Question

a person had 14,000 invested in two accounts, one paying 9% simple interest and one paying 10% simple interest. how much was invested in each account if the interest in the one year is $1339?

Answer 1

Let x and y be the amounts invested in the one paying 9% and the one paying 10% respectively.

x + y = 14000 ------------------------------------------(1)

`\text{ simple interest = }\frac{\text{ principal}*\text{ rate}*\text{ time}}{100}`

For the one paying 9% simple interest

`\text{ simple interest =}(x*9*1)/(100)=(9x)/(100)`

and

for the one paying 10% simple interest

`\text{ simple interest =}(y*10*1)/(100)=(10y)/(100)`

Hence,

`\text{ total interest = }(9x)/(100)+(10y)/(100)=1339`

Therefore,

`\begin{gathered} (9x+10y)/(100)=1339 \\ \Rightarrow9x+10y=133900---------------------(2) \end{gathered}`

From equation (1), isolating x, we have

x = 14000 - y --------------------------------(3)

Substituting equation (3) into equation (2), we have

`\begin{gathered} 9(14000-y)+10y=133900 \\ \text{ Hence} \\ 126000-9y+10y=133900 \\ \Rightarrow y=133900-126000=7900 \\ \end{gathered}`

Substituting the y = 7900 into equation (3), we have

x = 14000 - 7900 = 6100

Hence,

he invested $6100 in the account with 9% simple interest

and $7900 in the account with 10% simple interest