AIOU 1429 Solved Assignment Autumn 2025

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ALLAMA IQBAL OPEN UNIVERSITY

(Department of Mathematics)

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WARNING

1. Plagiarism or hiring of ghost writer(s) for solving the assignment(s) will debar the student from award of degree/certificate if found at any stage.

2. Submitting assignment(s) borrowed or stolen from other(s) as one's own will be penalized as defined in the "Aiou Plagiarism Policy".

Assignment Submission Schedule
6 Credit Hours	Due Date	3 Credit Hours	Due Date
Assignment 1	15-12-2025	Assignment 1	08-01-2026
Assignment 2	08-01-2026
Assignment 3	30-01-2026	Assignment 2	20-02-2026
Assignment 4	20-02-2026

Course: Business Mathematics (1429)	Semester: Autumn-2025
Level: BA / B.Com / BBA

Total Marks: 100	Pass Marks: 50

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ASSIGNMENT No. 1

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Q1(a). Find the sample space for choosing an odd number from 1 to 15 at random. ▶

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Step 1: Identify the range of numbers
The numbers are from \(1\) to \(15\).

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Step 2: List all odd numbers in this range
The odd numbers between \(1\) and \(15\) are \(1, 3, 5, 7, 9, 11, 13, 15\).

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Step 3: Write the sample space
The sample space \(S\) is the set of all possible outcomes when choosing an odd number from \(1\) to \(15\).

\[ S = {1, 3, 5, 7, 9, 11, 13, 15} \]

The answer is \(S = \{1, 3, 5, 7, 9, 11, 13, 15\}\)

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Q1(b). What is the difference between mutually exclusive events and collectively exhaustive events? ▶

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Definition of Mutually Exclusive Events
Two or more events are called mutually exclusive if they cannot occur at the same time. In other words, the occurrence of one event prevents the occurrence of the other. For example, when tossing a coin, getting heads and getting tails are mutually exclusive events because both cannot happen together.

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Definition of Collectively Exhaustive Events
Events are collectively exhaustive if at least one of them must occur when an experiment is performed. This means the set of all events covers all possible outcomes. For example, in a coin toss, the events "heads" and "tails" are collectively exhaustive because one of them must occur.

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Key Difference
Mutually exclusive events cannot occur together, while collectively exhaustive events ensure that one of the possible outcomes must occur. Some events can be both mutually exclusive and collectively exhaustive, such as the outcomes of a fair coin toss.

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Q1(c). The probability that an applicant for pilot school will be admitted is 0.5. If three applicants are selected at random, what is the probability that:
i. All three will be admitted.
ii. None will be admitted.
iii. Only one will be admitted. ▶

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Given Information
The probability that an applicant is admitted is \( p = 0.5 \). The number of applicants selected is \( n = 3 \). The situation follows a binomial distribution.

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Formula
The probability of exactly \( k \) admissions out of \( n \) applicants is given by \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \]

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(i) All three will be admitted
Here \( k = 3 \). \[ P(X = 3) = \binom{3}{3} (0.5)^3 (1 - 0.5)^{0} \] \[ P(X = 3) = 1 \times (0.5)^3 = 0.125 \]

0.125

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(ii) None will be admitted
Here \( k = 0 \). \[ P(X = 0) = \binom{3}{0} (0.5)^0 (1 - 0.5)^3 \] \[ P(X = 0) = 1 \times 1 \times (0.5)^3 = 0.125 \]

0.125

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(iii) Only one will be admitted
Here \( k = 1 \). \[ P(X = 1) = \binom{3}{1} (0.5)^1 (1 - 0.5)^2 \] \[ P(X = 1) = 3 \times 0.5 \times (0.5)^2 = 3 \times 0.125 = 0.375 \]

0.375

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Q2(a). Define a random variable. What is the difference between a discrete random variable and a continuous random variable? ▶

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Definition of a Random Variable
A random variable is a variable that takes on different numerical values based on the outcome of a random experiment. It assigns a number to each possible outcome of that experiment.

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Discrete Random Variable
A discrete random variable can take only a finite or countable number of possible values. Examples include the number of heads in three coin tosses or the number of students present in a class.

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Continuous Random Variable
A continuous random variable can take any value within a given range or interval. Its values are uncountable and are often measured rather than counted. Examples include the height of students or the time taken to complete a race.

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Key Difference
A discrete random variable deals with countable outcomes, while a continuous random variable deals with measurable outcomes that can take infinitely many possible values within an interval.

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Q2(b). The fire chief for a small volunteer fire department has compiled data on the number of false alarms called in each day for the past 360 days. Construct the probability distribution for this study. ▶

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Given Information
The fire chief collected data for 360 days on the number of false alarms received per day. Let the random variable \( X \) represent the number of false alarms in a day.

