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6. The main difference between mortgage and amortization are A. The time period in which the debt is repaid B. The amount of money to be borrowed C. Mortgage is investement but amortization is loan repayment D. A and B are the answer E. A and C are the answer 7. What will the amount if the size of each payment is birr100 payable at the end of each semi- annual for one year and the interest rate is 4% A. Br200 B. Br194 C. Br123 D. Br202 E. none 8. Which of the following term is both Components and Assumptions of linear programming model? A. Objective function B Decision variables C.Non-negativity D. Divisibility E. Linearity 9. Assume that a person is invested Br.4,000 at 5% simple interest for 3 years.then the total simple interest will. br520 B. br6000 C. br 60 D. br600 E. none 10.How much money you have to deposit in an account today at 5% simple interest rate if you are to receive Br.3,000 as an amount in 10 years? A br2000 B. br1200 C. br1500 D. br900 E. br500 III. Work out: show all necessary steps (18 pts) I. "Gosh" Automobile produces two different types of Car. These are G1 and G2. G1 requires 9 hours to assemble and 3 hours to package. While G2 requires 12 houres to assemble and 3 hours to package.The company has 108 hours of assembly and 30 hours of packaging time available.Profits are birr65 for G1 and birr75 for G2, both per unit (10pts). Required: a. Formulate the Linear programming model for the problem b. The possible coordinate points will C. Determine how many cars of each type to make in order to maximize total profit d. the maximuim profit and the slack amount will e. Interpret your result 2. Ato Amare purchased a house for Br.50,000 He made an amount of down payment and pay monthly Br.600 to retire the mortgage for 20 years at an annual interest rate of rate of 24% compounded 24% monthly (8pts). Required. Find the mortgage, down payment, interest charged and percentage of the down payment to the selling price.

Pergunta

6. The main difference between mortgage and amortization are
A. The time period in which the debt is repaid
B. The amount of money to be borrowed
C. Mortgage is investement but amortization is loan repayment
D. A and B are the answer
E. A and C are the answer
7. What will the amount if the size of each payment is
birr100 payable at the end of each semi-
annual for one year and the interest rate is
4% 
A. Br200
B. Br194
C. Br123
D. Br202
E. none
8. Which of the following term is both Components and Assumptions of linear programming model?
A. Objective function B Decision variables C.Non-negativity D. Divisibility E. Linearity
9. Assume that a person is invested Br.4,000 at 5%  simple interest for 3 years.then the total simple
interest will.
br520 B. br6000 C. br 60 D. br600 E. none
10.How much money you have to deposit in an account today at 5%  simple interest rate if you are to
receive Br.3,000 as an amount in 10 years?
A br2000
B. br1200
C. br1500 D.
br900
E. br500
III. Work out: show all necessary steps (18 pts)
I. "Gosh" Automobile produces two different types of Car. These are G1 and G2. G1 requires 9 hours
to assemble and 3 hours to package. While G2 requires 12 houres to assemble and 3 hours to
package.The company has 108 hours of assembly and 30 hours of packaging time available.Profits
are birr65 for G1 and birr75 for G2, both per unit (10pts).
Required:
a. Formulate the Linear programming model for the problem
b. The possible coordinate points will
C. Determine how many cars of each type to make in order to maximize total profit
d. the maximuim profit and the slack amount will
e. Interpret your result
2. Ato Amare purchased a house for Br.50,000 He made an amount of down payment and pay
monthly Br.600
to retire the mortgage for 20 years at an annual interest rate of rate of 24% compounded 24% 
monthly (8pts).
Required.
Find the mortgage, down payment, interest charged and percentage of the down payment to
the selling price.

