5. At a certain school, students have access to two sports: football and basketball. One-fifth of the football players play basketball, and one-seventh of the basketball players play football. Exactly 110 students are involved in only one sport. How many students are involved in both sports?

## 53rd Mathematical Competition for High School Students in Slovenia

Selection Competition, April 1, 2009

## Problems for 3rd Year Students
II/5. 1. method. Let the number of football players be denoted by $n$, the number of basketball players by $k$, and the number of students who engage in both sports by $d$. One-fifth of the football players also play basketball, so $\frac{n}{5}=d$. One-seventh of the basketball players also play football, so $\frac{k}{7}=d$. From this, it follows that $n=5d$ and $k=7d$.

The number of basketball players who do not play football is $k-d$, and the number of football players who do not play basketball is $n-d$. Therefore, the number of students who engage in only one sport is $(n-d) + (k-d)$, which is 110. Thus, $110 = n + k - 2d$. We get $110 = 10d$ or $d = 11$. Therefore, 11 students engage in both sports.

Introduction of notation 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=309&width=431&top_left_y=508&top_left_x=1418)
Equation $\frac{k}{7}=d$

Writing $n=5d$ and $k=7d$

2. method. Let $x$ be the number of students who only play basketball, $y$ be the number of students who only play football, and $z$ be the number of students who play both sports.

The total number of football players is $y + z$. Since one-fifth of the football players play basketball, it follows that $z = \frac{1}{5}(y + z)$, or $y = 4z$. The total number of basketball players is $x + z$, and one-seventh of them play football, which is $z$, so $z = \frac{1}{7}(x + z)$, or $x = 6z$.

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=323&width=443&top_left_y=1392&top_left_x=1412)

Since the number of students who engage in exactly one sport is $x + y$, we have $x + y = 110$, from which it follows that $110 = 6z + 4z = 10z$, so $z = 11$. Therefore, 11 students engage in both sports.

## Introduction of notation

1 point

Equation $z = \frac{1}{5}(y + z)$ (or equivalent) 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=75&width=1636&top_left_y=1930&top_left_x=216)

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=54&width=1631&top_left_y=1989&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=57&width=1633&top_left_y=2036&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-12.jpg?height=54&width=1631&top_left_y=2089&top_left_x=221)