Let \( x \) be the total number of swim club members and \( y \) be the total number of diving club members. We know from the problem statement:
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The total number of members is 55: \[ x + y = 55 \]
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One-third of the swim team members are seniors: \[ \text{Seniors in swim team} = \frac{1}{3}x \]
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One-fifth of the diving team members are seniors: \[ \text{Seniors in diving team} = \frac{1}{5}y \]
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The total number of seniors in the two clubs is 13: \[ \frac{1}{3}x + \frac{1}{5}y = 13 \]
Now we have the two equations:
- \( x + y = 55 \)
- \( \frac{1}{3}x + \frac{1}{5}y = 13 \)
Next, we'll solve these equations. Let's start with the first equation: \[ y = 55 - x \]
Now, substitute this value of \( y \) into the second equation: \[ \frac{1}{3}x + \frac{1}{5}(55 - x) = 13 \]
Next, simplify this equation. Distributing \( \frac{1}{5} \): \[ \frac{1}{3}x + \frac{55}{5} - \frac{1}{5}x = 13 \] \[ \frac{1}{3}x + 11 - \frac{1}{5}x = 13 \]
Now, isolate the terms with \( x \): \[ \frac{1}{3}x - \frac{1}{5}x = 13 - 11 \] \[ \frac{1}{3}x - \frac{1}{5}x = 2 \]
To combine the fractions, find a common denominator, which is 15: \[ \frac{5}{15}x - \frac{3}{15}x = 2 \] \[ \frac{2}{15}x = 2 \]
Now, multiply both sides by 15: \[ 2x = 30 \] \[ x = 15 \]
Now, substitute \( x = 15 \) back into the first equation to find \( y \): \[ 15 + y = 55 \] \[ y = 55 - 15 = 40 \]
Thus, the total number of members in each club is:
- Swim club members: \( x = 15 \)
- Diving club members: \( y = 40 \)
Now let's verify:
- Seniors in swim: \( \frac{1}{3} \cdot 15 = 5 \)
- Seniors in diving: \( \frac{1}{5} \cdot 40 = 8 \)
- Total seniors: \( 5 + 8 = 13 \), which is correct.
Finally, we determine that the diving club has \( 40 - 15 = 25 \) more members than the swim club.
Summary:
- Total swim club members: \( 15 \)
- Total diving club members: \( 40 \)
- The diving club has \( 25 \) more members than the swim club.
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