Question 331505
Let x=test1
   x+8=test2
    88=test3
jim had a b average so 80<=(t1+t2+t3)/3<=89 the avg of his test
   substitution gives us

80<=(x+x+8+88)/3<=89; 80<=(2x+96)3<=89; multiply by 3
240<=2x+96<=267; subtract 96
240-96<=2x<=267-96;  144<=2x<=171; now divide by 2
77<=x<=85.5
if jim first test was between 77 and 85.5 inclusive