Assessment math

Question

Larry Mitchell invested part of his $26,000 advance at 6% annual simple interest and the rest at 2% annual simple interest. If his total yearly

interest from both accounts was $800, find the amount invested at each rate

The amount invested at 6% is $

The amount invested at 2% is $

Answer 1

Hi,

Use the following simple interest Formula:

           I  =  P r t  

      Where :    

                I = Interest Amount

                p = Principal Amount

                r = annual interest rate in decimal

                t = time in years

Let the amount invested at  6%    =  $x

           Amount invested at 2%    =  $26000 - $x

Interest from 6 % :  Substitute  P = $x , r  = 0.06, and  t = 1 in the formula    

               

I₆ =   x ( 0.06 ) 1  =  0.06 x                 

Interest from 2% : Substitute P = $(26000 - x), r  = 0.02,  and t = 1 in the formula

       

    I₂ =  (26000 – x ) ( 0.02 )   =  520 – 0.02x

Total Interest is 800:    I₆  + I₂ = 800

                                 0.06 x + 520 – 0.02x = 800

                                                      0.04x = 280

                                                             x = 7000

The amount invested at 6% =  $x = $7,000

The amount invested at 2%  =  $26000 -  $x  

                              = $26000 – $7000

                              =  $19,000

Hope this is helpful!