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06 Jun 2017, 11:39
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
62% (02:29) correct
38%
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wrong
based on 443
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All of the arrangements made in a flower shop yesterday contained roses, tulips, or lilies, or some combination of the three types. 32 of the arrangements made yesterday contained roses, 42 contained tulips, and 41 contained lilies. 16 of the arrangements made yesterday contained roses and tulips, 18 contained lilies and tulips, and 8 contained roses and lilies. If 5 of the arrangements made yesterday contained all three types, how many total arrangements were made?
A. 63
B. 68
C. 78
D. 83
E. 115
A. 63
B. 68
C. 78
D. 83
E. 115
06 Jun 2017, 11:54
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Given data
P(R) = 32 | P(T) = 42 | P(L) = 41
P(R and T) = 16
P(L and T) = 18
P(R and L) = 8
P(R and T and L) = 5
Since P(R and T), P(L and T) and P(R and L) include P(R and T and L) their actual values are as follows:
P(R and T) = 11
P(L and T) = 13
P(R and L) = 3
P(Any 2) = 11 + 13 + 3 = 27
Also, since P(only R) contains P(R and T), P(R and L) and P(R and T and L), the actual value is 13
P(Only R) = 32-11-3-5
Similarly, we can deduce P(Only T) = 13 and P(Only L) = 20
P(Total) = P(Only R) + P(Only T) + P(Only L) + P(Any 2) + P(All 3) = 13 + 13 + 20 + 27 + 5 = 78
Therefore, 78(Option C) is total arrangements that could be made in the flower shop.
P(R) = 32 | P(T) = 42 | P(L) = 41
P(R and T) = 16
P(L and T) = 18
P(R and L) = 8
P(R and T and L) = 5
Since P(R and T), P(L and T) and P(R and L) include P(R and T and L) their actual values are as follows:
P(R and T) = 11
P(L and T) = 13
P(R and L) = 3
P(Any 2) = 11 + 13 + 3 = 27
Also, since P(only R) contains P(R and T), P(R and L) and P(R and T and L), the actual value is 13
P(Only R) = 32-11-3-5
Similarly, we can deduce P(Only T) = 13 and P(Only L) = 20
P(Total) = P(Only R) + P(Only T) + P(Only L) + P(Any 2) + P(All 3) = 13 + 13 + 20 + 27 + 5 = 78
Therefore, 78(Option C) is total arrangements that could be made in the flower shop.
29 May 2018, 00:40
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Bunuel
We have a formula for 3 sets.
The answer is C.
