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AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023

AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023 – Business Mathematics is a fundamental aspect of any business, providing tools for analysis and decision-making. As part of the Allama Iqbal Open University (AIOU) course code 1429, students delve into the intricacies of Business Mathematics. In this blog post, we will explore Assignment 1, spanning the first four chapters, covering five questions with a breakdown of each part.

Question 1: Probability in Quality Control

The question begins with a comprehensive table detailing acceptable and unacceptable products across three types. From calculating the probability of selecting a product type at random to conditional probabilities, the question tests students on their understanding of basic probability concepts.

Q. 1 (a)

A sample of 800 parts has been selected from three product lines and inspected by the quality control department. The given table summarizes the results of the inspection. If a part is selected at random from this sample, what is the probability that:

The part is of the product 1 type
The part is unacceptable?
The part is an acceptable unit of product 3?
The part is an unacceptable unit of product 1?
The part is acceptable, given that the selected part is a unit of product 2?
The part is product 1, given that the selected part is acceptable?
The part is product 3, given that the selected part is unacceptable?

(b) What is the meant by the sample space for an experiment? Write the sample spaces for the given experiments:

i) A day in June is chosen for a picnic.

ii) A coin is tossed and a die is rolled

iii) The manager of a small company must decide whether to buy office space of lease it.

Question 2: Probability Distributions and Family Structures

This question shifts the focus to probability distributions and family structures. Students are tasked with finding probabilities related to body types, such as overweight, and formulating a probability distribution for the number of boys in families with three children.

Q. 2 (a) A study on body types gave the following results: 45% were short, 25% were short and overweight, and 24% were tall and not overweight. Find the probability that a person is:

i) Overweight;

ii) Short, but not overweight;

iii) Tall and overweight.

(b) Find a formula for the probability distribution of the number of boys in families with three children assuming equal probabilities for boys and girls.

Question 3: Visualization and Random Variables

This question combines the graphical representation of a plane equation with assessing the validity of a given probability distribution for a random variable. It challenges students to apply mathematical concepts in visualizing equations and evaluating probability distributions.

Q. 3 (a) Sketch the plane representing 2x + 3y + 4z = 12

(b) If a random variable can take the values 0, 1, 2 and 3. Then check that whether the following satisfy the condition of being a probability distribution.

P(X=0) = 0.18 P(X=1) = 0.20 P(X=2) = 0.38 P(X=3) = 0.24

Question 4: Intervals, Equations, and Interpretations

Students are required to sketch various intervals on the number line, showcasing their understanding of different interval notations. The second part involves working with equations, converting them to slope-intercept form, identifying slopes and y-intercepts, and interpreting the meaning of slopes in a real-world context.

Q. 4 (a) Sketch the following intervals:

i) (–3, 0) ii) [–5, –3] iii) [–5, –1] iv) (5, 10] v) [–2, 3}

(b) Given the equation

i) Rewrite the equation in slope intercept form

ii) Identify the slope and y– intercept

iii) Interpret the meaning of the slope

Question 5: Inequalities and Graphical Solutions

The final question focuses on solving second-degree inequalities algebraically and graphically. Students tackle quadratic inequalities and then verify their solutions graphically. This question integrates both analytical and graphical problem-solving skills.

Q. 5 (a) Solve the following second degree inequality:

i) x 2 + 4x – 12 ≤ 0 ii) 5x 2 – 13x – 6 ≥ 0

(b) Solve graphically and check your answer algebraically.

–x + 3y = 2

4x – 12 = –8

Conclusion: AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023

AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023 provides students with a holistic review of probability, probability distributions, graphical representations, and inequalities. The questions are designed to reinforce theoretical concepts while challenging students to apply their mathematical knowledge to practical scenarios. By mastering these assignments, students are better equipped to navigate the mathematical intricacies of the business world.

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Last date of submission of AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023 is 01-04-2024.

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There are 5 questions in AIOU Course Code 1429 Business Mathematics Solved Assignment 1 Autumn 2023.

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Aiou code 1429 business mathematics assignments spring and autumn.

