AIOU 9424 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: Quantitative Reasoning (9424)	Semester: Autumn-2025
Level: BS

Total Marks: 100	Pass Marks: 50

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ASSIGNMENT No. 1

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Q1(a). Which fundamental concepts and methods in mathematics did the Babylonians introduce and develop? ▶

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Fundamental Concepts and Methods Introduced by the Babylonians

The Babylonians were among the earliest civilizations to develop systematic mathematical practices. Their contributions were practical yet advanced for their time, and many of their ideas continue to influence modern mathematics. Their work was primarily recorded on clay tablets using cuneiform script, which has given historians and mathematicians valuable insight into their techniques and problem-solving approaches.

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Sexagesimal Number System

One of the most significant achievements of Babylonian mathematics was their adoption of the sexagesimal, or base-60, number system. Unlike the decimal system, base-60 offered great flexibility because 60 is divisible by many numbers such as 2, 3, 4, 5, and 6. This made fractional calculations easier. The legacy of this system still exists today in our division of time into 60 seconds per minute and 60 minutes per hour, and in the measurement of angles with 360 degrees in a circle.

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Place Value System

The Babylonians developed an early place value system where the value of a symbol depended on its position. This was a revolutionary step compared to other ancient systems like Egyptian hieroglyphics. Though initially they lacked a symbol for zero, their positional system laid the groundwork for later numerical systems. Eventually, they introduced a placeholder symbol, which helped avoid ambiguities in large numbers.

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Mathematical Tables

The Babylonians created extensive mathematical tables on clay tablets to simplify calculations. These included multiplication tables, tables of squares, cubes, reciprocals, and square roots. These resources functioned much like early calculators, enabling scribes and scholars to solve complex problems more efficiently.

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Quadratic Equations

Babylonian mathematicians could solve quadratic equations, often applying methods similar to what we now recognize as the quadratic formula. They used these techniques for practical problems, such as calculating the dimensions of fields or the sides of geometric figures. Their ability to solve such equations shows a highly developed form of algebra.

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Pythagorean Triples

Tablets such as Plimpton 322 demonstrate that the Babylonians were familiar with Pythagorean triples long before Pythagoras. These sets of integers satisfied the relation a² + b² = c², suggesting that they had significant knowledge of right triangles and geometric properties, which were applied in surveying and construction.

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Practical Geometry

Geometry played a practical role in Babylonian society. They used formulas to approximate areas and volumes, such as calculating the area of a trapezoid or the volume of a cylinder. They also made approximations for π, sometimes taking it as 3 or as the fraction 25/8. These methods were essential for architecture, land measurement, and irrigation projects.

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Algebraic Problem Solving

Babylonian texts reveal strong algebraic thinking. They worked with equations involving one or two variables and often solved simultaneous equations. These were not abstract exercises but connected to real-life problems, such as distributing land, calculating shares, or trade-related computations.

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Astronomical Applications

Babylonian mathematics was closely tied to astronomy. They used their base-60 system to make accurate astronomical calculations, including predictions of lunar eclipses and planetary movements. This precise use of mathematics for celestial observations heavily influenced later Greek and Islamic astronomy.

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Fractions and Reciprocals

The Babylonians made extensive use of fractions and reciprocals. Unlike Egyptians, who mainly used unit fractions, Babylonians had a more versatile approach due to their base-60 system, which allowed simple representation of many fractions. This ability was particularly important in trade, measurements, and dividing resources.

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Legacy of Babylonian Mathematics

The mathematical innovations of the Babylonians formed the basis of later mathematical traditions. Their systematic methods, use of place value, and integration of mathematics into practical and astronomical contexts made their work foundational for Greek, Indian, and later Islamic mathematics. Even today, remnants of their sexagesimal system survive in our timekeeping and angle measurement.

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Conclusion

The Babylonians were pioneers in mathematics, introducing fundamental concepts such as the sexagesimal system, place value, quadratic solutions, practical geometry, and astronomical mathematics. Their ability to merge practical needs with systematic problem-solving reflects their ingenuity. The impact of their mathematical system is still visible in modern life, making their contributions timeless and universal.

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Q1(b). From A to B, Aryan travels 1 km. He travels back at a pace of three kilometers per hour faster than he did in the beginning. The travel takes 0.5 hours in total. Determine the pace for the initial and subsequent portions of the trip. ▶

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Problem Restatement

Aryan travels 1 km from A to B at some initial speed \( v \) km/h. On the return trip, he travels 1 km back from B to A at a faster speed of \( v + 3 \) km/h. The total travel time is 0.5 hours. We need to determine both speeds.

