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M21-24

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M21-24  [#permalink]

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New post 16 Sep 2014, 01:11
3
7
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

37% (01:12) correct 63% (01:24) wrong based on 126 sessions

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Re M21-24  [#permalink]

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New post 16 Sep 2014, 01:11
1
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E
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Re: M21-24  [#permalink]

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New post 01 Jul 2015, 06:13
Hi,

I think the anser is pretty logic, simple and fast. But I have one doubt... what if I have to solve with combinatorics formulas "C", wich would be the right solution?

Thanks a lot.

Regards.

Luis Navarro
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Re: M21-24  [#permalink]

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New post 01 Jul 2015, 06:19
luisnavarro wrote:
Hi,

I think the anser is pretty logic, simple and fast. But I have one doubt... what if I have to solve with combinatorics formulas "C", wich would be the right solution?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


This is not combinations question, it's a so called worst case scenario, min/max, question.

Check other Worst Case Scenario Questions from our Special Questions Directory to practice.

Hope it helps.
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Re: M21-24  [#permalink]

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New post 01 Jul 2015, 07:36
1
Bunuel wrote:
luisnavarro wrote:
Hi,

I think the anser is pretty logic, simple and fast. But I have one doubt... what if I have to solve with combinatorics formulas "C", wich would be the right solution?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


This is not combinations question, it's a so called worst case scenario, min/max, question.

Check other Worst Case Scenario Questions from our Special Questions Directory to practice.

Hope it helps.


Thanks a lot¡¡¡

Regards.

Luis Navarro Looking for 700
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Re: M21-24  [#permalink]

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New post 01 Jul 2015, 07:38
2
luisnavarro wrote:
Bunuel wrote:
luisnavarro wrote:
Hi,

I think the anser is pretty logic, simple and fast. But I have one doubt... what if I have to solve with combinatorics formulas "C", wich would be the right solution?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


This is not combinations question, it's a so called worst case scenario, min/max, question.

Check other Worst Case Scenario Questions from our Special Questions Directory to practice.

Hope it helps.


Thanks a lot¡¡¡

Regards.

Luis Navarro Looking for 700


Please use KUDOS button instead of "thank you" messages. Thank you!
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Re: M21-24  [#permalink]

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New post 12 Sep 2017, 08:05
Hi Bunuel,

The question has asked for the least number of balls that can be extracted.

For that reason I chose 18 as an answer. Can you please explain the solution in detail and why 18 isn't right.
15 balls (without any yellow balls), then 3 more to include at least 3 yellow and that makes the total as 18.

Please correct me what is wrong with my logic.

Thanks!
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Re: M21-24  [#permalink]

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New post 12 Sep 2017, 08:10
shalini14 wrote:
Hi Bunuel,

The question has asked for the least number of balls that can be extracted.

For that reason I chose 18 as an answer. Can you please explain the solution in detail and why 18 isn't right.
15 balls (without any yellow balls), then 3 more to include at least 3 yellow and that makes the total as 18.

Please correct me what is wrong with my logic.

Thanks!


We want to ensure that at least three pills of each color are among those extracted. 18 is not correct answer because we can have 10 red pills and then 8 yellow pills and we still won't have three pills of 3 different colors.

For more check other Worst Case Scenario Questions from our Special Questions Directory to practice.

Hope it helps.
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Re: M21-24  [#permalink]

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New post 27 Aug 2018, 04:15
hi bunuel,
irrespective of what as been asked (least or greatest),we have to consider worst case scenario right?
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Re: M21-24  [#permalink]

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New post 22 Sep 2018, 03:50
Bunuel wrote:
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E



buneul .... why cant least number of balls be 10 +5 +1? least number of balls to pick up would be 16 ( in best case and 22 in worst case).

so why are you taking the worst case,when question is about least?


thanks
sd
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Re: M21-24  [#permalink]

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New post 23 Sep 2018, 04:05
sdgmat89 wrote:
Bunuel wrote:
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E



buneul .... why cant least number of balls be 10 +5 +1? least number of balls to pick up would be 16 ( in best case and 22 in worst case).

so why are you taking the worst case,when question is about least?


thanks
sd


Explained here: https://gmatclub.com/forum/m21-184278.html#p1924687
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Re: M21-24  [#permalink]

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New post 06 Apr 2019, 11:01
Bunuel wrote:
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E




Hi Xylan

Can you please see what am i missing here??


