Question

A total of $13,000 is invested in two accounts. One pays 7.5% simple interest and the other pays 8.5% simple interest. If the total interest earned from both accounts total $1,035 for one year. How much was invested in the account having 8.5% simple interest?

Answer 1

ANSWER :

The amount invested in the account having 8.5% simple interest is $6000

EXPLANATION :

The simple interest formula is :

$I=Prt$

where I = interest

P = initial investment

r = interest rate

t = time in years

From the problem, a total of $13,000 is invested in two accounts.

Let x = amount invested in 7.5%

y = amount invested in 8.5%

The interests in 1 year (t = 1)

$\begin{gathered} I=Prt \\ I_1=x(0.075)(1)=0.075x \\ I_2=y(0.085)(1)=0.085y \end{gathered}$

The interests are 0.075x and 0.085y.

The total interest earned for both accounts is $1035

So the equation will be :

$0.075x+0.085y=1035$

Another equation is the sum of x and y which is $13000

$\begin{gathered} x+y=13000 \\ x=13000-y \end{gathered}$

We need to find the amount invested that pays 8.5% simple interest which is y.

Substitute x = 13000 - y to the first equation :

$\begin{gathered} 0.075x+0.085y=1035 \\ 0.075(13000-y)+0.085y=1035 \\ 975-0.075y+0.085y=1035 \\ 0.01y=1035-975 \\ 0.01y=60 \\ y=\frac{60}{0.01} \\ y=6000 \end{gathered}$