Linear equation, please help solving

[pearls_precious]

In a particular area, there are 5% more male than female. If total population is 75, find the no. of each.
Answer is 30 females and 45 Males. I need to know what will the equation.

 

[JeffM]

In a particular area, there are 5% more male than female. If total population is 75, find the no. of each.
Answer is 30 females and 45 Males. I need to know what will the equation.

By the way, the answer makes no sense. Please check the problem. I suspect you will find that 5% should read 50%.

Name your unknowns first.

\(\displaystyle m = percentage\ of\ males.\)

\(\displaystyle f = percentage\ of\ females.\)

\(\displaystyle x = number\ of\ males.\)

\(\displaystyle y = number\ of\ females.\)

Now write down as mathematical equations what you know about your unknowns from general knowledge or the statement of the problem.

You have four unknowns so you need four equations. I'll give you one

\(\displaystyle x = \dfrac{75m}{100}.\)

What are the other three?

Once you have four equations, start substituting until you are down to a single equation; then solve.

That approach will let you solve almost every word problem.

 

[pearls_precious]

Thanks Jeff, but can you please help me solve it fully. somehow, i cant solve this one.

 

[JeffM]

Thanks Jeff, but can you please help me solve it fully. somehow, i cant solve this one.

Please check the problem is it 5% or 50%. That may be your problem.

Also given my hint on an equation for x, what might you guess is the equation for y?

 

[pearls_precious]

yeah I get the equation of y ...
This is what I am looking at

m + f = 100 where m & f is male & female percentage
m = f + 5%

replacing first equation,

f + 5 + f = 100%
2f = 95%
f = 42.5 %

No of females = 75 * 42.5%
= 35.6 = 36

No. of males = 39 males

 

[JeffM]

yeah I get the equation of y ...
This is what I am looking at

m + f = 100 where m & f is male & female percentage
m = f + 5%

replacing first equation,

f + 5 + f = 100%
2f = 95%
f = 42.5 %

No of females = 75 * 42.5%
= 35.6 = 36

No. of males = 39 males

Well you are doing the algebra almost correctly, but half of 95% is not 42.5%. But I have asked you twice to check the problem statement. The problem will make sense to you if the 5% is really 50%. (Of course, some times the book has a misprint)

 

[pearls_precious]

This is the question, it is from an assignment, it could be wrong. But also, if we use 50%, I still dont get the answer.

 

[JeffM]

This is the question, it is from an assignment, it could be wrong. But also, if we use 50%, I still dont get the answer.

\(\displaystyle m = 150\%\ of\ f \implies m = 1.5f.\)

\(\displaystyle m + f = 100 \implies 1.5f + f = 100 \implies 2.5f = 100 \implies f = \dfrac{100}{2.5} = \dfrac{1000}{25} = 40.\)

So the percentage of females is 40%.

The percentage of males is then 1.5 * 40% = 60%.

\(\displaystyle y = \dfrac{75f}{100} = \dfrac{75 * 40}{100} = \dfrac{3000}{100} = 30.\)

\(\displaystyle x = \dfrac{75m}{100} = \dfrac{75 * 60}{100} = \dfrac{4500}{100} = 45.\)

 

[pearls_precious]

why did we do 150% of f?

\(\displaystyle m = 150\%\ of\ f \implies m = 1.5f.\)

\(\displaystyle m + f = 100 \implies 1.5f + f = 100 \implies 2.5f = 100 \implies f = \dfrac{100}{2.5} = \dfrac{1000}{25} = 40.\)

So the percentage of females is 40%.

The percentage of males is then 1.5 * 40% = 60%.

\(\displaystyle y = \dfrac{75f}{100} = \dfrac{75 * 40}{100} = \dfrac{3000}{100} = 30.\)

\(\displaystyle x = \dfrac{75m}{100} = \dfrac{75 * 60}{100} = \dfrac{4500}{100} = 45.\)

 

[pearls_precious]

Why we doing m = 1.5f

m = 1.5f [1]
m + f = 75; m = 75 - f [2]

Substitute [2] in [1]: 1.5f = 75 - f
Solve for f.
If you can't, you need classroom help.

 

[JeffM]

Why we doing m = 1.5f

Because m is 50% more than f. 50% = 0.5 so m = f + 0.5f = 1.5f