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 Business Mathematics Past Paper Spring 2025 Solved | Step-by-Step Solutions by Asif Brain Academy

For Part (a), we are given the quadratic equation y² – 3y – 5 = 0. This is in the standard form ay² + by + c = 0 where a = 1, b = -3, and c = -5. To solve this equation, we use the quadratic formula which states that y = [-b ± √(b² – 4ac)] / 2a. Substituting the values, we get y = [3 ± √((-3)² – 4(1)(-5))] / 2(1) = [3 ± √(9 + 20)] / 2 = [3 ± √29] / 2. Therefore, the two roots are y = (3 + √29)/2 and y = (3 – √29)/2. To determine the nature of these roots, we examine the discriminant which is Δ = b² – 4ac = 9 + 20 = 29. Since the discriminant is greater than zero and is not a perfect square, the roots are real, distinct, and irrational.

For Part (b), we are given the inequality 2x² + 4x + 8 < 0. We begin by simplifying the inequality by dividing both sides by 2. Since 2 is positive, the direction of the inequality remains unchanged, giving us x² + 2x + 4 < 0. Next, we find the discriminant of the quadratic expression x² + 2x + 4, which is Δ = (2)² – 4(1)(4) = 4 – 16 = -12. Because the discriminant is negative and the coefficient of x² is positive (1 > 0), the quadratic opens upward and is always positive for all real values of x. This means that x² + 2x + 4 > 0 for every real number x, and consequently the inequality x² + 2x + 4 < 0 has no solution. When representing this on the real line, we show an empty solution set, often denoted by the symbol ∅, indicating that there are no real numbers that satisfy the given inequality.

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