SOLUTION: A teacher invests $12,000 in accounts and earns $$870 in interest for the year. Part of the money was invested at 6.38% and the remainder at 7.83%, both simple interest. How much w
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Question 1197548: A teacher invests $12,000 in accounts and earns $$870 in interest for the year. Part of the money was invested at 6.38% and the remainder at 7.83%, both simple interest. How much was invested at each rate?
A setup for a solution by a standard algebraic method....
x = amount invested at 6.38%
12000-x = amount invested at 7.83%
The total interest was $870:
The solution is straightforward; but the numbers are ugly. Using a calculator might be a good idea for doing the actual arithmetic.
An informal solution.... The numbers are still ugly; but in the end you get the answer with less work.
(1) Use a calculator to find that $870 interest on $12000 is an interest rate of 7.25%.
(2) Using a number line and simple arithmetic, determine that 7.25% is 87/145 = 3/5 of the way from 6.38% to 7.83%.
(3) That means 3/5 of the total was invested at the higher rate.
ANSWER: 3/5 of $12,000, or $7200, at 7.83%; the other $4800 at 6.38%.
CHECK (use a calculator): .0638(4800)+.0783(7200)=870