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Step 1: Express the travel times

The time taken to travel from A to B is:

\[ t_1 = \frac{1}{v} \]

The time taken to return from B to A is:

\[ t_2 = \frac{1}{v+3} \]

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Step 2: Total time equation

The total time for the trip is 0.5 hours, therefore:

\[ t_1 + t_2 = 0.5 \]

\[ \frac{1}{v} + \frac{1}{v+3} = 0.5 \]

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Step 3: Simplify the equation

Multiply through by \( v(v+3) \) to remove denominators:

\[ (v+3) + v = 0.5 \cdot v(v+3) \]

\[ 2v + 3 = 0.5v^2 + 1.5v \]

Multiply both sides by 2 to remove the decimal:

\[ 4v + 6 = v^2 + 3v \]

Rearranging gives:

\[ v^2 - v - 6 = 0 \]

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Step 4: Solve the quadratic equation

The quadratic can be factorized as:

\[ (v - 3)(v + 2) = 0 \]

So the solutions are:

\[ v = 3 \quad \text{or} \quad v = -2 \]

Since speed cannot be negative, we take:

\[ v = 3 \, \text{km/h} \]

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Step 5: Find the return speed

The return speed is:

\[ v + 3 = 3 + 3 = 6 \, \text{km/h} \]

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The speed from A to B was \( 3 \, \text{km/h} \). The speed from B to A was \( 6 \, \text{km/h} \).

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Q2(a). Consider two candidates who appeared for the selection of the post of senior scientist. Both persons were thoroughly examined by a panel of examiners. Person A got 80 points out of 100, while Person B got 70 points out of 100. Scientist A got the position. Explain how numbers helped the panel of examiners to decide for selection. ▶

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Introduction

In any evaluation process, numbers play a crucial role in ensuring fairness and objectivity. When two candidates appear for the same position, their performance is often assessed through a scoring system that provides a measurable basis for comparison. This makes the selection process transparent and evidence-based.

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Performance Evaluation Through Scores

In the given situation, both Person A and Person B appeared for the selection of the post of senior scientist. Their performances were thoroughly examined by a panel of experts. After evaluation, Person A secured 80 points out of 100, while Person B scored 70 points out of 100. These numbers represent the overall assessment of their skills, knowledge, and suitability for the role.

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Numerical Comparison of Candidates

The scores provide a direct comparison between the two candidates. Person A’s score of 80 shows a higher level of achievement compared to Person B’s score of 70. The difference of 10 points reflects that Person A demonstrated greater competence, accuracy, and command over the subject matter. Without numbers, the judgment could have been subjective, but numerical values made the comparison clear and straightforward.

In mathematical form, we can write:

\[ \text{Person A's Score} = \frac{80}{100} = 0.8 \, (80\%) \]

\[ \text{Person B's Score} = \frac{70}{100} = 0.7 \, (70\%) \]

Thus, Person A outperformed Person B by: \[ 0.8 - 0.7 = 0.1 = 10\% \]

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Role of Objectivity in Decision-Making

Numbers remove personal bias from the selection process. The panel could rely on the numerical scores instead of personal impressions or preferences. This ensured that the decision to select Person A was based purely on merit, as reflected by the higher score. The use of a scoring system guarantees transparency and fairness.

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Why Person A Was Selected

Since Person A obtained 80 points, which is significantly higher than Person B’s 70 points, the numbers clearly indicated that Person A performed better overall. The panel used these results to select the candidate who demonstrated superior knowledge and skill. The numerical evidence provided a reliable justification for choosing Person A as the senior scientist.

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Conclusion

The use of numbers in evaluation ensures that decisions are logical, fair, and evidence-based. In this case, the numerical difference between the scores of Person A and Person B guided the panel to select Person A. Thus, numbers acted as a decisive tool, ensuring meritocracy and fairness in the selection process.

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Q2(b). In this case, the person who sold the pen for Rs. 96 made the same amount of money as the watch's cost price. If he sells it for twice as much as the percentage profit he made on it before, what will the new price be? ▶

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Problem Restatement

A person sold a pen for Rs. 96. The money he received from selling the pen is equal to the cost price of a watch. He made a certain profit percentage on the pen. If he sells the pen again but this time at twice the earlier percentage profit, what will be the new selling price?

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Step 1: Let the cost price of the pen be x

The selling price of the pen is 96, so the profit made is:

\[ \text{Profit} = 96 - x \]

The profit percentage is:

\[ \text{Profit \%} = \frac{96 - x}{x} \times 100 \]

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Step 2: New profit percentage condition

If the pen is sold at twice the earlier percentage profit, then the new selling price will be:

\[ \text{New SP} = x \Bigg(1 + \frac{2(96 - x)}{x}\Bigg) \]

Simplifying,

\[ \text{New SP} = x + 2(96 - x) = 192 - x \]

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Step 3: Interpretation

The new selling price depends on the cost price of the pen (x). Without knowing x exactly, the formula remains:

\[ \text{New SP} = 192 - x \]

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Step 4: Numerical Example

Suppose the cost price of the pen was Rs. 80.

Then the original profit is:

\[ 96 - 80 = 16 \]

The profit percentage is:

\[ \frac{16}{80} \times 100 = 20\% \]

If the profit percentage is doubled, it becomes 40%. Therefore, the new selling price is:

\[ 80 + \frac{40}{100} \times 80 = 80 + 32 = 112 \]

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In general, the new selling price of the pen is \(192 - x\), where \(x\) is the cost price of the pen. For example, if the cost price is Rs. 80, then the new selling price will be Rs. 112.

