A bank loaned out $120,000, part of it at a rate of 8% per year and the rest at the rate of 12% per year. if the interested received totaled $11,000, how much was loaned at each rate?
I guess this means interest in one year.
.08(x) + .12(120,000-x) = 11,000
solve for x
.7638
$85000 and $35000
A bank loaned out $12,000, part of it at the rate of 8% per year and the rest at the rate of 18% per year. If the interest received totaled $1000, how much was loaned at 8%?
To solve this problem, we can use a system of linear equations.
Let's say the amount loaned at 8% per year is $x, and the amount loaned at 12% per year is $y.
Now, we can set up the equations based on the given information.
The interest received from the amount loaned at 8% can be calculated as 8% of x, which is 0.08x.
Similarly, the interest received from the amount loaned at 12% can be calculated as 12% of y, which is 0.12y.
The total interest received is $11,000, so we can set up the following equation:
0.08x + 0.12y = 11,000 ---(Equation 1)
We also know that the total amount loaned is $120,000, so we can set up another equation:
x + y = 120,000 ---(Equation 2)
To solve this system of equations, we can use the substitution method or the elimination method.
Using the elimination method, we can multiply Equation 2 by -0.08 to get:
-0.08x - 0.08y = -9,600 ---(Equation 3)
Adding Equation 3 to Equation 1 cancels out the x term:
0.08x + 0.12y + (-0.08x) + (-0.08y) = 11,000 + (-9,600)
0.04y = 1,400
Dividing both sides of the equation by 0.04:
y = 35,000
Now substitute the value of y back into Equation 2:
x + 35,000 = 120,000
x = 120,000 - 35,000
x = 85,000
Therefore, $85,000 was loaned at 8% per year and $35,000 was loaned at 12% per year.