Question

A person invested $6100 for 1 year, part at 7%, part at 10%, and the remainder at 14%. The total annual income from these investments was $713. The amount of money invested at 14% was $700 more than the amounts invested at 7% and 10% combined. Find the amount invested at each rate.

Answer 1

Total investment = 6100

Time of investment = 1 year

First part at 7%

Second part 10%

Third part 14%

Total annual income = 713

Third part (14%) is equal to 700 + First part (7%) + Second part (10%)

Simple interest formula: A = P(1 + rt), where A is the final investment value, P is the principal, r is the interest rate and t is the time

If we say that F is the invested part at 7%, S at 10% and T at 14%, we can write:

Equation1: F + S + T = 6100 ==> T = 6100 - S - T

Equation 2: F(0.07) + S(0.1) + T(0.14) = 713 ==> 0.07F + 0.1S + 0.14T = 713

Equation 3: T = 700 + F + S

Answer:

F = 1100

S = 1600

T = 3400

Equation 1: F = 6100 - S - T

Equation 2: F = (713 - 0.1S - 0.14T)/0.07 = 10185.71429 - 1.428571429S - 2T

6100 - S - T = 10185.71429 - 1.428571429S - 2T

1.428571429S - S - T + 2T = 10185.71429 - 6100

0.428571429S + T = 4085.71429

T = 6100 - S - T

T = 700 + F + S

6100 - S - T = 700 + F + S

5400 = 2F + 2S = 2(F + S)

F + S = 2700

0.428571429S + T = 4085.71429

T = 700 + F + S

0.428571429S + 700 + F + S = 4085.71429

F + 1.428571429S = 4085.71429 - 700 = 3385.71429

F + S = 2700 ==> S = 2700 - F

F + 1.428571429S = 4085.71429 - 700 = 3385.71429

F + 1.428571429(2700 - F) = 3385.71429

F + 3857.142858 - 1.428571429F = 3385.71429

3857.142858 - 3385.71429 = 0.428571429F

471.428568/0.428571429 = F = 1100

S = 2700 - 1100 = 1600

T = 700 + F + S = 700 + 1100 + 1600 = 3400