Allama Iqbal Open University (AIOU) presents a variety of programs to empower University students with data and expertise . Amongst these, Business Mathematics (Code 1429) stands out as a vital topic that gives a basis for understanding mathematical ideas within the context of enterprise and economics. The assignments for this course, provided in each of the Spring and Autumn semesters, function as a bridge between ideas and sensible software.

AIOU Code 1429 Business Mathematics :

AIOU Code 1429, Business Mathematics , is designed to equip University students with the mathematical instruments essential for decision-making within the enterprise world. From fundamental arithmetic operations to superior statistical evaluation , the course covers a large spectrum of mathematical ideas related to numerous enterprise disciplines.

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Business Statistics (1430)
Commercial Geography (1428)
Principle of Management (1427)
Basics of Information Communication Technology (1431)
Compulsory English-II (1424)
Compulsory English-I (1423)
Business Communication (1416)
Introduction to Business Finance (1415)

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Business Mathematics Code 1429 Free Solved Assignments of AIOU

Subject : Business Mathematics

Code : 1429

Level : Bachelors (BA / B.Com)

Semester : Spring 2013

No. of Assignments : 1st and 2nd

Note : These assignments are out-dated, but I shared because these are helpful for exam preparations of Spring 2013. We will updated Autumn 2013 Solved Assignment of 1429 (According to Pattern of AIOU- Allama Iqbal Open University) soon on the same site.

Keybook is also available for Free, If you want to download free key-book of 1429, kindly contact us via email, or comment below.Best of Luck for your exam.

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English-II Code 1424 AIOU Free Solved Assignment No.1 Autumn 2013

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AIOU Course Code 1429-1 Solved Assignment Spring 2022

Course: Business Mathematics (1429)

dOWNLOAD button is given at the end of post

Semester: Spring, 2022

ASSIGNMENT No. 1

(a)

The part is of the product 1 type

180 /800 = 0.225

The part is unacceptable

If unacceptable so the calculation not possible.

The part is an acceptable unit of product 3

230 / 800 = 0.2875

The part is an unacceptable unit of product 1

800 / 180 = 4.44

The part is acceptable, given that the selected part is a unit of product 2

390 / 800 = 0.4875

The part is product 1, given that the selected part is acceptable

570 / 800 = 0.7125

The part is product 3, given that the selected part is unacceptable

800 / 390 = 2.051

(b) What is the difference between the states of statistical independence and statistical dependence?

When two events are dependent events, one event influences the probability of another event. A dependent event is an event that relies on another event to happen first. Dependent events in probability are no different from dependent events in real life: If you want to attend a concert, it might depend on whether you get overtime at work; if you want to visit family out of the country next month, it depends on whether or not you can get a passport in time. More formally, we say that when two events are dependent, the occurrence of one event influences the probability of another event.

Simple examples of dependent events:

Robbing a bank and going to jail.
Not paying your power bill on time and having your power cut off.
Boarding a plane first and finding a good seat.
Parking illegally and getting a parking ticket. Parking illegally increases your odds of getting a ticket.
Buying ten lottery tickets and winning the lottery. The more tickets you buy, the greater your odds of winning.
Driving a car and getting in a traffic accident.

An independent event is an event that has no connection to another event’s chances of happening (or not happening). In other words, the event has no effect on the probability of another event occurring. Independent events in probability are no different from independent events in real life. Where you work has no effect on what color car you drive. Buying a lottery ticket has no effect on having a child with blue eyes.

When two events are independent, one event does not influence the probability of another event.
Owning a dog and growing your own herb garden.
Paying off your mortgage early and owning a Chevy Cavalier.
Winning the lottery and running out of milk.
Buying a lottery ticket and finding a penny on the floor (your odds of finding a penny does not depend on you buying a lottery ticket).
Taking a cab home and finding your favorite movie on cable.
Getting a parking ticket and playing craps at the casino.
Here are more formal ways to quantify dependent or independent events. You’ll come across these formulas in basic probability .
P(A|B) = P(A).
P(B|A) = P(B)

The probability of A, given that B has happened, is the same as the probability of A. Likewise, the probability of B, given that A has happened, is the same as the probability of B. This shouldn’t be a surprise, as one event doesn’t affect the other. You can use the following equation to figure out probability for independent events: P(A∩B) = P(A) · P(B).

