College B Sports Participation Analysis

This problem involves analyzing student participation in sports at College B, specifically focusing on Hockey and Football. We need to determine what percentage of students who played Hockey actually played only Hockey.

Interpreting Sports Participation Data

Let's define the variables based on the information provided:

• Let $T$ represent the total number of students in College B.
• Let $H_{only}$ represent the number of students who participated only in Hockey.
• Let $F_{only}$ represent the number of students who participated only in Football.
• Let $H \cap F$ represent the number of students who participated in both Hockey and Football.

The problem states that every student participated in at least one sport. This means the total number of students $T$ is the sum of students in these three categories:

$T = H_{only} + F_{only} + (H \cap F)$

Sports Participation Data Interpretation

We are given the following key pieces of information:

• 36% of the students participated in both sports:

  $H \cap F = 0.36 \times T$

• The number of students who participated only in Football ($F_{only}$) is 120% more than those who participated only in Hockey ($H_{only}$). This can be written as:

  $F_{only} = H_{only} + (1.20 \times H_{only})$

  $F_{only} = 2.20 \times H_{only}$

• The total number of students who participated in only one sport is 960:

  $H_{only} + F_{only} = 960$

Calculating Only One Sport Participants

We can now use the relationships derived to find the exact number of students in each category.

Substitute the expression for $F_{only}$ (from the second point) into the equation for the total students participating in only one sport (the third point):

$H_{only} + (2.20 \times H_{only}) = 960$

Combine the terms involving $H_{only}$:

$3.20 \times H_{only} = 960$

Solve for $H_{only}$:

$H_{only} = \frac{960}{3.20} = \frac{9600}{32}$

$H_{only} = 300$

Now, calculate the number of students who played only Football ($F_{only}$):

$F_{only} = 2.20 \times H_{only} = 2.20 \times 300$

$F_{only} = 660$

We can verify this: $H_{only} + F_{only} = 300 + 660 = 960$, which matches the given information.

Total Students Calculation

We know that the total number of students $T$ is the sum of those playing only Hockey, only Football, and both sports:

$T = H_{only} + F_{only} + (H \cap F)$

Substitute the calculated values and the given relationship for $H \cap F$:

$T = 960 + (0.36 \times T)$

To find the total number of students $T$, rearrange the equation:

$T - 0.36 \times T = 960$

$0.64 \times T = 960$

$T = \frac{960}{0.64} = \frac{96000}{64}$

$T = 1500$

So, the total number of students in College B is 1500.

Both Sports Participants Calculation

Now we can calculate the number of students who participated in both Hockey and Football:

$H \cap F = 0.36 \times T = 0.36 \times 1500$

$H \cap F = 540$

Total Hockey Participants Calculation

The total number of students who participated in Hockey ($H_{total}$) includes those who played only Hockey and those who played both sports:

$H_{total} = H_{only} + (H \cap F)$

$H_{total} = 300 + 540$

$H_{total} = 840$

Required Percentage Calculation

The question asks for the percentage of students who participated in Hockey that participated in only Hockey. This is calculated as:

Percentage = $\frac{H_{only}}{H_{total}} \times 100\%$

Substitute the values we found:

Percentage = $\frac{300}{840} \times 100\%$

Simplify the fraction:

$\frac{300}{840} = \frac{30}{84} = \frac{5}{14}$

Now, calculate the percentage:

Percentage = $\frac{5}{14} \times 100\% = \frac{500}{14}\%$

Simplify the fraction further:

Percentage = $\frac{250}{7}\%$

Convert the improper fraction to a mixed number:

$\frac{250}{7} = 35 \frac{5}{7}$

Therefore, the percentage is $35\frac{5}{7}\%$.