SOLUTION: Sam invested $2050, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $54 more than twice the income from the 6% investment. How much
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Question 1140509: Sam invested $2050, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $54 more than twice the income from the 6% investment. How much did he invest at each rate? Answer by ikleyn(52786) (Show Source):
Let x = amount at 8%, in dollars.
Then the amount at 6% is the rest (2050-x) dollars.
The interest from the 8% amount is 0.08*x dollars.
The interest from the 6% amount is 0.06*(2050-x) dollars.
Your equation is
interest at 8% - 2*(interest at 6%) = 54 dollars, or
0.08*x - 2*0.06*(2050-x) = 54 dollars.
From the equation, express x and calculate the answer
x = = 1500.
Answer. The amount at 8% is $1500; the rest $2050-$1500 = $550 is the amount at 6%.
Check. 0.08*1500 - 2*0.06*550 = 54 dollars. ! Correct !
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It is a typical and standard problem on investment.
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).
