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?
12
17
18
23
25
OA is E
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rakeshd347
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theCodeToGMAT
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Considering worst case...
10 + 12 + 3 = 25
Answer [spoiler]{E}[/spoiler]?
10 + 12 + 3 = 25
Answer [spoiler]{E}[/spoiler]?
Last edited by theCodeToGMAT on Mon Oct 21, 2013 6:44 am, edited 1 time in total.
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Uva@90
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Consider the worst case scenario where he removes 12 yellow pills(which contributes more) then 10 red pills then 3 blue pills.
i.e. 12+10+3 =25
SO Answer is E
Regards,
Uva.
i.e. 12+10+3 =25
SO Answer is E
Regards,
Uva.
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Hi rakeshd347,
Uva@90 has described this type of question in a great way: the "worst case scenario." In questions that ask about what it would take to "ensure" or "guarantee" an outcome, you have to consider the "worst case scenario."
That "worst case scenario" here would be if you maximized the number of 2 colors before you got 3 of the final color:
12 yellows + 10 reds......+ 3 blues = 25
GMAT assassins aren't born, they're made,
Rich
Uva@90 has described this type of question in a great way: the "worst case scenario." In questions that ask about what it would take to "ensure" or "guarantee" an outcome, you have to consider the "worst case scenario."
That "worst case scenario" here would be if you maximized the number of 2 colors before you got 3 of the final color:
12 yellows + 10 reds......+ 3 blues = 25
GMAT assassins aren't born, they're made,
Rich
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GMATGuruNY
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As noted above, this is a WORST-CASE SCENARIO problem.rakeshd347 wrote: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?
12
17
18
23
25
OA is E
To GUARANTEE that 3 of each color are chosen, we must determine the worst-case scenario: the greatest number of marbles that can be removed WITHOUT choosing at least 3 of each color.
If all of the yellow pills and all of the red pills are removed first, the number of pills removed = 12+10 = 22.
Since there are more yellow pills and red pills than blue pills, 22 is the maximum number of pills that can be removed without choosing at least 3 of each color.
To remove at least 3 of each color, 3 blue pills must now be chosen, yielding the following total:
22+3 = 25.
The correct answer is E.
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sahilchaudhary
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We have to take the worst case.
12+10+3 = 25.
So, the answer is E.
12+10+3 = 25.
So, the answer is E.
Sahil Chaudhary
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