SOLUTION: Hi there were 1518 boys and girls in a hall.when 2/3 of the boys and 1/5 of the girls left there were twice as many girls as boys remaining. How many girls were there at the

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Hi there were 1518 boys and girls in a hall.when 2/3 of the boys and 1/5 of the girls left there were twice as many girls as boys remaining. How many girls were there at the       Log On


   



Question 983474: Hi
there were 1518 boys and girls in a hall.when 2/3 of the boys and 1/5 of the girls left there were
twice as many girls as boys remaining.
How many girls were there at the beginning
how many boys were there at the beginning
thanks

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
let g = original no. of girls
let b = original no. of boys
:
there were 1518 boys and girls in a hall.
g + b = 1518
when 2/3 of the boys and 1/5 of the girls left there were twice as many girls as boys remaining.
That means that 4/5 of the girls and 1/3 of the boys remained, so we can say
4%2F5g = 2(1%2F3b)
4%2F5g = 2%2F3b
multiply by 15 to get rid of the denominators
3(4g) = 5(2b)
12g = 10b
Simplify divide by 2
6g = 5b
b = 6%2F5g
b = 1.2 g
in the 1st equation replace b with 1.2g
g + 1.2g = 1518
2.2g = 1518
g = 1518/2.2
g = 690 girls originally
then
1518 - 690 = 828 boys originally
:
:
Check this, by finding the remaining girls and boys
1/3 * 828 = 276 boys remain
4/5 * 690 = 552 girls
2 * 276 = 552 twice as many girls as boys