A certain firm has 28 lawyers in three states. These lawyers have lice

Question

A certain firm has 28 lawyers in three states. These lawyers have license either in only one state or all three states. If 10 have license in state X, 11 have license in Y, 13 have license in Z. How many lawyers have license only in state Z ?

A. 3
B. 6
C. 10
D. 11
E. 13

A. 3
B. 6
C. 10
D. 11
E. 13

yashikaaggarwal · Accepted Answer

Total => 28 (A+B+C+D) ----------(1)
These lawyers have license either in only one state or all three states. If 10 have license in state X, 11 have license in Y, 13 have license in Z.
A+D = 10
B+D = 11
C+D = 13
Adding all 3
A+B+C+3D = 34 ------------(2)

Subtracting equation 1 from equation 2
A+B+C+3D = 34
A+B+C+D = 28
----------------
2D = 6 => D = 3

Now to get only C
C+D = 13
C = 13-3 => 10

Answer is C

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chetan2u · Answer

X = Only x + All
Y = Only y + All
Z = Only z + All
X+Y+Z=Only x + All +Only y + All + Only z + All
10+11+13=(Only x +Only y + Only z + All) + 2*All
34=28+2*All
All=3
So only z=Z-All=13-3=10

sanjitscorps18 · Answer

X = 10
Y = 11
Z = 13

X  U  Y  U  Z = X + Y + Z - X ∩ Y - Y ∩ Z - Z ∩ X + X ∩ Y ∩ Z

Now no two cities share lawyers, hence X ∩ Y - Y ∩ Z - Z ∩ X = 0

As a result of this scenario, we'll have to tweak the original formula and subtract the 2 * X  ∩  Y  ∩  Z as the segments don't overlap between 2 cities hence, we cannot subtract twice the tri-overlap area.

Hence X  U  Y  U  Z = X + Y + Z - 0 - 2 * X ∩ Y ∩ Z
=> 28 = 10 + 11+ 13 - 2 * X ∩ Y ∩ Z
=> 6 = 2 * X ∩ Y ∩ Z
=> X ∩ Y ∩ Z = 3

In order to find lawyers working only in City Z, we have to subtract lawyers working in all 3 cities from the total lawyers from City Z

Z - X ∩ Y ∩ Z
=> 13 - 3 = 10

Option C

AnkurGMAT20 · Answer

Bunuel , this is a GMAT Prep Focus mock question, please tag it.

Bunuel · Answer

________________________
Added the tag. Thank you!

laughingcowgirl · Answer

Can someone explain why the formula "Total=A+B+C−(AnB+AnC+BnC)+AnBnC+Neither" cannot be used to solve this question?

Bunuel · Answer

Because each of AnB, AnC, and BnC also includes AnBnC, and thus the sum of AnB + AnC + BnC won't be 0. You should use the second formula for three overlapping sets:

Total = A + B + C - (sum \ of \ EXACTLY \ 2-group \ overlaps) - 2*(all \ three) + Neither .

28 = 10 + 11 + 13 - 0 - 2*(all three) + 0

(all three) = 3

Only Z = Z - (all three) = 13 - 3 = 10.

Answer: C.

For further insights into the distinctions between these two formulas, check out  ADVANCED OVERLAPPING SETS PROBLEMS .

Hope it helps.

cs87956 · Answer

Because it's either 1 or all 3 --> there's no option of 2 which the formula includes

DanTheGMATMan · Answer

Pretty standard 3 group overlapping sets problem:

https://www.youtube.com/watch?v=byXL7i8kabA

GMATinsight · Answer

10+11+13-2*(License in all states) = 28

i.e. (License in all states) = 3

Licence only in state Z = 13-3 = 10

Answer: Option C

NEYR0N · Answer

"These lawyers have license either in  only one state  or  all three states. "

This rules out “exactly two states.”
 It also rules out “no states.”

A certain firm has 28 lawyers in three states. These lawyers have lice

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A certain firm has 28 lawyers in three states. These lawyers have lice

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