(a) Unemployment statistics within a western state indicate that 6 percent of those eligible to work are unemployed. Suppose that an experiment is conducted where three persons are selected at random and their employment status is noted. If the random variable for this experiment is defined as the number of persons unemployed.

i) Construct the probability distribution for this experiment and determine the probability that
ii) None of three is unemployed

iii) Or two or more are employed

(b) Find a formula for the probability distribution of the number of boys in families with three children assuming equal probabilities for boys and girls.

All girls – 0.5*0.5*0.5 =0.125 (12.5%)
2Boys 1 Girl — 3C2 * 0.125 = 37.5%
At least 2 girls — 2Girls 1 boy + all girls = 37.5% +12.5% = 50%
2 Boys at most — All posibilities — P(all boys) = 100%– 12.5% =67.5%

(a) Sketch the plane representing 2x + y + 0z = 4

(b) The average size of farms in the United State increased from 100 acres in 1920 to 700 acres in 1980. Let y be the average size x years after 1900. In what year was the average size 400 acres?

Number of years = 1980 – 1920 = 60

Total growth in farm size between (1920-1980) = 700 acres

Average yearly growth in farm size (x) = (700 acres – 100 acres)/ 60 = 600/60 = 10 acres

10 acres X 30 years = 300 acres

1920 + 30years = 1950

(a) Sketch the following intervals.

ii) [–5, –3]

iii) [–5, –1)

iv) (5, 10]

(b) Solve:

i) |7x – 12 | = |4–3x|
ii) [5x – 4| £ –10

(a) The value of machine is expected to decrease at a liner rate over the time. Two data points indicate that the value of the machine at t=0 (time of purchase) is $18,000 and its value in 1 year will equal $14,500.

i) Determine the slope intercept equation (V = mt + k) which relates the value V of the machine to its age t.

The two ordered pairs are (0, 18000) and (1, 14500).

So, the equation will be

⇒ V – 18000 = – 3500t

⇒ V = – 3500t + 18000 (Answer)

ii) Interpret the meaning of the slope and V intercept.

Now, the slope in the above equation i.e. – 3500 is the rate of decrease of machine value in $ per year.

And the V-intercept 18000 gives the initial value of the machine.

iii) Solve for the t intercept and interpret its meaning.

The t-intercept will give

0 = – 3500t + 18000

⇒ t = 5.14 years.

This means the value of the machine will become zero after 5.14 years.

(b) Solve graphically and check your answer algebraically.

–x + 3y = 2

4x – 12 = –8

x=1 and y=1

Sunday, July 23

Business mathematics (1429) - spring - 2023 - assignment 1.

Business Mathematics (1429)

Q. 1 (a) The table below gives the probability that a person has life insurance in the indicated range.

Amount of Insurance None Less than @10,000 $10,000–$24,999 $25,000–$49,999$50,000–$99,999 $100,000– $1999,999 $200,000 or more

Probability 0.17 0.20 0.17 0.14 0.15 0.12 0.05

Find the probability that an individual has the following amounts of life insurance.

i) Less than @10,000

ii) $10,000 to $99,999

iii) $50,000 or more

iv) Less than $50,000 or $100,000 or more

(b) In a survey of 410 salespersons and 350 construction workers, it is found that 164 of the salespersons and 196 of the construction workers were overweight. If a person is selected at random from the group, what is the probability that:

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i) This person is overweight?

ii) This person is a salesperson, given that the person is overweight?

iii) This person is overweight, given that the person is a salesperson?

iv) This person is a construction worker, given that the person is not overweight.

v) This person is not overweight, given that the person is a construction worker.

