Chapter 1 Numbers, Quantification and Numerical Applications (Concepts)

Modulo Arithmetic

Modulo Arithmetic is a system of arithmetic for integers, often referred to as "clock arithmetic". In this system, numbers "wrap around" upon reaching a certain value called the modulus. This concept is fundamental in various areas, including number theory, computer science (especially in cryptography, hashing, and error detection codes), and even in everyday time calculations (like determining the hour on a clock).

At its core, modulo arithmetic is concerned with remainders obtained from integer division.

The Modulo Operator

The modulo operation, denoted by $a \pmod{m}$, calculates the remainder when an integer $a$ (the dividend) is divided by a positive integer $m$ (the modulus).

For any integer $a$ and any positive integer $m$, the process of integer division states that there exist unique integers $q$ (the quotient) and $r$ (the remainder) such that:

and the remainder $r$ must satisfy the condition:

The result of the modulo operation $a \pmod m$ is this unique remainder $r$.

In simpler terms, to find $a \pmod m$, you divide $a$ by $m$, and the result is the remainder, ensuring that the remainder is never negative and always less than the modulus $m$.

Examples of Modulo Calculation

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Properties of Modulo Arithmetic

Modulo arithmetic operations (addition, subtraction, multiplication) behave in a way that makes calculations simpler, especially with large numbers. The key principle is that if you replace a number with its remainder modulo $m$, the result of the operation modulo $m$ remains the same.

Let $a$ and $b$ be any integers, and $m$ be a positive integer. Let $a_m = a \pmod m$ and $b_m = b \pmod m$ be their respective remainders when divided by $m$.

1. Addition Property:

The sum of two integers modulo $m$ is equivalent to the sum of their remainders modulo $m$.

Derivation:

By the definition of modulo, $a = q_a m + a_m$ and $b = q_b m + b_m$ for some integers $q_a, q_b$, where $0 \leq a_m, b_m < m$.

Adding $a$ and $b$:

Let $Q = q_a + q_b$. This is an integer. So, $a + b = Qm + (a_m + b_m)$.

This equation shows that when $(a+b)$ is divided by $m$, the remainder is the same as the remainder when $(a_m + b_m)$ is divided by $m$.

2. Subtraction Property:

The difference of two integers modulo $m$ is equivalent to the difference of their remainders modulo $m$.

Derivation:

Using $a = q_a m + a_m$ and $b = q_b m + b_m$:

Subtracting $b$ from $a$:

Let $Q = q_a - q_b$. This is an integer. So, $a - b = Qm + (a_m - b_m)$.

This equation shows that when $(a-b)$ is divided by $m$, the remainder is the same as the remainder when $(a_m - b_m)$ is divided by $m$.

3. Multiplication Property:

The product of two integers modulo $m$ is equivalent to the product of their remainders modulo $m$.

Derivation:

Using $a = q_a m + a_m$ and $b = q_b m + b_m$:

Multiplying $a$ and $b$:

Let $Q = q_a q_b m + q_a b_m + a_m q_b$. This is an integer. So, $a \times b = Qm + a_m b_m$.

This equation shows that when $(a \times b)$ is divided by $m$, the remainder is the same as the remainder when $(a_m \times b_m)$ is divided by $m$.

4. Exponentiation Property:

For any integer $a$, a positive integer $m$, and a non-negative integer $k$, the property for exponentiation modulo $m$ is:

This means you can take the modulo before raising to the power. This property is particularly useful for calculating powers of large numbers modulo $m$ without having to compute the large power first. You can repeatedly apply the multiplication property.

Example: Find $7^{10} \pmod{5}$.

Calculating $7^{10}$ is $282475249$. $282475249 \pmod 5 = 4$ (since the last digit is 9, and $9 \pmod 5 = 4$).

Using the property: We can calculate $7 \pmod 5$ first.

So, $7^{10} \pmod 5 = (7 \pmod 5)^{10} \pmod 5 = 2^{10} \pmod 5$.

Now we calculate powers of 2 modulo 5:

The remainders repeat in a cycle of length 4: 2, 4, 3, 1.

To find $2^{10} \pmod 5$, we look at the exponent 10. The position in the cycle is $10 \pmod 4 = 2$. The second value in the cycle is 4.

Thus, $7^{10} \pmod 5 = 4$. This method avoids calculating the large number $7^{10}$.

