At the end of a year he has earned $960 in interest. How much is invested at each rate?

Question

Levi invests $11000 in two accounts, the first pays 3% simple interest, the second pays 12% simple interest. At the end of a year he has earned $960 in interest. How much is invested at each rate?

Answer 1

We can solve this by solving a system of 2 equations.

 

Let X = amount of money (principal) invested in account 1

Let Y = amount of money (principal) invested in account 2

 

Therefore:    X + Y = $11000     (Equation 1)

 

 

Amount of interest earned from account 1 will equal 0.03X  (principal times 3% interest rate)

Amount of interest earned from account 2 will equal 0.12Y  (principal times 12% interest rate)

 

We also know that the total interest earned is $960.

 

Therefore:   0.03X + 0.12Y = $960     (Equation 2)

 

Solve equation 1 for X:     X = 11000 - Y

 

Plug into equation 2:   0.03(11000 - Y) + 0.12Y = 960

 

Simplify:     330 - 0.03Y + 0.12Y = 960

                  330 + 0.09Y = 960

                  0.09Y = 630

                        Y = 7000          and X = 11000 - Y= 11000 - 7000 or   X = 4000

 

Amount invested in account 1 is $7000

Interest rate of account 1= 12%

 

Amount invested in account 2 = $4000

Interest rate of account 2 = 3%

 

Check answer:

     Total interest = $7000*(0.12) * $4000*(0.03) = $840 + $120 = $960   Answer checks!