Question

Miguel bought a desktop computer and a laptop computer. Before finance charges, the laptop cost $200 less than the desktop. He paid for the computers using two different financing plans. For the desktop the interest rate was 8% per year, and for the laptop it was 6.5% per year. The total finance charges for one year were $335. How much did each computer cost before finance charges? Desktop: Laptop:

Answer 1

Desktop = $2400

Laptop = $2200

Explanation;

The formula for the interest is:

`I=P[(1+r)^t-1]`

Where:

• I is the interest after t years

,

• r is the rate of compounding, in decimal

,

• t is the years of compounding

In this case, let's call L = price of the laptop, D = price of the desktop

Since the laptop costs $200 less than the desktop:

`D-200=L`

Also, if we call I_D = interest of the desktop and I_L interest of the laptop, we know that the total after one year is

`I_D+I_L=335`

Now, we can write the equations for both financing plans:

`\begin{gathered} I_D=D[(1+0.08)^t-1] \\ I_L=L[(1+0.065)^t-1] \end{gathered}`

So far we have:

`\begin{gathered} (1)\text{ }D-200=L \\ (2)\text{ }I_D+I_L=335 \\ (3)I_D=D[(1+0.08)^t-1] \\ (4)\text{ }I_L=L[(1+0.065)^t-1] \end{gathered}`

Now, we can write:

`I_L=335-I_D`

And replace in the equation (4):

`335-I_D=(D-200)[1.065^t-1]`

Since t = 1 (one year passed)

we can solve for I_D:

`I_D=335-(D-200)0.065`

And solve:

`\begin{gathered} I_D=335-0.065D+13 \\ I_D=348-0.065D \end{gathered}`

Now, we can equate this with equation (3):

`\begin{cases}I_D=348-0.065D{} \\ I_D=D[(1+0.08)^t-1]\end{cases}`

But, before, let's make some work on the equation 3. Snce t = 1:

`\begin{gathered} I_D=D[(1+0.08)^1-1] \\ I_D=0.08D \end{gathered}`

Now equate:

`348-0.065D=0.08D`

And solve for D:

`\begin{gathered} 348=0.08D+0.065D \\ 348=0.145D \\ . \\ D=(348)/(0.145)=2400 \end{gathered}`

The price of the desktop computer is $2400. Now we can find the price of the laptop using the equation (1):

`L=2400-200=2200`

The price of the laptop is $2200