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Josh has a big drawer full of 40 packets, each containing a marker

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Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 11:01
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Question Stats:

53% (02:09) correct 47% (02:20) wrong based on 76 sessions

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Josh has a big drawer full of exactly 40 packets, each containing a marker which is either black or blue or red in colour. All 40 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 40, 23 packets contain a black marker each. How many packets contain a blue marker?

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

(2) If Josh withdraws packets without looking at their contents, he needs to draw minimum 20 packets to ensure that he has exactly 8 markers of any single colour out of red, blue, black.
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Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 12:26
From the question stem, we are able to get the following details
1. Each sealed packet has a marker - Red, Blue, Or Black(which we can't see from outside)
2. 23 of the 40 packets have a black marker. Therefore, Blue + Red = 17

1. Statement 1 tells us that Blue < Red
Case 1: Red = 13, Blue = 4
Case 2: Red = 10, Blue = 7
Case 3: Red = 9, Blue = 8
We cannot come to a unique value for the number of packets containing Blue markers (Insufficient)

2. If he needs to withdraw 20 markers to have 8 of a color,
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
We cannot come to a unique value for the number of packets containing Blue markers (Insufficient)

Combining the information from both the statements,
the only option possible is Black = 23, Blue = 5, Red = 12 (Sufficient - Option C)
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Re: Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 13:56
pushpitkc wrote:
2. If he needs to withdraw 20 markers to have 8 of a color,
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 9, Blue = 8
Case 2: Red = 8, Blue = 9
We cannot come to a unique value for the number of packets containing Blue markers(Insufficient)


What if Josh selects 7 black, 7 blue, 6 red? In that case, he will not have 8 markers of any one color.

Here's what I did...
40 packets
23 of them have black markers
(# of packets with red markers) + (# of packets with blue markers) = 17
We're asked to find the # of packets with blue markers.

Statement 1
(# of packets with blue markers) < (# of packets with red markers)
We can have (7 blue, 10 red), or (6 blue, 11 red).
Insufficient

Statement 2
This means that when Josh selects 19 packets, he is not guaranteed to have 8 packets of any one color. In the worst case scenario, Josh will select 7 packets of one color, 7 packets of another, and 5 packets of another color (total selected = 19 packets). The 20th packet he selects MUST give him 8 packets of one color, thus one of the colors must have 5 packets. So we either have (5 red, 12 blue) or (12 red, 5 blue).
Insufficient

Combine Statements 1 & 2
Per Stmt 1: red < blue
Per Stmt 2: (5 red, 12 blue) or (12 red, 5 blue)
Combined, we get 5 red, 12 blue

Answer: C
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Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 22 Mar 2018, 14:11
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aserghe1 wrote:
pushpitkc wrote:
2. If he needs to withdraw 20 markers to have 8 of a color,
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 9, Blue = 8
Case 2: Red = 8, Blue = 9
We cannot come to a unique value for the number of packets containing Blue markers(Insufficient)


What if Josh selects 7 black, 7 blue, 6 red? In that case, he will not have 8 markers of any one color.

Here's what I did...
40 packets
23 of them have black markers
(# of packets with red markers) + (# of packets with blue markers) = 17
We're asked to find the # of packets with blue markers.

Statement 1
(# of packets with blue markers) < (# of packets with red markers)
We can have (7 blue, 10 red), or (6 blue, 11 red).
Insufficient

Statement 2
This means that when Josh selects 19 packets, he is not guaranteed to have 8 packets of any one color. In the worst case scenario, Josh will select 7 packets of one color, 7 packets of another, and 5 packets of another color (total selected = 19 packets). The 20th packet he selects MUST give him 8 packets of one color, thus one of the colors must have 5 packets. So we either have (5 red, 12 blue) or (12 red, 5 blue).
Insufficient

Combine Statements 1 & 2
Per Stmt 1: red < blue
Per Stmt 2: (5 red, 12 blue) or (12 red, 5 blue)
Combined, we get 5 red, 12 blue

Answer: C




Have corrected my solution. Thanks for noticing aserghe1

I mistook the question "he needs to draw minimum 20 packets to ensure that he has exactly
8 markers of any single color out of red, blue, black." and thought it meant that he could take
8 of any color. But as you rightly pointed out there is a possibility that he may not have 8 of
either color if I used the numbers for markers: Black - 23, Red - 8, Blue - 9
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Re: Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 13 May 2019, 20:34
Why are considering Combinations of (12 & 5) only ?
Why it can't be (11&6) or (10&7)
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Re: Josh has a big drawer full of 40 packets, each containing a marker  [#permalink]

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New post 29 Aug 2019, 12:44
pushpitkc wrote:
From the question stem, we are able to get the following details
1. Each sealed packet has a marker - Red, Blue, Or Black(which we can't see from outside)
2. 23 of the 40 packets have a black marker. Therefore, Blue + Red = 17

1. Statement 1 tells us that Blue < Red
Case 1: Red = 13, Blue = 4
Case 2: Red = 10, Blue = 7
Case 3: Red = 9, Blue = 8
We cannot come to a unique value for the number of packets containing Blue markers (Insufficient)

2. If he needs to withdraw 20 markers to have 8 of a color,
2 cases are possible such that 8 markers of either color can be taken out.
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
We cannot come to a unique value for the number of packets containing Blue markers (Insufficient)

Combining the information from both the statements,
the only option possible is Black = 23, Blue = 5, Red = 12 (Sufficient - Option C)


Not certain how statement 2 results to Options of 12R,5B or 12B, 5R. Can it not be 11R,6B or 11B,6R this option also agrees to the statement of 8 of any one color. What am i missing here ?
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Re: Josh has a big drawer full of 40 packets, each containing a marker   [#permalink] 29 Aug 2019, 12:44
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