = 2 and y = 4. It satisfies even number condition but x is an integer.
Statement 2 is insufficient.
From Statement 1 and Statement 2:
From \(X^2 = 2y\) we can say, \(X = \sqrt{2} * \sqrt{y}\)
And as, x*y is an even number. This implies x * y is even integer ( as even number cannot be a fraction ).
and we already know that y is an integer ( From statemnt 1 ).
We can rewrite x * y as \(\sqrt{2} * \sqrt{y} * y\)= even integer.
\(\sqrt{2y} * y\) = even integer.
\(\sqrt{2y}\) this should result in integer then only as a whole x * y will result as evn number.
This implies x should be an integer.
The correct choice is C.
