Question

A person invested $8100 for 1 year, part 5%, part at 11%, and the remainder at 14%. The total annual income from these investments was $942. The amount of money invested at 14% was $500 more than the amounts invested at 5% and 11% combined. Find the amount invested at each rate.

Answer 1

The formula is

`C=c(1+n* i)`

Where C is the total, c initial capital, n the number of years and i interest percentage

We can modify the formula to express the total won

`C=c_1(1+n* i_1)+c_2(1+n* i_2)+c_3(1+n* i_3)`

Where c1 is the par 5%, C2 the par at 11% and c3 the par at 14%

i1,i2 and i3 the percentage of interest applied to each part

as a total you must use the total earned plus the initial capital

so replacing

`\begin{gathered} 8100+942=c_1(1+1*0.05)+c_2(1+1*0.11)+c_3(1+1*0.14) \\ 9042=c_1(1.05)+c_2(1.11)+c_3(1.14) \end{gathered}`

this was our first equation

the next comes from: the sum of all inverted parts is 8100

so

`c_1+c_2+c_3=8100`

hitrd equation is from: The amount of money invested at 14% was $500 more than the amounts invested at 5% and 11% combined.

`c_3=c_1+c_2+500`

Solution of the equations

`\begin{gathered} 9042=c_1(1.05)+c_2(1.11)+c_3(1.14) \\ c_1+c_2+c_3=8100 \\ c_3=c_1+c_2+500 \end{gathered}`

we replace the third equation on the second

`\begin{gathered} c_1+c_2+(c_1+c_2+500)=8100 \\ 2c_1+2c_2=8100-500 \\ 2c_1=7600-2c_2 \\ \\ c_1=(7600-2c_2)/(2) \\ \\ c_1=3800-c_2 \end{gathered}`

now replace c3 and c1 on the first equation to find c2

`9042=(3800-c_2)(1.05)+c_2(1.11)+(c_1+c_2+500)(1.14)`

now replace c1 again

`9042=(3800-c_2)(1.05)+c_2(1.11)+(3800-c_2+c_2+500)(1.14)`

and find c2

`\begin{gathered} 9042=3990-1.05c_2+1.11c_2+(4300)(1.14) \\ 9042-3990=0.06c_2+4902 \\ 5052-4902=0.06c_2 \\ 150=0.06c_2_{} \\ \\ c_2=(150)/(0.06)=2500 \end{gathered}`

the value of c2 or the part at 11% is $2,500

now we can replace c2 in one of the equations we solved, for example this

`c_1=3800-c_2`

and find c1

`\begin{gathered} c_1=3800-2500 \\ c_1=1300 \end{gathered}`

the value of c1 or the part at 5% is $1,300

now we can repalce c1 and c2 on the equation

`c_1+c_2+c_3=8100`

and find c3

`\begin{gathered} 1300+2500+c_3=8100 \\ c_3=8100-1300-2500 \\ c_3=4300 \end{gathered}`

the value of c3 or the part at 14% is $4,300

the total values are

part at 11% is $2,500

part at 5% is $1,300

part at 14% is $4,300