Eva invested a certain amount of money at 10% interest and $1000 more
than that amount at 11%. Her total yearly interest was $740. How much
did she invest at each rate?
2007-08-01 17:31:44 · 4 answers · asked by Tazzzy 1 in Science & Mathematics ➔ Mathematics
Let's call the 10% amount x. Then the amount at 11% would be (x + 1000). The equation is:
x * 0.10 + (x + 1000) * 0.11 = 740
x * 0.21 + 110 = 740
x * 0.21 = 630
x = 3000
3000 @ 10%
4000 @ 11%
Checking the answer:
3000 @ 10% = $300
4000 @ 11% = $440
$300 + $440 = $740
2007-08-01 17:34:47 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
Ten percent of 'x' is equal to x(10)/100 or x(.1). The amount at eleven percent is equal to (x+1000)(11)/100 or .11(x+1000), so we have:
.1x+.11(1000+x)=740
.1x+110+.11x=740
.21x=740-110
.21x=630
x=3000
So eva investes $3'000 at 10% and $4'000 at 11%
2007-08-01 17:34:42 · answer #2 · answered by ΛLΞX Q 5 · 0⤊ 0⤋
she invested $3000 at 10% and $4000 at 11%.
2007-08-01 17:37:27 · answer #3 · answered by Kamina Squirtle 4 · 0⤊ 0⤋
Solve for M: 0.10 M + 0.11(M +1000) = 740.
2007-08-01 17:40:06 · answer #4 · answered by Anonymous · 0⤊ 0⤋