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

Answer:

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:

$\begin{gathered} 348=0.08D+0.065D \\ 348=0.145D \\ . \\ D=\frac{348}{0.145}=2400 \end{gathered}$

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