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Step 1: Represent the Data
Suppose the collected data is summarized as follows (example data for illustration):

Number of False Alarms (X)	Number of Days (f)
0	180
1	90
2	60
3	30

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Step 2: Compute Probabilities
The probability of each value of \( X \) is found using \[ P(X) = \frac{f}{360} \]

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X	f	P(X) = f / 360
0	180	\( \frac{180}{360} = 0.5 \)
1	90	\( \frac{90}{360} = 0.25 \)
2	60	\( \frac{60}{360} = 0.1667 \)
3	30	\( \frac{30}{360} = 0.0833 \)

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Step 3: Verify Total Probability
Check that the sum of all probabilities equals 1. \[ 0.5 + 0.25 + 0.1667 + 0.0833 = 1.0 \]

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Final Probability Distribution

X	P(X)
0	0.5
1	0.25
2	0.1667
3	0.0833

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This table represents the probability distribution for the number of false alarms per day.

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Q2(c). Construct the discrete probability distribution that corresponds to the experiment of tossing a fair coin three times. Suppose the random variable X equals the number of heads occurring in three tosses. What is the probability of two or more heads? ▶

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Given Information
A fair coin is tossed three times. Let the random variable \( X \) represent the number of heads obtained. The possible values of \( X \) are 0, 1, 2, and 3.

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Step 1: List the Sample Space
The sample space for three coin tosses is: \[ \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \] There are 8 equally likely outcomes.

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Step 2: Determine the Value of \( X \) for Each Outcome

S.No	Number of Heads (X)	Favorable Outcomes	Count
1	0	TTT	1
2	1	HTT, THT, TTH	3
3	2	HHT, HTH, THH	3
4	3	HHH	1

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Step 3: Construct the Probability Distribution
Each outcome has probability \( \frac{1}{8} \). \[ P(X = k) = \frac{\text{Number of favorable outcomes}}{8} \]

X	P(X)
0	\( \frac{1}{8} = 0.125 \)
1	\( \frac{3}{8} = 0.375 \)
2	\( \frac{3}{8} = 0.375 \)
3	\( \frac{1}{8} = 0.125 \)

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Step 4: Probability of Two or More Heads
We find \( P(X \ge 2) \). \[ P(X \ge 2) = P(X = 2) + P(X = 3) \] \[ P(X \ge 2) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = 0.5 \]

The answer is 0.5

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Q3(a). Solve the following first-degree equations:

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(a) \( 8x - 6 = 5x + 3 \)
(b) \( -15 + 35x = 8x - 9 \)
(c) \( (x + 9) - (-6 + 4x) + 4 = 0\) ▶

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(a) \( 8x - 6 = 5x + 3 \)

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Step 1: Write the given equation
We have the equation \( 8x - 6 = 5x + 3 \).

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Step 2: Move all terms containing \( x \) to one side
Subtract \( 5x \) from both sides:
\( 8x - 5x - 6 = 3 \)

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Step 3: Simplify the equation
\( 3x - 6 = 3 \)

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Step 4: Move the constant term to the other side
Add \( 6 \) to both sides:
\( 3x = 9 \)

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Step 5: Solve for \( x \)
Divide both sides by \( 3 \):
\( x = \frac{9}{3} \)

Final Answer: \( x = 3 \)

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(b) \( -15 + 35x = 8x - 9 \)

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Step 1: Write the given equation
We have the equation \( -15 + 35x = 8x - 9 \).

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Step 2: Move all terms containing \( x \) to one side
Subtract \( 8x \) from both sides:
\( -15 + 35x - 8x = -9 \)

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Step 3: Simplify the equation
\( -15 + 27x = -9 \)

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Step 4: Move the constant term to the other side
Add \( 15 \) to both sides:
\( 27x = 6 \)

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Step 5: Solve for \( x \)
Divide both sides by \( 27 \):
\( x = \frac{6}{27} \)

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Step 6: Simplify the fraction
\( x = \frac{2}{9} \)

The answer is \( x = \frac{2}{9} \)

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(c) \( (x + 9) - (-6 + 4x) + 4 = 0\)

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Step 1: Write the given equation
We have the equation \( (x + 9) - (-6 + 4x) + 4 = 0 \).

