a woman has $27,000. Part is invested at 7% interest and the rest is invested at 9% interest. If the total interest earned is $2110, how much does she invest at each rate?
.07 x + .09 (27000 - x) = 2110
- .02 x = 2100 - 2430 = - 330
x = 16,500
27000 - x = 10,500
part at 7% --- x
part at 9% ---- 27000-x
.07x + .09(27000-x) = 2110
times 100
7x + 9(27000-x) = 211000
solve for x
(why not invest all at 9% ? )
To solve this problem, let's assign variables to the unknown quantities. Let's say the amount invested at 7% is 'x' dollars, and the remainder (i.e., the amount invested at 9%) is equal to '27000 - x' dollars.
Now, we can set up an equation based on the given information. The interest earned from the 7% investment plus the interest earned from the 9% investment should equal $2110:
0.07x + 0.09(27000 - x) = 2110
To solve for 'x,' we can simplify and solve the equation step-by-step:
0.07x + 2430 - 0.09x = 2110 [Distribute 0.09 to 27000 - x]
-0.02x + 2430 = 2110 [Combine like terms]
-0.02x = 2110 - 2430 [Subtract 2430 from both sides]
-0.02x = -320 [Simplify]
x = -320 / -0.02 [Divide both sides by -0.02]
x = 16000
Therefore, the woman invested $16,000 at 7% interest and the remaining amount, $27,000 - $16,000 = $11,000, at 9% interest.