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Q3(a). Four shirts, four sets of pants, and two hats cost $560 in a store. Nine shirts, nine pairs of pants, and six hats total $1,290 in cost. How much does one shirt, one pair of pants, and one item cost altogether? ▶

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Problem Restatement

We are given that four shirts, four sets of pants, and two hats cost $560 in a store. Nine shirts, nine pairs of pants, and six hats cost $1,290. We need to find the cost of one shirt, one pair of pants, and one hat altogether.

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Step 1: Define the variables

Let the price of one shirt be \( S \), one pair of pants be \( P \), and one hat be \( H \).

From the given information we can write:

\[ 4S + 4P + 2H = 560 \]

\[ 9S + 9P + 6H = 1290 \]

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Step 2: Simplify the first equation

Divide the first equation by 2:

\[ 2S + 2P + H = 280 \quad (1) \]

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Step 3: Eliminate H

Subtract 6 times equation (1) from the second equation:

\[ 9S + 9P + 6H - 6(2S + 2P + H) = 1290 - 1680 \]

\[ 9S + 9P + 6H - (12S + 12P + 6H) = -390 \]

\[ -3S - 3P = -390 \]

\[ S + P = 130 \]

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Step 4: Solve for H

Substitute \( S + P = 130 \) into equation (1):

\[ 2(130) + H = 280 \]

\[ 260 + H = 280 \]

\[ H = 20 \]

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Step 5: Final Calculation

Now the total cost of one shirt, one pair of pants, and one hat is:

\[ S + P + H = 130 + 20 = 150 \]

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The cost of one shirt, one pair of pants, and one hat altogether is $150.

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Q3(b). If the Golden Ratio is 1.618 and you have a piece of cardboard that is 7/8 ft wide. How long should you cut, so it follows the Golden Ratio? ▶

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Problem Restatement

The Golden Ratio is approximately \( \phi = 1.618 \). We are given a piece of cardboard that is \( \tfrac{7}{8} \) ft wide. We need to find the length so that the dimensions follow the Golden Ratio.

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Step 1: Recall the Golden Ratio Formula

For an object to follow the Golden Ratio:

\[ \frac{L}{W} = \phi \]

where \( L \) is the length and \( W \) is the width.

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Step 2: Substitute the Width

The width is given as:

\[ W = \tfrac{7}{8} \, \text{ft} = 0.875 \, \text{ft} \]

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Step 3: Calculate the Length

Using the Golden Ratio:

\[ L = \phi \times W \]

\[ L = 1.618 \times 0.875 \]

\[ L = 1.41575 \, \text{ft} \]

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The required length of the cardboard should be approximately 1.416 ft in order to follow the Golden Ratio with a width of 7/8 ft.

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Q4(a). Using the example of the population death rate and birth rate. Find in how many years the population has increased to 11000. ▶

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Population Growth Problem

We want to determine in how many years the population will increase to 11,000, given birth and death rates. The growth of a population can be modeled using the formula:

\[ P(t) = P_0 \times \left(1 + \frac{r}{100}\right)^t \]

Here, \( P_0 \) is the initial population, \( r \) is the net growth rate (birth rate minus death rate), and \( t \) is the number of years.

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Step 1: Net Growth Rate

If the birth rate is \( b\% \) and the death rate is \( d\% \), then the net growth rate is:

\[ r = b - d \]

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Step 2: Condition for Population

The target population is 11,000. Thus, the equation becomes:

\[ 11000 = P_0 \times \left(1 + \frac{r}{100}\right)^t \]

Taking logarithms:

\[ t = \frac{\ln\left(\tfrac{11000}{P_0}\right)}{\ln\left(1 + \tfrac{r}{100}\right)} \]

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Step 3: Example with Values

Suppose the initial population is \( P_0 = 10000 \). If the birth rate is 3% and the death rate is 1%, then:

\[ r = 3 - 1 = 2\% \]

The equation becomes:

\[ 11000 = 10000 \times (1.02)^t \]

\[ 1.1 = (1.02)^t \]

Taking logarithms:

\[ t = \frac{\ln(1.1)}{\ln(1.02)} \]

\[ t \approx \frac{0.09531}{0.01980} \approx 4.81 \, \text{years} \]

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The population will grow from 10,000 to 11,000 in about 5 years at a net growth rate of 2% per year.

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Q4(b). What types of connections did prehistoric humans have with their surroundings, and how did their behavior impact the environment. ▶

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Introduction

Prehistoric humans lived in close connection with their surroundings. Their survival depended entirely on the resources provided by nature such as water, plants, animals, and shelter. Unlike modern societies, they did not have advanced technology or artificial systems to control their environment. Instead, their daily activities and lifestyle were shaped directly by the natural world around them. These early connections reveal how humans interacted with the environment in ways that were both sustainable and, at times, damaging.

