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Re: A total of 9 women and 12 men reside in the 21 apartments that are in [#permalink]
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statement 1: sufficient
If 4 of the women are students, the chance of randomly selecting 1 among the 21 apartments is 4/21

statement 2: sufficient
If 5 of the women are not students, then 9-5=4 of the women are students. Same as above.

Answer D!
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Re: A total of 9 women and 12 men reside in the 21 apartments that are in [#permalink]
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A total of 9 women and 12 men reside in the 21 apartments that are in a certain apartment building, one person to each apartment. If a poll taker is to select one of the apartments at random, what is the probability that the resident of the apartment selected will be a woman who is a student?

(1) Of the women, 4 are students.
(2) Of the women, 5 are not students.

chetan2u
In Statement 2, How can we be so sure that the remaining 4 women are students?
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Re: A total of 9 women and 12 men reside in the 21 apartments that are in [#permalink]
Bunuel wrote:

Tough and Tricky questions: Word Problems.



A total of 9 women and 12 men reside in the 21 apartments that are in a certain apartment building, one person to each apartment. If a poll taker is to select one of the apartments at random, what is the probability that the resident of the apartment selected will be a woman who is a student?

(1) Of the women, 4 are students.
(2) Of the women, 5 are not students.


Kudos for a correct solution.


Total Residents =21
No. of Women = 9
Let X be the no of women who are students.

Probability to select Woman P (Woman) = \(\frac{9}{12}\)
Probability of getting a Women who is student = \(\frac{P(Women)*X}{9}\)
P(WomanStudent) = \(\frac{9}{21}*\frac{X}{9}\)
P(WomanStudent) = X/21

Now X we can find from both statement 1 and 2 seperately

Therefore Option D
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Re: A total of 9 women and 12 men reside in the 21 apartments that are in [#permalink]
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Re: A total of 9 women and 12 men reside in the 21 apartments that are in [#permalink]
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