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# If x,y,and z are positive numbers,what is the value of the average (ar

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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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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.

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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Clumsy question right?

Like from 1 we get that the question is y=?

Then from 2 the info is something that has nothing to do with nothing.

So I chose E because the the question was a complete mess and I couldn't evaluate anything essentially.
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Clearly E since we have 3 variables and only 2 equations
moreover

1) x-y=y-z
y= (x+z) /2 we dont know x,y,z A,D gone

2) x^2-y^2 = z
(x-y) (x+y) =z. we dont know x,y,z B gone

put y = x+z /2 in 2nd equation
which will give quadratic but mean of x and z will be impossible to obtain... Insufficient C GONE

Remaining E is answer
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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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.

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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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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

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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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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 [#permalink]
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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.
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Note that if the question stated that x, y, and z were negative integers, then both statements together would be sufficient.

(1) $$x - y = y - z$$

$$x - 2y = -z$$

$$2y - x = z$$

(2) $$x^2 - y^2 = z$$

$$x^2 - y^2 = 2y - x$$

$$x^2 + x - y^2 - 2y = 0$$

$$x(x - 1) - y(y + 2) = 0$$

$$x = 0, -1$$ and $$y = 0, -2$$

However, because the question states that x, y, and z are positive numbers, we have infinite potential solutions for x and y.
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Probus wrote:

Aren't the two Statements Contradictory

What One says is that y is the average of x&z
So we can say x<y<z or z<y<x

So if we look at statement 2 we have

(x+y)(x-y)=z

Since z is positive then x>y
we can infer that y<x<z

Where am i going wrong?

This is not necessary
suppose x=6 and y=5 then z=11
thus y<x<z

but if x=2 and y=1 then z=2
then x=z=2
In any case this is not what the question is asking, in DS question you have to take every statement separately first, and if any of them is not the answer then you can check for C, in this particular statement whatever we do we can't get a specific value of x+z/2 therefore the answer should be E.
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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Dear chetan2u ,

Can t we get the median as a variable in A and get a sufficient answer?

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
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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prashant0099 wrote:
Dear chetan2u ,

Can t we get the median as a variable in A and get a sufficient answer?

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

No, x is just a variable. It has no value.
When we look for a value, it has to be a number.

If a question wants an answer in a variable, it will ask you to find median/values of an expression in terms of x. Generally in PS, you will come across such questions
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If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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gd93 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

Can't this approach be a little risky as sometimes the variables cancel out? So wouldn't it still be worth it to try and work it out to make sure that we are in fact still only left with variables?

You should always try to solve the equations when you have time.
But in case you are short of time and in the given scenarios as in this question, you can mark E.

There is no number in the equations, it just contains variable.
If one of the variables cancel out we will have the other as equal to 0 in most of the cases, but here none can be 0 as all variables are positive integers.
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If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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

Posted from my mobile device
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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

Posted from my mobile device

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.

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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
Hi Bunuel

Aren't the two Statements Contradictory

What One says is that y is the average of x&z
So we can say x<y<z or z<y<x

So if we look at statement 2 we have

(x+y)(x-y)=z

Since z is positive then x>y
we can infer that y<x<z

Where am i going wrong?
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Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
1. Says that the difference b/w the x&y and y&z is same. this means they are in AP. But we wont get "One" value of y(it is the mean of x & z as per the stmt 1). Changing the values of X & Z, you get different values of Y.

2. Second stmt merely says that (x+y)(x-y) = z but we need answer for (X+Z)/2. hence this stmt is insufficient.

Combining the 2 stmts you get 2 equations . ie. from st 1. y = (x+z)/2 and from st 2 (x+y)(x-y) = z

but again note that there are 3 unknowns and 2 equations. Hence you will still not get "One" value for (x+z)/2

Hope this helps

Probus wrote:
Hi Bunuel

Aren't the two Statements Contradictory

What One says is that y is the average of x&z
So we can say x<y<z or z<y<x

So if we look at statement 2 we have

(x+y)(x-y)=z

Since z is positive then x>y
we can infer that y<x<z

Where am i going wrong?
Re: If x,y,and z are positive numbers,what is the value of the average (ar [#permalink]
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