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07 Mar 2020, 01:33
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A
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If x, y, z are integers and xy+z is an odd integer then is x even?
1. xy+xz is even
2. y+xz is odd
Please help me with this question.
Posted from my mobile device
1. xy+xz is even
2. y+xz is odd
Please help me with this question.
Posted from my mobile device
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07 Mar 2020, 01:59
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Darknight5
xy+z is an odd integer
i.e.
Case 1: x = even, y = odd, z = odd
Case 2: x = Odd, y = even, z = odd
Case 3: x = even, y = even, z = odd
Case 4: x = Odd, y = Odd, z = even
Question: Is x = even?
Statement 1: xy+xz is even
x(y+z) = even
case 1 and Case 3 satisfy this statement
Both result in x = even hence
SUFFICIENT
Statement 2: y+xz = odd
Case 1 and case 2 both satisfy this statement but
Case 1 says x=even while case2 says x=odd hence
NOT SUFFICIENT
Answer: Option A
07 Mar 2020, 02:50
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I have got you 
If x, y, z are integers and xy+z is an odd integer then is x even? Yes/No
Constraint: xy+z=odd int.
1. xy+xz is even
x(y+z) = even
Now case 1: even •even=even
Case 2: even•odd=even
Case 3: odd•even=even
Make sure these cases make sense with the constraint in the question to know wether x is consistently even or otherwise
Case1: when x=even ,y=even ,z=even
Constraint: xy+z =odd —> even(even)+even =even (ruled out!)
Case1: when x=even ,y&z=odd
Constraint: xy+z=odd —> even(odd)+odd=odd (Yeah! ). .: x=even
Case2:when x=even,y=even ,z=odd
Constraint: even(even)+odd=odd (Yeah!) .: x=even
When x=even,y=odd, z=even
Constraint: even(odd)+even=even (ruled out!)
Case3: when x=odd,y=odd,z=odd
Constraint:odd(odd)+odd=even
Or odd(even)+even=even (ruled out since doesn’t follow Constraint)
.: x=EVEN (Sufficient)
2. y+xz is odd
Case1: odd+even =odd
Case2: even+odd=odd
So case1 xz =even —> x=even/odd
z=even/odd, y=odd
Constraint: even(odd)+even=even (ruled out!)
Or even(odd) +odd =odd (Yeah!)
.: x= even
Or odd(odd) +even =odd (Yeah!)
.: x=odd
Case 2 here y=even ,xz =odd
Constraint: odd(even) +odd =odd
Here x=odd
(Not sufficient)
So A like America
Hope you got it
Posted from my mobile device
If x, y, z are integers and xy+z is an odd integer then is x even? Yes/No
Constraint: xy+z=odd int.
1. xy+xz is even
x(y+z) = even
Now case 1: even •even=even
Case 2: even•odd=even
Case 3: odd•even=even
Make sure these cases make sense with the constraint in the question to know wether x is consistently even or otherwise
Case1: when x=even ,y=even ,z=even
Constraint: xy+z =odd —> even(even)+even =even (ruled out!)
Case1: when x=even ,y&z=odd
Constraint: xy+z=odd —> even(odd)+odd=odd (Yeah! ). .: x=even
Case2:when x=even,y=even ,z=odd
Constraint: even(even)+odd=odd (Yeah!) .: x=even
When x=even,y=odd, z=even
Constraint: even(odd)+even=even (ruled out!)
Case3: when x=odd,y=odd,z=odd
Constraint:odd(odd)+odd=even
Or odd(even)+even=even (ruled out since doesn’t follow Constraint)
.: x=EVEN (Sufficient)
2. y+xz is odd
Case1: odd+even =odd
Case2: even+odd=odd
So case1 xz =even —> x=even/odd
z=even/odd, y=odd
Constraint: even(odd)+even=even (ruled out!)
Or even(odd) +odd =odd (Yeah!)
.: x= even
Or odd(odd) +even =odd (Yeah!)
.: x=odd
Case 2 here y=even ,xz =odd
Constraint: odd(even) +odd =odd
Here x=odd
(Not sufficient)
So A like America
Hope you got it
Posted from my mobile device
