SOLUTION: A total of $3700 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $330, how much was invested at each rate?
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Question 779591: A total of $3700 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $330, how much was invested at each rate? Found 2 solutions by mananth, josgarithmetic:Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! Part I 6.00% per annum ------------- Amount invested =x
Part II 11.00% per annum ------------ Amount invested = y
3700
Interest----- 330.00
Part I 6.00% per annum ---x
Part II 11.00% per annum ---y
Total investment
x + 1 y= 3700 -------------1
Interest on both investments
6.00% x + 11.00% y= 330
Multiply by 100
6 x + 11 y= 33000.00 --------2
Multiply (1) by -6
we get
-6 x -6 y= -22200.00
Add this to (2)
0 x 5 y= 10800
divide by 5
y = 2160 - x)
Part I 6.00% $ 1540
Part II 11.00% $ 2160
You can put this solution on YOUR website! A total of $3700 was invested, part of it at 6% interest and the remainder at 11%.
If the total yearly interest amounted to $330, how much was invested at each rate?
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A total of $T was invested, part of it at c% interest and the remainder at g%.
If the total yearly interest amounted to $n, how much was invested at each rate?
(All values are as variables.)
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invested v at g% and therefore invested T-g at c%.
The amounts of earned interests were and .
Their sum n, was given, so .
Solve the equation for v.
, for ALL such examples which fit this same exact format.
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