If z=(x+y)/2xy and 0<y<z<1 which of the following must be true about

Since both x and y are between 0 and 1 i.e. fractions, we can represent

x = 1/a where a > 1
y = 1/b where b > 1

Therefore \(z = \frac{\frac{1}{a} +\frac{1}{b}}{\frac{2}{ab}}\)

Therefore \(z = \frac{a+b}{ab}*\frac{ab}{2}\)

Therefore \(z = \frac{a+b}{2}\) where a and b > 1

Therefore \(a+b > 2\) (since individually a and b both are greater than 1 hence addition will always be greater than 2)

Therefore \(z=\frac{a+b}{2} > 1 \)

Answer - E

Here's a different approach....

To narrow down the answer choices let's first test a pair of values

If \(0<y<x<1\), then it COULD be the case that y = 0.2 and x = 0.5
In this case, \(z = \frac{0.5+0.2}{2(0.5)(0.2)}=\frac{0.7}{0.2}=3.5\)
If it is possible for \(z = 3.5\), then we can ELIMINATE A and D, because they suggest that z CANNOT equal 3.5

At this point, I notice that some of the remaining answer choices suggest that z COULD equal 1.
So let's see if this is possible.
In other words, given the possible values of \(x\) and \(y\), is it possible that \(1 = \frac{(x+y)}{2xy}\)?
Let's first multiply both sides of the equation by \(2xy\) to get: \(2xy = x+y\)
Now subtract x and y from both sides of the equation: \(2xy - x - y= 0\)
Rewrite \(2xy\) as \(xy + xy\) to get: \(xy + xy - x - y= 0\)
Rewrite as: \(xy - x + xy - y= 0\)
Factor in parts to get: \(x(y - 1) + y(x - 1)= 0\)
At this point we might be able to see that there can be no solution to this equation. Here's why:

Since y < 1, we know that (y - 1) is NEGATIVE
Likewise, since x < 1, we know that (x - 1) is NEGATIVE
We also know that x and y are both POSITIVE.

So our equation, \(x(y - 1) + y(x - 1)= 0\), becomes (POSITIVE)(NEGATIVE) + (POSITIVE)(NEGATIVE) = 0
Simplify to get: NEGATIVE + NEGATIVE = 0
This is impossible.

Since we are unable to find a solution to be equation \(1 = \frac{(x+y)}{2xy}\), we can conclude that z cannot equal 1

This means we can ELIMINATE B and C, because they suggest that z CAN equal 1

By the process of elimination the correct answer is E.

Cheers,
Brent

If Z = (X + Y)/2XY and 0<X<Y<1, which of the following must be true of the range of Z?
We know x and y are fractions between 0 and 1.

Now just pick the extreme values for x and y to determine the range of z
Now lower limit :When x=0.98 ,y=0.99
Z= (0.98+0.99)/2(0.99)(0.98)
Z= 1.97/1.95 = 1.01 (2d.p)

Now upper limit :When x=0.01 ,y=0.02
Z= (0.01+0.02)/2(0.01)(0.02)
Z= 0.03/(0.0002) =150

In all those extreme cases z must be greater than 1
.: Z > 1
E like elephant

Posted from my mobile device
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First, we can manipulate the equation on z:

\(z = \frac{(x+y)}{2xy} = \frac{x}{2xy} + \frac{y}{2xy} =\frac{1}{2y} + \frac{1}{2x}\).

Next we would like to find the range of \(\frac{1}{2y}\) and \(\frac{1}{2x}\). We can try plugging in extremes to find a range on z.

Extremely small x and y: \(x = 0.02\) and \(y = 0.01\). We can see z will skyrocket thus we can conclude there is no upper bound for z.

Extremely big x and y: \(x = 0.999\) and \(y = 0.99\). We can use x = y = 1 to approximate and we get \(z = 1/2 + 1/2 = 1\). So z is at least 1.

This concludes we choose z > 1.

Ans: E
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