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23 May 2020, 04:10
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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For all integers a and b, the operation θ is defined by \(aθb = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?
(1) \(x^2 = y^2\)
(2) \(x = –y\)
DS20593
(1) \(x^2 = y^2\)
(2) \(x = –y\)
DS20593
23 May 2020, 05:58
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Bunuel
For all integers a and b, the operation θ is defined by \(θ = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?
(1) \(x^2 = y^2\)
(2) \(x = –y\)
PS20593
(1) \(x^2 = y^2\)
(2) \(x = –y\)
PS20593
xθy = (x + y)/2, which will be an integer as long as x + y is even.
If x^2 = y^2, then either x = y, in which case x + y = x + x = 2x, which is even, or -x = y, in which case x + y = x - x = 0, which is again even. So Statement 1 guarantees that x + y is even and is sufficient.
Statement 2 is a special case of Statement 1; it tells us that xθy = 0, so is also sufficient, and the answer is D.
I think there's a typo in the question, though -- the operation is not defined properly, as written. It should say:
"the operation θ is defined by \(aθb = \frac{a + b}{2}\)"
25 May 2020, 23:48
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For all integers a and b, the operation θ is defined by aθb=a+b2aθb=a+b2. If x and y are integers, is xθyxθy an integer?
(1) x^2=y^2
(2) x=–y
The average of 2 numbers will be an integer if only the sum of the two numbers is divisible by 2.
statement 1 : x^2 = y^2
x=2 y=2 (x+y)/2 =2=integer
x=-2 y=-2 (x+y)/2 =0= integer Suff
statement 2 :
(2) x=–y
its a subset of statement 1
Hence Suff
Answer D
(1) x^2=y^2
(2) x=–y
The average of 2 numbers will be an integer if only the sum of the two numbers is divisible by 2.
statement 1 : x^2 = y^2
x=2 y=2 (x+y)/2 =2=integer
x=-2 y=-2 (x+y)/2 =0= integer Suff
statement 2 :
(2) x=–y
its a subset of statement 1
Hence Suff
Answer D
