Phyllis invested $12,000, part at 14% annual interest and the remainder at 10%. Last year she earned $1632 in interest. How much money did she invest at each rate?
*Thanks a bunch!
2007-01-11 13:01:10 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
14x + 10(12000-x) = 163200 (alredy multiplied both sides by 10 to get rid of percentages)
4x + 120000 = 163200
4x = 43200
x = 10800
10800 @ 14% = 1512
1200 @ 10% = 120
1512+120 = 1632
tada!
2007-01-11 13:09:25 · answer #1 · answered by tomkat1528 5 · 0⤊ 0⤋
She invested $1200 at 10% and $10,800 at 14%.
Solve this equation: .14(12,000-x) + .1x = 1632
1680 - .14x +.1x = 1632
1680 -.04x =1632
-.04x = -48
x = 1200
12,000-1200 = 10,800
2007-01-11 13:10:13 · answer #2 · answered by crane 1 · 0⤊ 0⤋
Lets say X is the amount of money invested at 14% and Y the amount of money invested at 10%
X + Y = 12,000 ------> X=12,000 - Y
(14/100)*X + (10/100)*Y = 1632
0.14X + 0.1Y= 1632
replace X by (12,000 - Y)
0.14(12,000 - Y) + 0.1Y= 1632
1680 - 0.14Y + 0.1Y =1632
0.04Y = 48
Y = 48/0.04 =1200
X = 12,000 - Y = 12000 - 1200 = 10,800
2007-01-11 13:11:30 · answer #3 · answered by Antonio R 3 · 0⤊ 0⤋
at the same time as a variable is with information from itself there's a a million in the front of it. 7u+4u+1u upload your numbers mutually 7+4+a million to get 12 placed the variable on the end. note: If between the united stateshave an exponent, or there is yet another variable, that's distinct. ex: 7u+4u+v might want to be 11u+v 7u+4u^2+u might want to be 8+4u^2
2016-12-02 03:44:43 · answer #4 · answered by boshell 4 · 0⤊ 0⤋
If it's easy....then why don't you do it?
2007-01-11 13:03:23 · answer #5 · answered by Anonymous · 0⤊ 0⤋