A quick way to approach this question is to plug in your own numbers.
When looking at these two statements, Statement 2 is more definite, (the average of two variables equals a constant rather than another variable) so start with that Statement.
The average of b and c is 4. Substitute 5 and 3 for b and c, respectively, since those numbers are one greater and one less than 4 (and thus average 4).
b = 5
c = 3
Now that we have numbers for b and c, we very clearly don't know what a is. So Answer B is eliminated. Now, let's see about combining this with Statement 1.
Solve for a using the b and c from before. You can set up an equation for an average, which is the sum of the numbers divided by how many there are (2).
$$\frac{a+5}{2}=3$$
$$a+5=6$$
$$a=1$$
Now that you have all 3 values, you can calculate the average for a, b, and c:
$$\frac{1+5+3}{3}=\frac{9}{3}=3$$
So at this point, the average COULD be 3. But what if you switched the values of b and c in the first place?
b=3
c=5
The average of b and c is still 4. But now solve for a
$$\frac{a+3}{2}=5$$
$$a+3=10$$
$$a=7$$
Now, the average of the 3 numbers will be:
$$\frac{7+3+5}{3}=\frac{15}{3}=5$$
Furthermore, you can have a situation where b and c are both equal to 4, making a equal to 4, and the average of a, b, and c is 4.
The issue with this problem is that the average of b and c can be equal to 4 whether b is greater than c, b is less than c, OR b is equal to c. But, each of these scenarios impacts the value of a, which then impacts the value of the average of a, b, and c.
Statement 1 is not sufficient. Statement 1 and 2, taken together, are not sufficient either. Answer E is correct.