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Dependence on Nature for Survival

Prehistoric humans relied on hunting animals, fishing in rivers, and gathering wild fruits, nuts, and roots. This dependence created a deep awareness of seasonal changes, weather patterns, and migration routes of animals. For example, the availability of edible plants in different seasons determined where groups of people would settle or move. The connection to natural food resources made prehistoric humans keen observers of their surroundings, linking their survival to the land and ecosystems they inhabited.

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Use of Tools and Fire

The invention of simple stone tools allowed prehistoric humans to cut wood, hunt more effectively, and process food. The discovery of fire was another major turning point, providing warmth, protection, and a new way to cook food. However, fire also gave humans the ability to clear land by burning forests, which began to alter natural landscapes. While this helped in hunting and farming later on, it also contributed to environmental changes such as the loss of trees and habitats for certain species.

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Nomadic Lifestyle and Environmental Balance

In the early stages, humans were nomadic, moving from one place to another in search of food and shelter. This lifestyle helped prevent overuse of a single area, allowing the land to recover after people left. Their movements ensured that resources such as plants and animals were not exhausted in one location. In this way, the connection between humans and nature maintained a natural balance, as they took only what was needed and adapted to what was available.

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Impact of Hunting on Animal Populations

While hunting was essential for survival, it also had lasting effects on the environment. Over time, excessive hunting led to the extinction of some large animals, such as mammoths in certain regions. The disappearance of these animals changed ecosystems, as other species were affected by the loss of their predators or competitors. This shows that even without advanced technology, human behavior could significantly reshape the natural environment.

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Early Farming and Domestication

The shift from hunting and gathering to farming marked a major change in the human relationship with the environment. By domesticating plants and animals, prehistoric humans began to settle in one place, clearing land for agriculture. This practice increased food supply but also led to deforestation and soil exhaustion in some areas. Farming represented the first step toward controlling and modifying the environment for human needs, which had both positive and negative long-term effects.

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Cultural and Spiritual Connections

Prehistoric humans often viewed their surroundings as sacred. Natural features like rivers, mountains, and forests were seen as sources of life and spirituality. Cave paintings, carvings, and burial sites suggest that they respected animals and natural forces. This spiritual connection influenced how they used resources, sometimes encouraging practices that protected the environment and showed gratitude for nature’s gifts.

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Conclusion

Prehistoric humans had deep and direct connections with their environment, shaped by survival needs, spiritual beliefs, and daily life. Their behaviors such as hunting, using fire, and early farming had clear impacts on ecosystems, sometimes maintaining balance and sometimes leading to environmental changes. These early interactions remind us that humans have always influenced nature, and they highlight the importance of using resources wisely to avoid long-term damage. The story of prehistoric life teaches us that the way we interact with our surroundings continues to shape the future of our environment.

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Q5(a). A weighing scale has an uncertainty of a maximum+5 grams per 1 kg. A load of 105kg is weighed on this scale. Find what could be a possible error that the scale can give when this load is weighed? ▶

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Problem Explanation

We are given a weighing scale that has an uncertainty of maximum ±5 grams for every 1 kilogram measured. A load of 105 kilograms is weighed on this scale. We need to calculate the possible error the scale could give for this load.

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Step 1: Understanding the Uncertainty

The scale error is given as ±5 grams per 1 kilogram. This means for each kilogram measured, the scale could be off by as much as 5 grams.

\[ \text{Error per 1 kg} = \pm 5 \, \text{g} \]

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Step 2: Total Load

The load being weighed is:

\[ 105 \, \text{kg} \]

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Step 3: Total Possible Error

The possible error increases proportionally with the weight. Therefore, the total error is:

\[ \text{Total Error} = 105 \times 5 \, \text{g} \]

\[ \text{Total Error} = 525 \, \text{g} \]

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Step 4: Convert into Kilograms

Since 1 kilogram = 1000 grams, we convert:

\[ 525 \, \text{g} = 0.525 \, \text{kg} \]

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The possible error in weighing a 105 kg load on this scale is:

\[ \pm 525 \, \text{g} \quad \text{or} \quad \pm 0.525 \, \text{kg} \]

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Q5(b). If you want to retire in 30 years and you estimate that you will need 10 lacs rupees for retirement, how much money do you need to save each month if your investments earn an average annual return of 8%? ▶

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Problem Explanation

The goal is to calculate how much money must be saved each month in order to accumulate 10 lacs rupees in 30 years, given that the average annual return on investment is 8%. This is a future value of annuity problem.

