Is the positive integer x odd?

Question

Tough and Tricky questions: properties of numbers.

(1) x = y^2 + 4y + 6, where y is a positive integer.
(2) x = 9z^2 + 7z - 10, where z is a positive integer.

WoundedTiger · Accepted Answer

St 1: x = y^2 + 4y + 6, where y is positive integer
Can be written as x=y^2+4y+4+2 or x=(y+2)^2+2
Now if y=even than x=even
but if y=odd then x=odd..

we have 2 cases so Option A and D ruled out

St 2: x = 9z^2 + 7z - 10

If z=even then we have Even (9z^2)+Even(7z)- Even(10)=Even
if z=Odd, then we have Odd+Odd-Even =Even..
Clearly x is even whether z is odd or even..

Ans is B

desiwolverine · Answer

1) Y^2 + 4Y + 6 = O + E + E = O if Y = O
 = E + E + E = E if Y = E 
Not suff

2) 9Z^2 + 7Z - 10 = E + E - E = E if Z = O
 = O + O - E = E if Z = E
Suff

Hence B.

rhine29388 · Answer

Statement 1 states that
x =  y^2+4y+6  where y is a positive integer, check for both y as a even number and a odd number
We find that x can be odd/even depending on value of y so insufficient

Statement 2 states that
x =  9z^2+7z-10  where z is a positive integer, check for both z as a even number and a odd number
We find that x is always even which gives us a definite NO as the answer to given question

So correct answer - B

stonecold · Answer

Tremendous Question
Here we are told that x is an integer and we need to get if its odd or not
Lets look at statements 
Statement 1
Here if y is even => x is even
If y is odd => x is odd 
No clue of y is given 
hence not sufficient 
Statement 2
Here irrespective of z being even/odd => 9z^+7z will always be even
hence x=even+even => Even
Hence Sufficient to say that x is not odd

Hence B

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Is the positive integer x odd?

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Is the positive integer x odd?

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