The question asks for the least number of balls that can be extracted.


10 red balls
5 blue balls
and 3 yellow balls to include "at least three pills of each color" as the Q stem states

This makes a total of 18 balls !!



would appreciate your help!
Thanks
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M21-24  [#permalink]

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New post 06 Apr 2019, 22:30
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JIAA wrote:
Bunuel wrote:
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E


Hi Xylan

Can you please see what am i missing here??

The question asks for the least number of pills that can be extracted.

10 red pills
5 blue pills
and 3 yellow pills to include "at least three pills of each color" as the Q stem states

This makes a total of 18 pills !!

would appreciate your help!
Thanks


JIAA It's one of the coolest Qs that work on Logic rather than on Combinatorics/formulae.
It's okay to get it wrong in the learning stage. However, Do understand the finer nuances and what EXACTLY the Q is seeking from the test-taker.

Given:
    10 Red pills | 5 Blue pills | 12 Yellow Pills - Total pills: \(10 + 5 + 12 = 27\)
    Q-statement:
      What is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?
    Q-meaning: The LEAST number of pills one should pick such that the picker MUST have ALL the three pills of EACH color - focus on the capital case-words; they ARE contextually vital.

Let's solve it in the reverse-way to help you better understand:
    Let's say you pick all the 27 pills:
      Now, you would definitely have ALL the pills of EACH color, let alone at least 3 colors of each pill.
      Left-out pills: \(0\) -The pills which ARE NOT picked
    Now, Decrease the picks such that, you are still FOR SURE left with ALL the three pills of EACH color.
    Let's say you pick all the 26 pills:
      Left out pill: \(1\) - Can be anything from R, B, and Y. Worst case scenario: If the \(1\) left-out pill is B, you are STILL left with a minimum of \(4\) pills of B
    Now, this part is where it becomes interesting:
    Let's say you pick all the 25 pills:
      Left out pill: \(2\) - Can be anything from R, B, and Y. Worst case scenario: If the \(2\) left-out pills are B, you are STILL left with a minimum of \(3\) pills of B
    Another decreased-pick, i.e., Picking 24 pills MAY result in being having LESS THAN 3 pills of B, i.e., 2 pills of B in the picked-pill selection, If the \(3\) left-out pills are of B-color.

Since we need to BE SURE of at least 3 pills of EACH color, Picking \(>= 25\) pills is a SAFE bet.
Thus, if we PICK 25 pills, we can SURE that we will have AT LEAST 3 pills of EACH color:
    worst case scenario would be that 10R, 3B, and 12Y are picked: \(10 + 3 + 12 = 25\) - The least pick count

Another approach:
    Let's say, there is a highly unfortunate picker in the Casino of Las-Vegas, and he needs to be SURE of the minimum picks such that at least 3 pills of EACH color are present in order to win the GRAND prize.
      - A made-up story, Yet Contextually striking! ;)
    The unfortunate picker in order to CONFIRM his chances of winning thought of the worst case scenario, i.e., what-if all the initial picks gave him pills of identical color.
      Thus, the case where all his initial picks resulted in picking 12 Yellow ones
    Now, being aware of the unfortunateness, he continued to pick the other balls of the identical color, i.e., Next pick of 10 Red ones.
    The interesting scenario - After picking all the pills of R and Y, the picker is NOW sure that the next pick will DEFINITELY give him the blue ones.
    Now, knowing this, he can simply add 3 more to the initial picks to confirm the grand prize! :cool:
    \(12 + 10 + 3 = 25\) - The least pick count

Picking 18 does NOT guarantee that one would have at least 3 of each color.
18 may result from (12Y & 6R) OR (9Y & 9R). - Hence, incorrect.
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Re: M21-24  [#permalink]

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New post 07 Apr 2019, 04:57
Xylan wrote:
JIAA wrote:
Bunuel wrote:
Official Solution:

If a box contains 10 red pills, 5 blue pills, and 12 yellow pills, what is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?

