A man invest $12000 in two accounts.If one account pays 10% per year and the other pays 7% per year, how much was invested in each account if the total interest earned in the first year was $960?
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part at 10% --- x
part at 7% = 12000 - x
solve for x
.1x + .07(12000-x) = 960
yugiu
To solve this problem, let's assume that the amount invested in the account that pays 10% per year is x dollars.
Therefore, the amount invested in the account that pays 7% per year would be $12000 - x, as the total investment is $12000.
Next, let's calculate the interest earned from each account.
The interest earned from the account that pays 10% per year would be (10/100) * x = 0.1x dollars.
The interest earned from the account that pays 7% per year would be (7/100) * ($12000 - x) = 0.07($12000 - x) dollars.
According to the problem, the total interest earned in the first year was $960.
So, we can set up the equation:
0.1x + 0.07($12000 - x) = $960.
Now, let's solve this equation step by step to find the value of x.
0.1x + 0.07($12000 - x) = $960
0.1x + 840 - 0.07x = $960
0.03x + 840 = $960
0.03x = $960 - $840
0.03x = $120
To solve for x, we divide both sides of the equation by 0.03:
x = ($120) / (0.03)
x = $4000
So, $4000 was invested in the account that pays 10% per year, and ($12000 - $4000) = $8000 was invested in the account that pays 7% per year.
Therefore, the man invested $4000 in the account that pays 10% per year and $8000 in the account that pays 7% per year.