F.5 MATHS

(a) There are x sheeps and (2000-x) cows in a pasture. Since the number of sheep exceeds the no. of cows by at most 600 and at least 500.

500 <= x - (2000 - x) <= 600 or 2500 <= 2x <= 2600...(1)

Also 2000 - x <= 720 or x >= 1280....(2)

(b) Combine (1) and (2) :

1280 <= x <= 1300

So, the minimum no. of cows in the pasture is 700