Question

a person invested $6700 for one year, part at 8%, part at 10%, and the remainder at 12%. the total annual income from these investments was $716. the amount of money invest at 12% was $300 more than the amount at 8% and 10% combined. find the amount invested at each rate.

Answer 1

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500

Explanation:

step 1

Let

x-----> the amount invested at 8% rate

y-----> the amount invested at 10% rate

z-----> the amount invested at 12% rate

`z=(x+y)+300` ----> equation A

`x+y+z=6,700` ----> equation B

substitute equation A in equation B

`x+y+(x+y+300)=6,700`

`2x+2y=6,400`

`x+y=3,200` -----> equation C

we know that

The simple interest formula is equal to

`I=P(rt)`

where

I is the Final Interest 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

`t=1\ year\\ P=\$6,700\\ I=\$716`

substitute in the formula above

`716=x(0.08)+y(0.10)+z(0.12)`

substitute equation A

`716=x(0.08)+y(0.10)+(x+y+300)(0.12)`

`716=0.08x+0.10y+0.12x+0.12y+36`

`716=0.20x+0.22y+36`

`0.20x+0.22y=680` -----> equation D

step 2

Solve the system of equations

x+y=3,200 -----> equation C

`0.20x+0.22y=680` -----> equation D

Solve the system by graphing

The solution is the point (1,200,2,000)

see the attached figure

Find the value of z

`z=(x+y)+300`

`z=(1,200+2,000)+300=3.500`

therefore

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500