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Re: At a certain university, 60% of the professors are women, and 70% of [#permalink]
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abhi758 wrote:
Quote:
therefore, women tenured + women professors + men tenured = 90%

men tenured = 30%


Not clear with the above part of the explanation??


well there are four options:

women tenured-men tenured
women not tenured-men not tenured

they are asking for professors that are women, tenured or both
that means that they want everyone except men not tenured
but women make up 60% (both tenured and not tenured) that leaves 30% tenured men professors
90%-60%
men tenured = 30%
women tenured = 40%
total tenured = 70%
women not tenured = 30%
total women = 60%
total men = 40%

and if you wanted to calculate men-not tenured = 10%
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Re: At a certain university, 60% of the professors are women, and 70% of [#permalink]
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another way to think about it:

they are asking for professors who are women, tenured or both

Total women = 60%
Total tenured = 70%
Total both = 130%

But adding both means we are double-counting women-tenured
130-90 = 40% are women tenured or the number being double-counted
That leaves 30% men tenured
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Re: At a certain university, 60% of the professors are women, and 70% of [#permalink]
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With problems like this were you are comparing two qualities, I think a simple table would be the easiest way to solve:

It was also easier to assume that there was a total of 10 people in the university. This is easier to deal with than percentages for me:

---------------------WOMEN----------MEN
TENURED---------------A---------------B--------7
NOT TENURED----------C---------------D--------3
-----------------------6----------------4-------10

Since 90% of the professors are women or tenured or both, this means that A+B+C = 9

So D = 10-(A+B+C) = 1, which means B is equal to (4-1) = 3.

Since we want the percentage of men tenured, it would be B/(B+D) or 3/4 = 75%
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Re: At a certain university, 60% of the professors are women, and 70% of [#permalink]
Quote:
A question from magoosh, i guess a Sub-600 level question.


Its not a very difficult question I agree, but is it really below level 600?

I wonder if there is anyway to truly determine the level of a question.
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At a certain university, 60% of the professors are women, an [#permalink]
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At a certain university, 60% of the professors are women, and 70% of the professors are tenured. If 90% of the professors are women, tenured, or both, then what percent of the men are tenured?

A. 25
B. 37.5
C. 50
D. 62.5
E. 75
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Re: Question [#permalink]
Bunuel, thanks. So "both" in the statement means all groups except men+nottenured?
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Re: Question [#permalink]
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mainhoon wrote:
Bunuel, thanks. So "both" in the statement means all groups except men+nottenured?


Yes, question means {Women/Tenured}+{Women/Not Tenured}+{Men/Tenured}, so all except {Men/Not Tenured}.

I agree here with you that the wording of the question is a little bit odd.
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Re: Question [#permalink]
Bunuel, thanks for the clarification. I have another request. In another post, there was a question about 4 numbers and 4 people and what was the probability for different scenarios. You had solved it using the combination approach for a variety of cases. I am trying to do the same but with the fractional approach (meaning picking each scenario and multiplying them).. Can you please check that post and reply? I don't know how to link this to that post..

It said something about 4 numbers and 4 people.. And what is the probability they will all select different numbers.. You answer was 4!/4^4. I wanted to arrive at the same answer using the fractional approach - I mean picking each case and multiplying them through.. Am not able to for this simple problem..

I say the first person can be selected in 4C1 ways and he can select any of the 4 numbers = 4/4
Second person can be selected in 3C1 ways and he can select any of the remaining 3 numbers = 1/3
Third in 2C1 and he can select 1/2
The fourth is left to select the last one.. Multiply them all together and I get
4 * 4/4 * 3 *1/3 * 2*1/2 - Makes no sense!
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Re: ven daigram [#permalink]
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shrive555 wrote:
At a certain university, 60% of the professors are women, and 70% of the professors are tenured. If 90% of the professors are women, tenured or both, then what percent of the men are tenured?