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Step 1: Formula for Future Value of Annuity

The formula for the future value of an ordinary annuity is:

\[ FV = P \times \frac{(1 + r)^n - 1}{r} \]

Where:

\(FV\) = Future Value (10,00,000 rupees)
\(P\) = Monthly savings (what we want to find)
\(r\) = Monthly interest rate
\(n\) = Total number of months

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Step 2: Convert Annual Interest Rate to Monthly Rate

The annual return is 8%. To get the monthly rate:

\[ r = \frac{0.08}{12} = 0.0066667 \, \text{per month} \]

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Step 3: Calculate Number of Months

Since retirement is in 30 years, the total months are:

\[ n = 30 \times 12 = 360 \, \text{months} \]

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Step 4: Substitute Values into Formula

\[ 10,00,000 = P \times \frac{(1 + 0.006667)^{360} - 1}{0.006667} \]

First calculate the growth factor:

\[ (1.006667)^{360} \approx 11.935 \]

Now subtract 1:

\[ 11.935 - 1 = 10.935 \]

Divide by \(r = 0.006667\):

\[ \frac{10.935}{0.006667} \approx 1640.25 \]

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Step 5: Solve for Monthly Savings

\[ P = \frac{10,00,000}{1640.25} \approx 609.8 \]

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The required monthly saving is about:

\[ \text{Rs. 610 per month (approximately)} \]

This means if you invest Rs. 610 every month for 30 years at an 8% annual return, you will accumulate around 10 lacs rupees for retirement.

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ASSIGNMENT No. 2

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Q1(a). A grain storage facility is a hemispherical shell with a radius of thirty meters. What is the volume of the building? How much paint is needed to cover the outside of the building completely? ▶

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Problem Statement

A grain storage facility is a hemispherical shell with radius \(r=30\) metres. Find (a) the volume of the building, and (b) how much paint is needed to cover the outside of the building completely. In the paint part we will assume the painted area is the outer curved surface of the hemisphere (the flat circular base lies on the ground and is not painted). We will also give example paint quantities for common paint coverages.

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Volume of a Hemisphere

The volume of a full sphere is \(\tfrac{4}{3}\pi r^3\). For a hemisphere (half a sphere) the volume is half of that, so

\[ V_{\text{hemisphere}}=\frac{1}{2}\cdot\frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3. \]

Substitute \(r=30\) m:

\[ V=\frac{2}{3}\pi(30)^3=\frac{2}{3}\pi\cdot 27{,}000=18{,}000\pi\ \text{m}^3. \]

Numerically, using \(\pi\approx3.14159265\):

\[ V\approx 18{,}000\times3.14159265 \approx 56{,}548.67\ \text{m}^3. \]

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Exterior Surface Area to be Painted

The curved surface area of a hemisphere (not including the base) is half the surface area of a sphere, i.e.

\[ A_{\text{curved}}=\frac{1}{2}\cdot 4\pi r^2=2\pi r^2. \]

With \(r=30\) m:

\[ A=2\pi(30)^2=2\pi\cdot900=1{,}800\pi\ \text{m}^2. \]

Numerically:

\[ A\approx 1{,}800\times3.14159265 \approx 5{,}654.87\ \text{m}^2. \]

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Paint Required (examples)

Paint requirements depend on the paint coverage (how many square metres one litre of paint covers). Common approximate coverages are 8–12 m\(^2\) per litre depending on paint and number of coats. Using the exterior area \(A\approx5654.87\ \text{m}^2\):

For coverage 10 m\(^2\)/L (typical rule of thumb):

\[ \text{Litres needed}=\frac{5654.87}{10}\approx 565.49\ \text{L}. \]

For coverage 12 m\(^2\)/L:

\[ \text{Litres needed}\approx\frac{5654.87}{12}\approx 471.24\ \text{L}. \]

For coverage 8 m\(^2\)/L (thicker coat / poorer spread):

\[ \text{Litres needed}\approx\frac{5654.87}{8}\approx 706.86\ \text{L}. \]

It is wise to allow about 10% extra for wastage, overlaps and additional coats. So for the 10 m\(^2\)/L case buy roughly \(565.49\times1.1\approx622\) L (round up to a convenient package size).

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Summary (rounded)

Volume of the hemispherical storage building:

\(V=18{,}000\pi\ \text{m}^3 \approx 56{,}548.7\ \text{m}^3\)

Exterior curved surface area to be painted:

\(A=1{,}800\pi\ \text{m}^2 \approx 5{,}654.9\ \text{m}^2\)

Example paint required (approximate, before wastage):

At 10 m\(^2\)/L: about 565.5 L. With 10% spare: about 622 L.

Adjust the final quantity according to the actual paint coverage indicated by the manufacturer and the number of coats you intend to apply.

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Q1(b). Write the following sentences in the form if p, then q. Find p and q without any trouble.
1. A resident of Karachi is a resident of Pakistan.
2. Daily exercise is a sufficient condition for being healthy.
3. Being bald is sufficient for being male.
4. Whenever it rains, I get wet.
5. There is no doubt that she is smart if she is an art expert. ▶

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Introduction

We are asked to rewrite each of the given sentences in the logical form "If p, then q". After that, we need to identify p (the hypothesis) and q (the conclusion). This allows us to see the structure of conditional statements clearly.

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1. A resident of Karachi is a resident of Pakistan.