(a) To find the probabilities for the given amounts of life insurance, we need to sum up the probabilities for the corresponding ranges.

i) Probability of having Less than @10,000 life insurance:

This corresponds to the probability in the range "Less than @10,000," which is 0.20.

ii) Probability of having $10,000 to $99,999 life insurance:

This corresponds to the sum of probabilities in the ranges "$10,000–$24,999," "$25,000–$49,999," and "$50,000–$99,999":

0.17 + 0.14 + 0.15 = 0.46

iii) Probability of having $50,000 or more life insurance:

This corresponds to the sum of probabilities in the ranges "$50,000–$99,999," "$100,000–$1999,999," and "$200,000 or more":

0.15 + 0.12 + 0.05 = 0.32

iv) Probability of having Less than $50,000 or $100,000 or more life insurance:

This corresponds to the sum of probabilities in the ranges "Less than @10,000," "$10,000–$24,999," "$25,000–$49,999," and "$100,000–$1999,999," and "$200,000 or more":

0.20 + 0.17 + 0.14 + 0.12 + 0.05 = 0.68

(b) Let's define the following events:

A: Selecting a salesperson

B: Selecting a construction worker

C: Selecting an overweight individual

Number of salespersons (n(A)) = 410

Number of construction workers (n(B)) = 350

Number of overweight salespersons (n(A ∩ C)) = 164

Number of overweight construction workers (n(B ∩ C)) = 196

i) Probability of selecting an overweight person (C):

P(C) = (Number of overweight individuals) / (Total number of individuals)

P(C) = (n(A ∩ C) + n(B ∩ C)) / (n(A) + n(B))

P(C) = (164 + 196) / (410 + 350)

P(C) = 360 / 760

P(C) ≈ 0.474

ii) Probability of selecting a salesperson, given that the person is overweight (A|C):

P(A|C) = (Probability of selecting a salesperson and overweight) / (Probability of being overweight)

P(A|C) = (n(A ∩ C) / n(A)) / (n(A ∩ C) + n(B ∩ C)) / (n(A) + n(B))

P(A|C) = (164 / 410) / (360 / 760)

P(A|C) ≈ 0.452

iii) Probability of selecting an overweight person, given that the person is a salesperson (C|A):

P(C|A) = (Probability of selecting a salesperson and overweight) / (Probability of being a salesperson)

P(C|A) = (n(A ∩ C) / n(A)) / (n(A) / (n(A) + n(B)))

P(C|A) = (164 / 410) / (410 / (410 + 350))

P(C|A) ≈ 0.4

iv) Probability of selecting a construction worker, given that the person is not overweight (B|C'):

P(B|C') = (Probability of selecting a construction worker and not overweight) / (Probability of not being overweight)

P(B|C') = (n(B) - n(B ∩ C)) / (n(A) + n(B) - (n(A ∩ C) + n(B ∩ C))) / (n(A) + n(B) - (n(A ∩ C) + n(B ∩ C)))

P(B|C') = (350 - 196) / (410 + 350 - (164 + 196)) / (410 + 350 - (164 + 196))

P(B|C') ≈ 0.692

v) Probability of selecting a person who is not overweight, given that the person is a construction worker (C'|B):

P(C'|B) = (Probability of selecting a construction worker and not overweight) / (Probability of being a construction worker)

P(C'|B) = (n(B) - n(B ∩ C)) / (n(B) / (n(A) + n(B)))

P(C'|B) = (350 - 196) / (350 / (410 + 350))

P(C'|B) ≈ 0.485

Note: The results are approximated to three decimal places.

Q. 2 (a) Obtain a probability distribution of the sum of spots when a pair of dice is rolled. (8)

(b) The continuous random variable X has the density function (12)

i) Show that P (0 < X < 2) = 1

ii) Find P(X < 1.2)

(a) Probability distribution of the sum of spots when a pair of dice is rolled:

When two dice are rolled, each die has 6 sides, numbered from 1 to 6. To find the probability distribution of the sum of spots, we need to calculate all the possible sums and their corresponding probabilities.

The possible sums of spots when rolling two dice are:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.

The probability of each sum can be calculated as follows:

- For a sum of 2, there is only one way to get it (rolling a 1 on both dice). So, the probability is 1/36.

- For a sum of 3, there are two ways to get it (1 and 2 or 2 and 1). The probability is 2/36 or 1/18.

- For a sum of 4, there are three ways to get it (1 and 3, 2 and 2, or 3 and 1). The probability is 3/36 or 1/12.

