Question

A membership at Gym A costs $50 for 5 months. A membership at Gym B down the street costs $40 for 3 months. You write two equations in the form of y=kx
to try and figure out which membership would be cheaper for a year. What is the value of k
for the cheaper membership?

Answer 1

To determine which gym membership is cheaper for a year, we need to find the total cost for each gym when considering their respective membership terms.

**For Gym A:**
The cost is $50 for 5 months. To find the cost per month (k), we can express the cost for one month as follows:

\[
y_A = k_A x
\]

where:
- \( k_A = \frac{50}{5} = 10 \) (cost per month for Gym A)
- \( x \) is the number of months

The cost equation for Gym A over 12 months (or 1 year) would be:

\[
y_A = 10x
\]
where \( x \) would be 12 months.

Now, to calculate the total cost for a year:

Since Gym A has an effective 5-month cycle, we solve:

\[
\text{Total for a year at Gym A} = \left(\frac{12}{5}\right) \times 50 = 2.4 \times 50 = 120
\]

**For Gym B:**
The cost is $40 for 3 months. We express the cost for one month (k) similarly:

\[
y_B = k_B x
\]

where:
- \( k_B = \frac{40}{3} \) (approximately $13.33 per month for Gym B)
- \( x \) is the number of months

The cost equation for Gym B over 12 months would be:

\[
y_B = \left(\frac{40}{3}\right)x
\]

Now, to calculate the total cost for a year:

\[
\text{Total for a year at Gym B} = \left(\frac{12}{3}\right) \times 40 = 4 \times 40 = 160
\]

Now we compare the yearly totals for both gyms:
- Total for Gym A for one year: $120
- Total for Gym B for one year: $160

Since Gym A is cheaper, the value of \( k \) for the cheaper membership (Gym A) is:

\[
\boxed{10}
\]