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
Two equations. x+y=56000 .07x=.05y Solve both of these equations simultaneously and it will be the answer. x+(.07/.05 x)=56000
$14,693.28
3125
d 2.5 years
2500
You invested $15,000 in two accounts paying 6% and 8% annual interest, respectively.
Two equations. x+y=56000 .07x=.05y Solve both of these equations simultaneously and it will be the answer. x+(.07/.05 x)=56000
2.5 years
$14,693.28
3125
d 2.5 years
3125
2500
2500
If you invested 7580 and after 5 years you have 3126.75 then the annual interest rate is negative. It is -16.23%.
Let P be the amount of invested money. Then, .08P = 336 P = 336/.08 = 4,200
Kate invested 4500.