Question

When the members of a family discussed where their annual reunion should take​ place, they found that out of all the family​ members, 11 would not go to a​ park, 12 would not go to a​ beach, 15 would not go to the family​ cottage, 7 would go to neither a park nor a​ beach, 8 would go to neither a beach nor the family​ cottage, 8 would go to neither a park nor the family​ cottage, 5 would not go to a park or a beach or to the family​ cottage, and 1 would go to all three places. What is the total number of family​ members?

Answer 1

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family​ cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$.

8 would go to neither a park nor the family​ cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family​ cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a​ beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family​ cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a​ beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a​ park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.