Question

The swim and diving clubs at Riverdale High School have a total of 55 members and no student is a member of both teams. One-third of the swim team members are seniors and One-fifth of the diving team members are seniors. If there are 13 seniors in the two clubs, how many members does each club have? Let x represent the total number of swim club members and let y represent the total number of diving club members.

The equation that represents the total number of members is
.

The equation that represents the total number of seniors is
.

The diving club has
more members than the swim club.

Answer 1

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:

1. The total number of members is 55:
\[
x + y = 55
\]

2. One-third of the swim team members are seniors:
\[
\text{Seniors in swim team} = \frac{1}{3}x
\]

3. One-fifth of the diving team members are seniors:
\[
\text{Seniors in diving team} = \frac{1}{5}y
\]

4. 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:
1. \( x + y = 55 \)
2. \( \frac{1}{3}x + \frac{1}{5}y = 13 \)

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.