MHB -ratio to ratio club membership

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Ratio
AI Thread Summary
The discussion revolves around determining the initial number of members in a club where 60% are men and 40% are women. After 45 new members joined, the number of female members increased by 50%, while the number of male members became 1.5 times the original count. The calculations reveal that there were initially 90 members, consisting of 54 men and 36 women. The final member count after the new additions is confirmed to be 135. The mathematical reasoning confirms that the initial ratios and conditions hold true throughout the problem.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
60% of the members of our club are men and the rest were women. When 45 new members joined the club, the number of a female members increased by 50% and the number of men was 1 1/2 times as many as before. how many club members were there initially?

ok from the $ \dfrac{60\%}{40\%}$ ration we have $\dfrac{3}{2}=\dfrac{m}{w}$

by changing the ratio $\dfrac{3+1.5(3)}{2+1}=\dfrac{15}{6}=\dfrac{5}{2}=\dfrac{m}{w}$$\dfrac{m}{m-45}=\dfrac{5}{2}\quad m=75$

howeve when I tried to appy this it didn't jive

thot this would be slam dung but :mad::mad::mad:
 
Mathematics news on Phys.org
The statements, “increased by 50%“ and “1.5 times as many as before” say the same thing, so the final ratio is unchanged from the 3/2 initial ratio. The ratio of m/w in the 45 that join is also 3/2.

Of the 45 new members, 27 are men and 18 are women

therefore, there were initially 54 men and 36 women

54+27 = 1.5(54)

36+18 = 1.5(36)
 
karush said:
60% of the members of our club are men and the rest were women. When 45 new members joined the club, the number of a female members increased by 50% and the number of men was 1 1/2 times as many as before. how many club members were there initially?

ok from the $ \dfrac{60\%}{40\%}$ ration we have $\dfrac{3}{2}=\dfrac{m}{w}$

by changing the ratio $\dfrac{3+1.5(3)}{2+1}=\dfrac{15}{6}=\dfrac{5}{2}=\dfrac{m}{w}$$\dfrac{m}{m-45}=\dfrac{5}{2}\quad m=75$

howeve when I tried to appy this it didn't jive

thot this would be slam dung but :mad::mad::mad:
I hope you mean "slam dunk"!

Let N be the number of members
"60% of the members of our club are men and the rest were women."
So there are .6N men and .4N women in the club.

"When 45 new members joined the club, the number of a female members increased by 50% and the number of men was 1 1/2 times as many as before."

As you have been told "increased by 50%" and and "1 1/2 times as many" are the same thing. There were .6N men and 50% of that is .3N so now there are .6N+ .3N= .9N men. There were .4N women and1/2 of that is .2N so now there are .6N women. There were 45 new member so now there are a total of N+ 45 members.

So .9N+ .6N= 1.5N= N+ 45.

Subtracting N from both sides .5N= 45. Dividing by .5 (the same as multiplying both sides by 2) N= 90.

There were, originally, 90 members.

Check: There were 90 members and 60% of them were men so there were 54 men. 40% of the members are women so there were 36 women. Half of 54 is 27 and half of 36 is 18. So the number of men increased by 27 to 54+ 27= 81 and the number of women increased by 18 to 36+ 18= 54. There are now 81+ 54= 135 members. Yes, 135- 90= 45 new members.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top