Course Content
AIOU 1429 Past Paper Spring 2025 – Question #01 (Topic: Probability Distribution and Histogram)
In this lesson, we will analyze the frequency distribution of ocean storms over fifty years. You will learn how to convert a frequency table into a probability distribution and then visualize the data using a histogram. This is a fundamental skill in business statistics and is directly taken from the AIOU 1429 past paper (Spring 2025). By the end of this lesson, you will be able to: Compute probabilities from given frequencies. Construct a probability distribution table. Draw and interpret a histogram for a discrete probability distribution.
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AIOU 1429 Past Paper Spring 2025 – Question #01 (Topic: Probability Distribution and Histogram)
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AIOU 1429 Past Paper Spring 2025 – Question #02 (Topic: Stationary Points, Maxima and Minima)
In this lesson, we solve Question #02 from the AIOU 1429 Business Mathematics past paper (Spring 2025). This question focuses on stationary points and maxima/minima of cubic functions. You will learn how to: Find stationary points by setting the first derivative equal to zero. Determine stationary values (function values at stationary points). Investigate whether a stationary point is a local maximum or minimum using the second derivative test. Two separate videos cover part (a) and part (b) in detail.
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Question 2 (a) Part: Stationary Points and Stationary Values
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Question 2 (b) Part: Maxima and Minima Investigation
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AIOU 1429 Past Paper Spring 2025 – Question #03 (Topic: Quadratic Equations and Inequalities)
In this lesson, we solve Question #03 from the AIOU 1429 Business Mathematics past paper (Spring 2025). This question has two parts: Part (a): Solving a second-degree quadratic equation and determining the nature of its roots using the discriminant. Part (b): Solving a quadratic inequality and representing the solution on the real number line. These concepts are fundamental in algebra and are widely used in business mathematics for break-even analysis, profit maximization, and other optimization problems.
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Question #03: Quadratic Equations and Inequalities
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AIOU 1429 Past Paper Spring 2025 – Question #04 (Topic: Horizontal Tangents and System of Linear Equations)
In this lesson, we solve Question #04 from the AIOU 1429 Business Mathematics past paper (Spring 2025). The question has two parts: Part (a): Finding the point where the tangent line to a given quadratic function is horizontal. This involves computing the derivative, setting it equal to zero, and finding the corresponding coordinates. Part (b): Solving a system of three linear equations in three variables using matrices. This requires setting up the augmented matrix and using row operations or matrix inversion to find the solution. Two separate videos cover part (a) and part (b) in detail.
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Question #04 (Part a) – Finding the Point Where Tangent is Horizontal
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Question #04 (Part b) – Solving System of Linear Equations Using Matrices
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AIOU 1429 Past Paper Spring 2025 – Question #05: Probability – Coin Toss and Card Draw
Part (a): We toss a fair coin three times. Each toss has two equally likely outcomes: head (H) or tail (T). The total number of possible outcomes is 2³ = 8. These outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. We want the probability of getting at least one head. Instead of counting all outcomes with at least one head directly, it is easier to use the complement rule. The complement of "at least one head" is "no heads," which means all tails. There is only one outcome with all tails: TTT. So the probability of no heads is 1/8. Therefore, the probability of at least one head is 1 – 1/8 = 7/8. Part (b): We draw one card from a standard deck of 52 playing cards. All cards are equally likely. i) Red card: In a standard deck, there are 26 red cards (13 hearts and 13 diamonds). So the probability of drawing a red card is 26/52 = 1/2. ii) Diamond card: There are 13 diamonds in the deck. Thus the probability of drawing a diamond is 13/52 = 1/4. These simple examples illustrate the basic probability formula: P(event)= number of favorable outcomes total number of possible outcomesP(event)= total number of possible outcomes number of favorable outcomes ​Key Takeaways The complement rule: P(at least one)=1−P(none) P(at least one)=1−P(none). For a fair coin, each toss has two outcomes, so total outcomes = 2ⁿ for n tosses. A standard deck has 52 cards: 26 red, 26 black; 13 cards in each suit (hearts, diamonds, clubs, spades). Probability is always a number between 0 and 1.
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Question #05: Basic Probability – Coin Toss and Card Draw
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AIOU 1429 Past Paper Spring 2025 – Question #06 (Intersection of Lines and Linear Equations)
In this lesson, we solve Question #06 from the AIOU 1429 Business Mathematics past paper (Spring 2025). The question consists of two parts: Part (a): Finding the point of intersection of two given linear equations. This involves solving a system of two equations in two variables using substitution or elimination method. Part (b): Deriving a linear equation for Celsius temperature (C) as a function of Fahrenheit (F) when the slope and C-intercept are provided. This demonstrates the application of the slope-intercept form in a real-world context. A single video covers both parts with detailed step-by-step solutions.
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Question #06: Intersection of Lines and Linear Temperature Equation
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AIOU 1429 Business Mathematics Past Paper Spring 2025 Solved | Step-by-Step Solutions by Asif Brain Academy

Part (a): We are given two linear equations: 3x + 2y = 3 and 2x – 6y = 1, and we need to find their point of intersection if it exists. The intersection point is the ordered pair (x, y) that satisfies both equations simultaneously. We can solve this system using the elimination method. First, we multiply the first equation by 3 to make the coefficients of y opposites: 3(3x + 2y) = 3(3) gives 9x + 6y = 9. The second equation is 2x – 6y = 1. Now we add the two equations together: (9x + 6y) + (2x – 6y) = 9 + 1, which simplifies to 11x = 10. Therefore, x = 10/11. To find y, we substitute x = 10/11 into the first original equation: 3(10/11) + 2y = 3, which gives 30/11 + 2y = 3. Converting 3 to 33/11, we have 30/11 + 2y = 33/11. Subtracting 30/11 from both sides gives 2y = 3/11, so y = 3/22. Thus the point of intersection is (10/11, 3/22). We can verify by substituting into the second equation: 2(10/11) – 6(3/22) = 20/11 – 18/22 = 20/11 – 9/11 = 11/11 = 1, which confirms our solution.

Part (b): We are told that Celsius (C) and Fahrenheit (F) are related by a linear equation. The slope is given as 5/9, and the C-intercept is given as -100/9. The C-intercept means the value of C when F = 0. In the slope-intercept form of a linear equation, we write C = mF + b, where m is the slope and b is the C-intercept. Substituting the given values, we get C = (5/9)F + (-100/9). Therefore, the linear equation is C = (5/9)F – 100/9. This is the familiar temperature conversion formula rearranged. We can also write it as C = (5F – 100)/9 or multiply both sides by 9 to get 9C = 5F – 100, which is another standard form of the Celsius-Fahrenheit relationship.

Key Takeaways

  • The intersection point of two lines is the solution to the system of their equations.

  • The elimination method involves making coefficients of one variable opposites and adding equations.

  • The slope-intercept form of a line is y = mx + b, where m is slope and b is the y-intercept.

  • In the context of temperature conversion, C = (5/9)(F – 32) can be rearranged to C = (5/9)F – 160/9, but here the given intercept differs, indicating a different scaling or possibly a misprint; however, we use the values as provided.

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