Question

the greens are moving. their real - estate agent located 78 houses listed for sale in their price range. of those houses listed for sale, 46 had a finished basement, 53 had a three - car garage, 34 had a finished basement and a three - car garage. complete parts a) through c). a) how many had a finished basement but not a three - car garage? (type a whole number.) b) how many had a three - car garage but not a finished basement? (type a whole number.) c) how many had either a finished basement or a three - car garage? (type a whole number.)

Explanation:

Step1: Define sets

Let $A$ be the set of houses with a finished basement, $|A| = 46$, and $B$ be the set of houses with a three - car garage, $|B| = 53$. The total number of houses is $n = 78$. The number of houses that have both is not given directly, but we can use set - theoretic concepts.

Step2: Find number for part a

We want to find the number of houses that have a finished basement but not a three - car garage. Using the formula $|A\cap\overline{B}|=|A|-|A\cap B|$. First, we assume we know the number of houses with both from the context (if not, we can't solve precisely). Let's assume we use the principle that $|A\cup B|=|A| + |B|-|A\cap B|$. We know that the number of houses with either a finished basement or a three - car garage or both is related to the total number of houses. But if we assume we want to find $|A\cap\overline{B}|$, and we know that the number of houses with both is say $x$. Then $|A\cap\overline{B}|=|A|-x$. If we assume no other information about the intersection, we note that the number of houses with a finished basement is 46. Let's assume the number of houses with both is 12 (a made - up value for illustration purposes, as the full data about intersection is not given clearly in the problem statement). Then $|A\cap\overline{B}|=46 - 12=34$.

Step3: Find number for part b

We want to find the number of houses that have a three - car garage but not a finished basement, i.e., $|B\cap\overline{A}|$. Using the formula $|B\cap\overline{A}|=|B|-|A\cap B|$. Again, assuming $|A\cap B| = 12$, then $|B\cap\overline{A}|=53 - 12 = 41$.

Step4: Find number for part c

We want to find the number of houses that have either a finished basement or a three - car garage. Using the formula $|A\cup B|=|A|+|B|-|A\cap B|$. Assuming $|A\cap B| = 12$, then $|A\cup B|=46 + 53-12=87$. But this is wrong as the total number of houses is 78. There is an error in our assumptions. Let's start over.
Let's use a Venn - diagram approach. Let the number of houses with both be $x$.
We know that $|A| = 46$, $|B| = 53$ and the total number of houses $n=78$.
The number of houses with only a finished basement is $a=46 - x$, the number of houses with only a three - car garage is $b = 53 - x$ and the number of houses with both is $x$ and the number of houses with neither is $y$. So $a + b+x + y=78$.
We know that $|A|=46$, $|B| = 53$.
a) The number of houses with a finished basement but not a three - car garage:
We know that the number of houses with a finished basement is 46. Let the number of houses with both be $x$. The number of houses with a finished basement but not a three - car garage is $46-(46 + 53-78)=31$.
b) The number of houses with a three - car garage but not a finished basement:
The number of houses with a three - car garage is 53. The number of houses with a three - car garage but not a finished basement is $53-(46 + 53-78)=25$.
c) The number of houses with either a finished basement or a three - car garage:
Using the formula $|A\cup B|=|A|+|B|-|A\cap B|$, where $|A\cap B|=46 + 53-78 = 21$. So $|A\cup B|=46+53 - 21=78$.

Answer:

a) 31
b) 25
c) 78