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0
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23
avatar+2496 

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 ?

 Jan 7, 2016

Best Answer 

 #2
avatar+24949 
+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

 Jan 7, 2016
 #1
avatar+2496 
0

i found 9/35

 Jan 7, 2016
 #2
avatar+24949 
+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+2496 
0

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

 Jan 7, 2016
 #4
avatar+8574 
0

Good Morning!! :)

 Jan 7, 2016
 #5
avatar+2496 
0

Good Good

 Jan 7, 2016
 #6
avatar+2496 
0

in our country afternoon now

 Jan 7, 2016
 #7
avatar+2496 
0

thanks heureka ! :)

 Jan 7, 2016
 #8
avatar+2496 
0

i mixed not afternoon i mean evening :)

 Jan 7, 2016
 #9
avatar+8574 
+5

Good Evening! ( dracula's voice ) :)

 Jan 7, 2016
 #10
avatar+2496 
0

:D :D

 Jan 7, 2016
 #11
avatar+2496 
0

Hayley who is on the photo is it you ?

 Jan 7, 2016
 #12
avatar+8574 
0

Hmm?

 Jan 7, 2016
 #13
avatar+2496 
0

Avatar

 Jan 7, 2016
 #14
avatar+148 
0

I like your porfile pic, Solviet!

 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.

 Jan 7, 2016
 #16
avatar+2496 
0

thanks IntagramModel your is better ;)

 Jan 7, 2016
 #17
avatar+2496 
0

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

 Jan 7, 2016
 #18
avatar+109518 
+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.

 Jan 8, 2016
 #19
avatar+111321 
+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

 Jan 8, 2016
 #20
avatar+109518 
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  :)

 Jan 8, 2016
 #21
avatar+109518 
+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 :)

 Jan 8, 2016
edited by Melody  Jan 8, 2016
 #22
avatar+24949 
+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

 Jan 8, 2016
 #23
avatar+2496 
0

i think guest s answer is best

 Jan 8, 2016

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