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Step 2: Remove parentheses
Subtracting a negative changes the signs:
\( x + 9 + 6 - 4x + 4 = 0 \)

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Step 3: Combine like terms
\( (x - 4x) + (9 + 6 + 4) = 0 \)
\( -3x + 19 = 0 \)

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Step 4: Move the constant term to the other side
Subtract \( 19 \) from both sides:
\( -3x = -19 \)

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Step 5: Solve for \( x \)
Divide both sides by \( -3 \):
\( x = \frac{-19}{-3} \)

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Step 6: Simplify the signs
\( x = \frac{19}{3} \)

The answer is \( x = \frac{19}{3} \)

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Q4(a). Solve the following quadratic equations using the quadratic formula:

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(i) \( 4x^2 + 3x - 1 = 0 \)
(ii) \( 4t^2 + 64 = 0 \) ▶

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(i) \( 4x^2 + 3x - 1 = 0 \)

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Step 1: Write the given quadratic equation
We have the equation \( 4x^2 + 3x - 1 = 0 \).

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Step 2: Identify the coefficients
\( a = 4 \), \( b = 3 \), and \( c = -1 \).

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Step 3: Write the quadratic formula
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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Step 4: Substitute the values of \( a \), \( b \), and \( c \)
\[ x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-1)}}{2(4)} \]

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Step 5: Simplify under the square root
\[ x = \frac{-3 \pm \sqrt{9 + 16}}{8} \]

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Step 6: Simplify further
\[ x = \frac{-3 \pm \sqrt{25}}{8} \]

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Step 7: Take the square root of 25
\[ x = \frac{-3 \pm 5}{8} \]

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Step 8: Find the two values of \( x \)
For the plus sign: \[ x = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4} \]
For the minus sign: \[ x = \frac{-3 - 5}{8} = \frac{-8}{8} = -1 \]

The answers are \( x = \frac{1}{4} \) and \( x = -1 \)

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(ii) \( 4t^2 + 64 = 0 \)

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Step 1: Write the given quadratic equation
We have the equation \( 4t^2 + 64 = 0 \).

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Step 2: Move the constant term to the other side
\( 4t^2 = -64 \)

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Step 3: Divide both sides by 4
\( t^2 = -16 \)

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Step 4: Take the square root of both sides
\( t = \pm \sqrt{-16} \)

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Step 5: Simplify using imaginary numbers
\( t = \pm 4i \)

The answers are \( t = 4i \) and \( t = -4i \)

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Q4(b). A railing is to enclose a rectangular area of 1800 square feet. The length of the plot is twice the width. How much railing must be used? ▶

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Step 1: Write the given information
The area of the rectangular plot is \( 1800 \, \text{ft}^2 \).
The length is twice the width.

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Step 2: Let the width be \( x \)
Then the length will be \( 2x \).

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Step 3: Write the formula for the area of a rectangle
\( \text{Area} = \text{Length} \times \text{Width} \)
\( 1800 = 2x \times x \)

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Step 4: Simplify the equation
\( 1800 = 2x^2 \)

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Step 5: Solve for \( x \)
Divide both sides by 2:
\( x^2 = 900 \)
Take the square root of both sides:
\( x = 30 \)

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Step 6: Find the length
\( \text{Length} = 2x = 2(30) = 60 \)

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Step 7: Find the perimeter (total railing needed)
\( \text{Perimeter} = 2(\text{Length} + \text{Width}) \)
\( \text{Perimeter} = 2(60 + 30) = 2(90) = 180 \)

The answer is \( 180 \, \text{feet of railing} \)

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Q5(a). Solve the linear equation \(y = 2x + 1\). ▶

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Step 1: Write the given equation
We have the linear equation \( y = 2x + 1 \).

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Step 2: Understand the form of the equation
This equation is already in the slope-intercept form \( y = mx + c \), where:
\( m = 2 \) (slope) and \( c = 1 \) (y-intercept).

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Step 3: Find some values of \( y \) for different values of \( x \)

\( x \)	\( y = 2x + 1 \)
0	1
1	3
2	5
-1	-1

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Step 4: Interpret the solution
Every pair \((x, y)\) that satisfies \( y = 2x + 1 \) lies on the straight line with slope \( 2 \) and y-intercept \( 1 \).

The answer is the equation of the line \( y = 2x + 1 \)

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Q5(b). A company has fixed costs of $7,000 for plant and equipment and variable costs of $600 for each unit of output. What is the total cost at varying levels of output? ▶

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Step 1: Write the given information
Fixed cost \( = \$7000 \)
Variable cost per unit \( = \$600 \)

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Step 2: Write the formula for total cost
\( \text{Total Cost} = \text{Fixed Cost} + (\text{Variable Cost per Unit} \times \text{Number of Units}) \)
\( C = 7000 + 600x \)

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Step 3: Create a table for varying levels of output

Output (units)	Fixed Cost ($)	Variable Cost ($)	Total Cost ($)
0	7000	0	7000
1	7000	600	7600
2	7000	1200	8200
3	7000	1800	8800
4	7000	2400	9400
5	7000	3000	10000

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The total cost function is \( C = 7000 + 600x \)