- For a sum of 5, there are four ways to get it (1 and 4, 2 and 3, 3 and 2, or 4 and 1). The probability is 4/36 or 1/9.

- For a sum of 6, there are five ways to get it (1 and 5, 2 and 4, 3 and 3, 4 and 2, or 5 and 1). The probability is 5/36.

- For a sum of 7, there are six ways to get it (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, or 6 and 1). The probability is 6/36 or 1/6.

- For a sum of 8, there are five ways to get it (2 and 6, 3 and 5, 4 and 4, 5 and 3, or 6 and 2). The probability is 5/36.

- For a sum of 9, there are four ways to get it (3 and 6, 4 and 5, 5 and 4, or 6 and 3). The probability is 4/36 or 1/9.

- For a sum of 10, there are three ways to get it (4 and 6, 5 and 5, or 6 and 4). The probability is 3/36 or 1/12.

- For a sum of 11, there are two ways to get it (5 and 6 or 6 and 5). The probability is 2/36 or 1/18.

- For a sum of 12, there is only one way to get it (rolling a 6 on both dice). So, the probability is 1/36.

(b) The continuous random variable X has the density function:

The problem mentions that X is a continuous random variable with a given density function, but the actual density function is missing from the question. Without the specific density function, we cannot perform the requested calculations (P(0 < X < 2) and P(X < 1.2)).

If you have the density function for X, please provide it, and I'd be happy to help you with the calculations. Alternatively, if you have any other questions or need assistance with different topics, feel free to ask!

Q. 3 In a certain marketplace the demand and supply functions for a commodity are as follows: D : p = 100–5q (20)

S : p = 20 + 4q

i) What are the initial equilibrium price and quantity?

ii) Assume an imaginative advertising campaign shifts the demand function two places to the right. Sketch the initial demand and supply functions. Sketch the new demand function. Find the new equilibrium price and quantity.

iii) Assume that a tax of $1 per unit is levied on the seller. What will be the effect on the supply function? Depict the situation graphically. Determine the new equilibrium price and quantity.

(a) Equilibrium price and quantity:

To find the equilibrium price and quantity, we need to set the demand and supply functions equal to each other and solve for the value of q (quantity) and p (price).

Demand function: D: p = 100 - 5q

Supply function: S: p = 20 + 4q

Setting D equal to S:

100 - 5q = 20 + 4q

Now, solve for q:

100 - 20 = 4q + 5q

q = 80/9 ≈ 8.89

Now, substitute the value of q back into either the demand or supply function to find the equilibrium price (p):

p = 100 - 5(8.89) ≈ 55.56

So, the initial equilibrium price is approximately $55.56, and the initial equilibrium quantity is approximately 8.89 units.

(b) Imaginative advertising campaign:

An imaginative advertising campaign shifts the demand function two places to the right. This means that the demand function will change as follows:

New demand function: D': p = 100 - 5(q - 2)

To sketch the initial demand and supply functions, we plot them on a graph with the quantity (q) on the x-axis and the price (p) on the y-axis. The initial demand function (D) and supply function (S) will be represented as straight lines. The new demand function (D') will also be a straight line but shifted two places to the right.

Now, we need to find the new equilibrium price and quantity using the new demand function (D') and the original supply function (S).

Setting D' equal to S:

100 - 5(q - 2) = 20 + 4q

100 - 5q + 10 = 20 + 4q

90 - 5q = 20 + 4q

q = 90/9 = 10

Substitute the value of q back into the new demand function (D') to find the equilibrium price (p):

p = 100 - 5(10 - 2) = 100 - 40 = 60

So, the new equilibrium price is $60, and the new equilibrium quantity is 10 units.

(d) Effect of a tax on the supply function:

If a tax of $1 per unit is levied on the seller, the supply function will change. The new supply function (S') will be given as follows:

New supply function: S': p = 20 + 4q - 1

This is because for each unit sold, the seller now has to pay $1 in tax, which reduces the effective price they receive.

To depict the situation graphically, we plot both the original supply function (S) and the new supply function (S') on the same graph. The new supply function (S') will be parallel to the original supply function (S) but shifted down by 1 unit.

