SOLUTION: A bank loaned out $13,500, part of it at the rate of 9% annual interest, and the rest at 13% annual interest. The total interest earned for both loans was $1,635.00. How much was l

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Question 1158944: A bank loaned out $13,500, part of it at the rate of 9% annual interest, and the rest at 13% annual interest. The total interest earned for both loans was $1,635.00. How much was loaned at each rate?
Found 3 solutions by Amily_2190, MathTherapy, greenestamps:
Answer by Amily_2190(27) About Me  (Show Source):
You can put this solution on YOUR website!
Lets say that
x= amount loaned at the rate of 9% annual interest
y=amount loaned at the rate of 13% annual interest
The two loans add up to $13,500, so this means that x%2By=%2413500
x=13,500-y
y=13,500-x
You know how much the annual rate of interest is for each loan and the total interest that is earned, so you can create this equation:
($13,500-y)(.09)+($13,500-x)(.13)=$1,635

Now, we need to solve for x and y.
-First, distribute the second equation to get
1215-.09y%2B1755-.13x=1%2C635 and simplify it to -.09y-.13x=-1335
Equation 1......x%2By=%2413%2C500
Equation 2......-.09y-.13x=-1335

Now that you have two equations, use elimination. Multiply the first equation by .09 to cancel out the y values.
(.09)(x+y)=(.09)(13,500)
.09x+.09y=1215....(Add this to the second equation)

.09x+.09y=1215
+(-.09y-.13x=-1335)
-0.04x=-120
x=%243000
Now substitute x back into the first equation to find Y
x+y=13,500
3,000+y=13,500
y=%2410500

Answer: A bank loaned $3000 at the rate of 9% annual interest and $10500 at 13% annual interest.
Hope this helped!


Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!

A bank loaned out $13,500, part of it at the rate of 9% annual interest, and the rest at 13% annual interest. The total interest earned for both loans was $1,635.00. How much was loaned at each rate?
Let the amount invested at 9% be N
Then amount invested at 13% = 13,500 - N
We then get: .09N + .13(13,500 - N) = 1,635
.09N + 1,755 - .13N = 1,635
.09N - .13N = 1,635 - 1,755
- .04N = - 120
Amount invested at 9%, or
Now, do you think you can find the amount invested at 13%?

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a non-algebraic alternative for solving "mixture" problems like this.

$13,500 all at 9% would yield $1215 interest.
$13,500 all at 3% would yield $1755 interest.
The actual interest was $1635.

On a number line (if you want to think of it that way), 1635 is 420/540 = 7/9 of the way from 1215 to 1755.

Therefore, 7/9 of the total was invested at 13% (and the other 2/9 of it at 9%).

2/9 of $13,500 is $3000; 7/9 is $10,500.

ANSWER: $10,500 at 13%; $3000 at 9%.

CHECK:
.13(10,500)+.09(3000) = 1365+270 = 1635