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

Answer:

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

Step-by-step 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