Original: A resident of Karachi is a resident of Pakistan.

If p, then q form: If a person is a resident of Karachi, then that person is a resident of Pakistan.

Here, \(p\): "A person is a resident of Karachi."
\(q\): "That person is a resident of Pakistan."

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2. Daily exercise is a sufficient condition for being healthy.

Original: Daily exercise is a sufficient condition for being healthy.

If p, then q form: If a person exercises daily, then that person is healthy.

Here, \(p\): "A person exercises daily."
\(q\): "That person is healthy."

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3. Being bald is sufficient for being male.

Original: Being bald is sufficient for being male.

If p, then q form: If a person is bald, then that person is male.

Here, \(p\): "A person is bald."
\(q\): "That person is male."

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4. Whenever it rains, I get wet.

Original: Whenever it rains, I get wet.

If p, then q form: If it rains, then I get wet.

Here, \(p\): "It rains."
\(q\): "I get wet."

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5. There is no doubt that she is smart if she is an art expert.

Original: There is no doubt that she is smart if she is an art expert.

If p, then q form: If she is an art expert, then she is smart.

Here, \(p\): "She is an art expert."
\(q\): "She is smart."

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Q2(a). Individuals who visited an Islamabad hospital on a single day were prescribed the following amounts of antibiotics (A), painkillers (P), and blood pressure medications (BP):

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• A only 12
• P only 22
• BP only 8
• P and A only 24
• P and BP only 16
• BP and A only 15
• All three 2
• Outside (none) 2

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i. Sketch the Venn diagram with three circles to summarize the table results.
ii. How many patients took BP medicines or antibiotics?
iii. How many patients specifically took BP medicine but did not take any pain medicine?
iv. What proportion of patients used painkillers at all?
v. How many patients used antibiotics and blood pressure medications but not painkillers?
vi. How many patients used pain relievers, antibiotics, or blood pressure medication? ▶

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i. Sketch the Venn diagram with three circles to summarize the table results.

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ii. How many patients took BP medicines or antibiotics?
Counts that include A or BP are A only 12, BP only 8, A and P only 24, P and BP only 16, A and BP only 15, All three 2. Sum = 12 + 8 + 24 + 16 + 15 + 2 = 77 patients.

iii. How many patients specifically took BP medicine but did not take any pain medicine?
Regions with BP and without P are BP only 8 and A and BP only 15. Sum = 8 + 15 = 23 patients.

iv. What proportion of patients used painkillers at all?
Patients using P are P only 22, A and P only 24, P and BP only 16, All three 2. Total using P = 22 + 24 + 16 + 2 = 64. The overall total of all people counted is 101. Proportion = 64 / 101 ≈ 0.6337 which is about 63.37 percent.

v. How many patients used antibiotics and blood pressure medications but not painkillers?
That is the A and BP only region: 15 patients.

vi. How many patients used pain relievers, antibiotics, or blood pressure medication?
That is everyone who used at least one of the three. Total of all regions except none = 101 - 2 = 99 patients.

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Q2(b). In triangle \(\triangle ABC \) angle \(A\) is twice the measure of angle \(B\), and angle \(C\) is 40 degrees. Find the measures of angles \(A\) and \(B\). ▶

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1. Let the measure of \(\angle B = B\).
2. Then \(\angle A = 2B\).
3. The sum of angles in a triangle is \(180^\circ\), so:

   \[\angle A + \angle B + \angle C = 180\]

4. Substitute \(\angle A = 2B\) and \(\angle C = 40^\circ\):

   \[2B + B + 40 = 180\]

5. Simplify:

   \[3B + 40 = 180\]

   \[3B = 140\]

   \[B = \frac{140}{3} = 46\frac{2}{3}^\circ\]

6. Find \(\angle A\):

   \[\angle A = 2B = 2 \times \frac{140}{3} = \frac{280}{3} = 93\frac{1}{3}^\circ\]

7. Check \(\angle C\):

   \[\angle C = 40^\circ\]

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• \(\angle A = 93\frac{1}{3}^\circ\)
• \(\angle B = 46\frac{2}{3}^\circ\)
• \(\angle C = 40^\circ\)

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• \(\angle A = 93.33^\circ\)
• \(\angle B = 46.67^\circ\)
• \(\angle C = 40^\circ\)

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Q3(a). Solve the inequality \(u^2 \le 3u + 4\). ▶

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Step 1: Bring all terms to one side:

\[ u^2 - 3u - 4 \le 0 \]

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Step 2: Factor the quadratic:

\[ u^2 - 3u - 4 = (u - 4)(u + 1) \]

So the inequality becomes:

\[ (u - 4)(u + 1) \le 0 \]

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Step 3: Find the critical points:

\[ u - 4 = 0 \implies u = 4 \]

\[ u + 1 = 0 \implies u = -1 \]

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Step 4: Test intervals between the critical points:

1. \(u < -1\): \((u - 4)(u + 1) = (-)(-) = +\) → does not satisfy \(\le 0\)

