Question

A bank loaned out $37,500, part of it at the rate of 11% annual interest, and the rest at 9% annual interest. The total interest earned for both loans was $3,745.00. How much was loaned at each rate?

Answer 1

Answer:

The Loan amount at 11 % interest rate is $ 18,500   And

The Loan amount at 9 % interest rate is $ 19,000

Step-by-step explanation:

Given as :

Let The loan given at interest rate of 11 % annual  = $ x

And The loan given at interest rate of 9 % annual = ($37,500 - $ x)

Total interest fro both loan = $3745

I.e  CI 1 + CI 2 = $3745

Now, From compound interest method :

$Amount= Principal\times (1 + \frac{Rate}{100})^{Time}$

$A 1 = 1\times (1 + \frac{11}{100})^{1}$

Or,  A 1 = $ 1.11 x

Similarly

$A 2 = ( 37,500 -x ) \times (1 + \frac{9}{100})^{1}$

Or, A 2 = 1.09 × ($37,500 - $ x)

∵  Compound Interest = Amount - principal

Or, $ 3745 = CI 1 + CI 2

Or,  $ 3745 = ($ 1.11 x - $ x) + ( 1.09 × ($37,500 - $ x) -  ($37,500 - $ x) )

Or,  $ 3745 = $ .11 x+  ($37,500 - $ x) ( .09 )

Or,  $ 3745 = $ .02 x + $ 3375

or,  0.02 x =  $ 3745 - $ 3375

∴   x = $ $\frac{370}{.02}$

SO,  x = $ 18,500

And $ 37500 - x = $ 19,000

Hence The Loan amount at 11 % interest rate is $ 18,500   And

           The Loan amount at 9 % interest rate is $ 19,000   Answer