a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how far from the pivot must a 25 N weight be hung to balance the metre rule horizontally? ​

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

Distance from the pivot =  3 cm

it has to be b, about 97 percent.

i don't understand. explain it clearly