(18) Out of 120 students in SSS 3, [tex]$\frac{5}{8}$[/tex] like football and [tex]$\frac{2}{5}$[/tex] like swimming. Every student like at least one of these sports. (a) What fraction of the students in [tex]$SSS _3$[/tex] like both subjects? (b) What fraction of the students like football but not swimming? (c) What fraction of the students like swimming only?
Answer (2)
Calculate the number of students who like football: 8 5 × 120 = 75 .
Calculate the number of students who like swimming: 5 2 × 120 = 48 .
Find the number of students who like both: 120 = 75 + 48 − n ( F ∩ S ) ⟹ n ( F ∩ S ) = 3 . Then, the fraction is 120 3 = 40 1 .
Determine the fractions for football only and swimming only: 120 75 − 3 = 5 3 and 120 48 − 3 = 8 3 .
The fractions are: 40 1 , 5 3 , 8 3 .
Explanation
Analyze the problem Let's analyze the problem. We have 120 students in total. We know the fraction of students who like football and the fraction who like swimming. We also know that every student likes at least one of the sports. We need to find the fraction of students who like both sports, the fraction who like football but not swimming, and the fraction who like swimming only.
Set up the equations Let F be the set of students who like football, and S be the set of students who like swimming. Let n ( F ) be the number of students who like football, n ( S ) be the number of students who like swimming, and n ( F ∩ S ) be the number of students who like both. We are given that n ( F ) = 8 5 × 120 and n ( S ) = 5 2 × 120 .
Since every student likes at least one sport, n ( F ∪ S ) = 120 .
We know that n ( F ∪ S ) = n ( F ) + n ( S ) − n ( F ∩ S ) . Therefore, 120 = 8 5 × 120 + 5 2 × 120 − n ( F ∩ S ) .
Calculate the number of students who like both sports First, let's calculate the number of students who like football: n ( F ) = 8 5 × 120 = 75 Next, let's calculate the number of students who like swimming: n ( S ) = 5 2 × 120 = 48 Now we can substitute these values into the equation: 120 = 75 + 48 − n ( F ∩ S ) 120 = 123 − n ( F ∩ S ) n ( F ∩ S ) = 123 − 120 = 3 So, 3 students like both football and swimming.
Calculate the fraction of students who like both sports The fraction of students who like both sports is: 120 n ( F ∩ S ) = 120 3 = 40 1 Therefore, 40 1 of the students like both football and swimming.
Calculate the fraction of students who like football but not swimming The number of students who like football but not swimming is: n ( F ) − n ( F ∩ S ) = 75 − 3 = 72 The fraction of students who like football but not swimming is: 120 72 = 60 36 = 10 6 = 5 3 Therefore, 5 3 of the students like football but not swimming.
Calculate the fraction of students who like swimming only The number of students who like swimming only is: n ( S ) − n ( F ∩ S ) = 48 − 3 = 45 The fraction of students who like swimming only is: 120 45 = 24 9 = 8 3 Therefore, 8 3 of the students like swimming only.
Final Answer In conclusion: (a) The fraction of students who like both sports is 40 1 .
(b) The fraction of students who like football but not swimming is 5 3 .
(c) The fraction of students who like swimming only is 8 3 .
Examples
Understanding sports preferences in a school can help in planning extracurricular activities. For example, knowing the fraction of students who like both football and swimming can guide the school in organizing combined sports events or allocating resources effectively. This type of analysis can also be used in marketing to target specific groups with tailored advertisements or promotions, ensuring that resources are used efficiently and that the needs and interests of the students are met.
In summary, 40 1 of the students like both football and swimming, 5 3 like football but not swimming, and 8 3 like swimming only.
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