Question 308911: Not exactly sure where this type of equation fits in but I am in need of obtaining assistance.
A college student invested part of a $25,000 inheritance at 7% interest and the rest at 6%. If his annual interest is $1,671 how much did he invest at 6%? If the answer is $8,000, explain how you would solve the problem
Found 2 solutions by scott8148, Earlsdon:
Answer by scott8148(6628)   (Show Source):
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! Well, the answer is not $8,000 but I will explain how I would solve this problem anyway.
First, I would compute the amount of interest earned on each of the two amounts, $x (invested at 6% per annum) and $(25,000-x) (invested at 7% per annum).
Then I would add these two amounts and set the sum equal the annual interest earned ($1,671.00).
Finally, I would solve for x, the amount invested at 6%.
Here are the algebraic steps:
Change the percentages to their decimal equivalents. (6% = 0.06) and (7% = 0.07).
0.06x+0.07(25000-x) = 1671 Simplify and solve for x.
0.06x+1750-0.07x = 1671 Combine the like-terms.
-0.01+1750 = 1671 Subtract 1750 from both sides.
-0.01x = -79 Finally, divide both sides by -0.01
x = 7900
$7,900 was invested at 6% per annum.
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