Question

You invested 7000 between two accounts paying 6% and 8% annual interest, respectively. If the total interest eared for the year was 480, how much wasinvested at each rate?

Answer 1

Let the principal between the two accounts be x and y

x = principal invested at 6%

y = principal invested at 8%

7000 was invested in the two accounts.

Principal

$x+y=7000\ldots\ldots...\ldots\ldots....\ldots..\ldots\text{.}(1)$

Interest

$\begin{gathered} I=\frac{PR}{100} \\ \text{For first account with principal (x)} \\ I=x\times\frac{6}{100} \\ I=0.06x \\ \\ \text{For second account with principal (y)} \\ I=y\times\frac{8}{100} \\ I=0.08y \\ \\ \text{The total interest earned for the year was 480} \\ 0.06x+0.08y=480\ldots\ldots\ldots\ldots\ldots\ldots\ldots.\mathrm{\cdot}..(2) \end{gathered}$

Solving equations (1) and (2) simultaneously,

$\begin{gathered} x+y=7000\ldots\ldots\ldots\ldots......(1)\text{ x6} \\ 6x+6y=42000\ldots\ldots..\ldots\ldots....\ldots\ldots\ldots\ldots......\mathrm{}(1) \\ \\ 0.06x+0.08y=480\ldots\ldots..\ldots..\ldots\text{.}(2)\text{ x100} \\ 6x+8y=48000\ldots.\ldots\ldots\ldots..\ldots\ldots....\ldots\ldots...(2) \\ \\ \\ Subtract\text{ ing equation (1) from (2);} \\ 6x-6x+8y-6y=48000-42000 \\ 2y=6000 \\ \text{Dividing both sides by 2,} \\ y=\frac{6000}{2} \\ y=3000 \end{gathered}$

Substituting y in equation (1) to get the value of x

$\begin{gathered} x+y=7000 \\ x+3000=7000 \\ x=7000-3000 \\ x=4000 \end{gathered}$

Therefore, the principal invested at 6% is 4000 and the principal invested at 8% is 3000.