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10 Aug 2023, 12:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
73% (01:46) correct
27%
(01:57)
wrong
based on 1291
sessions
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A certain law firm has 28 lawyers and has offices in three states, X, Y, and Z. Every lawyer in the firm is either licensed in all three states or is licensed in only one of the three states. If 10 lawyers are licensed in State X, 11 are licensed in State Y, and 13 are licensed in State Z, how many lawyers are licensed only in State Z ?
A. 3
B. 7
C. 8
D. 10
E. 25
A. 3
B. 7
C. 8
D. 10
E. 25
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10 Aug 2023, 13:16
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Bunuel
...A certain law firm has 28 lawyers...
Given:
\(x + y + z + p= 28\) --(1)
...10 lawyers are licensed in State X...
\(x + p = 10\) --(2)
...11 are licensed in State Y...
\(y + p = 11\) --(3)
...13 are licensed in State Z..
\(z + p = 13\) --(4)
(2) + (3) + (4) =
\(x + y + z + 3p = 10 + 11 + 13 = 34\)
\(x + y + z + 3p = 34\) -- (5)
Subtract (1) from (5)
\(2p = 34 - 28 = 6\)
\(p = 3\)
From (4)
\(z + p = 13 - 3 = 10\)
Option D
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14 Aug 2023, 03:57
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Bunuel
To solve this problem, we can use one of the 3-overlapping-sets formulas:
Total # of Unique Elements = # in (Group X) + # in (Group Y) + # in (Group Z) - # in (Groups of Exactly Two) – 2[# in (Group of Exactly Three)] + # in (Neither)
Total # of Unique Elements = 28
# in (Group X) = 10
# in (Group Y) = 11
# in (Group Z) = 13
# in (Groups of Exactly Two) = 0
# in (Neither) = 0
So, we have:
28 = 10 + 11 + 13 – 0 - 2[# in (Group of Exactly Three)] + 0
2[# in (Group of Exactly Three)] = 6
# in (Group of Exactly Three) = 3
Since each of the 13 lawyers licensed in State Z is either licensed in all three states or is licensed only in State Z, we have:
# only in State Z = 13 – 3 = 10
Answer: D
