A lender offers a choice between two loans. For loan A the lender charges 12% a year compounded 12 times a year. For loan B the lender charges 12.5% a year and compounds it once a year. Is loan A or loan B a cheaper loan for the borrower. Please show work!!!
Loan A:
P = Po(1+r)^n.
Po = $100.
r = 0.12/12mo. = 0.01/mo.
P = 100(1.01)^12 = $$112.68
Loan B:
P = 100(1.125)^1 = $112.50
The amount of loan can be any value as long as the loans are equal.
To determine which loan is cheaper for the borrower, we need to compare the effective annual interest rates (EAR) of both Loan A and Loan B. The EAR takes into account how often the interest is compounded and provides a standard basis for comparison.
Let's start by calculating the EAR for Loan A, which has an interest rate of 12% compounded 12 times a year (monthly compounding).
Step 1: Convert the annual interest rate to a decimal:
12% = 0.12
Step 2: Divide the annual interest rate by the compounding frequency:
0.12 / 12 = 0.01
Step 3: Add 1 to the result:
1 + 0.01 = 1.01
Step 4: Raise the previous result to the power of the compounding frequency (12):
(1.01)^12 ≈ 1.126825
Step 5: Subtract 1 from the previous result and multiply by 100 to get the equivalent EAR as a percentage:
(1.126825 - 1) * 100 ≈ 12.68%
So, for Loan A, the effective annual interest rate (EAR) is approximately 12.68%.
Now let's calculate the EAR for Loan B, which has an interest rate of 12.5% compounded once a year (annual compounding).
Step 1: Convert the annual interest rate to a decimal:
12.5% = 0.125
Step 2: Add 1 to the result:
1 + 0.125 = 1.125
Step 3: Subtract 1 from the previous result and multiply by 100 to get the equivalent EAR as a percentage:
(1.125 - 1) * 100 ≈ 12.5%
So, for Loan B, the effective annual interest rate (EAR) is approximately 12.5%.
Now, comparing the two EARs, we can see that Loan A has an EAR of 12.68%, while Loan B has an EAR of 12.5%. Since the effective annual interest rate for Loan B is lower, it means that Loan B is the cheaper loan for the borrower.