Question

toward the middle of the harvesting season, peaches for canning come in three types, early, late, and extra late, depending on the expected date of ripening. during a certain week, the data to the right were recorded at a fruit delivery station. complete parts (a) through (d) below.
29 trucks went out carrying early peaches; 66 carried late peaches; 51 carried extra late peaches; 22 carried early and late; 30 carried late and extra late; 4 carried early and extra late; 2 carried all three; 8 carried only figs (no peaches at all).
(a) how many trucks carried only late variety peaches?
trucks
(type a whole number.)

Explanation:

Step1: Use the principle of inclusion - exclusion

Let \(E\) be the set of trucks carrying early peaches, \(L\) be the set of trucks carrying late peaches, and \(X\) be the set of trucks carrying extra - late peaches. We know \(n(E)=29\), \(n(L) = 66\), \(n(X)=51\), \(n(E\cap L)=22\), \(n(L\cap X)=30\), \(n(E\cap X)=4\), and \(n(E\cap L\cap X)=2\).

Step2: Calculate the number of trucks carrying only late peaches

The number of trucks carrying only late peaches can be found by subtracting the number of trucks that carry late peaches along with other types from the total number of trucks carrying late peaches. The number of trucks that carry late peaches and other types is \((n(E\cap L)-n(E\cap L\cap X))+(n(L\cap X)-n(E\cap L\cap X))+n(E\cap L\cap X)\).
The number of trucks carrying only late peaches \(=n(L)-[(n(E\cap L)-n(E\cap L\cap X))+(n(L\cap X)-n(E\cap L\cap X))+n(E\cap L\cap X)]\).
First, \(n(E\cap L)-n(E\cap L\cap X)=22 - 2=20\), \(n(L\cap X)-n(E\cap L\cap X)=30 - 2 = 28\).
Then \(n(L)-[(n(E\cap L)-n(E\cap L\cap X))+(n(L\cap X)-n(E\cap L\cap X))+n(E\cap L\cap X)]=66-(20 + 28+2)\).
\(=66 - 50=16\).

Answer:

16