Question

A total of $4000 was invested, part of it at 8% interest and the remainder at 11%. if the total yearly interest amounted to $365, how much was invested at each rate?

Answer 1

$1500 was invested at 11% $2500 was invested at 8% Assuming simple interest for each investment, we have the following expressions 0.11x = interest on 11% investment. (x = amount invested at 11%) 0.08(4000-x) = interest on 8% investment Adding the 2 expressions together and setting the sum to 365 gives 0.11x + 0.08(4000-x) = 365 Now solve for x by first distributing the 0.08 0.11x + 320 - 0.08x = 365 Subtract 320 from both sides and combine x's 0.03x = 45 Divide both sides by 0.03 x = 1500 So $1500 was invested at 11% and (4000-1500) = 2500 was invested at 8%