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

Question

In a tin can, there is a certain number of pencils, 40 percent without erasers, and 35 percent without points. If of the pencils 1/6 have no erasers and no points, what fractional part of the pencils have both points and erasers?

A) 11/12
B) 7/12
C) 5/12
D) 1/3
E) 1/4

There is a seemingly simple problem but I have got struck at a subtle point. Could someone post a solution so that I can compare it with the official explanation. Thanks

Vips0000 · Answer

Form a double matrix like shown, using some smart number (any multiple of 6 with enough 0s) eg 600.

from this, fraction that we want is = 250/600 = 5/12

Ans C it is.

Amateur · Answer

n(aUb)=n(a)+n(b)-n(anb)
so total without erasers and points will be 0.75-1/6 which will be 7/12. 
So pencils with both erasers and points will be 1-7/12 which should be 5/12
C

g3kr · Answer

Can you explain more clearly

Amateur · Answer

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

PareshGmat · Answer

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

CrackverbalGMAT · Answer

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

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

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