SOLUTION: In a church choir the ratio of males to females is 2:3. On one Sunday service ten male members were absent and six new female members joined the choir as guests for the day. If on
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Question 1156580: In a church choir the ratio of males to females is 2:3. On one Sunday service ten male members were absent and six new female members joined the choir as guests for the day. If on this day the ratio of males to females was 1:3, how many regular choir members does the choir have? Found 2 solutions by Theo, Edwin McCravy:Answer by Theo(13342) (Show Source):
when 10 males are missing and 6 females are added, the ratio becomes 1/3.
you get (m-10)/(f+6) = 1/3.
you have two equations that need to be solved simultaneously.
these two equations can be solved simultaneously in various ways. i chose to make them into standard inear equations and then add or subtract one equation from the other to narrow down the unknown variables to one so they can be solved.
first equation is m/f = 2/3.
multiply both sides of this equation by f to get m = 2/3 * f.
multiply both sides of this equation by 3 to get 3m = 2f.
subtract 2f from both sides of this equation to get 3m - 2f = 0.
second equation is (m-10)/(f+6) = 1/3.
multiply both sides of this equation by (f+6) to get m-10 = 1/3 * (f+6).
multiply both sides of this equation by 3 to get 3 * (m-10) = f + 6.
simplify to get 3m - 30 = f + 6.
subtract f from both sides of this equation and add 30 to both sides of this equation and combine like terms to get 3m - f = 36.
your two equations that need to be solved simultaneously are now:
3m - 2f = 0
3m - f = 36
subtract the first equation from the second to get f = 36.
go back to the original ratio of m/f = 2/3 and replace f with 36 to get m/36 = 2/3.
multiply both sides of this equation by 36 to get m = 36 * 2/3.
simplify to get m = 24.
you have m = 24 and f = 36.
24/36 is the same as 2/3, so the original ratio is correct.
(24 - 10) / (36 + 6) = 14/42.
divide both numerator and denominator by 14 to get 1/3.
the modified ratio is also correct.
this confirms the values of m and f are good.
your solution is that the choir has 24 males and 36 females as regular members.
that's a total of 60.
On one Sunday service ten male members were absent and six new female members joined the choir as guests for the day.
Then the ratio was
If on this day the ratio of males to females was 1:3,
how many regular choir members does the choir have?
Solve this system:
Cross multiply both equations:
From the second equation, substitute (3M-36) for F
in the first equation, and solve to find M. Then substitute
what you get for M in 3M-36=F to find F.
I'll give you only the answer for the number of females, F=36.
You find the number of men and add them to find the number of
regular choir members.
Oh crap! The other tutor already gave you the answer, darn it!
Edwin