SOLUTION: An investment adviser invested $14,000 in two accounts. One investment earned 8% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of
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Question 1127076: An investment adviser invested $14,000 in two accounts. One investment earned 8% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of interest earned for 1 year was $757. How much was invested in each account? Answer by ikleyn(52770) (Show Source):
0.08x + 0.025(14000-x) = 757.
Solve for x, which is the amount invested at 8%.
x = = (one click in my MS Excel) = 7400.
Answer. 7400 dollars were invested at 8%. The rest, 14000-7400 = 6600 dollars were invested at 2.5%.
Check. 0.08*7400 + 0.025*6600 = 757 dollars. ! 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).
