SOLUTION: Aimee calhoun invested $12500 part at 7% simple interest and part at 6% simple interest for a period of 1 year. how much was invested at each rate if each account earned the same i

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Question 112653This question is from textbook ELEMENTARY & iNTERMEDiATE
: Aimee calhoun invested $12500 part at 7% simple interest and part at 6% simple interest for a period of 1 year. how much was invested at each rate if each account earned the same interest? This question is from textbook ELEMENTARY & iNTERMEDiATE

Found 2 solutions by josmiceli, chitra:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
.07x+=+.06%2812500+-+x%29
.07x+=+750+-+.06x
.13x+=+750
x+=+5769.23
12500+-+x+=+6730.75
$5769.23 was invested @ 7%, and $6730.75 was invested @ 6%

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
The total amount invested = $12500

Let x be the amount invested in the first part at the rate of 7% for a duration of 1 yr

Therefore Simple Interest(SI) = PTR/100

where P = principle T = time and R = rate

SI = x * 1 * (7/100) -----------------------(1)

Let the remaining amount be (12,500 - x) invested at 6% for a duration of 1 yr

SI = (12,500 - x) * 1 * (6/100) -------------------(2)

It is told that the interest earned by both the investments is the same.

so equating EQN'S (1) and (2) we get:

x+%2A+7%2F100 = (12,500 - x)6%2F100

Solving for x we get:

7x = (12,500 - x)6

==> 7x = 75,000 - 6x

Adding 6x on both sides we get:

13x = 75,000

==> x+=+75%2C000%2F13

==> x = 5769.230 was invested at 7%

So, now the other value is got by:


==> 12,500 - 5769.230 = 6730.769 was invested at 6%

thus the solution.