Now, to determine the new equilibrium price and quantity with the tax, we set D equal to S':

100 - 5q = 20 + 4q - 1

100 - 20 + 1 = 4q + 5q

q = 81/9 = 9

Substitute the value of q back into the new supply function (S') to find the equilibrium price (p):

p = 20 + 4(9) - 1 = 20 + 36 - 1 = 55

So, the new equilibrium price with the tax is $55, and the new equilibrium quantity is 9 units.

That concludes the explanation for the question. If you have any specific queries or need further clarification on any part of the question, feel free to ask!

Q. 4 (a) Sketch the graph of each of the following linear functions: (12)

i) Graph passes through the point (2,–1) with slope of 3.

ii) Graph passes through the point (–3,–2) with slope of –1.

iii) Graph passes through the point (2, 4) with slope of 0.

iv) Graph passes through the point (5, 0) with undefined slope.

(b) Solve the following simultaneous linear equations by graphical method:

6x + 4y = 5 (8) 2x + 3y =3

(a) Sketching the graphs of linear functions:

To sketch the graph of each linear function, we'll use the slope-intercept form of a linear equation, which is given by y = mx + b, where m is the slope and b is the y-intercept (the value of y when x = 0).

i) Graph passes through the point (2, -1) with a slope of 3:

The equation of the line can be written as y = 3x + b. To find the value of b, we substitute the coordinates (2, -1) into the equation:

-1 = 3(2) + b

So, the equation of the line is y = 3x - 7. Now, let's plot the graph:

ii) Graph passes through the point (-3, -2) with a slope of -1:

The equation of the line can be written as y = -x + b. To find the value of b, we substitute the coordinates (-3, -2) into the equation:

-2 = -(-3) + b

So, the equation of the line is y = -x - 5. Now, let's plot the graph:

iii) Graph passes through the point (2, 4) with a slope of 0:

When the slope is 0, the line is horizontal, and the equation becomes y = b (where b is the y-coordinate of the point the line passes through). In this case, y = 4, so the equation of the line is y = 4. Let's plot the graph:

iv) Graph passes through the point (5, 0) with an undefined slope:

When the slope is undefined, the line is vertical, and the equation becomes x = a (where a is the x-coordinate of the point the line passes through). In this case, x = 5, so the equation of the line is x = 5. Let's plot the graph:

(b) Solving simultaneous linear equations by the graphical method:

To solve the simultaneous equations 2x + 3y = 3 and 6x + 4y = 5, we'll graph both equations and find the point of intersection, which represents the solution.

First, rewrite the equations in slope-intercept form (y = mx + b):

1) 2x + 3y = 3

3y = -2x + 3

y = (-2/3)x + 1

2) 6x + 4y = 5

4y = -6x + 5

y = (-6/4)x + 5/4

y = (-3/2)x + 5/4

Now, let's plot both graphs:

The solution to the system of equations is the coordinates of the point where the two lines intersect. By observing the graph, we can estimate the point of intersection to be approximately (1, 1/3).

So, the solution to the simultaneous equations is x ≈ 1 and y ≈ 1/3.

Q. 5 (a) Suppose that the demand and price for a certain brand of shampoo are related by

(20)

Where p is price in dollars and q is demand.

i) Find the price for a demand of: 0 units ; 8 units

ii) Find the demand for the shampoo at a price of: $6, $11, $16

iii) Graph :

Suppose the price and supply of the shampoo are related by,

Where q represents the supply, and p the price.

iv) Find the supply when the price is : $0, $10, $20

v) Graph on the same axes used for part (iii)

vi) Find the equilibrium supply

vi) Find the equilibrium price.

(a) The given relationship between demand (q) and price (p) for the shampoo is:

We are given the demand function, and we need to find the corresponding prices and the supply function.

i) To find the price for a demand of 0 units and 8 units, we can use the demand function.

For demand (q) of 0 units:

p = 40 - 3(0)

p = 40 dollars

For demand (q) of 8 units:

p = 40 - 3(8)

p = 40 - 24

p = 16 dollars

ii) To find the demand for the shampoo at a price of $6, $11, and $16, we need to rearrange the demand function to solve for q.

