SOLUTION: Sam invested $2000, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $20 more than twice the income from the 6% investment. How much
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-> SOLUTION: Sam invested $2000, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $20 more than twice the income from the 6% investment. How much
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Question 1140269: Sam invested $2000, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $20 more than twice the income from the 6% investment. How much did he invest at each rate? Answer by ikleyn(52781) (Show Source):
Let x = amount at 8%, in dollars.
Then the amount at 6% is the rest (2000-x) dollars.
The interest from the 8% amount is 0.08*x dollars.
The interest from the 6% amount is 0.06*(2000-x) dollars.
Your equation is
interest at 8% - 2*(interest at 6%) = 20 dollars, or
0.08*x - 2*0.06*(2000-x) = 20 dollars.
Simplify this equation, express x and calculate the answer
0.08x - 2*0.06*2000 + 0.12x = 20,
x = = 1300.
Answer. The amount at 8% is $1300; the rest $2000-$1300 = $700 is the amount at 6%.
Check. 0.08*1300 - 2*0.06*700 = 20 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).