A. 12
B. 17
C. 18
D. 23
E. 25

In the worst possible scenario, one will draw 22 pills without drawing a single blue pill. To make sure that there will be 3 blue pills among the extracted, one has to draw 3 more than 22 (i.e. 25).

Answer: E


Hi Xylan

Can you please see what am i missing here??

The question asks for the least number of pills that can be extracted.

10 red pills
5 blue pills
and 3 yellow pills to include "at least three pills of each color" as the Q stem states

This makes a total of 18 pills !!

would appreciate your help!
Thanks


JIAA It's one of the coolest Qs that work on Logic rather than on Combinatorics/formulae.
It's okay to get it wrong in the learning stage. However, Do understand the finer nuances and what EXACTLY the Q is seeking from the test-taker.

Given:
    10 Red pills | 5 Blue pills | 12 Yellow Pills - Total pills: \(10 + 5 + 12 = 27\)
    Q-statement:
      What is the least number of pills one must extract from the box to ensure that at least three pills of each color are among those extracted?
    Q-meaning: The LEAST number of pills one should pick such that the picker MUST have ALL the three pills of EACH color - focus on the capital case-words; they ARE contextually vital.

Let's solve it in the reverse-way to help you better understand:
    Let's say you pick all the 27 pills:
      Now, you would definitely have ALL the pills of EACH color, let alone at least 3 colors of each pill.
      Left-out pills: \(0\) -The pills which ARE NOT picked
    Now, Decrease the picks such that, you are still FOR SURE left with ALL the three pills of EACH color.
    Let's say you pick all the 26 pills:
      Left out pill: \(1\) - Can be anything from R, B, and Y. Worst case scenario: If the \(1\) left-out pill is B, you are STILL left with a minimum of \(4\) pills of B
    Now, this part is where it becomes interesting:
    Let's say you pick all the 25 pills:
      Left out pill: \(2\) - Can be anything from R, B, and Y. Worst case scenario: If the \(2\) left-out pills are B, you are STILL left with a minimum of \(3\) pills of B
    Another decreased-pick, i.e., Picking 24 pills MAY result in being having LESS THAN 3 pills of B, i.e., 2 pills of B in the picked-pill selection, If the \(3\) left-out pills are of B-color.

Since we need to BE SURE of at least 3 pills of EACH color, Picking \(>= 25\) pills is a SAFE bet.
Thus, if we PICK 25 pills, we can SURE that we will have AT LEAST 3 pills of EACH color:
    worst case scenario would be that 10R, 3B, and 12Y are picked: \(10 + 3 + 12 = 25\) - The least pick count

Another approach:
    Let's say, there is a highly unfortunate picker in the Casino of Las-Vegas, and he needs to be SURE of the minimum picks such that at least 3 pills of EACH color are present in order to win the GRAND prize.
      - A made-up story, Yet Contextually striking! ;)
    The unfortunate picker in order to CONFIRM his chances of winning thought of the worst case scenario, i.e., what-if all the initial picks gave him pills of identical color.
      Thus, the case where all his initial picks resulted in picking 12 Yellow ones
    Now, being aware of the unfortunateness, he continued to pick the other balls of the identical color, i.e., Next pick of 10 Red ones.
    The interesting scenario - After picking all the pills of R and Y, the picker is NOW sure that the next pick will DEFINITELY give him the blue ones.
    Now, knowing this, he can simply add 3 more to the initial picks to confirm the grand prize! :cool:
    \(12 + 10 + 3 = 25\) - The least pick count

Picking 18 does NOT guarantee that one would have at least 3 of each color.
18 may result from (12Y & 6R) OR (9Y & 9R). - Hence, incorrect.



Super Helpful! Thanks a ton!
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Re: M21-24   [#permalink] 07 Apr 2019, 04:57
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