A person invested ​$6700 for 1​ year, part at 5​%, part at 10​%, and the remainder at 14​%.

Question

A person invested ​$6700 for 1​ year, part at 5​%, part at 10​%, and the remainder at 14​%. The total annual income from these investments was ​$753. The amount of money invested at 14​% was ​$700 more than the amounts invested at 5​% and 10​% combined. Find the amount invested at each rate.

Answer 1

.05x + .10y +.14z = 753, x+y+z = 6,700, z=700+x+y, y = 6700 -z -x =6700-(700+x+y) -x = 6000-2x-y

2y = 6000 -2x, y =3000 -x

05x+ .10y + .14(700+x+y) = 753

5x +10y +14(700+ x+y) = 75300

19x +24y = 75300- 700(14) = 75,300 -2,800 = 72,500

19x +24y = 72500

19x +24(3000-x) = 72,500

x-5 = 72,500 - 72000 = 500

x = $495 invested at 5%

y = 3000-495 = $2,505 invested at 10%

z = 700+495+2,505 = $3,700 invested at 14%