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1. xy + xz is an even integer - SUFFICIENT Given: xy + z is odd ...(i) xy + xz is even ...(ii)

subtracting (ii) from (i) we get xz - z, which should be odd (* since odd - even = odd) => z(x-1) is odd => both z and (x-1) is odd => since (x-1) is odd, x must be even.

2. y + xz is an odd integer -INSUFFICIENT Given: xy + z is odd ...(i) y + xz is odd ...(ii)

subtracting (ii) from (i) we get xy + z - y - xz = (x-1)(y-z) , which should be even => either (x-1) is even or (y-z) is even ....insufficient to determine
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You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.
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Re: GMAT Prep...How much time did u take to solve this one ?? [#permalink]

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27 Sep 2009, 08:58

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IanStewart wrote:

You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.

Fr St2, y+xz odd xy+z odd => y+zx+xy+z even => y(x+1)+z(x+1) even => (y+z)(x+1) event x+1 can be odd or even means that x can be even or odd, insuff

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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19 Mar 2014, 18:41

Given condition: xy + z = odd implies either xy = odd (x =odd and y = odd) and z = even or xy = even (x or y can be odd and even respectively and vice versa) and z = odd

condition 1:

xy + xz = even; Implies x(y+z) = even which again implies the following:

i) x even and y+z = odd - where again y or z can be odd and even respectively and vice versa ii) x odd and y +z = even - where again y and z has to be both odd or both even

inconclusive

condition 2:

y + xz = odd

again inconclusive 1 + 2: Add xy + z + y + xz = odd + odd implies: (x + 1)(y+z) = even and x (y+z) is also even according to 2.. so y + z = even <y and z both even or y + z both odd>, x can be odd or even but by 1 xy + z = odd which means y and z both odd, so x is even.

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer (2) y + xz is an odd integer

Odd/Even questions can be usually solved quite easily if one tries some operations with the statements

We want to know if x is even integer

We are given that xy+z is odd

Statement 1

xq + xz is even

Subtracting

z(x+1) is odd

Therefore, x+1 should be odd and x should be even

Sufficient

Statement 2

Not sufficient

Answer is A

Just my 2c

Cheers J

I am not clear with the red part.

When you subtract \(xy + z=odd\) from \(xy+xz=even\) you'll get: \(xz-z=even-odd=odd\) --> \(z(x-1)=odd\). For the product of two integers to be odd, both of them must be odd --> \(z\) and \(x-1\) are odd. If \(x-1=odd\), then x must be even: \(x-1=x-odd=odd\) --> \(x=odd+odd=even\).

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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21 Mar 2014, 19:46

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Odd(O) Even (E) given: x,y,z integers xy+z=O so only the following scenarios can fulfill the constraints a) EO+O b) EE+O c) OE+O d) OO+E

question: x=E?

1) x(y+z)=E i. (E)(O+O) --> fits scenario a -->yes, x can be even ii. (O)(E+E) --> n/a - doesn't fit any scenarios iii. (O)(O+O) --> n/a - doesn't fit any scenarios

stop testing, x can't be odd, sufficient

2) y+xz = O i. E+(O)(O) --> fits scenario a -->yes, x can be even ii. O+(E)(E) --> n/a - doesn't fit any scenarios iii. O+(O)(E) --> fits scenario d -->no, x can be odd

If x, y, z are integers and xy+z is an odd integer, is x an even integer.

1. xy +xz is an even integer

2. y+ xz is an odd integer

Statement 1

xy+xz is even xy+z is odd

Subtract the two. xy+xz-xy-z = xz-z

xz-z will be odd (even - odd will always result in odd)

z(x-1) will be odd

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

Sufficient.

Statement 2

y+xz is odd xy+z is odd

Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z).

(x+1)(y+z) will be even (odd+odd is even)

Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result) or x+1 can be odd and y+z can be even (Even * odd will result in an even result) or both (x+1) and (y+z) can be even (Even * even will result in an even result)

So x+1 can be both odd and even. In other words, x can be both even and odd.

