Question

Suppose that at the end of a month, a customer owes a bank K1500. In the middle of the month, the customer pays Kx to the bank where x<1000, and at the end of the month, the bank adds interest at a rate of 4% of the remaining amount still owed. This process continues every month until the money owed is repaid in full.

(i) Find the value of x for which the customer still owes 1500 at the start of every month.

(ii) Find the value of x for which the whole amount owed is paid off exactly after the second payment.

(iii) Show that the value of x for which the whole amount owed is paid off exactly after the (n+1)th payment is given by x = 15000 (r)^(n) (r - 1) / (r)^(n + 1), where r=1.04.

A. For (i), x = 600; For (ii), x = 400; For (iii), x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1)

B. For (i), x = 800; For (ii), x = 300; For (iii), x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1)

C. For (i), x = 500; For (ii), x = 200; For (iii), x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1)

D. For (i), x = 700; For (ii), x = 350; For (iii), x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1)

Answer 1

To find the value of x for which the customer still owes K1500 at the start of every month, the value of x is 1560. To find the value of x for which the whole amount owed is paid off exactly after the second payment, the value of x is 1620. The value of x for which the whole amount owed is paid off exactly after the (n+1)th payment is x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1).

Step-by-step explanation:

To find the value of x for which the customer still owes 1500 at the start of every month, we need to solve the equation:

K1500 + K1500*0.04 = Kx

Simplifying, we get:

1500 + 60 = x

x = 1560

So, the value of x for which the customer still owes 1500 at the start of every month is 1560.

To find the value of x for which the whole amount owed is paid off exactly after the second payment, we need to solve the equation:

K1500 + K1500*0.04 + K1500*0.04 = Kx

Simplifying, we get:

1500 + 60 + 60 = x

x = 1620

So, the value of x for which the whole amount owed is paid off exactly after the second payment is 1620.

For the third part, we can prove the given equation using mathematical induction. Let's assume the equation x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1) is true for some positive integer n. We need to show that it is also true for n+1.

Using the assumption, we have:

x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1)

Multiplying both sides by (1.04), we get:

1.04x = 15000 (1.04)^(n+1) (1.04 - 1) / (1.04)^(n + 1)

Simplifying, we get:

1.04x = 15000 (1.04 - 1)

1.04x = 15000 * 0.04

1.04x = 600

x = 600

So, the given equation x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1) holds true for n+1. Therefore, the value of x for which the whole amount owed is paid off exactly after the (n+1)th payment is given by x = 15000 (1.04)^(n) (1.04 - 1) / (1.04)^(n + 1), where r=1.04.