If x, y, and z are integers, with x

Question

Tough and Tricky questions: Algebra.

If  x ,  y , and  z  are integers, with  x \lt y \lt z , what is the average (arithmetic mean) of  x ,  y , and  z ?

(1)  (x + y)z = 5

(2)  x + z \lt 3

Kudos for a correct  solution .

DangerPenguin · Answer

1. I can only think of 1 possibility that work for statement 1. x=0,y=1,z=5 in which case the avg would be 3. I'm going to lean towards this statement being true.

2. Knowing that both equations must be correct and also agree with each other, I now know that there is another possibility for statement 1 since what I came up with does not fit this equation. Regardless, this equation is not sufficient by itself because there could be infinite values for x + z to be less than 3. Statement 2 is insufficient.

3. Now that I know there is another combo of #s (besides the ones I came up with in step 1) that must work for statement 1, I know that statement 1 is also insufficient. Down to answer choices C and E. After a little more thought I came up with If x=-3, y=-2, z=-1, the avg is -2. This combo of #s also fits statement 1 and we now know it's insufficient.

4. Both equations combined tell us that the only combo of #s we can use is x=-3,y=-2,z=-1 for an avg of -2.

Answer C!

Loham · Answer

The way I see it, statement 1 implies that the sum of x,y,z has to be 6 (x+y=1 and z=5, or x+y=5 and z=1).

I cant see any value in statement two.

So I guess the answer is A.

santorasantu · Answer

I tried the problem in the following way:

from statement 1:
(x+y)z = 5, as the product is positive, (x+y) and z should be both positive or negative, if positive, then, (x+y+z) = 6, we get AM as 2. but if (x+y) = -5 and z = -1, we get (x+y+z) = -6, or AM = -2. two different answers, so NSF.

from statement 2, 
NSF as no info on y.

statement 1 + 2, 
x+y and z should be negative as the sum of x+z is less than 3, as from statement 1, if x+y = 1 and z = 5, then the sum will be greater than 3, the other possibility z = 1 and x+y = 5 is not possible as z>y>x.
so, AM = -2, sufficient.
my answer

Bunuel · Answer

Official Solution:

If  x ,  y , and  z  are integers, with  x \lt y \lt z , what is the average (arithmetic mean) of  x ,  y , and  z ? 
 
Since the average of  x ,  y , and  z  is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of  x + y + z ?" We also note the restrictions on the possible values of  x ,  y , and  z  - the variables must be integers in ascending order from  x  to  z  (not necessarily consecutive). Moreover, they must be different integers, since the inequality  x \lt y \lt z  indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.
 
Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set  x = 0 ,  y = 1 , and  z = 2  meets all conditions  (x + z = 2 \lt 3 , all variables are integers and in ascending order), with  x + y + z = 3 . Another set ( x = -1 ,  y = 0 , and  z = 1 ) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.
 
Statement (1): INSUFFICIENT. The equation states that  x + y  (which must be an integer) multiplied by  z  (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that  x \lt y \lt z , we can construct the only sets that work:
 
 x + y = 1  and  z = 5  (There's no way to assign  z = 1  and  x + y = 5  while preserving  x \lt y \lt z .)
 
 x = 0 ,  y = 1 ,  z = 5 ,  sum = 6 
 
 x = -1 ,  y = 2 ,  z = 5 ,  sum = 6 
 
 x = -2 ,  y = 3 ,  z = 5 ,  sum = 6 
 
 x = -3 ,  y = 4 ,  z = 5 ,  sum = 6 
 
 x + y = -5  and  z = -1 
 
 x = -3 ,  y = -2 ,  z = -1 ,  sum = -6 
 
Since there are 2 possible sums, this statement is insufficient.
 
Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of  x + z  for each case, keeping only the cases in which  x + z  is less than 3. Two cases remain.
 
Case 1:  x = -3 ,  y = 4 ,  z = 5 ,  x + z = 2 \lt 3 ,  x + y + z = 6 
 
Case 2:  x = -3 ,  y = -2 ,  z = -1 ,  x + z = -4 \lt 3 ,  x + y + z = -6 
 
Since the two cases yield different sums, we cannot determine a single value for that sum.
 
Answer: E.

santorasantu · Answer

aah missed to look deeper into the cases from statement A, thanks for the explanation Bunuel

exc4libur · Answer

x,y,z = integers

(1) (x+y)z=5   insufic

x,y,z: -3,-2,-1…-5*-1=5
x,y,z: 0,1,5…-1*5=5
x,y,z: -3,4,5…1*5=5

(2) x+z<3   insufic

(1&2)    insufic

x+z<3: -3+-1=-4<3
x+z<3: -3+5=2<3

Ans (E)

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If x, y, and z are integers, with x < y< x, what is the average

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