The computer club had ¾ as many members as the drama club. After half the drama club members left to join the computer club, the computer club had 27 more members than the drama club. How many members did the drama club have at first?
Because I don’t know and I am supposed to get help that’s what this website is about not being rude to ppl that need help.
oh, wow - a whole hour. Have you tried to solve it while you were waiting and whining?
c = 3/4 d
c + d/2 = 27 + d/2
clearly, c=27
so d = 4/3 * 27 = 36
check:
18 of the 36 drama members left to join the 27 computer club
27+18 = 45 = 36/2 + 27
To solve this problem, let's break it down step by step.
Let's assume the number of members in the drama club at first was D.
According to the problem, the computer club had ¾ as many members as the drama club. So, the number of members in the computer club at first was ¾D.
After half the drama club members left to join the computer club, the number of members in the computer club increased. So, the new number of members in the computer club is (¾D + (1/2) * D).
Also, the problem tells us that the computer club had 27 more members than the drama club after this change. So, we can set up the equation:
(¾D + (1/2) * D) = D + 27
To solve this equation, let's first simplify both sides:
3/4D + 1/2D = D + 27
Now, let's get rid of the fractions by multiplying the entire equation by 4:
3D + 2D = 4D + 108
Combine like terms:
5D = 4D + 108
Subtract 4D from both sides:
D = 108
Therefore, the drama club had 108 members at first.