Long word problem

Question

You invested $20,000 in two accounts paying 8% and 9% annual interest, respectively. If the total interest earned for the year was $1620, how much was invested at each rate?

Answer 1

Pa = account A present worth ($) 

 

Pb = account B present worth ($) 

 

Simple Interest A = Ia = 8% or 0.08

 

Simple Interest B = Ib = 9% or 0.09

 

Interest Amount = P*I*t

 

Let A = interest amount for A = Pa*Ia*t

 

      B = interest amount for B = Pb*Ib*t

 

Given:  t = 1

 

1) Pa + Pb = $20,000   or   Pa = 20,000 - Pb

 

2) A + B = $1620

 

Substituting function in equation 2):

 

Pa*Ia*t + Pb*Ib*t = 1620       

 

Insert given values:  t = 1 yr and Ia = 0.08   and    Ib = 0.09

 

Pa*(0.08)*(1) + Pb*(0.09)*(1) = 1620

 

(0.08)*Pa + (0.09)*Pb = 1620

 

Substitute Pa = 20,000 - Pb           

 

0.08*(20,000 - Pb) + (0.09)*Pb = 1620     

 

1600 - (0.08)*Pb + (0.09)*Pb = 1620       (combine like terms and solve for Pb)

 

Pb = (1620 - 1600)/0.01 = 2000

 

Pa = 20,000 - Pb = $20,000 - $2000 = $18,000

 

checking,      1620

 

($18,000)*(0.08)*(1) + ($2000)*(0.09)*(1) = $1440 + $180 = $1620       checks out

 

Therefore, $18,000 was invested at 8% interest   =  $1440

          and $2000 was invested at 9% interest      = $  180

                                                                            --------

                                                                            $1620 total return after 1 yr