How far to the left of the pivot must a 3.6 kg cat stand to keep the seesaw balanced? express your answer to two significant figures and include the appropriate units?

The solution for this problem is:
This is just a statics problem since no one of the components of the system is touching. The bowl and the smaller cat will have a habit of to twist the seesaw one way, while the weightier cat will twist it the other way.But since the weight of the cat and bowl of tuna is not given, we will assume that:Weight of the cat is 5.4kgAnd the weight of the tuna bowl is 2.5kgLength of the seesaw is 4m, but since the cat and the tuna bowl is in each half, it will be 2 m.(5.4 x 2) = 10.8kg/m. torque. (2.5 x 2) m= 5kg/m. for bowl. 
Difference = (10.8 - 5) = 5.8kg/m. 
(5.8/3.8) = 1.5 meters left of the pivot.

Ithink the answer would be a