This is only possible when both z and (x-1) are odd.

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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31 Jan 2017, 18:03

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(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that z-xz = z(1-x) is odd. Thus 1 - x is odd, so x must be even. SUFF

(2) SUbtracting, we see that xy - y + z - xz = y(x - 1) - z(x - 1) =(y - z)(x - 1) is even. If y - z is odd, x - 1 is even and thus x is odd. On the other hand, if y - z is even, x is even. NOT SUFF

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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14 Mar 2017, 23:52

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer ->z(x-1) = odd so X= even

(2) y + xz is an odd integer ->(Y+Z)(X+1)= even,{given xy + z=odd,either XY =odd and Z =even,satisfies in statement 2 so X can be odd} { Now XY= even and Z=odd, either X/Y =even or X&Y =even,As both wont satisfy in statement 2 so X cannot be even} Hence i got answer D

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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15 Mar 2017, 00:07

sreenu7464 wrote:

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer ->z(x-1) = odd so X= even

(2) y + xz is an odd integer ->(Y+Z)(X+1)= even,{given xy + z=odd,either XY =odd and Z =even,satisfies in statement 2 so X can be odd} { Now XY= even and Z=odd, either X/Y =even or X&Y =even,As both wont satisfy in statement 2 so X cannot be even} Hence i got answer D

In (2), I see no problem if x is even, y is odd and z is odd so xy+z is odd and y+xz is odd.
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Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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26 May 2017, 08:25

Solved this problem by plugging numbers: Given: xy + z = odd integer

Now, plug numbers that satisfy the above condition: [ I plugged numbers by considering the scenarios : (i) xy = even and z = odd (ii) xy = odd and z = even]

X Y Z 2 1 3 1 2 3 2 4 3 3 1 2 1 3 2 3 5 2

Use the above values in given conditions to check whether X can be odd or even:

Statement 1: xy+xz = even => x(y+z) = even substitute values from above table and identify the values that satisfy statement 1 You will get : x = 2 and corresponding (y,z) as (1,3) and (4,3). Rest of cases does not satisfy statement 1. Hence we can say that X is even and cannot be odd SUFFICIENT

Statement 2: y+xz = odd substitue values in statement 2 to identify values that satisfy it 1 + 2*3 = 7 (x=2) 2 + 1*3 = 5 (x=1) This means x can be both odd and even. Hence INSUFFICIENT

Re: If x, y and z are integers and xy + z is an odd integer, is [#permalink]

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25 Dec 2017, 08:58

Xy+z = 5 Xy+xz = 2 by subtracting Z – xz = 3 or z( 1-x) = 3 , so 1-x = odd, therefore x is even. Sufficient xy+z = 5 y+xz =3 by adding y(x+1) + z(x+1) = 8 or (x+1)(y+z) = 8, here x+1 can be odd or even and hence x can be even or odd. NS

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer (2) y + xz is an odd integer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x and y) and 1 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

Conditions 1) & 2) xy + xz : even ⇒ xy:even, xz=even | xy:odd, xz:odd y + xz : odd ⇒ y: odd, xz=even | y: even, xz: odd xy + z: odd ⇒ xy: even, z:odd | xy:odd, z:even ⇒ x:even, y:odd, z:odd Both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1) xy + xz : even ⇒ xy:even, xz=even | xy:odd, xz:odd xy + z: odd ⇒ xy: even, z:odd | xy:odd, z:even ⇒ x:even, y:?, z:odd The condition 1) is sufficient since we can determine x is even.

Condition 2) y + xz : odd ⇒ y: odd, xz=even | y: even, xz: odd xy + z: odd ⇒ xy:odd, z:even | xy: even, z:odd x = 2, y = 1, z = 1 | x = 1, y = 2, z = 1 The condition 2) is not sufficient.

Therefore, A is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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