Chapter 8.3, Problem 20Q

(a)

To determine

To Find: The location for the first stop.

Answer to Problem 20Q

The first stop must be post office.

Explanation of Solution

Given:

The distance from the school to the post is $376\text{yards}$ .

The distance between the post office and the house is $929\text{yards}$ .

The distance from the grocery store to the house is 513 yards.

Calculation:

Consider the location of the house is H, school is S, grocery store is G and the post office is P.

From the given data it is clear that the first stops must be post office at it is close to the school.

(b)

To determine

To Find: The distance between the post office and the grocery sotre.

Answer to Problem 20Q

The required distance is of $416\text{yards}$ .

Explanation of Solution

Since, the distance between the house and the grocery is 513 yards and the tot the post it is 929 yards. Then, the distance between the post office and the store is $929-513=416\text{yards}$ .

(c)

To determine

To Find: The distance between the school and the house.

Answer to Problem 20Q

The required distance is $1305\text{yards}$ .

Explanation of Solution

Since, the distance of the post office is 376 yards from the school and 929 yards from the house this suggest that the total distance between the house and the school is,

  $929+376=1305\text{yards}$ .

(d)

To determine

To Find: The time taken to cover the distance from home to school when the speed is of 75 yards per minute.

Answer to Problem 20Q

The required time is $7.4\text{minutes}$ .

Explanation of Solution

Since the speed is 75 yards per minute.

Then, the time taken to cover the distance of 1305 yards is,

  $\frac{1305}{75}=17.4\text{minutes}$