Question

A person invested ​$7600 for 1​ year, part at 6​%, part at 9​%, and the remainder at 13​%. The total annual income from these investments was ​$818. The amount of money invested at 13​% was ​$1200 more than the amounts invested at 6​% and 9​% combined. Find the amount invested at each rate.

Answer 1

Explanation:

Let X be the amount invested at 6%, Y be the amount invested at 9%, and Z be the amount invested at 13%.

From the problem, we know that:

X + Y + Z = 7600 ---(1) (the total amount invested is $7600)

0.06X + 0.09Y + 0.13Z = 818 ---(2) (the total income from the investments is $818)

Z = X + Y + 1200 ---(3) (the amount invested at 13% is $1200 more than the amounts invested at 6% and 9% combined)

We can use equation (3) to substitute for Z in equations (1) and (2), then solve for X and Y as follows:

X + Y + (X + Y + 1200) = 7600

2X + 2Y = 6400

X + Y = 3200

0.06X + 0.09Y + 0.13(X + Y + 1200) = 818

0.06X + 0.09Y + 0.13X + 0.13Y + 156 = 818

0.19X + 0.22Y = 662

Using the system of equations X + Y = 3200 and 0.19X + 0.22Y = 662, we can solve for X and Y to get:

X = 800

Y = 2400

Substituting back into equation (3), we get:

Z = X + Y + 1200 = 4400

Therefore, the amounts invested at 6%, 9%, and 13% were $800, $2400, and $4400 respectively.