In a tin can, there is a certain number of pencils, 40

Venn diagrams..... Consider two samples a, b which overlap..... we have the formula I wrote above....
so if you consider no erasers to be sample space a, and no points to be sample space b....
when they say 0.4 had no erasers, it means they also contain a few which donot have points too.... likewise when they say 0.35 had no points they contain pencils which didnot have erasers too.... so pencils without erasers and points are represented in both the cases above... so if you add no erasers and no points samples, think about it, you are adding pencils without erasers and points twice.... So subtract quantity (1/6) from what you got by adding 0.35+0.4=0.75. So on the whole pencils with no erasers only +pencils with no points only +pencils with no both erasers and points = 0.75-1/6=7/12....
so 7/12 are pencils which are defective on the whole.... but you want good pencils... anything apart from defective pencils are good right... Remember set theory, sum of all samples=1...
So good pencils will be 1-7/12 which is 5/12