Question

last year Charlie had $20,000 to invest. he invested some of it in an account that paid 10% simple interest per year , and he invested the rest in an account that paid 9% simple interest per year. after one year he received a total of $1830 in interest. how much did he invest in each account

Answer 1

`He\text{ invested \$3,000 in the first account and \$17,000 in the second account}`

Here, we want to know the amount invested in each of the two accounts

Let the amount invested in the first account be $x while the amount invested in the second account be $y

From the question;

`x\text{ + y = 20,000}`

Now let us work with the interests;

`\begin{gathered} \text{For the first account, we have the interest as;} \\ 10\text{ percent of x = 0.1x} \\ \\ \text{For the second account, we have the interest as } \\ 9\text{ percent of y which is 0.09y} \\ \\ \text{Adding both; } \\ \\ 0.1x\text{ + 0.09y = 1830} \end{gathered}`

From the first equation, we can have;

`x\text{ = 20000-y}`

Substitute this into the equation of the interest;

`\begin{gathered} 0.1(20000-y)\text{ + 0.09y = 1830} \\ 2000-0.1y+0.09y\text{ = 1830} \\ \\ 0.1y-0.09y\text{ = 2000-1830} \\ 0.01y\text{ = 170} \\ \\ y\text{ = }(170)/(0.01) \\ \\ y\text{ = 17,000} \end{gathered}`

From above,

`\begin{gathered} x\text{ = 20000-y} \\ \\ x\text{ = 20000-17000} \\ \\ x\text{ = \$3000} \end{gathered}`