sana.noor wrote:If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd
(2) (y+2)!/x! is greater than 2
OA is C
Statement 1: (y+2)!/x! = odd
Let (y+2)!/x! = 1, implying that y+2 = x.
If x=3, then y=1.
Result: y is ODD.
If x=4, then y=2.
Result: y is EVEN.
INSUFFICIENT.
Statement 2: (y+2)!/x! is greater than 2
If (y+2)!/x! = 3, then 3 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 3*2*1.
x! = 2*1.
In this case, y+2 = 3, so y=1.
Result: y is ODD.
If (y+2)!/x! = 4, then 4 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 4*3*2*1
x! = 3*2*1.
In this case, y+2 = 4, so y=2.
Result: y is EVEN.
INSUFFICIENT.
Statements combined:
If (y+2)!/x! = 3, then y is ODD, as shown above.
If (y+2)!/x! = 5, then 5 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 5*4*3*2*1
x! = 4*3*2*1.
In this case, y+2 = 5, so y=3.
Result: y is ODD.
If (y+2)!/x! = 7, then 7 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 7*6*5*4*3*2*1
x! = 6*5*4*3*2*1.
In this case, y+2 = 7, so y=5.
Result: y is ODD.
The cases above illustrate that y must be ODD.
SUFFICIENT.
The correct answer is
C.
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