Question

Frank the trainer has two solo workout plans that he offers his clients: Plan A and Plan B. Each client does either one or the other (not both). On Wednesday there were 3 clients who did Plan A and 8 who did Plan B. on Thursday there were 5 clients who did Plan A and 2 who did Plan B. Frank trained his Wednesday clients for a total of 7 hours and his Thursday clients for a total of 6 hours. How long does each of the workout plans last?

Answer 1

Let's say first that duration of plan A is a, and duration for plan B is b.

We can now establish 2 equations. One for Wednesday and one for Thursday.

On Wednesday, frank trained clients for a total of 7 hours, being 3 clients with plan A, and 8 with plan B. Then:

`3\cdot a+8\cdot b=7`

On Thursday, frank trained clients for 6 hours, being 5 clients with plan A, and 2 clients with plan B. Then:

`5\cdot a+2\cdot b=6`

Now we have a system of 2 equations and 2 unknowns:

`\begin{gathered} 3a+8b=7 \\ 5a+2b=6 \end{gathered}`

It will be easy to solve the system by reduction method. We can eliminate terms 8b and 2b mmultiplying the second equation by -4:

`\begin{gathered} -4\cdot(5a+2b)=-4\cdot6 \\ -20a-8b=-24 \end{gathered}`

Now, the system is:

`\begin{gathered} 3a+8b=7 \\ -20a-8b=-24 \end{gathered}`

Combining two equations (adding them), we have:

`\begin{gathered} (3-20)a+(8-8)b=7-24 \\ -17a+0b=-17 \\ -17a=-17 \\ a=(-17)/(-17) \\ a=1 \end{gathered}`

Now, we know that plan A lasts 1 hour.

We can use any equation to solve for b, knowing that a = 1.

`\begin{gathered} 5a+2b=6 \\ 5\cdot(1)+2b=6 \\ 5+2b=6 \\ 2b=6-5 \\ 2b=1 \\ b=(1)/(2) \end{gathered}`

Now, we know plan B lasts half an hour (1/2 hour).