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Re: If x,y,and z are positive numbers,what is the value of the average (ar
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01 Jul 2017, 03:00
If x, y, and z are positive numbers, what is the value of the average (arithmetic mean) of x and z ?
All are positive integers.
Both statement (1) and statement (2) do not give us any values, for x, y or z and hence, we will not be able to arrive at a certain value for the average of x & z.
Hence, Answer is E _________________
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Re: If x,y,and z are positive numbers,what is the value of the average (ar
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03 Jul 2017, 03:38
1
We need to find (x+z)/2
Statement 1. x-y = y-z . this means z, y, x are in Arithmetic Progression (increasing order also).. And x+z = 2y or (x+z)/2 = y. But we don't know the value of y, so insufficient.
Statement 2. Clearly insufficient.
Combining the two statements: Let the three AP terms: z = z, y = z+d, x = z+2d Now from second statement we are given that x^2 - y^2 = z or (z+2d)^2 - (z+d)^2 = z Solving we get 2zd + 3d^2 = z.. But this is NOT sufficient to find the value of z/d (thus that of y too) so the question cannot be answered.
Re: If x,y,and z are positive numbers,what is the value of the average (ar
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31 Aug 2017, 10:32
1
Statement (1) can be simplified and looked at conceptually. No need to pick numbers.
(x + z)/2 = y
Since we have no idea what y is, statement (1) is insufficient.
Statement (2) tempts us to use the difference of squares common equation.
(x + y)(x – y) = z
Since x, y and z are all positive, we know x – y must be positive, or x – y > 0. So x > y. But this is about all we can get with this approach, and this is something we could have observed from x^2 – y^2 = z, since z is positive.
The fact that there are no values in x^2 – y^2 = z is a clue that there are many possible combinations of x, y and z that could work. In picking numbers, I always like to start with simple numbers that fit the situation. So since x > y, here are some numbers that fit:
x = 2 y = 1 z = 2^2 – 1 = 3 (x + z)/2 = 2.5
OR
x = 3 y = 2 z = 3^2 – 2^2 = 5 (x + z)/2 = 4
Since we get two possible values for (x + z)/2, statement (2) is insufficient.
For (1) and (2) together, here’s an alternative to picking numbers. Notice that (1) tells us that y is the midpoint of x and z. Since (2) tells us that x > y, we know that z < y < x, and they are all evenly spaced. I like the idea of representing the spacing as d, so y = z + d, and x = z + 2d. Therefore, (2) gives us:
(z + 2d)^2 – (z + d)^2 = z z^2 + 4dz + 4d^2 – z^2 – 2dz – d^2 = z 2dz + 3d^2 = z 3d^2 = z – 2dz = z(1-2d) z = 3d^2/(1-2d)
In order to keep z positive, we must have d < ½, but as long as we do that, z could be lots of numbers, and therefore y and x could also be lots of numbers. Since (1) showed that our question is equivalent to y = ?, (1) and (2) together are insufficient.
Re: If x,y,and z are positive numbers,what is the value of the average (ar
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17 Nov 2017, 11:37
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AbdurRakib wrote:
If x, y, and z are positive numbers, what is the value of the average (arithmetic mean) of x and z ?
(1) x - y = y - z
(2) x^2 - y^2 = z
We need to determine (x + z)/2.
Statement One Alone:
x - y = y - z
Simplifying the equation, we have:
x - y = y - z
x + z = 2y
Since x + z = 2y, we have:
(x + z)/2 = 2y/2 = y
However, since we do not know the value of y, we cannot determine the average. Statement one alone is not sufficient to answer the question.
Statement Two Alone:
x^2 - y^2 = z
This does not provide enough information to determine the average of x and z. For example, if x = 3 and y = 2, then z = 5, and the average of x and z would be 4. However, if x = 4 and y = 2, then z = 12, and the average of x and z would be 8. Statement two alone is not sufficient to answer the question.
Statements One and Two Together:
From statement one, we see that z = 2y - x, and from statement two, we have z = x^2 - y^2. Thus, we have 2y - x = x^2 - y^2.
Notice that the equation above has two variables; thus, there are infinitely many solutions. That is, we won’t have a unique value for x or y, and hence we don’t have a unique value for z, either. The two statements together are still not sufficient to answer the question.
Answer: E
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Re: If x,y,and z are positive numbers,what is the value of the average (ar
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19 Nov 2017, 12:48
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1
Hi All,
We're told that X, Y and Z are POSITIVE numbers. We're asked for the average of X and Z. This question can be solved with a bit of math and TESTing VALUES.
1) X - Y = Y - Z
We can simplify this equation... X + Z = 2Y
IF.... X=1, Z=1, Y=1, then the answer to the question is... (1+1)/2 = 1 IF.... X=2, Z=4, Y=3, then the answer to the question is... (2+4)/2 = 3 Fact 1 is INSUFFICIENT
2) X^2 - Y^2 = Z
IF.... X=2, Y=1, Z=3, then the answer to the question is... (2+3)/2 = 2.5 IF.... X=3, Y=1, Z=8, then the answer to the question is... (3+8)/2 = 5.5 Fact 2 is INSUFFICIENT
Combined, we can combine the two given equations.... X + Z = 2Y X^2 - Y^2 = Z into.... X + (X^2 - Y^2) = 2Y X + X^2 = Y^2 + 2Y
Given that last equation - and since all 3 variables are POSITIVE - as X gets bigger, Y will ALSO get bigger. Since X could become a huge number (and Z is positive so it cannot 'offset' the increase in X'), the answer to the question "what is the average of X and Z?" will vary. Combined, INSUFFICIENT
Re: If x,y,and z are positive numbers,what is the value of the average (ar
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25 Nov 2018, 13:38
lolmenot wrote:
chetan2u wrote:
AbdurRakib wrote:
If x, y, and z are positive numbers, what is the value of the average (arithmetic mean) of x and z ?
(1) x - y = y - z
(2) x^2 - y^2 = z
Hi,
The info and observations from the Q.. 1) all numbers are >0. 2) there is no numeric value at all in the options/statements.
In the given circumstances, you cannot find the value. ANSWER at the best will in variables .
So E
From 1) don't we understand that Y is the middle value as it's equal numbers away from X and Z.
Example x,y,z is 10,8,6
Shouldn't A be the correct answer
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Hi lolmenot,
In your example (X = 10, Y = 8, Z = 6), the answer to the question is (10+6)/2 = 8. What if you used different values though? Would the answer always be 8 or could it be anything else? If the answer to the question changes, then Fact 1 is insufficient.
close
72% (01:35) correct 