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Congruence Modulo

Building upon the concept of modulo arithmetic, Congruence Modulo m provides a formal way to describe the relationship between two integers that have the same remainder when divided by a specific positive integer $m$. This idea is foundational in number theory and has numerous applications in fields where patterns repeat cyclically.

Definition of Congruence Modulo m

Two integers $a$ and $b$ are said to be congruent modulo m, where $m$ is a positive integer (called the modulus), if they have the same remainder when divided by $m$.

Mathematically, this relationship is denoted as:

This notation is read as "$a$ is congruent to $b$ modulo $m$".

An equivalent and often more useful definition of congruence modulo $m$ is that $a$ and $b$ are congruent modulo $m$ if and only if their difference, $(a - b)$, is divisible by $m$.

Here, $m \mid (a - b)$ means that $(a - b)$ is a multiple of $m$, i.e., $a - b = k \times m$ for some integer $k$.

Equivalence of the Definitions:

Let's show that the two definitions are equivalent:

Part 1: If $a \equiv b \pmod m$ (same remainder), then $m \mid (a - b)$.

If $a$ and $b$ have the same remainder $r$ when divided by $m$, then by the definition of the modulo operation (Formula (i) from the previous section on Modulo Arithmetic), we can write:

Subtracting the second equation from the first:

Since $q_1$ and $q_2$ are integers, $q_1 - q_2$ is also an integer. Let $k = q_1 - q_2$.

This means that $a - b$ is a multiple of $m$, which is the definition of $m \mid (a - b)$.

Part 2: If $m \mid (a - b)$, then $a \equiv b \pmod m$ (same remainder).

If $m \mid (a - b)$, then $a - b = km$ for some integer $k$. This can be written as $a = b + km$.

Let the remainder when $b$ is divided by $m$ be $r_b$. So, $b = q_b m + r_b$, where $0 \leq r_b < m$.

Substitute this expression for $b$ into the equation for $a$:

Let $Q = q_b + k$. This is an integer. So, $a = Qm + r_b$.

Since $0 \leq r_b < m$, by the uniqueness of the remainder in integer division, $r_b$ is the remainder when $a$ is divided by $m$.

Thus, $a$ and $b$ have the same remainder $r_b$ when divided by $m$. By definition, $a \equiv b \pmod m$.

The two definitions are equivalent.

Relationship with the Modulo Operation

The definition of congruence modulo $m$ is directly related to the result of the modulo operation.

$a \equiv b \pmod m$ if and only if $a \pmod m = b \pmod m$.

This is simply restating the first definition using the result of the modulo operation.

Example: $17 \pmod 5 = 2$ and $12 \pmod 5 = 2$. Since their remainders are the same, $17 \equiv 12 \pmod 5$.

Let's check using the difference definition: $17 - 12 = 5$. Is 5 divisible by 5? Yes, $5 = 1 \times 5$. So, $5 \mid (17 - 12)$, confirming $17 \equiv 12 \pmod 5$.

Properties of Congruence Modulo m

Congruence modulo $m$ defines a relationship between integers. This relationship is an equivalence relation on the set of integers, meaning it satisfies the following three properties for any integers $a, b, c$ and a positive integer $m$:

1. Reflexive Property:

Any integer $a$ is congruent to itself modulo $m$.

This is true because $a - a = 0$. Since $m > 0$, $0$ is divisible by $m$ ($0 = 0 \times m$). By the definition $m \mid (a-b)$, we have $m \mid (a-a)$, so $a \equiv a \pmod m$.

2. Symmetric Property:

If $a$ is congruent to $b$ modulo $m$, then $b$ is congruent to $a$ modulo $m$.

Proof:

3. Transitive Property:

If $a$ is congruent to $b$ modulo $m$, and $b$ is congruent to $c$ modulo $m$, then $a$ is congruent to $c$ modulo $m$.

Proof:

Since congruence modulo $m$ is reflexive, symmetric, and transitive, it partitions the set of integers into disjoint equivalence classes. Each class consists of all integers that have the same remainder when divided by $m$.

Arithmetic Properties of Congruences

Congruences behave well with respect to addition, subtraction, and multiplication. These properties are direct consequences of the properties of the modulo operator discussed in the previous section.

If $a, b, c, d$ are integers and $m$ is a positive integer such that $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then:

1. Addition Property:

The sum of congruent numbers is congruent modulo $m$.

This follows from $(a+c) \pmod m = ((a \pmod m) + (c \pmod m)) \pmod m$ and $(b+d) \pmod m = ((b \pmod m) + (d \pmod m)) \pmod m$. Since $a \pmod m = b \pmod m$ and $c \pmod m = d \pmod m$, their sums modulo $m$ will be the same.

