No idea if this approach is correct, this seems like an extremely hard question:

\(\frac{85n+x}{n+2}=81\)

Where 85n is the sum of the prior test results, and x is the sum of the last 2 test taken.

\(85n+x=81n+162\)

\(4n=162-x\)

\(n=\frac{162-x}{4}\)

Notice that n must be at least 1, so 162-x>=1

\(162-x>=1\)

\(161>=x\)

As x is the sum of both tests, the average is x/2

\(80.5>=\frac{x}{2}\)

Since x is close to the boundary, lets check x/2=80

If x/2=80, x=160. But 162-160 is 2/4, which is not an integer, which is not valid for n.

x/2 must therefore be less than 80

QB>QA

B.