Mrs. K is paid at a reduced rate for contracts completed

Let's call the original, unreduced prices of the 3 contracts - \(x, y\) and \(z\).

The reduction would then be: \(0.1x, 0.1y\) and \(0.15z\).

The percentage reduction may then be expressed as:

\(100 *\frac{(0.1x + 0.1y + 0.15z)}{(x + y + z)}\)

This is the value we need to establish as being on either side of 11% to get a sufficient answer.

Let's play around with it a bit to reduce the expression. Splitting \(0.15z\) into \(0.1z + 0.05z\), and factoring out \(0.1\) gives us:

\(100* \frac{[0.1(x+y+z) + 0.05z]}{(x + y + z)}\)

This further reduces to:

\(10 + \frac{5z}{(x+y+z)}\)

So if we are able to establish, that the second term in this equation is always less than 1 or always more than 1, we have an answer.

Let's look at statement 1. We have a value here for \(z\), which gives us a clear idea of the numerator of the term.
We also have a minimum value of \(x\) and \(y\). We can further arrive at the inference that \((x+y)\) is atleast \(2400\). Let's look at that value of the term:

\(\frac{(5 * 550)}{(2400 + 550)}\)

= \(\frac{55}{59}\)

This is obviously less than 1, which means the total percent will be some value between 10% and 11%.

If we continue to increase \(x\) and \(y\), over \(1200\), it should be clear that since the numerator only depends on the value of \(z\), an increase in the denominator \((x + y + z\)) only further reduces the value of this term, bringing the total percent closer to 10%.

Thus statement (1) is sufficient to conclude that the percentage will always be less than 11%.

Let's look at statement (2). All we have is a value of \(x\). Since nothing is indicated about \(y\) and more importantly \(z\), we can not conclude whether the term is less than 1 or more than 1. INSUFFICIENT.

Hope that clears things up, and that you guys find weighted averages to be simpler now!

EXTRA NOTE:
When would \(\frac{5z}{(x+y+z)} = 1\)?
This would be when\(x + y = 4z\)
i.e. when \(z\) = 20% of \((x + y + z)\) or 25% of \((x +y)\)