Question

Mike has $6000 to invest. He invested part of his total into an account earning 6% interest and the rest in an account earning 11% interest. If at the end of the year, he earned $460.00 in interest, how much money was invested in each account?

Answer 1

Answer:$4000 was invested into the account earning 6% interest.

$2000 was invested into the account earning 11% interest.

Step-by-step explanation:

Let x represent the amount invested into the account earning 6% interest.

Let y represent the amount invested into the account earning 11% interest.

Mike has $6000 to invest. He invested part of his total into an account earning 6% interest and the rest in an account earning 11% interest. This means that

x + y = 6000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the account earning 6% interest, the interest would be

I = (x × 6 × 1)/100 = 0.06x

Considering the account earning 11% interest, the interest would be

I = (x × 11 × 1)/100 = 0.11y

If at the end of the year, he earned $460.00 in interest, it means that

0.06x + 0.11y = 460 - - - - - - - - - - -1

Substituting x = 6000 - y into equation 1, it becomes

0.06(6000 - y) + 0.11y = 460

360 - 0.06y + 0.11y = 460

- 0.06y + 0.11y = 460 - 360

0.05y = 100

y = 100/0.05 = $2000

x = 6000 - y = 6000 - 2000

x = $4000