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). (b) how many carried only extra late? 19 trucks (type a whole number.) (c) how many carried only one type of peach? 40 trucks (type a whole number.) (d) how many trucks (in all) went out during the week? 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 \(EL\) be the set of trucks carrying extra - late peaches.
We know \(n(E)=29\), \(n(L) = 66\), \(n(EL)=51\), \(n(E\cap L)=22\), \(n(L\cap EL)=30\), \(n(E\cap EL)=4\), \(n(E\cap L\cap EL)=2\), and the number of trucks carrying only figs \(n(F) = 8\).

Step2: Find the number of trucks carrying only early peaches

\(n(\text{only }E)=n(E)-n(E\cap L)-n(E\cap EL)+n(E\cap L\cap EL)=29 - 22-4 + 2=5\)

Step3: Find the number of trucks carrying only late peaches

\(n(\text{only }L)=n(L)-n(E\cap L)-n(L\cap EL)+n(E\cap L\cap EL)=66-22 - 30+2=16\)

Step4: Find the total number of trucks carrying peaches

We use the formula \(n(E\cup L\cup EL)=n(E)+n(L)+n(EL)-n(E\cap L)-n(L\cap EL)-n(E\cap EL)+n(E\cap L\cap EL)\)
\[

$$\begin{align*} n(E\cup L\cup EL)&=29+66 + 51-22-30-4 + 2\\ &=98 \end{align*}$$

\]

Step5: Find the total number of trucks

The total number of trucks that went out is the sum of the number of trucks carrying peaches and the number of trucks carrying only figs.
So \(n(\text{total})=n(E\cup L\cup EL)+n(F)=98 + 8=106\)

Answer:

106