A pet care service, Canine Care, charges a flat monthly registration fee of 25 plus 15 per… — ACT

A pet care service, Canine Care, charges a flat monthly registration fee of $25 plus$ 15 per dog walk. A competing service, Paws & Claws, charges a flat monthly registration fee of $45 plus$ 10 per dog walk. If a client's total monthly cost with Canine Care is $30 more than their total monthly cost with Paws & Claws for the same number of walks, how many dog walks did the client's dog receive during that month?

Answer: 10 walks

Answer

The client's dog received 10 dog walks during the month.

The correct answer is 10. By translating the verbal descriptions into algebraic expressions, we find that Canine Care's monthly cost is

15w + 25

and Paws & Claws' monthly cost is

10w + 45

. Setting up the equation where Canine Care is

30

more than Paws & Claws gives

15w + 25 = 10w + 45 + 30

. Simplifying the right side results in

15w + 25 = 10w + 75

. Subtracting

10w

from both sides gives

5w + 25 = 75

. Subtracting

25

from both sides gives

5w = 50

, and dividing by

5

yields

w = 10

Step-by-Step Solution

Define the variable and write the cost expression for Canine Care.

15w + 25

, where

w

is the number of walks.

To represent the total monthly cost of Canine Care algebraically based on the flat fee and per-walk rate.

Write the cost expression for Paws & Claws.

10w + 45

, where

w

is the number of walks.

To represent the total monthly cost of Paws & Claws algebraically based on the flat fee and per-walk rate.

Set up the equation relating the two costs.

15w + 25 = 10w + 45 + 30

The problem states the cost of Canine Care is $30 more than the cost of Paws & Claws.

Simplify the equation and solve for

w

5w = 50 \Rightarrow w = 10

Combine like terms and isolate the variable

w

to find the number of dog walks.

Key Concept

Translating and Solving Algebraic Word Problems

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