Question 662795
I interpret the wording as meaning that 80 and 90 are both a B.
I'll solve based on that.
 
WITH ROUNDING:
If we are going to be realistic, we have to consider rounding.
I expect that an average of 79.5 would be rounded up to 80, and would be enough for a B.
I expect that an average of 90.5 would be rounded up to 91, and that would be enough for an A,
but any average of less than 90.5 (even a 90.4) would be a B.
So I am starting from
{{{79.5<=average<90.5}}}
I am assuming they calculate the average as the sum of all tests grades divided by the number of tests.
If the score in the last test is {{{x}}}, the average will be
{{{average=(92+66+72+88+x)/5=(318+x)/5}}}
Now my equation is
{{{79.5<=(318+x)/5<90.5}}}
Multiplying all sides of the signs times 5, we get an equivalent inequality
{{{79.5*5<=318+x<90.5*5}}} --> {{{397.5<=318+x<452.5}}}
Subtracting {{{318}}} from all sides of the signs, we an the equivalent inequality
{{{397.5<=318+x<452.5}}} --> {{{397.5-318<=318+x-318<452.5-318}}} --> {{{79.5<=x<132}}}
So there is no possibility of getting an A, and a score of {{{highlight(80)}}} in the fifth test is needed to get an A as the final grade.
 
WITH ROUNDING:
If the average has to be exactly 80.0 for a B,
then {{{80<=(318+x)/5}}} --> {{{400<=318+x}}} --> {{{82<=x}}}, and an {{{highlight(82)}}} in the fifth test is needed for a B.