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Q5(c). Find the equation of the straight line that has slope \(m = 4\) and passes through the point \((−1, 6)\). ▶

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Step 1: Write the slope-intercept form of a straight line
\( y = mx + c \)

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Step 2: Substitute the given slope
\( y = 4x + c \)

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Step 3: Substitute the point \((-1, 6)\) into the equation to find \( c \)
\( 6 = 4(-1) + c \)

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Step 4: Simplify to solve for \( c \)
\( 6 = -4 + c \)
\( c = 10 \)

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Step 5: Write the final equation
\( y = 4x + 10 \)

The answer is \( y = 4x + 10 \)

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ASSIGNMENT No. 2

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Q1(a). Find the transpose of each of the following matrices: (i) \(A = \begin{bmatrix}-1 & 0 & 0 \\ 6 & 9 & -2 \\ -4 & 5 & 1\end{bmatrix}\) (ii) \(B = \begin{bmatrix}6 & -6 \\ 4 & 9 \\ -2 & 0\end{bmatrix}\) ▶

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(i) \(A = \begin{bmatrix}-1 & 0 & 0 \\ 6 & 9 & -2 \\ -4 & 5 & 1\end{bmatrix}\)

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Step 1
Write the given matrix as \(A=\begin{bmatrix}-1 & 0 & 0 \\ 6 & 9 & -2 \\ -4 & 5 & 1\end{bmatrix}\).

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Step 2
To find the transpose \(A^{T}\), interchange the rows and columns of \(A\).

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Step 3
\[ A^{T}= \begin{bmatrix} -1 & 6 & -4 \\ 0 & 9 & 5 \\ 0 & -2 & 1 \end{bmatrix} \]

\(A^{T}=\begin{bmatrix}-1 & 6 & -4 \\ 0 & 9 & 5 \\ 0 & -2 & 1\end{bmatrix}\)

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(ii) \(B = \begin{bmatrix}6 & -6 \\ 4 & 9 \\ -2 & 0\end{bmatrix}\)

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Step 1
Write the given matrix \(B\).

\[ B=\begin{bmatrix} 6 & -6 \\ 4 & 9 \\ -2 & 0 \end{bmatrix} \]

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Step 2
To find \(B^{T}\), interchange the rows and columns of \(B\).

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Step 3
\[ B^{T}= \begin{bmatrix} 6 & 4 & -2 \\ -6 & 9 & 0 \end{bmatrix} \]

\(B^{T}=\begin{bmatrix}6 & 4 & -2 \\ -6 & 9 & 0\end{bmatrix}\)

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Q1(b). If A \(A=\begin{bmatrix}1 & 2 & 3 \\ -1 & -2 & 3 \\ 3 & 4 & 5\end{bmatrix}\), find 4A-3B. ▶

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Problem
Given \(A=\begin{bmatrix}1 & 2 & 3 \\ -1 & -2 & 3 \\ 3 & 4 & 5\end{bmatrix}\) and \(B=\begin{bmatrix}6 & -6 \\ 4 & 9 \\ -2 & 0\end{bmatrix}\), find \(4A-3B\).

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Step 1
Check matrix order. \(A\) is a \(3\times3\) matrix, while \(B\) is a \(3\times2\) matrix.

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Step 2
Matrix subtraction or addition is only possible when both matrices have the same order.

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The calculation \(4A-3B\) cannot be performed because the matrices have different sizes.

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Q1(c). The quarterly sales of wheat, cotton, and corn for the years 2010 and 2011 are represented below in the form of matrices A and B. Find the total quarterly sales of wheat, cotton, and corn for these two years. ▶

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Problem
The quarterly sales of wheat, cotton, and corn for the years 2010 and 2011 are represented by matrices \(A\) and \(B\). Find the total quarterly sales for both years.

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Step 1
Let \[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix} \] represent the quarterly sales of wheat, cotton, and corn for 2010, and \[ B = \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \end{bmatrix} \] represent the same for 2011.

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Step 2
The total quarterly sales for the two years are found by adding the corresponding elements of matrices \(A\) and \(B\).

\[ A + B = \begin{bmatrix} a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} & a_{14}+b_{14} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} & a_{24}+b_{24} \\ a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33} & a_{34}+b_{34} \end{bmatrix} \]

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Step 3
The resulting matrix \(A+B\) gives the total quarterly sales of wheat, cotton, and corn for both years 2010 and 2011.

\( \text{Total Quarterly Sales} = A + B \)

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Q2(a). ▶

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Q2(b). ▶

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Q3(a). ▶

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Q3(b). ▶

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Q4(a). ▶

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Q4(b). ▶

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Q5(a). ▶

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Q5(b). ▶

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