For a price (p) of $6:

6 = 40 - 3q

3q = 40 - 6

q = 34/3 ≈ 11.33 units

For a price (p) of $11:

11 = 40 - 3q

3q = 40 - 11

q = 29/3 ≈ 9.67 units

For a price (p) of $16:

16 = 40 - 3q

3q = 40 - 16

q = 24/3 = 8 units

iii) Graphing the demand function:

To graph the demand function, we plot points on a graph with price (p) on the y-axis and demand (q) on the x-axis. We can find additional points by substituting different values of q into the demand function and then connecting the points to create the graph.

iv) To find the supply when the price is $0, $10, and $20, we need to use the supply function.

The supply function is given as q = 10p - 150.

For a price (p) of $0:

q = 10(0) - 150

q = -150 units (Note: Negative quantity doesn't make sense in this context)

For a price (p) of $10:

q = 10(10) - 150

q = 100 - 150

q = -50 units (Negative quantity doesn't make sense)

For a price (p) of $20:

q = 10(20) - 150

q = 200 - 150

q = 50 units

v) Graphing the supply function:

To graph the supply function, we plot points on a graph with price (p) on the y-axis and supply (q) on the x-axis. Similar to the demand function, we find additional points by substituting different values of p into the supply function and then connect the points to create the graph.

vi) Finding the equilibrium supply and price:

The equilibrium occurs when the demand and supply are equal, i.e., when the quantity demanded equals the quantity supplied.

Set the demand function equal to the supply function:

40 - 3q = 10p - 150

Now, solve for the equilibrium price (p):

10p = 40 + 3q + 150

10p = 190 + 3q

p = (190 + 3q)/10

Substitute the equilibrium price (p) into either the demand or supply function to find the equilibrium supply (q).

Let's assume the equilibrium price (p) is $x:

q = 10x - 150

The equilibrium occurs when the quantity demanded (40 - 3q) is equal to the quantity supplied (10x - 150).

40 - 3q = 10x - 150

Solve for x (the equilibrium price):

40 + 150 = 10x + 3q

190 = 10x + 3q

x = (190 - 3q)/10

So, the equilibrium supply (q) is (10x - 150) and the equilibrium price (p) is (190 - 3q)/10.

Note: The explanation and calculations above should give you a good understanding of the problem. If you need any further details or have specific questions, feel free to ask! Dear Student,

Code No 1429 Business Mathematics , AIOU Old Papers

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Business Mathematics (1429) B.A / B.Com / BBA Autumn 2023 Solved Assignments

April 5, 2023 Sarfraz Allma Iqbal Open University 0

Business Mathematics (1429) BA/B.Com/BBA Autumn 2023 Solved Assignments

Note: We Can Not Post The Solution Of The Assignment Because AIOU is Not Allowed To Share The Solution Of the Assignment. We Are Just Sharing The Question of Assignment .

Any student can download AIOU solved assignment from here. Click on the concerned level for the download assignment. The University has introduced a wide range of undergraduate, graduate, Master’s, M.Phil . and Doctoral programs. The University also offers basic functional courses for illiterates and semi-literates. These programs have given distinction to the University in relation to other educational institutions in Pakistan as it caters to the needs of all age groups and levels of education. A summary of the Existing Programs is as follows:

As we know Allama Iqbal Open Univesity Pakistan offers many different types of courses. Allama Iqbal Open University AIOU tutor is a federal Govt. that provides quality education to all Pakistani students. Moreover, Allama Iqbal Open University provides distance learning education. So, AIOU is based on a semester system as known as Spring and Autumn. But the Spring semester starts from July to the end of the year means December. Moreover, the Autumn semester starts from December to July.

Assignment Updates 2023

The idea to make and Develop a separate page for the AIOU Solved assignment was taken from AIOU’s old papers. We make this section to be understandable to a normal user. So, this Section will automatically be updated as the AIOU semester change from autumn to spring. The assignment will automatically be added for the spring semester when AIOU asks for spring-solved assignments or autumn-solved assignments. So, Hope you all will enjoy this new section for the assignment. If you are facing any problems, then feel free to comment below.

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