6. The main difference between mortgage and amortization are A. The time period in which the debt is repaid B. The amount of money to be borrowed C. Mortgage is investement but amortization is loan repayment D. A and B are the answer E. A and C are the answer 7. What will the amount if the size of each payment is birr100 payable at the end of each semi- annual for one year and the interest rate is 4% A. Br200 B. Br194 C. Br123 D. Br202 E. none 8. Which of the following term is both Components and Assumptions of linear programming model? A. Objective function B Decision variables C.Non-negativity D. Divisibility E. Linearity 9. Assume that a person is invested Br.4,000 at 5% simple interest for 3 years.then the total simple interest will. br520 B. br6000 C. br 60 D. br600 E. none 10.How much money you have to deposit in an account today at 5% simple interest rate if you are to receive Br.3,000 as an amount in 10 years? A br2000 B. br1200 C. br1500 D. br900 E. br500 III. Work out: show all necessary steps (18 pts) I. "Gosh" Automobile produces two different types of Car. These are G1 and G2. G1 requires 9 hours to assemble and 3 hours to package. While G2 requires 12 houres to assemble and 3 hours to package.The company has 108 hours of assembly and 30 hours of packaging time available.Profits are birr65 for G1 and birr75 for G2, both per unit (10pts). Required: a. Formulate the Linear programming model for the problem b. The possible coordinate points will C. Determine how many cars of each type to make in order to maximize total profit d. the maximuim profit and the slack amount will e. Interpret your result 2. Ato Amare purchased a house for Br.50,000 He made an amount of down payment and pay monthly Br.600 to retire the mortgage for 20 years at an annual interest rate of rate of 24% compounded 24% monthly (8pts). Required. Find the mortgage, down payment, interest charged and percentage of the down payment to the selling price.

Solução

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GustavoMestre · Tutor por 5 anos

