SOLUTION: There were twice as many boys as girls in a Swimming Club. After 40 boys and 15 girls left the club, the number of girls who remained was 3/5 of the number of boys who remained in
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Question 1201509: There were twice as many boys as girls in a Swimming Club. After 40 boys and 15 girls left the club, the number of girls who remained was 3/5 of the number of boys who remained in the club. How many children were there in the club at first? Found 2 solutions by math_tutor2020, ikleyn:Answer by math_tutor2020(3816) (Show Source):
x = number of girls initially
2x = number of boys initially
40 boys leave
2x becomes 2x-40
15 girls leave
x becomes x-15
"number of girls who remained was 3/5 of the number of boys who remained in the club" tells us that
x - 15 = (3/5)*(2x-40)
5(x - 15) = 3(2x-40)
5x - 75 = 6x-120
5x-6x = -120+75
-x = -45
x = 45 girls initially
which leads to
2x = 2*45 = 90 boys initially
45+90 = 135 is the final answer.
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Check:
90 boys at first
40 boys leave
90-40 = 50 boys left
45 girls at first
15 girls leave
45-15 = 30 girls left
ratio of girls remaining to boys remaining is 30/50 = 3/5, which confirms the answer is correct.
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There were twice as many boys as girls in a Swimming Club.
After 40 boys and 15 girls left the club, the number of girls who remained
was 3/5 of the number of boys who remained in the club.
How many children were there in the club at first?
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Originally x girls ans 2x boys.
The number x is unknown, and our goal is to find x.
Agter After 40 boys and 15 girls left the club,
(x-15) girls and (2x-40) boys remained.
The problem dictates us this proportion
= 3/5.
To solve it, first cross multiply
5(x-15) = 3*(2x-40).
Next simplify and find x
5x - 75 = 6x - 120
120 - 75 = 6x - x
45 = x.
ANSWER. Originally, there were 45 girls and 2*45 = 90 boys.
The total was 45 + 90 = 135 children.