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Daniel has a drawer full of 59 packets, each containing a marker

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Daniel has a drawer full of 59 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 07:37
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26% (02:25) correct 74% (02:14) wrong based on 51 sessions

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Daniel has a big drawer full of exactly 59 packets, each containing a marker which is either black or blue or red in colour. All 59 packets are sealed and are completely identical from outside. It is not possible to know which colour marker is inside unless the packet is fully opened. Out of 59, 23 packets contain a black marker each. How many packets contain a blue marker?

(1) If Daniel withdraws packets without looking at their contents, he needs to draw minimum 57 packets to ensure that he has ALL marbles of any single colour out of red, blue, black.

(2) Drawer has less packets containing red markers than those containing blue markers.
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Re: Daniel has a drawer full of 59 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 08:03
amanvermagmat wrote:
Daniel has a big drawer full of exactly 59 packets, each containing a marker which is either black or blue or red in colour. All 59 packets are sealed and are completely identical from outside. It is not possible to know which colour marker is inside unless the packet is fully opened. Out of 59, 23 packets contain a black marker each. How many packets contain a blue marker?

(1) If Daniel withdraws packets without looking at their contents, he needs to draw minimum 57 packets to ensure that he has ALL marbles of any single colour out of red, blue, black.

(2) Drawer has less packets containing red markers than those containing blue markers.



Total - 59 packets
Black - 23 packets
so R + Blue = 59-23=36
Blue = ?

(1) If Daniel withdraws packets without looking at their contents, he needs to draw minimum 57 packets to ensure that he has ALL marbles of any single colour out of red, blue, black.
Nothing much, this can be easily deducted ..
to ensure that you have complete colour of any one type will always MEAN leaving ONE each of other two.. so 59-2=57
insuff

(2) Drawer has less packets containing red markers than those containing blue markers.
R < Blue
insuff

combined
R<Blue ..
Insuff

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Daniel has a drawer full of 59 packets, each containing a marker  [#permalink]

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New post 19 Apr 2018, 22:23
Hi chetan2u,

shouldn't the answer be C?
First two statements are insufficient.

To count minimum number of marbles to have marbles of all possible color is (Highest + next highest + the third highest + ........ + second lowest + 1)
Combining,
Let us say Black is least of all, in which case, minimum no of marbles to draw to ensure we have marbles of all color = Blue + Red + 1 = 57 => Blue + Red = 56 (which is not possible as given sum is 59 and black is 23)
so Black is not lowest of all,
now from statement 2, Red < Blue
So Red must be lowest of all.

so, Black + Blue + 1 = 57 => 23 + Blue + 1 = 57 => Blue = 33, are we not getting number of blues?

Am i missing anything?

Please help.

Thanks
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Re: Daniel has a drawer full of 59 packets, each containing a marker  [#permalink]

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New post 19 Apr 2018, 22:53
hellosanthosh2k2 wrote:
Hi chetan2u,

shouldn't the answer be C?
First two statements are insufficient.

To count minimum number of marbles to have marbles of all possible color is (Highest + next highest + the third highest + ........ + second lowest + 1)
Combining,
Let us say Black is least of all, in which case, minimum no of marbles to draw to ensure we have marbles of all color = Blue + Red + 1 = 57 => Blue + Red = 56 (which is not possible as given sum is 59 and black is 23)
so Black is not lowest of all,
now from statement 2, Red < Blue
So Red must be lowest of all.

so, Black + Blue + 1 = 57 => 23 + Blue + 1 = 57 => Blue = 33, are we not getting number of blues?

Am i missing anything?

Please help.

Thanks


Hello

So we are talking about combining the two statements.

There are 23 black markers, rest 36 contain some red and some blue, but red markers are less, so red must be least in number. Lets say there are 'x' red markers, so '36-x' is the number of blue markers. And x < 36-x so we have x < 18.

Now, to get marbles of any one single colour, we first take the worst case scenario. Worst case scenario would be if we get 22 black markers, 'x-1' red markers and '36-x-1' blue markers (basically we have all but one marble of each color). And in this worst case scenario, now whichever next marker comes up will complete the objective - of having all marbles of any one particular colour. So lets add up 22, x-1 and 36-x-1 and 1:- this will give us the minimum number required.
22 + x-1 + 36-x-1 + 1 = 57.
So irrespective of the value of x, answer is coming to be 57 only (this answer is now independent of the number of red and blue markers).

You can check, lets say there are 23 black, 17 red, 19 blue. Then minimum no of markers = 22+16+18 + 1 = 57
Lets check now with 23 black, 10 red, 26 blue. Then minimum no of markers = 22+9+25 + 1 = 57 again.

So we cannot determine the number of red or blue markers based on this info.
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Re: Daniel has a drawer full of 59 packets, each containing a marker  [#permalink]

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New post 19 Apr 2018, 23:31
Hi amanvermagmat,

Thanks for reply.
Yes i misread the statement 1 as minimum number of markers required to get atleast one marker of each color and ended up with C.

Now understood as, minimum number of markers required to ensure to get all markers of one color
(worst case) Black - 1 + Blue - 1 + Red - 1 + 1 = 57
=> Blue + Read = 36

even with both statements, we can't get the number of blues as there are many possible values for Blue > Red and satisfy Blue + Red = 36

Clean E
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Re: Daniel has a drawer full of 59 packets, each containing a marker &nbs [#permalink] 19 Apr 2018, 23:31
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