Part of $6,000 is invested at 9%, another part at 11%, and the remainder at 12% yearly interest. The total yearly income from the three investments is $650. The sum of the amounts invested at 9% and 11% equals the amount invested at 12%. How much is invested at each rate?
A) $2,000 at 9%, $1,000 at 11%, and $3,000 at 12%
B) $1,500 at 9%, $0 at 11%, and $3,500 at 12%
C) $3,000 at 9%, $0 at 11%, and $3,000 at 12%
D) $1,500 at 9%, $1,000 at 11%, and $3,500 at 12%
E) $2,500 at 9%, $2,000 at 11%, and $2,000 at 12%

Question

Quizplus · Accepted Answer

To solve this, set up equations based on the information given. Let x, y, and z represent the amounts invested at 9%, 11%, and 12%, respectively. We have three pieces of information:1. $x + y + z = 6000$ (the total amount invested)2. $0.09x + 0.11y + 0.12z = 650$ (the total interest earned)3. $x + y = z$ (the sum of the amounts invested at 9% and 11% equals the amount invested at 12%)Substituting the third equation into the first gives $2z = 6000$, so $z = 3000$. Substituting $z = 3000$ into the second equation allows us to solve for $x$ and $y$, leading to $x = 2000$ and $y = 1000$. Therefore, $2,000 is invested at 9%, $1,000 at 11%, and $3,000 at 12%.

Part of $6,000 Is Invested at 9%, Another Part at 11