Question

Sam invests $6,000 in two bank accounts. One of the accounts pays 8% interest per year, and the other account pays 10% interest per year. If the total interest earned on the investments is $560 after one year, how much money was invested in each account?​

Answer 1

Sam invested $2000 in the account that pays 8% interest and $4000 in the account that pays 10% interest.

Let's assume Sam invested x dollars in the account that pays 8% interest, and ($6000 - x$) dollars in the account that pays 10% interest.

The interest earned from the account that pays 8% interest in one year is:

I = PRT (P is principal, R is rate and T is time)

$I_1 =0.08x$

The interest earned from the account that pays 10% interest in one year is:

$I_2=0.10(6000 - x)$

According to the problem, the total interest earned is $560. Therefore, we can set up the equation:

$0.08x + 0.10(6000 - x) = 560$

Simplifying the equation:

$0.08x + 600 - 0.10x = 560\\-0.02x = -40$

Dividing both sides by -0.02:

$x = -40 \div (-0.02)\\x = 2000$

So, Sam invested $2000 in the account that pays 8% interest, and the remaining amount of $6000 - 2000 = 4000$ in the account that pays 10% interest.

Therefore, Sam invested $2000 in the account that pays 8% interest and $4000 in the account that pays 10% interest.

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