A stockbroker has money in three accounts. The interest rates on the three accounts are 10 % 10%, 11 % 11%, and 12 % 12%. If she has twice as much money invested at 11 % 11% as she does in 10 % 10%, three times as much at 12 % 12% as she has at 10 % 10%, and the total interest for the year is $ 136 $136, how much is invested at each rate?

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

63,081,754

Step-by-step explanation:

sixty-three million: 63 (the millions place)

eighty one thousand: 081 (thousands place)

seven hundred fifty-four (hundreds place)