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Question 472835: Sally invested $3000 in two parts. One part was invested at 8% interest and the second part at 10% interest.
Her total earnings were $256 dollars.
How much was invested at 8% and how much was invested at 10%
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! We don't know how much money Sally invested at 8%. So let's call that amount X.
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The remainder of the $3000 she had available to invest at 10%. So, the amount she had left from the $3000 to invest at 10% was $3000 - X.
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The amount X was invested at 8% so the return that it earned is 8% times X, or equivalently 0.08 * X.
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The amount left to invest at 10% was $3000 - X so the return that it earned is 10% times 3000 - X, or 0.10*(3000 - X). Doing the distributed multiplication, it earned 300 -0.10*X
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When you add together the two earnings it is equal to $256. In equation form this is:
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0.08X + 300 - 0.10X = 256
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First, get rid of the 300 on the left side by subtracting 300 from both sides of the equation:
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0.08X + 300 - 300 - 0.10X = 256 - 300
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When you do this subtraction you are left with:
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0.08X - 0.10X = -44
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Combine the two terms on the left side and you have:
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-0.02X = -44
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Solve for X by dividing both sides by the coefficient (multiplier) of X which is -0.02:
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X = -44/-0.02 = 2200
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So now we know that Sally invested $2200 at 8%. The rest of her money is $3000 minus the $2200 which is $800, and it was invested at 10%
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Check:
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$2200 times 8% = $2200 * 0.08 = $176
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and
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$800 times 10% = $800 * 0.01 = $80
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So the total earnings is $176 + $80 = $256
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This is exactly the amount that the problem said it should be and, therefore, the answer that we got is correct. $2200 was invested and 8% and $800 was invested at 10%.
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Hope that this helps you to understand the problem and the procedure to follow in solving it.
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