2. \(-1 \le u \le 4\): \((u - 4)(u + 1) = (-)(+) = -\) → satisfies \(\le 0\)

3. \(u > 4\): \((u - 4)(u + 1) = (+)(+) = +\) → does not satisfy \(\le 0\)

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Step 5: Include the endpoints because the inequality is \(\le 0\):

\[ u = -1 \quad \text{and} \quad u = 4 \]

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\[ -1 \le u \le 4 \]

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Q3(b). The air duct in a stadium measures 40 feet in length and 18 inches in radius on a circular cross-section. What is the volume of the duct and how many square feet of paint are needed to paint the outside of the duct? ▶

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Air Duct Dimensions and Problem Statement

The problem involves an air duct in a stadium that measures 40 feet in length and has a circular cross-section with a radius of 18 inches. The objective is to determine the volume of the duct and the area of its surface that needs to be painted. To solve this problem, we will use the formulas for the volume and surface area of a cylinder.

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Understanding the Cylinder

The air duct is cylindrical in shape. A cylinder is defined as a three-dimensional figure with two parallel circular bases connected by a curved lateral surface. The height of the cylinder corresponds to the length of the duct, while the radius of the circular base is given in inches. Recognizing the duct as a cylinder allows us to apply the formulas for volume and surface area accurately.

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Converting Units

The length of the duct is given in feet, and the radius is in inches. For consistency in calculations, we convert the radius from inches to feet. Since 1 foot equals 12 inches, the radius in feet is 18 ÷ 12 = 1.5 feet. Ensuring consistent units is critical for accurate results in mathematical calculations.

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Formula for Volume

The volume V of a cylinder is calculated using the formula V = πr²h, where r is the radius and h is the height or length of the cylinder. This formula allows us to determine the internal space of the duct, which is essential in understanding the air-carrying capacity of the duct system.

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Calculating Volume

Substitute the known values: r = 1.5 feet and h = 40 feet. First, square the radius: r² = 1.5 × 1.5 = 2.25. Then multiply by the height: 2.25 × 40 = 90. Multiply this result by π to get the volume: V = 90 × π ≈ 282.74 cubic feet. This gives the total volume of air the duct can contain.

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Formula for Lateral Surface Area

The lateral surface area A of a cylinder is calculated using the formula A = 2πrh. This represents the area of the curved surface connecting the two circular bases. Determining this area allows us to calculate how much paint is needed for the exterior of the duct.

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Calculating Lateral Surface Area

Using the given radius r = 1.5 feet and height h = 40 feet, compute 2 × r × h = 2 × 1.5 × 40 = 120. Multiply by π to get the lateral surface area: A = 120 × π ≈ 376.99 square feet. This is the area of the duct that needs painting if only the curved surface is considered.

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Including the Ends

If the circular ends of the duct are also to be painted, the area of each end is given by πr². Calculate the area of one end: π × 1.5² = π × 2.25 ≈ 7.07 square feet. Multiply by 2 for both ends: 7.07 × 2 ≈ 14.14 square feet. Adding this to the lateral surface area: 376.99 + 14.14 ≈ 391.13 square feet. This represents the total paintable surface area of the duct.

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Summary of Calculations

The calculations yield the following results: the volume of the duct is approximately 282.74 cubic feet, the lateral surface area to be painted is approximately 376.99 square feet, and the total surface area including the ends is approximately 391.13 square feet. These calculations provide precise information for both the design and maintenance of the duct.

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Step-by-Step Verification of Volume

To verify, the steps are: radius in feet r = 1.5, height h = 40, r² = 2.25, r² × h = 2.25 × 40 = 90, 90 × π = 282.74 cubic feet. This confirms the volume calculation is correct and consistent with the standard formula for a cylinder.

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Step-by-Step Verification of Surface Area

For the lateral surface area: 2 × r × h = 2 × 1.5 × 40 = 120, multiply by π: 120 × 3.1416 ≈ 376.99 square feet. For the ends: π × r² = π × 2.25 ≈ 7.07 square feet per end, total for two ends: 7.07 × 2 ≈ 14.14, total surface area: 376.99 + 14.14 ≈ 391.13 square feet. This step-by-step verification ensures no computational error.

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Practical Implications of Volume

The volume indicates how much air the duct can carry. A volume of approximately 282.74 cubic feet allows engineers to determine the air flow required to ventilate stadium areas efficiently. It also informs the specifications for fans and pressure requirements to maintain consistent airflow throughout the duct system.

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Practical Implications of Surface Area

The surface area calculation informs the amount of paint needed. Lateral surface area only: 376.99 square feet. Including ends: 391.13 square feet. Accurate surface area ensures proper estimation of paint, reducing waste and optimizing cost. It also helps in maintenance planning and protective coating application for the duct.

