Question

a)Paul was due to make loan payments of $1,870 7 months ago, $1,951 5 months ago, and $1,232 in 9 months. Instead, he is to make a single payment today. If money is worth 7.61%, what is the size of the replacement payment?

***(Final answer should be rounded to two decimal places.)

b)

After the COVID-19 pandemic was over, Malika decided to go on a vacation and borrowed $2,520. If interest was charged on the loan at 5.2%, how much interest would she have to pay in 236 days?

***(Round to two decimal places otherwise Blackboard will mark as incorrect.)

c) Interest of $99 is earned at 7.05% on a deposit of $4,398 in how many days?

d)

Alley opened a bank account on April 16, 2020 and deposited $1,443. She earned simple interest of 1.41%. How much interest was earned and paid into Alley’s account on November 1, 2020?

***(Round answer to two decimals, otherwise Blackboard will mark as incorrect.)

Answer 1

a)

Now the payments which were due few months ago will add more interest till today and payments which are due in future already consider interest payments hence need to be discounted back to find present value.

Assuming simple interest the interest on first payment will be

SI = (PRT)/100 = (1870 * 7.61 * (7/12)) / 100 = 83.01

Hence amount paid today will be 1870 + 83.01 = 1953.01

Similarly for 2nd payment

SI = (PRT)/100 = (1951 * 7.61 * (5/12)) / 100 = 61.86

Hence amount paid today will be 1951+ 61.86 = 2012.86

Now the third payment which will happen in future contains both Principal and interest and hence we can write

SI + P = 1232

(PRT)/100 + P = 1232

(P * 7.61 * (9/12))/100 + P = 1232

0.057P + P = 1232

P = 1165.56

Hence adding all 3 present values we will get Size of replacement payment

= 1953.01 + 2012.86 + 1165.56 = 5131.43