Question

Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles. She can ride 9 mph faster than she can walk. How fast can she walk? Using r r as your variable to represent the rate at which she walks, write an equation using the information as it is given above that can be used to solve this problem.

Answer 1

`(48)/(r+9)=(12)/(r)`

Explanation:

Let r represent Linda's walking rate.

We have been given that Linda can ride 9 mph faster than she can walk, so Linda's bike riding rate would be
`t+9` miles per hour.

`\text{Time}=\frac{\text{Distance}}{\text{Rate}}`

We have been given that Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles.

`\text{Time while riding}=(48)/(r+9)`

`\text{Time taken while walking}=(12)/(r)`

Since both times are equal, so we will get:

`(48)/(r+9)=(12)/(r)`

Therefore, the equation
`(48)/(r+9)=(12)/(r)` can be used to solve the rates for given problem.

Cross multiply:

`48r=12r+108`

`48r-12r=12r-12r+108`

`36r=108`

`(36r)/(36)=(108)/(36)`

`r=3`

Therefore, Linda's walking at a rate of 3 miles per hour.

Linda's bike riding rate would be
`t+9\Rightarrow 3+9=12` miles per hour.

Therefore, Linda's riding the bike at a rate of 12 miles per hour.