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

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amanvermagmat
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Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   22 Mar 2018, 11:01
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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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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   22 Mar 2018, 13:56
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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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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   02 May 2021, 22:01
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Anush56 wrote:
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.



Think of it this way -

Stmnt 2 tells you that you need to withdraw 20 markers to ensure that you have at least 8 of the same colour.

So this means that you could pick 19 markers and still not have 8 of the same colour. But when you pick the 20th, you MUST have 8 of a colour.

So then, worst case scenario, how many of what colour markers could this 19 be such that the 20th has to ensure that you have 8 markers of 1 colour, no matter which marker you pick next?
We would assume that this happens in case we have 7 markers of all colours so that when we pick the next marker, it gives us 8 of one of the colours but that is not possible since we have only 19 markers. This means we must have total only 5 markers of one colour and 7 of the other two to make up 19. Now the leftover markers would only be of the other two colours. So whichever marker we pick next, it will give us 8 of one of the two other colours.

Since we know there are 23 black markers, it means we have only 5 markers of either red or blue.
If we have 5 red markers, then we have total 12 blue markers.
If we have 5 blue markers, then we have total 12 red markers.
Either case is possible.
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Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   22 Mar 2018, 12:26
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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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Josh has a big drawer full of 40 packets, each containing a marker [#permalink] 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] New post   02 May 2021, 19:32
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Anush56 wrote:
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.


From the second statment you can conclude that red or blue must be 5.
If red or blue could be 6 then the second statment would be false because choosing 20 packets the combination black:7, red:7 and blue:6 would be possible and you does not have 8 markers of a single coluor.
We can also conclude that red or blue can not be 4 because the minimum would be 19 packets and not 20 as the statment says.
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] 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] 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] New post   02 May 2021, 06:37
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   03 May 2021, 02:14
VeritasKarishma wrote:
Anush56 wrote:
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.



Think of it this way -

Stmnt 2 tells you that you need to withdraw 20 markers to ensure that you have at least 8 of the same colour.

So this means that you could pick 19 markers and still not have 8 of the same colour. But when you pick the 20th, you MUST have 8 of a colour.

So then, worst case scenario, how many of what colour markers could this 19 be such that the 20th has to ensure that you have 8 markers of 1 colour, no matter which marker you pick next?
We would assume that this happens in case we have 7 markers of all colours so that when we pick the next marker, it gives us 8 of one of the colours but that is not possible since we have only 19 markers. This means we must have total only 5 markers of one colour and 7 of the other two to make up 19. Now the leftover markers would only be of the other two colours. So whichever marker we pick next, it will give us 8 of one of the two other colours.

Since we know there are 23 black markers, it means we have only 5 markers of either red or blue.
If we have 5 red markers, then we have total 12 blue markers.
If we have 5 blue markers, then we have total 12 red markers.
Either case is possible.


Thank you so much ! This helps a lot. I find Min/ Max problems always tricky and addition when the question relates to maximizing or minimizing a certain quantity makes it all the more stressful.
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   05 May 2021, 12:51
Prompt:
Let Black=B, Blue=E, Red=R

B+E+R=40
B=23

This gives us

E+R = 17

We need to find E

1) E < R. Insufficient. (E,R) can be (1,16),(2,15) and so on.
2) Start with worst case scenario, you end up with 7 of each color. If this is the case, then you need to draw 22 pens (7 of each color, to guarantee next marker will be the 8th of a color). We however only need to draw 20. This means we we have either:

7+ = 7 or more

7+, 6, 6 (draw 12 from the two colors with 6 markers each, 7 from the 7+ color, and you're guaranteed 8th of the 7+ color on the 20th draw)
or
7+, 7+, 5 (draw 5 of the color with 5 markers, 7 from each of the 7+, then 20th marker is 8th of either of the 7+ colors)

Black is 23, so it's one of the 7+
This means 7+,6,6 combination is impossible because 6+6= 12 !=17 that we need for E+R

(E,R) can be (12,5) or (5,12)

2) is insufficient

Combine 1) and 2) and we see (E,R) = (5,12) and we have 5 blue markers.

Answer: C
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   08 Sep 2022, 08:00
KarishmaB wrote:
Anush56 wrote:
Hello VeritasKarishma
Can you please help me out with this question ?
I'm not understanding how from the second condition we can assume the below
Case 1: Red = 12, Blue = 5
Case 2: Red = 5, Blue = 12
I saw a similar question in Veritas question bank.



Think of it this way -

Stmnt 2 tells you that you need to withdraw 20 markers to ensure that you have at least 8 of the same colour.

So this means that you could pick 19 markers and still not have 8 of the same colour. But when you pick the 20th, you MUST have 8 of a colour.

So then, worst case scenario, how many of what colour markers could this 19 be such that the 20th has to ensure that you have 8 markers of 1 colour, no matter which marker you pick next?
We would assume that this happens in case we have 7 markers of all colours so that when we pick the next marker, it gives us 8 of one of the colours but that is not possible since we have only 19 markers. This means we must have total only 5 markers of one colour and 7 of the other two to make up 19. Now the leftover markers would only be of the other two colours. So whichever marker we pick next, it will give us 8 of one of the two other colours.

Since we know there are 23 black markers, it means we have only 5 markers of either red or blue.
If we have 5 red markers, then we have total 12 blue markers.
If we have 5 blue markers, then we have total 12 red markers.
Either case is possible.


Why Is this case not possible Red=9, blue 8 which adds to 17 and balance 3 markers are picked up from 23 black markers?
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink] New post   23 Sep 2023, 03:39
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Re: Josh has a big drawer full of 40 packets, each containing a marker [#permalink]
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