Find the amount invested at each rate

Question

Larry Mitchell invested part of his $17000 advance at 5% annual simple interest and the rest 6% annual simple interest. If his total yearly interest from both accounts was $870, find the amount invested at each rate.

Answer 1

Average interest rate = 100 x 870/17000

Amount at 5% = (17000)(6 - 100 x 870/17000) = $15,000

Amount at 6% = $2,000

 

Answer 2

Interest = Principle*Rate*Time

 

When you split the principle into two parts and invest each part at a different rate:

 

Interest = (Part of Principle)*(One Rate)*(Time) + (Rest of the Principle)*(Other Rate)*(Time)

 

Let x = the amount of the Principle invested at 5%.  The amount invested at 6% is $17,000 - x.

 

$870 = (x)*(0.05)*(1 year) + ($17,000 - x)*(0.06)*(1 year)

 

$870 = 0.05x + $1020 - 0.06x

 

-$150 = -0.01x

 

Solve for x.

Answer 3

Kim:

 

Let the "rest"  of his investment be$ P

 

Interest on $17000 = (17000)(0.05) = 850.

Interest on $P        = (0.06) P

 

Total interest          = 850 + (0.06) P

 

This total is given to be $870.  Solve for P, 850 + (0.06)P = 870.  That is P = 20/(0.06) = $333.33 (to 2 decimals).

 

Answer: 17000.00 at 5%

                333.33 at 6% ( to 2 decimals)

 

Dr. G.