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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
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16 Sep 2014, 01:11
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: M2124
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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 Looking for 700



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Re: M2124
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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: M2124
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01 Jul 2015, 07:36
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: M2124
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01 Jul 2015, 07:38
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: M2124
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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: M2124
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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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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: M2124
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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: M2124
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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/m21184278.html#p1924687
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Re: M2124
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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 XylanCan 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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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 XylanCan 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 testtaker. Given: 10 Red pills  5 Blue pills  12 Yellow Pills  Total pills: \(10 + 5 + 12 = 27\) Qstatement:
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? Qmeaning: 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 casewords; they ARE contextually vital. Let's solve it in the reverseway 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. Leftout 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 \(1\) leftout 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 \(2\) leftout pills are B, you are STILL left with a minimum of \(3\) pills of B Another decreasedpick, i.e., Picking 24 pills MAY result in being having LESS THAN 3 pills of B, i.e., 2 pills of B in the pickedpill selection, If the \(3\) leftout pills are of Bcolor. 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 LasVegas, 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 madeup story, Yet Contextually striking! The unfortunate picker in order to CONFIRM his chances of winning thought of the worst case scenario, i.e., whatif 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! \(12 + 10 + 3 = 25\)  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: M2124
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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 XylanCan 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 testtaker. Given: 10 Red pills  5 Blue pills  12 Yellow Pills  Total pills: \(10 + 5 + 12 = 27\) Qstatement:
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? Qmeaning: 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 casewords; they ARE contextually vital. Let's solve it in the reverseway 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. Leftout 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 \(1\) leftout 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 \(2\) leftout pills are B, you are STILL left with a minimum of \(3\) pills of B Another decreasedpick, i.e., Picking 24 pills MAY result in being having LESS THAN 3 pills of B, i.e., 2 pills of B in the pickedpill selection, If the \(3\) leftout pills are of Bcolor. 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 LasVegas, 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 madeup story, Yet Contextually striking! The unfortunate picker in order to CONFIRM his chances of winning thought of the worst case scenario, i.e., whatif 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! \(12 + 10 + 3 = 25\)  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!










