Normally, we have three tables:
i) Base Tolerances Table
ii) Base Deviations of Shafts Table
iii) Base Deviations of Holes Table
all are in micrometer and are of DIN-ISO 286 T1.
Steps to determine the max and min tolerance:
From 60P6, we know that it refers to a hole with inner diameter 60 mm.
In the Base Tolerances Table, search for the nominal size range that the inner diameter of the hole lies in. So it should be (50-80) mm.
Then refer to the base tolerance scale IT and look for the grade number 6. You will obtain the corresponding base tolerance (with respect to the nominal range) of 19.
After that, refer to Base Deviations of Holes Table. Look for the capital letter P column and nominal range 50-65 mm row. You will see a value of -32. This is the upper deviation EI. The lower deviation EI can be found by EI=ES-IT=(-32-19) \mu m=-51
Because base tolerance IT is 6, which is up to IT 7, a \delta has to added. In this case \delta =6 \mu m. So ES_{corrected}=-32+\delta =-32+6=-26 \mu m. So as EI_{corrected}=-45 \mu m.
Hence, the maximum tolerance is 60 mm + (-26 micrometer) = 59.974 mm while the minumum tolerance is 60 mm + (-45 micrometer) = 59.955 mm
And that's it
EDIT: Latex problem solved