25
37.5
50
62.5
75

i made a ven diagram for this question but messed up "If 90% of the professors are women, tenured or both," here :(


You make a ven diagram for Woman and Tenured. and put their number as x,y, and z ( Take y as Woman and tenured)
Consider the total number of professor is 100.

According to the question,
x+y=60
y+z=70
x+y+z=90 ( Woman,Tenured or Both) ( It means 10 Professors are left out who are not woman and not tenured )

Now z is the number of professors who are not woman but tenured.
z=90-60=30

So total number of Men professors = 30 + 10(left out)=40

Percent of men who are tenured = (30/40)*100 = 75%

Answer is E.

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Re: ven daigram [#permalink]
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shrive555 wrote:
Quote:
According to the question,
x+y=60
y+z=70


cud'nt get this part :?:



X is the number of professor who are women only ,y is the number of professor who are woman and tenured both ,whereas z is the number of professor who are men and tenured .
60% of the professors are women, means they can just woman or (woman and tenured too).
Hence x+y=60

Same way,70% of the professors are tenured means they can be woman tenured or gents tenured

Hence y+z=70
[img]
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[/img]

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Re: ven daigram [#permalink]
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pzazz12:
When you have so many variations, try making a table as shown in the attachment.
When you put 60% as women, you get 40% are men.
When you put 70% as tenured, you get 30% as not tenured.
Now draw a venn diagram since you know 90% are women or tenured or both.
Check out the attachment. Calculations involved are almost negligible.
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Ten.pdf [95.74 KiB]
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Re: ven daigram [#permalink]
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For people who might be more at ease with OVER-LAPPING SETS TABLE from MGMAT - see attachment.
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Re: At a certain university, 60% of the professors are women [#permalink]
gmatgambler wrote:
At a certain university, 60% of the professors are women, and 70% of the professors are tenured. If 90% of the professors are women, tenured, or both, then what percent of the men are tenured?


A)25

B)37.5

C)50

D)62.5

E)75



Assume there are a hundred professors in all. that means there are 60 women and 40 men. and that there are 70 tenured and hence 30 NTed professors.
Now its said that 90 of the 100 are either tenured or women. that means non tenured men = 10. now there are 40 men that means 30 of them are tenured.
Thus 30 of 40 or 75% is tenured.
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Re: At a certain university, 60% of the professors are women, and 70% of [#permalink]
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Total Prof = 100

W: Women
M: Men
T: Tenured
NT: Not tenured

W(T) +W(NT) + M(T) + M(NT) = 100
Given,
W(T) + W(NT) = 60
M(T) + W(T)= 70
W(T) + W(NT) + M(T) = 90

Using all three equations, M(T)= 30
& M(NT)= 10

Desired percentage = M(NT) / M
=30/40
=75%
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Re: At a certain university, 60% of the professors are women, an [#permalink]
let,
W= women
m=men
t= tenured
u= untenured
total professor=100

W or t = 90 = w + t - w and t
W and t = 60+70-90=40

....... w..... m..................total
t...... 40... 70-40=30........ 70
u......................................30
total. ..60...100-60=40..... 100

30/40*100=75%

ANS:E
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Re: At a certain university, 60% of the professors are women, an [#permalink]
can some body explain me the logic of both?
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Re: At a certain university, 60% of the professors are women, an [#permalink]
KarishmaB wrote:
pzazz12:
When you have so many variations, try making a table as shown in the attachment.
When you put 60% as women, you get 40% are men.
When you put 70% as tenured, you get 30% as not tenured.
Now draw a venn diagram since you know 90% are women or tenured or both.
Check out the attachment. Calculations involved are almost negligible.


Dear Karishma,

Can you please explain me the logic behind the Venn Diagram regarding Women and Tenured? I understand that we consider 90% total, then 60% as for Women and 70% for Tenured. But, how we get the overlap of 40%?
Nevertheless, that's the tricky moment of that problem!
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