Q:

Make the Average smaller then the Median

This is a homework question but it is worded in a way that says "Find the answer" and not "Is there an answer". Heuristic knowledge on the course and how it's written indicates an answer as well so my friends and I are extremely curious if we're missing some angle. 
The question:
We have a sample (1.0, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3.0) and, changing only one value, make the average smaller than the median. 
So the two approaches we started off with is either increasing the median (1.8) or decreasing the average (1.98). 
Following the increase of the median, the limits we have on the median are 1.6 to 2.0. Increasing the median to 2 with the smallest increase in number (to keep the average low) would be 1.8 to 2.0. This however makes the average = the median (actually the answer to the next question in the exercise, making us further believe there must be an answer for the question). 
This means we can't change the median to anything to solve this problem. The only other option we have is drop the average. 
I'm on a phone so if someone could write this better, that would be great. 
(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11)/11 < x6
x1+x2+x3+x4+x5+x7+x8+x9+x10+x11 < 10x6
We know x6 has to be 1.8 since we can't change the median. So that means;
20 < 18
We need to drop the average by more than 2 to get it lower than the median. We can only drop numbers on the right side of the median down to 1.8. This is at most a change of 3 - 1.2.
Additionally the left side of the median can only drop by 1.6.
Through this exhaustion of deduction, we figure the only way to get the average lower is to drop a number into the negative. But that seems like a bit of a cheat. Is it possible to do this without going into negative?

A:

There are no solutions without dropping one of the numbers below 0. Besides your arithmetic solution to show this, you can run a simulation in R.
The following simulation tries adding various numbers between -3 and 3 (at .05 intervals) to each item in your list, then compares the mean to the median.
x <- c(1.0, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3.0)

# change each of the numbers by some value between -3 and 3
# in this case, -3, -2.95, -2.98, -2.97, etc.
changes <- seq(-3, 3, .05)

num_tries <- length(changes) * length(x)
i <- seq_len(num_tries)

# create a matrix of 11 by all the possible tries
m <- replicate(num_tries, x)
m[cbind(i %% 11 + 1, i)] <- m[cbind(i %% 11 + 1, i)] + rep(changes, each = 11)

# remove any containing negatives
m <- m[, colSums(m < 0) == 0]

# are there any cases where the mean equals the median?
sum(apply(m, 2, mean) == apply(m, 2, median))
# Result: yes, 1, the one you already found 

# are there any cases where the mean is less than the median?
sum(apply(m, 2, mean) < apply(m, 2, median))
# Result: no, the answer is 0.

It's OK if you don't know R and can't follow this example, but it's just more evidence that you're not missing an arithmetic trick. With the numbers as you give them, you can't change one value to make the mean < median.
The only answers to the problem as you state them involve sending a number negative: for example, changing the 1.0 to -2.0.

