Question

1. a mail carrier delivered mail on two different routes: the morning route and the afternoon route. last month, he delivered the morning route 16 times and the afternoon route 12 times for a total distance of 141 miles. this month, he delivered the morning route 10 times and the afternoon route 15 times for a total distance of 123.75 miles. what is the distance of the morning route?
2. there are 200 adults and 300 children at a zoo. the zoo makes a total of $7000 from the entrance fees, and the cost for an adult and a child to attend is $30. write a system of equations that will determine the cost of both adult and children’s tickets.

Explanation:

Problem 2

Step1: Define variables

Let \( x \) be the distance of the morning route (in miles) and \( y \) be the distance of the afternoon route (in miles).

Step2: Set up equations

Last month: \( 16x + 12y = 141 \)
This month: \( 10x + 15y = 123.75 \)

Step3: Simplify equations

Multiply the first equation by \( 5 \): \( 80x + 60y = 705 \)
Multiply the second equation by \( 4 \): \( 40x + 60y = 495 \)

Step4: Subtract equations

Subtract the second new equation from the first: \( (80x + 60y) - (40x + 60y) = 705 - 495 \)
Simplify: \( 40x = 210 \)

Step5: Solve for \( x \)

\( x = \frac{210}{40} = 5.25 \)

Step1: Define variables

Let \( a \) be the cost of an adult ticket (in dollars) and \( c \) be the cost of a child ticket (in dollars).

Step2: Set up equations

From the total number of people and total cost: \( 200a + 300c = 7000 \)
From the cost of one adult and one child: \( a + c = 30 \)

Answer:

The distance of the morning route is \( 5.25 \) miles.

Problem 3