Bunuel
Of the 180 judges appointed by a certain President, 30 percent were women and 25 percent were from minority groups. If 1/9 of the women appointed were from minority groups, how many of the judges appointed were neither women nor from minority groups?
A. 75
B. 81
C. 87
D. 93
E. 99
Attachment:
judicialapptees..png [ 21.68 KiB | Viewed 6327 times ]
Order in which I filled the chart
1) 180 judges total
2)
# of women = 30% of 180 = (.3 * 180) =
54 women
3) # of men: (180 - 54) = 126 men
4) # of minority group persons
= 25% of 180 = (.25 * 180) = 45 members of minority groups
5) # of persons NOT in minority groups: (180 - 45) = 135
6)
# of women appointed who were from minority groups**
\(\frac{1}{9}\) of the women appointed were from minority groups
\((\frac{1}{9})\) * 54 = 6 of the women appointed were from minority groups
7) # of men appointed who WERE from minority groups
(45 - 6) = 39
8)
ANSWER:# of men who were NOT from minority groups = (??)
# of appointees were neither women nor from minority groups
(126 - 39) =
87Answer C
Am I missing something?
I have serious doubts about this answer and/or
what seems like inconsistency between
how to pick an answer (choose the intersection box)
and
how to calculate the # of women who are members of minority groups
(do NOT use the intersection box. If you enter "20" in that
intersection box, you'll get 101 as your answer. Not a choice.)
**Prompt states: "1/9 of the women appointed were from minority groups..."
In that statement, "of the women" controls the calculation.
• There are 54 women
\(\frac{1}{9}\) of those 54 women are from minority groups
\(\frac{1}{9}\) * 54 = 6
• Scenario
W = Women
A = Member of a minority group
X = all appointees
(1/9 of W were A) \(\neq\) (1/9 of X were W AND A)
Do not put (\(\frac{1}{9}\)*180) = 20 in the box in which members of minority groups and women intersect.