SOLUTION: two investments are made totaling $10,000. In one year these investments
yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and
the rest at 6 3/4%. How
Algebra.Com
Question 443238: two investments are made totaling $10,000. In one year these investments
yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and
the rest at 6 3/4%. How much more money is invested at 6 3/4%?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
two investments are made totaling $10,000. In one year these investments
yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and
the rest at 6 3/4%. How much more money is invested at 6 3/4%?
----
2-variable solution:
Quantity Eq. :: x + y = 10,000
Interest Eq. ::0.055x + 0.0675y = 650
-----------------------------------
Multiply 1st by 550
Multiply 2nd by 10000
-------------------------
550x + 550y = 550*10,000
550x + 675y = 6400000
----------------------------------------
Subtract and solve for "y":
125y = 9000000
y = $7200 (amt invested at 6 3/4%)
===================
Cheers,
Stan H.
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