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)

$\text{ simple interest = }\frac{\text{ principal}\times\text{ rate}\times\text{ time}}{100}$

For the one paying 9% simple interest

$\text{ simple interest =}\frac{x\times9\times1}{100}=\frac{9x}{100}$

and

for the one paying 10% simple interest

$\text{ simple interest =}\frac{y\times10\times1}{100}=\frac{10y}{100}$

Hence,

$\text{ total interest = }\frac{9x}{100}+\frac{10y}{100}=1339$

Therefore,

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

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