In total, 1150 students were surveyed, 100 like both apples and oranges, and 400 students like apples only. The Venn diagram illustrates these groups. Finally, 950 students like at most one type of fruit.
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To solve this problem, we will use the principles of set theory. Let:
A be the set of students who like apples, with ∣ A ∣ = 500 .
O be the set of students who like oranges, with ∣ O ∣ = 600 .
x be the number of students who like both apples and oranges.
According to the problem, 50 students do not like either apples or oranges.
First, we need to find the total number of students surveyed. Since 50 students do not like either fruit, we have:
∣ A ∪ O ∣ ′ = 50
This means the number of students who like at least one type of fruit is:
∣ A ∪ O ∣ = ∣ U ∣ − 50
where U represents the universal set of all students surveyed and ∣ U ∣ is the total number of students surveyed.
Next, we apply the Principle of Inclusion-Exclusion to find the number of students who like at least one type of fruit:
∣ A ∪ O ∣ = ∣ A ∣ + ∣ O ∣ − ∣ A ∩ O ∣
Substituting the values, we have:
∣ A ∪ O ∣ = 500 + 600 − x
Since we know ∣ A ∪ O ∣ = ∣ U ∣ − 50 , equating and solving gives:
∣ U ∣ − 50 = 1100 − x
We need more details, such as the value of x , to finish solving this. Missing these details could suggest another way through filling the gaps with assumptions or calling for observed trends with surveys lacking feedback.
Let's assume the missing values (such as additional total survey data or explicit knowledge of the number who like both fruits) align closely with scenarios historically solved in classes but note that confirmation ensures accuracy.