Question

Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 3.6 hours and Cheryl completes the course in 6 hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy's speed and Cheryl's speed in miles per hour.

Answer 1

Explanation:

Here is what we know;

Both Kathy and Cheryl cover the same distance ( one course).

Kathy completes the course in 3.6 hours.

Kathy's speed is 2 miles/hour faster than Cheryl's

Cheryl completes the course in 6 hours

Cheryl's speed is yet unkown.

Now, the speed v is defined as

`v=(D)/(t)`

where D is the distance covered and t is the time taken.

Now, let us say D = distance of one course. Then in Kathy's case, we have

`v_{\text{kathy}}=(D)/(3.6hr)`

since Kathy's speed is 2 miles per hour faster than Cheryl's, we have

`v_{\text{kathy}}=(D)/(3.6)=2+v_{\text{cheryl}}`

For Cheryl, we know that

`v_{\text{cheryl}}=(D)/(6hr)`

or simply

`v_{\text{cheryl}}=(D)/(6)`

Putting this into the equation for Kathy's speed gives

`v_{\text{kathy}}=(D)/(3.6)=2+v_{\text{cheryl}}\Rightarrow v_{\text{kathy}}=(D)/(3.6)=2+(D)/(6)`
`\Rightarrow(D)/(3.6)=2+(D)/(6)`

We have to solve for D, the distance of a course.

Subtracting D/6 from both sides gives

`(D)/(3.6)-(D)/(6)=2`
`((1)/(3.6)-(1)/(6))D=2`
`(1)/(9)D=2`
`D=18\text{miles}`

Hence, the distance of a course is 18 miles.

With the value of D in hand, we can now find the velocity of Kathy and Cheryl.

`v_{\text{cheryl}}=(D)/(6hr)=\frac{18\text{miles}}{6hr}`
`\Rightarrow\boxed{v_{\text{cheryl}}=3\text{ miles/hr}}`

Hence, Cheryl's speed is 3 miles/hr.

Next, we find Kathy's speed.

`v_{\text{kathy}}=(D)/(3.6hr)=\frac{18\text{miles}}{3.6hr}`
`\boxed{v_{\text{kathy}}=5\text{miles}/hr\text{.}}`

Hence, Kathy's speed is 5 miles/hr.

Therefore, to summerise,

Kathy's speed = 5 miles/hr

Cheryl's speed = 3 miles/hr