Janet invested $20,000, part at 11% and part at 10%. If the total interest at the end of the year is $2,110,?

Question

Janet invested $20,000, part at 11% and part at 10%. If the total interest at the end of the year is $2,110, how much did she invest at each rate

Answer 1

Let's say that she invested $x at 11%, and the remainder ($20,000 - x) at 10%. Then the equation is:

(x)(0.11) + (20000-x)(0.10) = 2110

0.11x + 2000 - 0.10x = 2110
0.01x = 110
x = 11,000

So...she invested $11,000 at 11%, and $9,000 at 10%.

Checking the result:
$11,000 @ 11% = $1,210 interest
$9,000 @ 10% = $900 interest

$1,210 + $900 = $2,110, which is the correct total interest amount.

Answer 2

$9000 at 10% = $900 & $11000 at 11% = $1210

Assuming interest was added at the end of the year.

Answer 3

hi, enable t = amt. invested at 10% enable w = amt. invested at 12% Then t + w = ten thousand and 10p.c.t + 12p.c.w = 1100 Now multiply the 2nd equation via one hundred to clean it of fractions. Giving us 10t + 12w = 110000 Now divide via 2 and we've 5t + 6w = 55000 ok multiply the 1st equation via -5 we've -5t -5w = -50000 upload the two above equations then we've w = $5000 Then t = $5000 desire This facilitates!!

Answer 4

amount invested rate
x 0.11
20000-x 0.10

0.11x+0.1(20000-x) = 2110
0.11x+2000-0.1x=2110
0.01x=110/0.01
x=11000
Janet invested 11000 at 11% and 9000 at 10%