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Mackey invested some money in an account paying 6% interest for one year. She invested $1000 more than this amount in an account paying 8.5%. How much did she invest in total if the total interest earned in the year was $737.50?

User Gplayer
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2 Answers

5 votes

$4500

$4500

$4500

$4500

User Bao HQ
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8.3k points
6 votes

Explanation:

Let's assume the amount Mackey initially invested in the 6% account is x dollars.

According to the given information, Mackey invested $1000 more than x in the 8.5% account, which means she invested (x + $1000) dollars in that account.

The total interest earned from the 6% account would be 0.06x dollars, and the total interest earned from the 8.5% account would be 0.085(x + $1000) dollars.

The total interest earned in the year is given as $737.50, so we can set up the equation:

0.06x + 0.085(x + $1000) = $737.50

Simplifying the equation:

0.06x + 0.085x + 0.085($1000) = $737.50

Combining like terms:

0.145x + $85 = $737.50

Subtracting $85 from both sides:

0.145x = $737.50 - $85

0.145x = $652.50

Dividing both sides by 0.145:

x = $652.50 / 0.145

x ≈ $4500

Therefore, Mackey initially invested approximately $4500 in the 6% account, and she invested $1000 more, which means she invested a total of $5500.

User Rajendra Thorat
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