Question 622654: How can $60,000 be invested, part at 7% annual simple interest and the remainder at 12% annual simple interest, so that the interest earned by the two accounts will be equal?
Answer by math-vortex(648)   (Show Source):
You can put this solution on YOUR website! Hi, there--
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We can use algebra to solve this.
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Let x be the amount of money invested at 7% simple interest.
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The money invested in both accounts sums up to $60,000, so the expression 60000-x represents the amount of money invested at 12% simple interest.
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Now express the interest rates in decimal form: 7% = 0.07 and 12% = 0.12
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The amount of interest earned on the the first account is the interest rate times the amount invested, or 0.07x.
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Likewise, the amount of interest earned on the second account is its interest rate times the amount invested, or 0.12*(60000-x).
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We want the interest earned on the two accounts to be equal, so we write an equation showing the two expressions to be equal.
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[interest earned at 7%] = [interest earned at 12%]
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Now simplify and solve for x.
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Add 0.12x to both sides of the equation.
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Divide both sides of the equation by 0.19.
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In the context of our problem x=37894.74 means that $37,894.74 should be invested in the account with 7% interest.
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The remainder should be invested in the account with 12% interest. Since 60,000-37,894.74= 22,105.26, we should invest $22,105.26 in the second account.
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The final step is to check that this allocation of the money actually earns the same amount of interest.
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$37,894.74 at 7% interest earns $2,652.63 since 0.07*37894.74=2652.63
$22,105.26 invested at 12% interest earns $2,652.63 since 0.12*22105.26=2652.63
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That's it. Feel free to email if you have questions about any part of this explanation.
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Ms.Figgy
math.in.the.vortex@gmail.com
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