average problem- any solun by plugging of nos

W,X,Y and Z are 4 variables used.

Average of X and Y is Z

Let X = 0, Y = 2. Therefore Z = 1

Average of Z and X is W

Z = 1 and X = 0. Therefore W = 1/2

You're asked (X-W)/(W-Y) = (0-1/2) / (1/2-2) = -1/2 / (-3/2) = [spoiler]1/3 [/spoiler]

The best approach to do this question is by plugging in numbers

Let us say x = 40 and y = 20
Avg of x and y = z = (20+40)/2 = 30

Avg of z and x = w = (20+30)/2 = 35

(x-w)/(w-y) = (40-35)/(35-20) = 5/15 = 1/3

Option B

HI shankar.ashwin

As per your plugging of numbers it does not match with the question,

As per qn, W should be greater than Y, but in your plugging you got W=1/2 and Y=2 this is wrong.

Iam not sure whether the qn is correct r wrong. If the condition W>Y is not given in qn then your approach is correct.

So any one pls check if W> Y then we can get the same answer as 1/3.

The problem states W>Y,
Equation 1: X+Y/2=Z
Equation 2: Z+X/2=w

Solving Equation 1 & 2
X+Y=2Z Z+X=2w => Z=2W-X

X+Y=2(2W-X)
3X+Y=4W => X=(4W-Y)/3...Equation 3

We need to find (x-w)/(w-y)
Sub Equation 3 to solve

((4W-Y)/3 -W)/ (W-Y)
= 1/3

kishokbabu wrote:HI shankar.ashwin

As per your plugging of numbers it does not match with the question,

As per qn, W should be greater than Y, but in your plugging you got W=1/2 and Y=2 this is wrong.

Iam not sure whether the qn is correct r wrong. If the condition W>Y is not given in qn then your approach is correct.

So any one pls check if W> Y then we can get the same answer as 1/3.

The statement W>Y .. doesnt really play any part in the solution
Because either way we would get the same solution

W>Y would correspond to X>Y which further means X>W
So (X-W)/(W-Y)= +ve

W<Y corresponds to Y>X which further means X<W
(X-W)/(W-Y)= -ve/-ve = +ve [The magnitude remaining the same]

Sorry didnt notice that while reading. But as rijul pointed out, the solution remains the same, just pick up a set of numbers which satisfy W>Y, you would still get the same answer for this question