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x, y, and z are positive integers. The average (arithmetic m
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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. I and III only
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Originally posted by above720 on 29 Jan 2007, 21:44.
Last edited by Bunuel on 25 Aug 2013, 22:54, edited 1 time in total.
Renamed the topic and added the OA.




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Re: x, y, and z are positive integers. The average (arithmetic m
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25 Aug 2013, 23:00
above720 wrote: 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. I and III only The average (arithmetic mean) of x, y, and z is 11 > \(x+y+z=3*11=33\) z is two greater than x > \(z=x+2\) (from this we have that either both x and z are odd or both are even) > \(x+y+(x+2)=33\) > \(2x+y=31\) > \(even+y=odd\) > \(y=oddeven=odd\). As for x and z: if \(y=1=odd\), then \(x=15=odd\) and \(z=17=odd\) BUT if \(y=3=odd\), then \(x=14=even\) and \(z=16=even\). Therefore only II must be true. Answer: B.
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Re: Problem Solving  Odds and Evens
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02 Feb 2007, 00:18
X+Y+Z =33
Z=X+2
2X+2+Y=33
2(X+1)+Y=33
even + Y =odd
Therefore, Y must be odd
(B)
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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



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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.



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Re: Problem Solving  Odds and Evens
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25 Aug 2013, 21:16
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
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Re: x, y, and z are positive integers. The average (arithmetic m
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19 Apr 2014, 23:41
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)



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Re: x, y, and z are positive integers. The average (arithmetic m
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29 Nov 2015, 10:52
above720 wrote: 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. I and III only x + y + z = 33 x + y + (x+2) = 33 2x + y + 2 = 33 2x + y = 31 2x = 31 y x = \(\frac{31y}{2}\) 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 ODDHence answer is (B)
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Re: x, y, and z are positive integers. The average (arithmetic m
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26 Feb 2016, 14:44
Could someone explain this? Any great breakdowns of the datasufficiency type of questions?
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x, y, and z are positive integers. The average (arithmetic m
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26 Feb 2016, 19:40
leonidbasin1 wrote: Could someone explain this? Any great breakdowns of the datasufficiency 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.phpIt will be of help to you if you ask specific questions rather than asking such broad ended questions. Hope this helps.



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Re: x, y, and z are positive integers. The average (arithmetic m
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Re: x, y, and z are positive integers. The average (arithmetic m
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10 Apr 2017, 23:27
EVENODD RULES are being tested hereGiven: 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= ODDEVEN y= ODDx and z can be even or odd and its not possible to get a unique value for them. Thus only statement 2 is correct.
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Re: x, y, and z are positive integers. The average (arithmetic m
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18 Apr 2017, 16:32
above720 wrote: 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. I and III only 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
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Re: x, y, and z are positive integers. The average (arithmetic m
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13 Aug 2017, 10:41
Bunuel wrote: above720 wrote: 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. I and III only The average (arithmetic mean) of x, y, and z is 11 > \(x+y+z=3*11=33\) z is two greater than x > \(z=x+2\) (from this we have that either both x and z are odd or both are even) > \(x+y+(x+2)=33\) > \(2x+y=31\) > \(even+y=odd\) > \(y=oddeven=odd\). As for x and z: if \(y=1=odd\), then \(x=15=odd\) and \(z=17=odd\) BUT if \(y=3=odd\), then \(x=14=even\) and \(z=16=even\). Therefore only II must be true. 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 = 122= 10 and y=11(mean) ? In such case, x must be even and y is odd. Thank you.
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Re: x, y, and z are positive integers. The average (arithmetic m
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14 Aug 2017, 00:51
TaN1213 wrote: Bunuel wrote: above720 wrote: 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. I and III only The average (arithmetic mean) of x, y, and z is 11 > \(x+y+z=3*11=33\) z is two greater than x > \(z=x+2\) (from this we have that either both x and z are odd or both are even) > \(x+y+(x+2)=33\) > \(2x+y=31\) > \(even+y=odd\) > \(y=oddeven=odd\). As for x and z: if \(y=1=odd\), then \(x=15=odd\) and \(z=17=odd\) BUT if \(y=3=odd\), then \(x=14=even\) and \(z=16=even\). Therefore only II must be true. 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 = 122= 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.
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Re: x, y, and z are positive integers. The average (arithmetic m
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15 Aug 2017, 03:37
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 = 122= 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.
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Re: x, y, and z are positive integers. The average (arithmetic m
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18 Aug 2017, 09:09
above720 wrote: 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. I and III only 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
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