x, y, and z are positive integers. The average (arithmetic m

X, y, and z are positive integers. The average (arithmetic mean) of x, y, and z is 11. If z is two greater than x, which of the following must be true?

I. x is even
II. y is odd
III. z is odd

a. I only
b. II only
c. III only
d. I and II only
e. 1 and III only
2x+y+2 = 33 ie: 2x+y = 31

y is sure odd

my answer is B

Just to add...
Once you have established that I need not be true, you can eliminate II as well because:
z = x + 2.
Unless you know whether x is even or odd, you cannot know whether z is even or odd.

Hi Guys,

I got a very good qns. Try it.

X, y, and z are positive integers. The average (arithmetic mean) of x, y, and z is 11. If z is two greater than x, which of the following must be true?

I. x is even
II. y is odd
III. z is odd

a. I only
b. II only
c. III only
d. I and II only
e. 1 and III only


OA is B

Given that:

x+y+z = 33 and z=x+2

using both equations

2x(Even) + y = 31 (Odd)


Since the sum is Odd 'y' has to be odd. Hence , only options (B) and (D) to be verified.

To check for 'x'.

x+y+z=33

x+z= 33- Odd

x+z= Even

Even + Even = Even
Odd+Odd = Even

So (B)

x + y + z = 33

x + y + (x+2) = 33

2x + y + 2 = 33

2x + y = 31

2x = 31 -y

x = \(\frac{31-y}{2}\)

So, we can say y must be an odd no and x can be even / odd.

Further we have z = x + 2

Since x can be odd/even ; z will be odd if x is odd and z will be even if x is even

So, we can be definite only about y as ODD

Hence answer is (B)

Could someone explain this? Any great breakdowns of the data-sufficiency type of questions?

You need to note what is given to you and what is getting asked. Additionally, this is a MUST BE TRUE type of a question ---> correct option will be true for ALL cases.

Given: x,y,z are POSITIVE INTEGERS and the average of x,y,z=11

From average of x,y,z = 11----> (x+y+z)/3 = 11 ---> x+y+z=33 (odd)

Also, z=x+2 ---> if x=odd, then z=odd+even=odd (think of 2+3=5, odd) and if x=even, then z=even+2=even. (Remember that odd+odd=even, even + even = even, odd +even =odd)

Also as x+y+z=33, odd ---> you need to have 1 odd number to make the sum of 3 integers = odd. Thus, the only case possible is

x=even, y=odd and z=even.

Lets analyse the options:

I. x is even , not possible. Eliminate.
II. y is odd, yes, keep.
III. z is odd, not possible. Eliminate.

Thus, only II is must be true and hence B is the correct answer.

As for your DS related question, it is a very broad question that can not be answered in simpler terms. You need to practice to understand how to tackle DS problems. The concept remain the same as those for PS only the application part differs. Take a look at Manhattan GMAT's DS book and practice questions with DS tags at viewforumtags.php

It will be of help to you if you ask specific questions rather than asking such broad ended questions.

Hope this helps.

Here is my take on this one->

x+y+z=33
z=x+2=> z-x = 2 => Even

Hence x and z will have the same Even/Odd nature.
Hence y=> odd-eve=> odd
y is always odd
x and y can be both even or both odd

Hence B

EVEN-ODD RULES are being tested here

Given: x,y,z>0
x+y+z=33 (Average*total no. of terms=11*3=33)
z=x+2

x+y+x+2=33
2x+y=31

EVEN+y= ODD
y= ODD-EVEN
y= ODD

x and z can be even or odd and its not possible to get a unique value for them.

Thus only statement 2 is correct.

We are given that the average (arithmetic mean) of x, y, and z is 11.

Using the formula average = sum/number, we have:

11 = (x + y + z)/3

33 = x + y + z

We are also given that z is two greater than x, thus we now have:

x + y + (2 + x) = 33

2x + y = 31

Since 31 is odd and 2x is even, and the only way to get an odd sum is even + odd (or vice versa), we see that y MUST be odd. However, we do not know anything about x or z.

Answer: B

Hi Bunuel,
I have understood the solution. Just for getting better at concept, I would like to get my doubt clarified.
Since x,y and z are positive integers , 11 is the mean and z is two greater than x, why can't it imply that z is 12, x = 12-2= 10 and y=11(mean) ? In such case, x must be even and y is odd.

Thank you.

The point is that the question asks which of the following MUST be true, not COULD be true. So, while x = 10 = even, y = 11 = mean, and z = 12 = even is certainly possible, but this is only one of the cases, and x = even is true only for some of the cases (it COULD be true) but on the other hand y = odd is true for ALL cases (it MUST be true).

Hope it helps.

Hi Bunuel,
I have understood the solution. Just for getting better at concept, I would like to get my doubt clarified.
Since x,y and z are positive integers , 11 is the mean and z is two greater than x, why can't it imply that z is 12, x = 12-2= 10 and y=11(mean) ? In such case, x must be even and y is odd.

Thank you.[/quote]

The point is that the question asks which of the following MUST be true, not COULD be true. So, while x = 10 = even, y = 11 = mean, and z = 12 = even is certainly possible, but this is only one of the cases, and x = even is true only for some of the cases (it COULD be true) but on the other hand y = odd is true for ALL cases (it MUST be true).

Hope it helps.[/quote]

Yes! I get it now.. Thanks Bunuel. I went wrong in assuming x,y,z to be consecutive integers which is why I thought 10,11,12 is the only possible case.

Since the average (arithmetic mean) of x, y, and z is 11:

x + y + z = 33

Since z = x + 2:

x + y + x + 2 = 33

2x + y = 31

y = 31 - even number, so y must be odd. Thus, we only can determine that y must be odd.

Answer: B

x + y + z = 33
x + y + x + 2 = 33
2x + y = 31

Which must be true?

I. We know from the above that y must be odd. If x is even, then z is even. If x is odd, then z is odd. We can clearly get both outcomes. Statement I doesn't have to be true.

II. If 2x + y = 31, y MUST be odd.

III. From statement I, we know we don't have conclusive information.

Answer is B.

\(x + y + z = 33\)

Or, \(z = x + 2\)

Thus, \(x + y + z = 33\)

Or, \(x + y + x + 2 = 33\)

Or, \(2x + y = 31\)

Here , 2x must be even (Whether x is even/Odd) so y must be Odd, Answer must be (B)

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