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If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y 01 Apr 2020, 02:30
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62% (01:53) correct 38% (01:22) wrong based on 253 sessions

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If \(x\) and \(y\) are positive integers, is \(x\) even ?


(1) \(x^{2} + y^{2} = 98\)

(2) \(x = y\)


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If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y 01 Apr 2020, 02:38
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Bunuel
If \(x\) and \(y\) are positive integers, is \(x\) even ?


(1) \(x^{2} + y^{2} = 98\)

(2) \(x = y\)


Project DS Butler Data Sufficiency (DS3)


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Given: x and y are positive and Integers

Question: Is x even?

statement 1: \(x^{2} + y^{2} = 98\)

Perfect square \(x^2\). and \(y^2\) have to be one of {1, 4, 9, 16, 25, 36, 49, 64, 81}

i.e. \(x^2 = y^2 = 49\)

i.e. x = y = 7 i.e. answer to the question is NO

SUFFICIENT

Statement 2: x = y

Both x and y may be 2 (even) or both may be 7 (odd) hence

NOT SUFFICIENT

Answer: Option A
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If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y 01 Apr 2020, 02:45
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Bunuel
If \(x\) and \(y\) are positive integers, is \(x\) even ?

(1) \(x^{2} + y^{2} = 98\)
(2) \(x = y\)
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(1) \(x^{2} + y^{2} = 98\)
--> Only Possible value of (\(x\), \(y\)) = (\(7\), \(7\))
--> '\(x\)' is definitely not even --> Sufficient

(2) \(x = y\)
--> '\(x\)' can take infinite values (Even or Odd) --> Insufficient

Option A
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If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y 01 Apr 2020, 10:06
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Bunuel
If \(x\) and \(y\) are positive integers, is \(x\) even ?


(1) \(x^{2} + y^{2} = 98\)

(2) \(x = y\)


Project DS Butler Data Sufficiency (DS3)


For DS butler Questions Click Here

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

If x is an even integer, then we have \(x = 2k\) for an integer \(k\).
We have \(y^2 = 98 - x^2 = 98 - 4k^2 = 4(24-k^2) + 2\) from condition 1).
A square of an integer \(y^2\) can't have a remainder \(2\) when it is divided by \(4\) for the following reasoning.
Thus, \(x\) can not be even.

If \(y\) is an odd integer, \(y = 2a + 1\), then \(y^2 = (2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2+a) + 1\) and \(y^2\) has a remainder \(1\).
If \(y\) is an even integer, \(y = 2a\), then \(y^2 = (2a)^2 = 4a^2 = 4a^2 + 0\) and \(y^2\) has a remainder \(0\).

Condition 1) yields a unique answer 'no'.

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)

Since condition 2) does not yield a unique solution obviously, it is not sufficient.

Therefore, A is the answer.
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If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y 09 Apr 2021, 12:55
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If \(x\) and \(y\) are positive integers, is \(x\) even ?


(1) \(x^{2} + y^{2} = 98\)
This combination is possible only when x & y = 7
Sufficient

(2) \(x = y\)
Not sufficient

IMO A
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