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In a tin can, there is a certain number of pencils, 40 [#permalink]
15 Nov 2012, 08:39

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Question Stats:

69% (03:48) correct
31% (02:29) wrong based on 101 sessions

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

Re: Groups, Percents, Fractions [#permalink]
15 Nov 2012, 08:59

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mhadi wrote:

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

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

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

Re: In a tin can, there is a certain number of pencils, 40 [#permalink]
15 Nov 2012, 13:17

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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

Re: In a tin can, there is a certain number of pencils, 40 [#permalink]
15 Nov 2012, 13:39

Amateur wrote:

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

Re: In a tin can, there is a certain number of pencils, 40 [#permalink]
15 Nov 2012, 14:57

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g3kr wrote:

Amateur wrote:

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

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

Re: In a tin can, there is a certain number of pencils, 40 [#permalink]
19 Jun 2014, 09:55

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