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Conclusion

In conclusion, the calculations for the air duct yield a volume of approximately 282.74 cubic feet, a lateral surface area of approximately 376.99 square feet, and a total surface area including ends of approximately 391.13 square feet. These results are essential for practical applications such as ventilation design, material estimation, and maintenance planning. Using standard formulas for a cylinder and consistent units, the problem has been solved accurately with all relevant mathematical steps verified.

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Q4(a). Construct the truth table for the logical expression \((p \lor q) \lor (p \land \neg r)\)). Determine under what conditions the expression is true. ▶

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p	q	r	¬r	p ∧ ¬r	p ∨ q	(p ∨ q) ∨ (p ∧ ¬r)
T	T	T	F	F	T	T
T	T	F	T	T	T	T
T	F	T	F	F	T	T
T	F	F	T	T	T	T
F	T	T	F	F	T	T
F	T	F	T	F	T	T
F	F	T	F	F	F	F
F	F	F	T	F	F	F

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Conclusion:

The expression \((p \lor q) \lor (p \land \neg r)\) is true for all combinations except when both p and q are false, regardless of r. In interval terms, it is true whenever \(p = T\) or \(q = T\) or \(p = T\) and \(r = F\).

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Q4(b). The labeling on the insulin vial specifies a required storage temperature range between 36°F and 46°F. What is the equivalent temperature range on the Celsius scale? ▶

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Problem: The insulin vial specifies a storage temperature range between 36°F and 46°F. Convert this range to Celsius.

Solution:

The formula to convert Fahrenheit (F) to Celsius (C) is:

\[ C = \frac{5}{9} (F - 32) \]

Step 1: Convert the lower limit (36°F)

\[ C = \frac{5}{9} (36 - 32) = \frac{5}{9} \times 4 = \frac{20}{9} \approx 2.22^\circ C \]

Step 2: Convert the upper limit (46°F)

\[ C = \frac{5}{9} (46 - 32) = \frac{5}{9} \times 14 = \frac{70}{9} \approx 7.78^\circ C \]

Step 3: Final Celsius Range

\[ \boxed{2.22^\circ C \text{ to } 7.78^\circ C} \]

Thus, the insulin should be stored at approximately 2.2°C to 7.8°C.

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Q5. Rewrite each of the following logical statements into symbolic notation, then construct a truth table to determine whether each of the following is a tautology, an absurdity, or a contingency.
1. (𝑖𝑓 𝑝, 𝑡ℎ𝑒𝑛 𝑛𝑜𝑡 𝑝) 𝑜𝑟 (𝑖𝑓 𝑝, 𝑡ℎ𝑒𝑛 𝑞)
2. 𝑖𝑓 (𝑝 𝑎𝑛𝑑 ~𝑝), 𝑡ℎ𝑒𝑛 𝑟
3. 𝑖𝑓 𝑝, 𝑡ℎ𝑒𝑛 (𝑝 𝑜𝑟 𝑞
4. 𝑛𝑜𝑡 𝑞 𝑎𝑛𝑑 (𝑖𝑓 (𝑖𝑓 𝑝, 𝑡ℎ𝑒𝑛 𝑞), 𝑡ℎ𝑒𝑛 𝑛𝑜𝑡 𝑝)
5. 𝑖𝑓 𝑃, 𝑡ℎ𝑒𝑛 (𝑖𝑓 𝑞, 𝑡ℎ𝑒𝑛 𝑝)
6. 𝑞 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑞
▶

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Problem: Rewrite the following logical statements into symbolic notation and construct a truth table to determine whether each is a tautology, absurdity, or contingency.

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1. (if p, then not p) or (if p, then q)

Symbolic notation: \((p \to \neg p) \lor (p \to q)\)

p	q	p → ¬p	p → q	(p → ¬p) ∨ (p → q)
T	T	F	T	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

Classification: Contingency

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2. if (p and ¬p), then r

Symbolic notation: \((p \land \neg p) \to r\)

p	r	p ∧ ¬p	(p ∧ ¬p) → r
T	T	F	T
T	F	F	T
F	T	F	T
F	F	F	T

Classification: Tautology

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3. if p, then (p or q)

Symbolic notation: \(p \to (p \lor q)\)

p	q	p ∨ q	p → (p ∨ q)
T	T	T	T
T	F	T	T
F	T	T	T
F	F	F	T

Classification: Tautology

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4. ¬q and (if (if p then q), then ¬p)

Symbolic notation: \(\neg q \land ((p \to q) \to \neg p)\)

p	q	¬q	p → q	(p → q) → ¬p	¬q ∧ ((p → q) → ¬p)
T	T	F	T	F	F
T	F	T	F	T	T
F	T	F	T	T	F
F	F	T	T	T	T

Classification: Contingency

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5. if p, then (if q, then p)

Symbolic notation: \(p \to (q \to p)\)

p	q	q → p	p → (q → p)
T	T	T	T
T	F	T	T
F	T	F	T
F	F	T	T

Classification: Tautology

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6. q and not q

Symbolic notation: \(q \land \neg q\)

q	¬q	q ∧ ¬q
T	F	F
F	T	F

Classification: Absurdity

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