You invested 19000 in two accounts paying 2 percent and 6 percent annual interest If the total interest earned for the year was 620 how much was invested at each rate?

Answer 1

13000 at 2% and 6000 at 6%

Create 2 simultaneous equations and solve:

Let the amount invested at 2% be x and at 6% be y, then:

x + y = 19000

0.02x + 0.06y = 620

(Since 2% means 2/100 = 0.02; similarly 6% = 6/100 = 0.06). Divide the second equation by 0.02 (multiply by 100 and divide by 2):

x + y = 19000

x + 3y = 31000

Subtract first from second and solve:

2y = 12000

y = 6000

Substitute back into first and solve:

x + 6000 = 19000

x = 13000

Substitute solution back into (original) second equation to check:

0.02 x 13000 + 0.06 x 6000 = 260 + 360 = 620