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

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

### 4 Answers

- McFateLv 71 decade agoFavorite Answer
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.

- angieLv 51 decade ago
$9000 at 10% = $900 & $11000 at 11% = $1210

Assuming interest was added at the end of the year.

- Anonymous4 years ago
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!!

- 1 decade ago
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%