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18 Nov 2024, 01:30
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E
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Law firm has branches in X, Y, Z states. Each lawyer has either a license in one state or in all three states. There are 10, 11, 13 lawyers licensed in the 3 states respectively, and 28 lawyers in total in the firm. How many lawyers are only licensed in state Z?
A. 2
B. 4
C. 6
D. 8
E. 10
A. 2
B. 4
C. 6
D. 8
E. 10
18 Nov 2024, 03:19
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We can use the sum of exactly 2 groups formula here
\(Total = X+Y+Z - (sum of exactly 2 groups) -2*(all 3)+Neither\)
sum of exactly 2 groups and neither will be 0, bcoz "Each lawyer has either a license in one state or in all three states."
So, \(28 = 10+11+13 - 0 - 2*A + 0\)
All three = \(3\)
So, Only \(Z= 10\)
Another method would be to draw venn diagram.
Only X + Only Y + Only Z + All three = 28
Only \(X = 10- a\)
Only \(Y = 11- a\)
Only \(Z = 13- a\)
So, \(34- 2a = 28\)
and \(a=3\)
Only \(Z= 10\)
I think E.
Hope it helps.
\(Total = X+Y+Z - (sum of exactly 2 groups) -2*(all 3)+Neither\)
sum of exactly 2 groups and neither will be 0, bcoz "Each lawyer has either a license in one state or in all three states."
So, \(28 = 10+11+13 - 0 - 2*A + 0\)
All three = \(3\)
So, Only \(Z= 10\)
Another method would be to draw venn diagram.
Only X + Only Y + Only Z + All three = 28
Only \(X = 10- a\)
Only \(Y = 11- a\)
Only \(Z = 13- a\)
So, \(34- 2a = 28\)
and \(a=3\)
Only \(Z= 10\)
I think E.
Hope it helps.
General Discussion
18 Nov 2024, 08:31
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Total= x+y+z-(x&y)-(x&z)-(y&z)-2(x,y&z)
28=10+11+13-0-2(common in x,y&z)
Common in x,y&z= 6/2= 3
So, only lawyers licensed in state Z= 13-3= 10
Answer: E
28=10+11+13-0-2(common in x,y&z)
Common in x,y&z= 6/2= 3
So, only lawyers licensed in state Z= 13-3= 10
Answer: E
