Answer: he invested $200 at 10%, $400 at 11% and $600 at 12%
Step-by-step explanation:
Let x represent the amount invested in the account earning 10% interest.
Let y represent the amount invested in the account earning 11% interest.
Let z represent the amount invested in the account earning 12% interest.
If she has twice as much money invested at 11 % as she does in 10 %, it means that
y = 2x
If she has three times as much at
12 % as she has at 10 %, it means that
z = 3x
The interest from the first account is
x × 0.1 × 1 = 0.1x
The interest from the second account is
y × 0.11 × 1 = 0.11y
The interest from the third account is
z × 0.12 × 1 = 0.12z
If the total interest for the year is $136, it means that
0.1x + 0.11y + 0.12z = 136 - - - - - - - - 1
Substituting y = 2x and z = 3x into equation 1, it becomes
0.1x + 0.11(2x) + 0.12(3x) = 136
0.1x + 0.22x + 0.36x = 136
0.68x = 136
x = 136/0.68 = 200
y = 2x = 2 × 200
y = $400
z = 3x = 3 × 200
z = $600
