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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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Difficulty:   25% (medium)

Question Stats: 100% (01:02) correct 0% (00:00) wrong based on 13 sessions

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If x, y, and z are positive integers, is x+y divisible by 2?

1) x+z is divisible by 2

2) y+z is divisible by 2

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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Official Solution:

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, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):

There are two cases to consider.

Case 1: If x, y, and z are all even, then the answer is 'yes'

Case 2: If x, y, and z are all odd, then the answer is also 'yes'

Since no other cases are possible, both conditions are sufficient, when taken together.

Since this is an integer question (one of the key question areas), CMT

(Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)

Since it tells us nothing about y, condition 1) is not sufficient.

Condition 2)

Since it tells us nothing about y, condition 2) is not sufficient.

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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Can someone please Explain How both the statements taken together are sufficient since from the equations 1 and 2 we get they are either odd or even but we are not sure if they are even or odd , right??
Intern  B
Joined: 02 Jul 2018
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Anirudh134 wrote:
Can someone please Explain How both the statements taken together are sufficient since from the equations 1 and 2 we get they are either odd or even but we are not sure if they are even or odd , right??

From equation 1 & 2, we determine either x, y, z are all even OR are all odd. I think you got it till there.

In both the cases (all even or all odd), if you try calculating x+y/2, we can conclude that it is divisible by 2. Hence sufficient.
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Anirudh134 wrote:
Can someone please Explain How both the statements taken together are sufficient since from the equations 1 and 2 we get they are either odd or even but we are not sure if they are even or odd , right??

Anirudh134

If x, y, and z are positive integers, is x+y divisible by 2?

1) x+z is divisible by 2

2) y+z is divisible by 2

ST : 1 ) gives

X+Z is even number.

even + even = even or odd + odd = odd.

so both X and Z are either even or odd.

Question - Is X + Y even ? We can not tell at this point. Not sufficient.

Now ST : 2 ) gives

Y+Z is even number.

again even + even = even or odd + odd = odd.

so both Y and Z are either even or odd.

Question - Is X + Y even ? We can not tell this from only statement 2. Not sufficient.

Taking St-1 and St-2 together

Case I) - If from St-1 X and Z both even then Y and Z also even. All X, Y and Z even.

So X + Y even - Sufficient.

Case II) - If from St-1 X and Z both odd then Y and Z also odd. All X, Y and Z odd.

So X + Y = Odd + Odd = Even - Sufficient.

Does this help?
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I solved this in a little different manner, and i believe some people not good at picking values or going for stuff like even+odd+... will find it useful.
Stmt 1: x+z=2*b (insf because we dont know anything about y)
Stmt 2: y+z=2*a (insf because we dont know anything about x)

Stmt 1 + Stmt 2: We add the two sides of the two equations and we get: x+y+2z=2a+2b ===> we take 2z to the right side: x+y=2a+2b-2z ====> x+y=2(a+b-z) ===> divisible by 2 and Sufficient ===> (C). Re: M60-15   [#permalink] 11 Apr 2019, 03:25
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# M60-15

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