SOLUTION: Sam invested $1950, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $46 more than twice the income from the 6% investment. How much
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-> SOLUTION: Sam invested $1950, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $46 more than twice the income from the 6% investment. How much
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Question 1129828: Sam invested $1950, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $46 more than twice the income from the 6% investment. How much did he invest at each rate?
Let x be the amount invested at 8%, in dollars.
Then the amount invested at 6% is the rest, (1950-x) dollars.
The equation from the condition is THIS:
0.08x - 2*0.06*(1950-x) = 46.
0.08x + 2*0.06x = 46 + 2*0.06*1950
x = = (one click in my MS Excel) = 1400.
Answer. $1400 were invested at 8%, and the rest 1950-1400 = 550 dollars were invested at 6%.
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).
