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In a group of 50 people, at least 20 smoke cigar, at least 15 smoke ho
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23 Oct 2019, 06:05
23 Oct 2019, 06:05
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Difficulty:
95%
(hard)
Question Stats:
28% (02:51) correct
72%
(02:33)
wrong
based on 67
sessions
In a group of 50 people, at least 20 smoke cigar, at least 15 smoke hookah and at least 11 smoke cigarette. If 25 of the people in the group do not smoke, then at most how many smoke exactly one of cigar, hookah and cigarette?
(A) 7
(B) 25
(C) 4
(D) 14
(E) CBD
(A) 7
(B) 25
(C) 4
(D) 14
(E) CBD
In a group of 50 people, at least 20 smoke cigar, at least 15 smoke ho
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23 Oct 2019, 09:13
23 Oct 2019, 09:13
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From given, A≥20, B≥15, C≥11, N = 25
the general formula : Total = A+B+C - (II) + (III) + N
At least one = Exact one + Exact two + Exact three = 25
At least one = A+B+C -(II) + (III) = 25 ---> 25 + (II) - (III) = A+B+C ≥ 46 ---> (II)-(III) ≥ 21
Exact two = (II)-3(III)
Exact three = (III)
To maximize "Exact one", "Exact two" should be = 0, and (III) is minimum
so if (II)-3(III) = 0 , then 2(III) = (II)-(III) ≥ 21 and (III) ≥ 10.5 ---> (III) = 11
Exact (1) = At least (1) - Exact (II) - Exact (III) = 25 - 0 - 11 = 14
the general formula : Total = A+B+C - (II) + (III) + N
At least one = Exact one + Exact two + Exact three = 25
At least one = A+B+C -(II) + (III) = 25 ---> 25 + (II) - (III) = A+B+C ≥ 46 ---> (II)-(III) ≥ 21
Exact two = (II)-3(III)
Exact three = (III)
To maximize "Exact one", "Exact two" should be = 0, and (III) is minimum
so if (II)-3(III) = 0 , then 2(III) = (II)-(III) ≥ 21 and (III) ≥ 10.5 ---> (III) = 11
Exact (1) = At least (1) - Exact (II) - Exact (III) = 25 - 0 - 11 = 14
General Discussion
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Joined: 15 Feb 2018
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Re: In a group of 50 people, at least 20 smoke cigar, at least 15 smoke ho
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24 Oct 2019, 19:34
24 Oct 2019, 19:34
What is the question source?
