These three tables show the floor layout of 5 departments in an office building, the average number of daily trips between each pair of offices, and the distance between each pair of offices. What is the total cost of this layout if there is a $0.50 per foot transportation cost? Average number of trips between departments
$$\begin{array} { | l | c | c | c | c | c | }
\hline & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { A } & \cdots & 25 & 15 & 25 & 30 \
\hline \text { B } & & \cdots & 40 & 50 & 35 \
\hline \text { C } & & & \cdots & 15 & 15 \
\hline \mathrm { D } & & & & \cdots & 20 \
\hline \text { E } & & & & & \cdots \
\hline
\end{array}$$ $$\begin{array} { | l | l | l | }
\hline \mathrm { A } & \mathrm { B } & \mathrm { D } \
\hline & \mathrm { C } & \mathrm { E } \
\hline
\end{array}$$ Distance between departments (feet)
$$\begin{array} { | l | c | c | c | c | c | }
\hline & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { A } & \cdots & 20 & 25 & 40 & 45 \
\hline \text { B } & & \cdots & 5 & 20 & 25 \
\hline \text { C } & & & \cdots & 25 & 20 \
\hline \text { D } & & & & \cdots & 5 \
\hline \text { E } & & & & & - \
\hline
\end{array}$$
A)$460.00
B)$2300.00
C)$3037.50
D)$4785.50

Question

These three tables show the floor layout of 5 departments in an office building, the average number of daily trips between each pair of offices, and the distance between each pair of offices. What is the total cost of this layout if there is a $0.50 per foot transportation cost? Average number of trips between departments
$$\begin{array} { | l | c | c | c | c | c | } 
\hline & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { A } & \cdots & 25 & 15 & 25 & 30 \
\hline \text { B } & & \cdots & 40 & 50 & 35 \
\hline \text { C } & & & \cdots & 15 & 15 \
\hline \mathrm { D } & & & & \cdots & 20 \
\hline \text { E } & & & & & \cdots \
\hline
\end{array}$$ $$\begin{array} { | l | l | l | } 
\hline \mathrm { A } & \mathrm { B } & \mathrm { D } \
\hline & \mathrm { C } & \mathrm { E } \
\hline
\end{array}$$ Distance between departments (feet)
$$\begin{array} { | l | c | c | c | c | c | } 
\hline & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \
\hline \text { A } & \cdots & 20 & 25 & 40 & 45 \
\hline \text { B } & & \cdots & 5 & 20 & 25 \
\hline \text { C } & & & \cdots & 25 & 20 \
\hline \text { D } & & & & \cdots & 5 \
\hline \text { E } & & & & & - \
\hline
\end{array}$$
A)$460.00
B)$2300.00
C)$3037.50
D)$4785.50

Quizplus · Accepted Answer

To calculate the total cost, multiply the average number of daily trips between each pair of departments by the distance between them, and then by the transportation cost ($0.50 per foot). Sum these values for all department pairs:- A to B: $25 \times 20 \times 0.50 = 250$- A to C: $15 \times 25 \times 0.50 = 187.50$- A to D: $25 \times 40 \times 0.50 = 500$- A to E: $30 \times 45 \times 0.50 = 675$- B to C: $40 \times 5 \times 0.50 = 100$- B to D: $50 \times 20 \times 0.50 = 500$- B to E: $35 \times 25 \times 0.50 = 437.50$- C to D: $15 \times 25 \times 0.50 = 187.50$- C to E: $15 \times 20 \times 0.50 = 150$- D to E: $20 \times 5 \times 0.50 = 50$Adding these values gives a total cost of $250 + 187.50 + 500 + 675 + 100 + 500 + 437.50 + 187.50 + 150 + 50 = 3037.50$.

These Three Tables Show the Floor Layout of 5 Departments