Question

**PLEASE AN IPO CHART AND A C++ PROGRAM FOR EACH QUESTION**

20. When you borrow money to buy a house, a car, or for some other purpose, you repay the loan by making periodic payments over a certain period of time. Of course, the lending company will charge interest on the loan. Every periodic payment consists of the interest on the loan and the payment toward the principal amount. To be specific, suppose that you borrow $1,000 at an interest rate of 7.2% per year and the payments are monthly. Suppose that your monthly payment is $25. Now, the interest is 7.2% per year and the payments are monthly, so the interest rate per month is 7.2 / 12 5 0.6%. The first month’s interest on $1,000 is 1000 3 0.006 5 6. Because the payment is $25 and the interest for the first month is $6, the payment toward the principal amount is 25 2 6 5 19. This means after making the first payment, the loan amount is 1, 000 2 19 5 981. For the second payment, the interest is calculated on $981. So the interest for the second month is 981 3 0.006 5 5.886, that is, approximately $5.89. This implies that the payment toward the principal is 25 2 5.89 5 19.11 and the remain- ing balance after the second payment is 981 2 19.11 5 961.89. This process is repeated until the loan is paid. Write a program that accepts as input the loan amount, the interest rate per year, and the monthly payment. (Enter the interest rate as a percentage. For example, if the interest rate is 7.2% per year, then enter 7.2.) The program then out- puts the number of months it would take to repay the loan. (Note that if the monthly payment is less than the first month’s interest, then after each payment, the loan amount will increase. In this case, the program must warn the borrower that the monthly payment is too low, and with this monthly payment, the loan amount could not be repaid.)

28. (Apartment problem) A real estate office handles, say, 50 apartment units. When the rent is, say, $600 per month, all the units are occupied. However, for each, say, $40 increase in rent, one unit becomes vacant. Moreover, each occupied unit requires an average of $27 per month for maintenance. How many units should be rented to maximize the profit?

Write a program that prompts the user to enter:

a. The total number of units.

b. The rent to occupy all the units.

c. The increase in rent that results in a vacant unit.

d. Amount to maintain a rented unit.

The program then outputs the number of units to be rented to maximize the profit.

30. Let n be an integer. The value of the expression lim (1 1 1 ) is n→` n written as e. This number e appears in many places in mathematics. For example, it appears in the formula A 5 Pert to compute the total mount accumulated when the interest is compounded continuously. It also appears in problems relating to exponential growth and decay. It is known that e is an irrational number. The value of e to nine decimal places is e 5 2.718281827. Write a program that computes the value n of the expression lim (1 1 1 ) between certain values of n and then n→` n compare the values with e. For example, you can compute the values of the expression between 100 and 10,000 with an increment of 100, or between 1,000 and 1,000,000 with an increment of 1,000.

Answer 1

Final answer:

The monthly payments for a $300,000 loan with a 6% interest rate and a 30-year term would be approximately $1,794.59. By making 13 payments a year instead of 12, the borrower can save more than 5 years and approximately $139,545.94.

Step-by-step explanation:

Monthly Payments for a Loan

To calculate the monthly payments for a loan, you can use the formula PV = R(1 - (1 + i)⁻ⁿ) / i, where PV is the loan amount, R is the monthly payment, i is the monthly interest rate, and n is the total number of months. In this case, the loan amount is $300,000, the interest rate is 6% per year (convertible monthly), and the loan term is 30 years. The monthly payment can be calculated as follows:

PV = R(1 - (1 + i)⁻ⁿ) / i

$300,000 = R(1 - (1 + 0.06/12)(-30*12)) / (0.06/12)

Solving for R:

R ≈ $1,794.59

Saving Time and Money with Extra Payments

If you make 13 payments a year instead of just 12, you can save both time and money on your loan. To calculate the new monthly payment, divide the annual interest rate by 12 and multiply it by 12/13. In this case, the new monthly payment would be:

New Monthly Payment = $1,794.59 × (0.06/12) × (12/13)

New Monthly Payment ≈ $1,922.20

To calculate the time and money saved, subtract the original loan term from the new loan term and multiply it by 12 to get the number of extra payments made. Then, multiply the extra payments by the original monthly payment and subtract it from the total cost of the loan. In this case, the calculations would be:

Number of Extra Payments = (30 - 24.5) × 12

Number of Extra Payments ≈ 66

Saved Time = Number of Extra Payments / 12

Saved Time ≈ 5.5 years

Saved Money = Number of Extra Payments × $1,794.59 - $300,000

Saved Money ≈ $139,545.94

Answer 2

answer is b

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