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

Can you explain more clearly

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

Refer Venn Diagram (We require to find orange region)

Say total pencils = 100

Without erasers = 40

Without points = 35

So, without erasers & without points \(= \frac{100}{6} (Blue region)\)

Total pencils with defect

\(= 40 + 35 - \frac{100}{6}\)

\(= \frac{350}{6}\)

Total pencils without defect

\(= 100 - \frac{350}{6}\)

\(= \frac{250}{6}\)

\(Fraction = \frac{\frac{250}{6}}{100}\)

\(= \frac{5}{12}\)

Answer = C

CONCEPT:Fractions, Basics of Set Theory

SOLUTION:Given,40% + 35% =75% =¾ of the pencils either do not have erasers or do not have a point.
1/6 of the pencils have NO erasers and NO points.
Hence ¾ - 1/6 =7/12 of the pencils either do not have erasers or do not have point or both.
Thus, the fraction of the pencils that have both points and eraser= 1-7/12
=5/12 (C)

Hope this helps. Keep studying!
Devmitra Sen