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

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)   (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=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
and simplify it to
Equation 1......
Equation 2......

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=
Now substitute x back into the first equation to find Y
x+y=13,500
3,000+y=13,500
y=

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

Answer by MathTherapy(10550)   (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)   (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

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