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# help

0
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+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

#2
+19064
+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 ?

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

heureka  Jan 7, 2016
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#1
+2493
0

i found 9/35

Solveit  Jan 7, 2016
#2
+19064
+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 ?

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

heureka  Jan 7, 2016
#3
+2493
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we both wrong the answer is 3/5 because he asking from female i find it now :P

Solveit  Jan 7, 2016
#4
+8613
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Good Morning!! :)

Hayley1  Jan 7, 2016
#5
+2493
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Good Good

Solveit  Jan 7, 2016
#6
+2493
0

in our country afternoon now

Solveit  Jan 7, 2016
#7
+2493
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thanks heureka ! :)

Solveit  Jan 7, 2016
#8
+2493
0

i mixed not afternoon i mean evening :)

Solveit  Jan 7, 2016
#9
+8613
+5

Good Evening! ( dracula's voice ) :)

Hayley1  Jan 7, 2016
#10
+2493
0

:D :D

Solveit  Jan 7, 2016
#11
+2493
0

Hayley who is on the photo is it you ?

Solveit  Jan 7, 2016
#12
+8613
0

Hmm?

Hayley1  Jan 7, 2016
#13
+2493
0

Avatar

Solveit  Jan 7, 2016
#14
+148
0

I like your porfile pic, Solviet!

InstagramModel  Jan 7, 2016
#15
+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
+2493
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thanks IntagramModel your is better ;)

Solveit  Jan 7, 2016
#17
+2493
0

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

Solveit  Jan 7, 2016
#18
+91921
+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
+84066
+5

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

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

CPhill  Jan 8, 2016
#20
+91921
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
+91921
+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
+19064
+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 ?

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

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

heureka  Jan 8, 2016
#23
+2493
0

i think guest s answer is best

Solveit  Jan 8, 2016

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