Problema Solution
A person invested $6600 for 1 year, part at 7%, part at 10%, and the remainder at 14%. The total annual income from these investments was $772. The amount of money invested at 14% was $800 more than the amounts invested at 7% and 10% combined. Find the amount invested at each rate.
Answer provided by our tutors
interest = principal * rate * time
we take time to be equal to 1
let 'x' represent the amount invested at 7%, 'y' the amount invested at 10%, then '6600-x-y' represents the amount invested at 14%
we have two equations with two unknowns:
772 = 0.07x + 0.10y + 0.14(6600-x-y)
(6600-x-y) = x + y + 800
solving this system gives:
x = 1200
y = 1700
6600 - 1200 - 1700 = 3700
$1,200 was invested at 7%, $1,700 was invested at 10% and $3,700 was invested at 14%