+0  
 
0
321
23
avatar+2493 

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

Solveit  Jan 7, 2016

Best Answer 

 #2
avatar+18715 
+20

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

 

 

\(\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \\ \end{array}\)

 

laugh

heureka  Jan 7, 2016
Sort: 

23+0 Answers

 #1
avatar+2493 
0

i found 9/35

Solveit  Jan 7, 2016
 #2
avatar+18715 
+20
Best Answer

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

 

 

\(\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \\ \end{array}\)

 

laugh

heureka  Jan 7, 2016
 #3
avatar+2493 
0

we both wrong the answer is 3/5 because he asking from female i find it now :P

Solveit  Jan 7, 2016
 #4
avatar+8621 
0

Good Morning!! :)

Hayley1  Jan 7, 2016
 #5
avatar+2493 
0

Good Good

Solveit  Jan 7, 2016
 #6
avatar+2493 
0

in our country afternoon now

Solveit  Jan 7, 2016
 #7
avatar+2493 
0

thanks heureka ! :)

Solveit  Jan 7, 2016
 #8
avatar+2493 
0

i mixed not afternoon i mean evening :)

Solveit  Jan 7, 2016
 #9
avatar+8621 
+5

Good Evening! ( dracula's voice ) :)

Hayley1  Jan 7, 2016
 #10
avatar+2493 
0

:D :D

Solveit  Jan 7, 2016
 #11
avatar+2493 
0

Hayley who is on the photo is it you ?

Solveit  Jan 7, 2016
 #12
avatar+8621 
0

Hmm?

Hayley1  Jan 7, 2016
 #13
avatar+2493 
0

Avatar

Solveit  Jan 7, 2016
 #14
avatar+148 
0

I like your porfile pic, Solviet!

InstagramModel  Jan 7, 2016
 #15
avatar
+10

Why NOT use concrete numbers for ALL professors!

Let us take the total numbers of professors to be=700

4/7 X 700=400 male professors

3/7 X 700=300 Female professors.

But we have a ratio of 2:5- being those >50 years of age to those <= 50 years of age. Therefore,

2/(2+5) X 700=200 male and female professors > 50 years of age, and similarly,

5/(2+5) X 700=500 male and female professors <=50 years of age. But 1/5 of male professors are older than 50 years, therefore,

1/5 X 400=80 male professors older than 50, and it follows that:

200 - 80 =120 female profeesors older than 50. Hence we have:

400 - 80 =320 male professors less than 50, and,

300 - 120=180 female professors less than 50. Therefore, the percentage of female professors <= 50 years, out of the total number of female professors is:

180/300=60% or 3/5. But, as a percentage of the total faculty of 700, then we have:

180/700=~25.71% or 9/35.

Guest Jan 7, 2016
 #16
avatar+2493 
0

thanks IntagramModel your is better ;)

Solveit  Jan 7, 2016
 #17
avatar+2493 
0

Thanks everybody! the answer is 3/5 we all misread the question :)

Solveit  Jan 7, 2016
 #18
avatar+91049 
+15

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

 

Let 

T = total number of proffessors

M= number of male professors            M=(4/7) T

F = number of female professors         F = (3/7) T

 

MU (males under 50)

MQ (males over 50)

FU (females under 50)

FQ (females over 50)

 

(MQ+FQ)=(2/7)T

(MU+FU)=(5/7)T


\(\frac{MQ}{M} =\frac{ 1}{5}\\ \frac{MQ}{ (4/7)T} =\frac{ 1}{5}\\ MQ =\frac{ 1}{5}\times \frac{4T}{7}=\frac{4T}{35}\\ \)


\((MQ+FQ)=\frac{2T}{7}\\ \frac{4T}{35}+FQ=\frac{2T}{7}\\ \frac{4T}{35}+FQ=\frac{10T}{35}\\ FQ=\frac{10T-4T}{35}\\ FQ=\frac{6T}{35}\\ so\\ FU=F-FQ\\ FU=\frac{3T}{7}-\frac{6T}{35}\\ FU=\frac{15T}{35}-\frac{6T}{35}\\ FU=\frac{9T}{35}\\~\\ \frac{FU}{F}=\frac{9T}{35}\div \frac{3T}{7}\\ \frac{FU}{F}=\frac{9T}{35}\times \frac{7}{3T}\\ \frac{FU}{F}=\frac{3}{5}\)

 

 

So 3/5 of the female professors are under 50.

Melody  Jan 8, 2016
 #19
avatar+78750 
+5

Impressive, Melody......I like the way you did that.....!!!

 

It may take me a few minutes to wrap my head around it.....!!!!

 

 

cool cool cool

CPhill  Jan 8, 2016
 #20
avatar+91049 
0

I have a definite advantage with questions like this - my native language is English.

 It can make a world of difference when interpreting the nuances of language  :)

Melody  Jan 8, 2016
 #21
avatar+91049 
+5

Thanks Chris :)

 

I used Q to represent older because o is too easily confused with 0.

Q looks a bit like an O that is why I used it.

 

These types of questions are always much easier if you you letters that are relevant to the question.

 

I think Heureka's answer is very similarto mine.  He just misinterpreted the last little bit that was being asked for - a very easy error to make :)

 

Guest's answer is also very similar.  I think the way guest has assigned an easier to work with total is really good.  

It often makes it easier for people to think if there are less letters :)

Melody  Jan 8, 2016
edited by Melody  Jan 8, 2016
 #22
avatar+18715 
+5

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

 

\(\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \end{array} \)

 

\(\begin{array}{rcl} \hline \text{continuation}\\ \hline \\ \frac{ \text{female}_{\le50}}{f+m} &=& \frac{9}{35} \\ \text{female}_{\le50} &=& \frac{9}{35}\cdot (f+m) \quad | \quad :f \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (1+\frac{m}{f}) \\\\ \boxed{~ \begin{array}{rcl} \frac{m}{m+f} &=& \frac{4}{7} \\ \frac{f}{m+f} &=& 1- \frac{4}{7} = \frac{3}{7}\\ \frac{ \frac{m}{m+f} } { \frac{f}{m+f} } &=& \frac{ \frac{4}{7} } { \frac{3}{7} }\\ \frac{ m } { f } &=& \frac43 \end{array} ~}\\\\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (1+ \frac43) \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (\frac73) \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9\cdot 7}{35\cdot 3} \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{3}{5} \\ \end{array}\)

 

laugh

heureka  Jan 8, 2016
 #23
avatar+2493 
0

i think guest s answer is best

Solveit  Jan 8, 2016

20 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details