Question 25721
Let b = number of boys on the 1st day of school
let g = number of girls on the 1st day of school
so, b + g equal number of boys + number of girls
60% of (b + g) is boys on the 1st day
later on, 6 girls leave and 6 boys are added
That means (b + g) does not change, but boys are now 75% of (b + g)
so, on the 1st day,
(a).....b/(b + g) = .60
and later on,
(b).....(b + 6)/(b + g) = .75
solve the first equation for b
b = .6b + .6g
.4b = .6g
b = (3/2)g
now solve the second equation for g
.75(b + g) = b + 6
.75b + .75g = b + 6
.75g = .25b + 6
express as fractions
(3/4)g = (1/4)b + 6
3g = b + 24
g = (1/3)b + 8
substitute b = (3/2)g (from before)
g = (1/3)(3/2)g + 8
g = (1/2)g + 8
2g = g + 16
g = 16
now substitute this value in either of the two equations (a) or (b)
b/(b + 16) = 3/5
b = (3/5)(b + 16)
(5/3)b = b + 16
(2/3)b = 16
b = (3/2)16
b = 24
so, there are 16 girls and 24 boys on the first day of school
If I add 6 boys and subtract 6 girls like the problems says
are the boys 75% of (boys + girls)?
(24 + 6) /((24 + 6) +(16 - 6)) = .75
30/(30 + 10) = 30/40 = .75
answer checks