SOLUTION: dr e man invested $10,000, part at 8 1/2% per annum and the rest at 6 3/4% per annum in one year the amiunt earned at the 8 1/% was $14 more than twice the amount earned at the 6 3
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Question 1174215: dr e man invested $10,000, part at 8 1/2% per annum and the rest at 6 3/4% per annum in one year the amiunt earned at the 8 1/% was $14 more than twice the amount earned at the 6 3/4% how much was invested at each rate and how much interest was earned at each rate?
Note that the calculations are straightforward (although a bit ugly) once you have set up the problem; by far the most important part of solving the problem is converting the given information into an appropriate equation.
So pay close attention to that part of the solution below.
The total investment is $10,000; let x be the amount invested at 8.5% and ($10,000-x) be the amount invested at 6.75%. Note that is a typical strategy for starting on a problem where the sum of two numbers is given -- one number is x, and the other is the given sum minus x.
The interest from the first investment is 8.5% of x; the interest from the second is 6.75% of ($10,000-x).
The interest from the first investment was $14 more than twice the interest from the second:
Solve algebraically; or graph the two expressions on a graphing calculator and see that they intersect at x=6200.
So $6200 was invested at 8.5%, and the other $3800 was invested at 6.75%.
