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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60 [#permalink]
Can someone please provide a more detailed explanation to this one?
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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60 [#permalink]
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question asks for 10 pair of socks, 10 pairs irrespective of color (Red, Green, Black, and Blue).

trick here is to know for getting 1 pair we need 5 socks.

To get 1st pair we need atleast 5 socks --> 2 of same color and 3 of different
To get 2nd pair we need additional 2 socks --> 3 (left out from previous selection) + 2 new socks
To get 3rd pair we need additional 2 socks and so on.

For each pair we need additional 2 socks. After 1st pair we need 9 more pairs so total socks to be drawn = 5 + 2*9 = 23

Answer : B
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A drawer in a darkened room contains 100 red socks, 80 green socks, 60 [#permalink]
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These problems are about considering the worst case scenario.

How can I not get pairs?

The first four socks can be red, green, black, blue

They are also of the form ODD, ODD, ODD, ODD (1 of each)

The fifth sock has to form a pair. Let's assume we get a blue pair.

So now we have red, green, black, blue, blue,

and are of the form ODD, ODD, ODD, EVEN


How can I not get a pair on the 6th selection? I have to get a blue (there was an even number so adding one won't form a pair)

So now we have red, green, black, blue, blue, blue.

We are back to ODD, ODD, ODD, ODD

No matter what I get on the 7 selection it will form a pair and the form will be

ODD, ODD, ODD, EVEN

To maintain this "worst case scenario" the 8th selection must be of the same color as the 7th (be the even one).

The 9th must necessarily form another pair.

So (x,y) represent (pairs, selections).

(1,5) (2,7) (3,9)....(10,y)

We can great an equation y=2x+3. Substituting 10 for x we obtain 23.
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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60 [#permalink]
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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60 [#permalink]
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