replayyyy wrote:Of the 60 animals in a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?
(1) the farm has more than twice as many cows as it has pigs
(2) the farm has more than 12 pigs.

P + C = 60 * 2/3 = 40

1. C > 2P
let P = 10 then C = 30
let P = 11 then C = 29
.. So multiple chances. Thus NS.

2. P > 12
let P = 13 then C = 27
let P = 14 then C = 26
let P = 15 then C = 25
.. So multiple chances. Thus NS.

Combining both only P = 13 then C = 27 will satisfy.
P = 13 2P = 26 < 27 =C thus satisfied.

PS: No more other chances try with 14 it will not satisfy A

replayyyy wrote:Of the 60 animals in a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?
(1) the farm has more than twice as many cows as it has pigs
(2) the farm has more than 12 pigs.

assign properties to problem and conditions 1, 2

problem c+p = 40
condition (1) c > 2p
condition (2) p > 12


solve inequalities (1) p < c/2; (2) 12 < p; 12 < p < c/2.
p must be integer (split pigs?) p=13
solve for c, c > 2p, c > 26, c = 27

plug into problem c + p = 40, 13 + 27 = 40

with both conditions the problem can be solved.

p.s. MGMAT tips for DS