Part of $7,000 is invested at 10%, another part at 11%, and the remainder at 12% yearly interest. The total yearly income from the three investments is $795. The sum of the amounts invested at 10% and 11% equals the amount invested at 12%. How much is invested at each rate?
A)$1,000 at 10%, $2,500 at 11%, and $3,500 at 12%
B)$500 at 10%, $2,000 at 11%, and $4,000 at 12%
C)$1,500 at 10%, $2,000 at 11%, and $3,500 at 12%
D)$500 at 10%, $2,500 at 11%, and $4,000 at 12%
E)$1,500 at 10%, $3,000 at 11%, and $3,000 at 12%

Question

Quizplus · Accepted Answer

Let x be the amount invested at 10%, y be the amount invested at 11%, and z be the amount invested at 12%.From the given information, we have:x + y + z = 70000.10x + 0.11y + 0.12z = 795y = zSubstituting y = z into the first equation, we get:x + 2z = 7000Substituting y = z and x + 2z = 7000 into the second equation, we get:0.10x + 0.11z + 0.12z = 7950.10x + 0.23z = 795Solving for x, we get:x = 795 - 0.23zSubstituting x = 795 - 0.23z into the equation x + 2z = 7000, we get:795 - 0.23z + 2z = 70001.77z = 6205z = 3500Substituting z = 3500 into the equation y = z, we get:y = 3500Substituting z = 3500 into the equation x = 795 - 0.23z, we get:x = 795 - 0.23(3500)x = 1000Therefore, the amount invested at 10% is $1000, the amount invested at 11% is $3500, and the amount invested at 12% is $3500.

Part of $7,000 Is Invested at 10%, Another Part at 11