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Difficulty:
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Question Stats:
74% (01:22) correct
26%
(01:09)
wrong
based on 126
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If x and y are integers, is the sum of x and y even?
(1) \(x^2\) = \(y^2\)
(2) \(x^3\) = \(y^3\)
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(1) \(x^2\) = \(y^2\)
(2) \(x^3\) = \(y^3\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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find out x+y= even or not.
1. x^2=y^2 , by rearranging, we can write this as (x+y)(x-y)=0; x+y=0 and 0 is even number so sufficient.
2. x^3-y^3 =0 by rearranging we do not about x+y so not sufficient.
Hence answer is A.
1. x^2=y^2 , by rearranging, we can write this as (x+y)(x-y)=0; x+y=0 and 0 is even number so sufficient.
2. x^3-y^3 =0 by rearranging we do not about x+y so not sufficient.
Hence answer is A.
Posts: 1,776
Updated on: 14 Apr 2016, 20:14
Originally posted by OptimusPrepJanielle on 14 Apr 2016, 01:09.
Last edited by OptimusPrepJanielle on 14 Apr 2016, 20:14, edited 1 time in total.
Last edited by OptimusPrepJanielle on 14 Apr 2016, 20:14, edited 1 time in total.
Bunuel
x + y = even?
Statement 1: x^2 = y^2
Case 1:
Either x = y (x = 2 and y = 2)
or x = -y (x = 2 and y = -2)
Hence x + y = 2x or 0
Yes, the sum is even
Sufficient
Statement 2: x^3 = y^3
x = y (x = 2 and y = 2 OR x = -2 and y = -2)
Therefore x + y = 2x
Yes, the sum is even.
Sufficient
Option D
Posts: 1,776
lipsi18
Hi lipsi18,
In the statement 1, either of (x+y) or (x-y) can be zero.
Hence we cannot say that x + y = 0
subhamoycomputer
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Is it B ?
If so, will provide my line of reasoning.
If so, will provide my line of reasoning.
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subhamoycomputer
Hi subhamoycomputer,
The correct option here should be D.
However do post your reasoning as to why you thing option A is insufficient.
If x and y are integers, is the sum of x and y even?
(1) x^2 = y^2
(2) x^3 = y^3
x + y even?
Statement 1:
x^2 = y^2. This implies x = y or x =-y.
Either way x+y will be even (0 or 2x).
Statement 2:
X^3 = y^3.
Implies x = y.
So x + y is 2x, which is even.
So each statement is individually sufficient.
Answer D.
(1) x^2 = y^2
(2) x^3 = y^3
x + y even?
Statement 1:
x^2 = y^2. This implies x = y or x =-y.
Either way x+y will be even (0 or 2x).
Statement 2:
X^3 = y^3.
Implies x = y.
So x + y is 2x, which is even.
So each statement is individually sufficient.
Answer D.
Here in both the cases we can say that both x and y must have the same odd/even nature
hence x+y=even
hence D
hence x+y=even
hence D
GMAT 1: 760 Q51 V42
Posts: 10,057
If x and y are integers, is the sum of x and y even?
(1) x^2 = y^2
(2) x^3 = y^3
There are 2 variables (x and y) in the original condition. In order to match the number of variables to the number of equations, we need 2 more equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer. Using both the condition 1) and the condition 2), we get 1)=2), from which we can obtain x=y=odd or x=y=even. The answer is always yes and the conditions are sufficient. Therefore, the correct answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
(1) x^2 = y^2
(2) x^3 = y^3
There are 2 variables (x and y) in the original condition. In order to match the number of variables to the number of equations, we need 2 more equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer. Using both the condition 1) and the condition 2), we get 1)=2), from which we can obtain x=y=odd or x=y=even. The answer is always yes and the conditions are sufficient. Therefore, the correct answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
Balajikarthick1990
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Bunuel
The Answer is D..Check for pairs (0,0) ...(1,1)....(-1,-1)
Balajikarthick1990
You are correct...
But for (1), check for |x|=|y|.. May be important in some other Q
x=1, y=-1 ; x=-1, y=1; x=1,y=1; x=-1, y=-1
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Let me understand, is 0 considered even?
Because if we use x = -y
Then the answer would be 0 for option A
Because if we use x = -y
Then the answer would be 0 for option A
douglasvg
hi,
0 is EVEN but it is neither NEGATIVE nor POSITIVE...
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Bunuel gmatbusters VeritasKarishma chetan2u
Could you share your approach?
Q is asking: Are both x and y even or odd?
For St 1, I took |x| = |y|
which is same as:
x = y and x = -y
or
x-y = 0
or
x + y = 0
Hence implying, both x and y have to be even/odd.
What is St 2 saying? How do I interpret odd powers of integer?
Could you share your approach?
Q is asking: Are both x and y even or odd?
For St 1, I took |x| = |y|
which is same as:
x = y and x = -y
or
x-y = 0
or
x + y = 0
Hence implying, both x and y have to be even/odd.
What is St 2 saying? How do I interpret odd powers of integer?
Posts: 17
1) x²=y²
So, x²-y²=0
So, (x+y)(x-y)=0
Either x+y=0 where 0 is even.
Or x-y=0 where 0 is even. If the difference of two integers is even, their sum is also even.
So statement (1) is sufficient.
(2) Cube of a number preserves the sign.
So x³=y³ means x=y, say x=y=a
So the sum will always be even =2a.
So statement (2) is sufficient.
Both are statements alone are sufficient so option D is the correct answer
Posted from my mobile device
So, x²-y²=0
So, (x+y)(x-y)=0
Either x+y=0 where 0 is even.
Or x-y=0 where 0 is even. If the difference of two integers is even, their sum is also even.
So statement (1) is sufficient.
(2) Cube of a number preserves the sign.
So x³=y³ means x=y, say x=y=a
So the sum will always be even =2a.
So statement (2) is sufficient.
Both are statements alone are sufficient so option D is the correct answer
Posted from my mobile device
Posts: 1,925
if \(x^ a = y ^a\), where a is an even integer
we get x = +/- y
But, if \(x^ b = y ^b\), where b is an odd integer
we get x = y
we get x = +/- y
But, if \(x^ b = y ^b\), where b is an odd integer
we get x = y
Good news, odd power are even easier to handle
Hence for Statement 2, sum of x and y = 2x or 2y = even
adkikani
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