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A drawer in a darkened room contains 100 red socks, 80 green socks, 60

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A drawer in a darkened room contains 100 red socks, 80 green socks, 60  [#permalink]

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New post 14 Mar 2019, 23:57
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Difficulty:

  65% (hard)

Question Stats:

46% (02:47) correct 54% (02:30) wrong based on 37 sessions

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TOUGH QUESTION:



A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair).


(A) 21

(B) 23

(C) 24

(D) 30

(E) 50

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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60  [#permalink]

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New post 15 Mar 2019, 01:46
My logic may be wrong. Luckily the question asks for 10 pairs (and not 10 pairs of the same colour). We just need the one combination which might disprove an answer. We can consider each answer in turn applying pretty much the same logic (each letter represents the colour of the sock and Bl. represents black socks):

A. He can pick out 19Bl., 1R and 1G. That's 9.5 pairs and short of the 10 pairs. At this point the youngster may consider turning the lights on, looking at the socks he is picking out or asking for help.
B. Same logic applies as above: 19Bl., 1R, 1G, 1B and the final one has to be any of Bl, B, R or G, making 10 pairs. It is necessary to think of mathematical counter-arguments: he could've equally have picked out 5G, 5B, 5R, 8Bl giving 10 pairs! Other combos? 6G, 6B, 6R and 2Bl giving 10 pairs. What if the first few socks were all of different colours then 1G, 1R, 1B, 20 Bl. 10 Pairs. I think it help picking out the extremes.
C. Again, same logic so as to test out the B. 1G, 1R, 1B and 21Bl works and at this point we probably would have exhausted most of our time on the question.
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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60  [#permalink]

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New post 16 Mar 2019, 01:53
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Bunuel wrote:

TOUGH QUESTION:



A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair).


(A) 21

(B) 23

(C) 24

(D) 30

(E) 50


so as to get 1 pair of socks we would require to pick atleast 5 socks ;similarly to get 2 pairs of socks 7 socks are required to be removed
for 3 pairs of socks 9 socks are required to be removed...
if we observe there is a pattern of 2p+3 ; p is no of desired pairs ; so here p = 10
we get 23
IMO B
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Re: A drawer in a darkened room contains 100 red socks, 80 green socks, 60  [#permalink]

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New post 23 Mar 2019, 00:24
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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New post 23 Mar 2019, 14:31
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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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New post 23 Mar 2019, 20:08
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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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A drawer in a darkened room contains 100 red socks, 80 green socks, 60   [#permalink] 23 Mar 2019, 20:08
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