SOLUTION: 14,600 is invested, part at 5% and the rest at 2%. If the interest earned from the amount invested at 5% exceeds the interest earned from the amount invested at 2% by $479.00, how
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-> SOLUTION: 14,600 is invested, part at 5% and the rest at 2%. If the interest earned from the amount invested at 5% exceeds the interest earned from the amount invested at 2% by $479.00, how
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Question 1153662: 14,600 is invested, part at 5% and the rest at 2%. If the interest earned from the amount invested at 5% exceeds the interest earned from the amount invested at 2% by $479.00, how much is invested at each rate? (Round to two decimal places if necessary.) Answer by ikleyn(52787) (Show Source):
Let x be the part invested at 5%.
Then the amount invested at 2% is (14600-x) dollars.
The 5% investment produces the interest of 0.05x dollars.
The 2% investment produces the interest of 0.02*(14600-x) dollars.
The condition says
0.05x - 0.02(14600-x) = 479.
Express x and calculate
x = = 11014.29.
ANSWER. $11014.29 was invested at 5%. The rest, (14600-11014.20) = 3585.71 dollars were invested at 2%.
CHECK. 0.05*11014.29 - 0.02*3585.71 = 479.00 dollars. ! Precisely correct !
Solved.
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It is a standard and typical problem on investments.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
