Question 106578: Janet invested $31,000, part at 10% and part at 1%. If the total interest at the end of the year is $1,390, how much did she invest at 10%?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the amount Janet invested at 10%, then ($31,000 -x) = the amount invested at 1%.
The interest can be written as:
x(10%) + ($31,000)(1%) = $1,390 Change the percentages to their decimal equivalents (10% = 0.1) and (1% = 0.01)
Now you can write the equation to solve for x.
0.1x + (0.01)($31,000 - x) = $1,390 Simplify and solve for x.
0.1x + $310 - 0.01x = $1,390
0.09x + $310 = $1,390 Subtract $310 from both sides.
0.09x = $1,080 Divide both sides by 0.09
x = $12,000 This is the amount Janet invested at 10%
|
|
|