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Let's address each question and problem step by step:<br /><br />### Question 6:<br />The main difference between mortgage and amortization are:<br />- **Mortgage**: A loan specifically used to purchase real estate.<br />- **Amortization**: The process of paying off debt with a fixed repayment schedule in regular installments over time.<br /><br />**Answer:** E. A and C are the answer<br /><br />### Question 7:<br />To find the amount if the size of each payment is $100 payable at the end of each semi-annual for one year and the interest rate is 4%, we need to calculate the future value of an ordinary annuity.<br /><br />Future Value (FV) of an ordinary annuity formula:<br />\[ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) \]<br /><br />Where:<br />- \( P \) = Payment amount per period ($100)<br />- \( r \) = Interest rate per period (4% annually, so 2% semi-annually or 0.02)<br />- \( n \) = Number of periods (2 periods in one year)<br /><br />\[ FV = 100 \times \left( \frac{(1 + 0.02)^2 - 1}{0.02} \right) \]<br />\[ FV = 100 \times \left( \frac{1.0404 - 1}{0.02} \right) \]<br />\[ FV = 100 \times 2.02 \]<br />\[ FV = 202 \]<br /><br />**Answer:** D. $Br202$<br /><br />### Question 8:<br />Components and Assumptions of linear programming model include:<br />- Objective function<br />- Decision variables<br />- Constraints<br />- Non-negativity<br />- Divisibility<br />- Linearity<br /><br />**Answer:** E. Linearity<br /><br />### Question 9:<br />Simple interest formula:<br />\[ I = P \times r \times t \]<br /><br />Where:<br />- \( P \) = Principal amount ($4000)<br />- \( r \) = Annual interest rate (5% or 0.05)<br />- \( t \) = Time in years (3 years)<br /><br />\[ I = 4000 \times 0.05 \times 3 \]<br />\[ I = 600 \]<br /><br />**Answer:** D. $br600$<br /><br />### Question 10:<br />To find the present value (PV) needed to receive $3000 in 10 years at 5% simple interest:<br /><br />Simple interest formula rearranged for principal:<br />\[ P = \frac{A}{1 + rt} \]<br /><br />Where:<br />- \( A \) = Amount received in the future ($3000)<br />- \( r \) = Annual interest rate (5% or 0.05)<br />- \( t \) = Time in years (10 years)<br /><br />\[ P = \frac{3000}{1 + 0.05 \times 10} \]<br />\[ P = \frac{3000}{1.5} \]<br />\[ P = 2000 \]<br /><br />**Answer:** A. $br2000$<br /><br />### Work out Problem 1:<br />#### a. Formulate the Linear Programming Model:<br />Let \( x_1 \) be the number of G1 cars produced and \( x_2 \) be the number of G2 cars produced.<br /><br />Objective Function:<br />Maximize Profit \( Z = 65x_1 + 75x_2 \)<br /><br />Subject to constraints:<br />\[ 9x_1 + 12x_2 \leq 108 \] (Assembly hours constraint)<br />\[ 3x_1 + 3x_2 \leq 30 \] (Packaging hours constraint)<br />\[ x_1 \geq 0 \]<br />\[ x_2 \geq 0 \]<br /><br />#### b. Possible Coordinate Points:<br />From the constraints, solve for boundary points:<br />1. \( 9x_1 + 12x_2 = 108 \)<br />2. \( 3x_1 + 3x_2 = 30 \)<br /><br />Solving these equations:<br />For \( 3x_1 + 3x_2 = 30 \):<br />\[ x_1 + x_2 = 10 \]<br />Possible points: (0, 10), (10, 0), (5, 5)<br /><br />For \( 9x_1 + 12x_2 = 108 \):<br />\[ 3x_1 + 4x_2 = 36 \]<br />Possible points: (0, 9), (12, 0), (4, 6)<br /><br />#### c. Determine Optimal Solution:<br />Evaluate the objective function at each feasible point:<br />- At (0, 9): \( Z = 65(0) + 75(9) = 675 \)<br />- At (12, 0): \( Z = 65(12) + 75(0) = 780 \)<br />- At (4, 6): \( Z = 65(4) + 75(6) = 260 + 450 = 710 \)<br /><br />Optimal solution: Produce 12 units of G1 and 0 units of G2 for maximum profit.<br /><br />#### d. Maximum Profit and Slack:<br />Maximum Profit: \( Z = 780 \)<br />Slack in assembly hours: \( 108 - (9 \times 12 + 12 \times 0) = 0 \)<br />Slack in packaging hours: \( 30 - (3 \times 12 + 3 \times 0) = 6 \)<br /><br />#### e. Interpretation:<br />The company should produce 12 units of G1 and no units of G2 to maximize profit, resulting in a maximum profit of $780. There will be no slack in assembly hours and 6 hours of slack in packaging time.<br /><br />### Work out Problem 2:<br />Given:<br />- House price: $50,000<br />- Monthly payment: $600<br />- Annual interest rate: 24% compounded monthly<br /><br />Monthly interest rate: \( \frac{24\%}{12} = 2\% \) or 0.02<br /><br />Using the mortgage formula:<br />\[ M = P \times \frac{r(1+r)^n}{(1+r)^n - 1} \]<br /><br />Where:<br />- \( M \) = Monthly payment ($600)<br />- \( r \) = Monthly interest rate (0.02)<br />- \( n \) = Total number of payments (20 years * 12 months/year = 240)<br /><br />Rearranging to solve for \( P \):<br />\[ P = \frac{M \times ((1+r)^n - 1)}{r(1+r)^n} \]<br /><br />\[ P = \frac{600 \times ((1+0.02)^{240} - 1)}{0.02(1+0.02)^{240}} \]<br /><br />Calculating:<br />\[ (1+0.02)^{240} \approx 54.349 \]<br /><br />\[ P = \frac{600 \times (54.349 - 1)}{0.02 \times 54.349} \]<br />\[ P = \frac{600 \times 53.349}{1.087} \]<br />\[ P \approx 29,448.66 \]<br /><br />Down payment:<br />\[ 50,000 - 29,448.66 = 20,551.34 \]<br /><br />Interest charged:<br />\[ Total paid = 600 \times 240 = 144,000 \]<br />\[ Interest = Total paid - Principal = 144,000 - 29,448.66 = 114,551.34 \]<br /><br />Percentage of down payment:<br />\[ \frac{20,551.34}{50,000} \times 100 \approx 41.1\% \]<br /><br />**Summary:**<br />- Mortgage: $29,448.66<br />- Down payment: $20,551.34<br />- Interest charged: $114,551.34<br />- Percentage of down payment: 41.1%
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