Question

Brad plans to invest a total of $6000, part of it in an account at 2% annual interest and the rest in an account at 8% annual interest. How much should he invest in each account so that the total interest in one year will be $192?

Answer 1

In an account at 2% invest
`\$4,800`

In an account at 8% invest
`\$1,200`

Explanation:

we know that

The simple interest formula is equal to

`A=P(1+rt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have two cases

First case

`t=1\ year\\ P=\$x\\ A=\$6,000+\$192=\$6,192\\r=0.02`

Second case

`t=1\ year\\ P=\$6,000-\$x\\ A=\$6,000+\$192=\$6,192\\r=0.08`

substitute both cases in the formula above

`\$6,192=\$x(1+(0.02*1))+(\$6,000-\$x)(1+(0.08*1))`

solve for x

`6,192=x(1.02)+(6,000-x)(1.08)`

`6,192=1.02x+6,480-1.08x`

`1.08x-1.02x=6,480-6,192`

`0.06x=288`

`x=\$4,800`

so

In an account at 2% invest
`\$4,800`

In an account at 8% invest
`\$1,200`