intermediate algebra

Question

It takes an older computer twice as long to send out a company's email as it does a newer computer. Working together, it takes the two computers 30 minutes to send out the email. How long will it take the older computer to send out the email on its own?

Answer 1

Marisa, these “shared work problems” just boil down into simple “system of two equations” problems, but the tricky part is recognizing the concepts and translating into math.

The fundamental formula is

Work = rate * time

As always, give names to all quantities.

Let Rs be the rate at which the old slow computer can do work, in work / time

Let Rf be the rate at which the new fast computer can do work

Let Ts be the time it takes the slow computer to do the work

Let Tf be the time it takes the fast computer to do the work

The problem states that Ts = 2Tf

It also says that working together, they get the job done in 30 minutes. Since Ts and Tf are both 30, and W (the amount of work to do) is 1 email job, this translates to

1 = 30Rs + 30Rf

finally let’s convert each R into a T, so that we can solve for Ts.

Because 1=RT, we can recognise

Rs = 1/Ts

Rf = 1/Tf

now substitute into the equation

1 = 30/Ts + 30/Tf

Lets solve for Ts. First we need to move the Tf term out of the way:

1 - 30/Tf = 30/Ts

then put the left hand side over a common denominator

(Tf - 30) / Tf = 30/Ts

Now invert both sides to get Ts on top

Tf / (Tf -30) = Ts / 30

Finally multiply by 30

Ts = 30Tf / (Tf-30)

and substitute what we already knew about Ts and Tf

(Ts = 2Tf, therefore Tf = Ts/2)

Ts = 30(Ts/2) / (Ts/2 - 30)

Multiply by the denominator

Ts( Ts/2 - 30) = 30(Ts/2)

multiply everything out

(Ts^2)/2 - 30Ts = 15Ts

multiply by denominator

Ts^2 - 60Ts = 30Ts

Collect terms

Ts^2 - 90 Ts = 0

Factor

Ts ( Ts - 90) = 0

The only realistic answer is Ts = 90

Now, to verify the solution, solve for Tf from

Ts = 2 Tf

Tf = 45

and put those values into the shared work equation

1 = 30/Ts + 30/Tf

1= 30/90 + 30/45

1 = 30/90 + 60/90 = 90/90

The key tricky part of these “shared work” problems is to be clear on Rates and Times, how to tell one from the other and how to convert one to the other.

If you have any questions, please ask.