2. Subtraction Property:

The difference of congruent numbers is congruent modulo $m$.

This follows similarly from the subtraction property of the modulo operator.

3. Multiplication Property:

The product of congruent numbers is congruent modulo $m$.

Proof (using the difference definition):

4. Exponentiation Property:

If $a$ is congruent to $b$ modulo $m$, then $a$ raised to the power of any positive integer $k$ is congruent to $b$ raised to the power of the same integer $k$, modulo $m$.

This property can be formally proven using mathematical induction, based on the multiplication property.

Base case ($k=1$): $a^1 \equiv b^1 \pmod m$, which is $a \equiv b \pmod m$. This is given.

Inductive step: Assume $a^k \equiv b^k \pmod m$ is true for some positive integer $k$. We need to show $a^{k+1} \equiv b^{k+1} \pmod m$.

We know $a \equiv b \pmod m$ (given) and $a^k \equiv b^k \pmod m$ (inductive hypothesis).

Using the Multiplication Property with $(a, b)$ and $(a^k, b^k)$, we have $a \cdot a^k \equiv b \cdot b^k \pmod m$.

This simplifies to $a^{k+1} \equiv b^{k+1} \pmod m$.

By the principle of mathematical induction, the property holds for all positive integers $k$.

Examples Using Congruence Properties

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Alligation and Mixture

In Applied Mathematics, problems involving the mixing of two or more substances with different properties (like price, concentration, density, etc.) to produce a mixture with a desired average property are commonly encountered. Alligation is a technique or rule that helps solve a specific type of mixture problem, particularly useful for finding the ratio in which two ingredients with known prices or concentrations should be mixed to obtain a mixture of a desired mean price or concentration. The term Mixture refers to the result of combining two or more distinct ingredients.

The Rule of Alligation

The Rule of Alligation provides a quick method to determine the ratio of the quantities of two ingredients that need to be mixed to obtain a mixture with a specific mean value of a property (like cost per unit, percentage concentration, etc.).

Suppose we have two ingredients:

• Ingredient 1: A cheaper ingredient with cost (or value of property) $C_1$ per unit.
• Ingredient 2: A dearer ingredient with cost (or value of property) $C_2$ per unit.

We want to mix these two ingredients to form a mixture with a mean cost (or value of property) $C_m$ per unit. It is always assumed that $C_1 < C_m < C_2$.

The rule of alligation states that the ratio of the quantity of the cheaper ingredient to the quantity of the dearer ingredient in the mixture is given by the ratio of the absolute differences between the mean value and the individual values, taken diagonally:

This rule is often visualized using the "alligation cross" or "alligation diagram":

The values $(C_2 - C_m)$ and $(C_m - C_1)$ represent the parts of the cheaper and dearer ingredients, respectively, in the ratio of mixing. The value $(C_2 - C_m)$ is placed diagonally opposite to $C_1$, and $(C_m - C_1)$ is placed diagonally opposite to $C_2$.

This rule is based on the principle of weighted averages.

Derivation of the Rule of Alligation

Let $Q_1$ be the quantity of the cheaper ingredient (with value $C_1$) and $Q_2$ be the quantity of the dearer ingredient (with value $C_2$).

When these quantities are mixed, the total quantity of the mixture is $Q_1 + Q_2$.

The total "value" (e.g., total cost, total amount of pure substance) in the mixture is the sum of the values contributed by each ingredient:

The mean value of the mixture, $C_m$, is the total value divided by the total quantity:

Our goal is to find the ratio $\frac{Q_1}{Q_2}$. Let's rearrange the equation to isolate the terms involving $Q_1$ and $Q_2$.

Multiply both sides by $(Q_1 + Q_2)$:

Distribute $C_m$ on the left side:

Group terms involving $Q_1$ on one side and terms involving $Q_2$ on the other side:

Factor out $Q_2$ on the left side and $Q_1$ on the right side:

Since $C_1 < C_m < C_2$, the terms $(C_m - C_2)$ and $(C_1 - C_m)$ are negative. To get positive differences as shown in the alligation diagram, we can rewrite the equation as:

Divide both sides by $-1$ and then rearrange to find the ratio $\frac{Q_1}{Q_2}$:

This derived ratio matches the Rule of Alligation. It's important that the values $C_1$, $C_2$, and $C_m$ refer to the same property (e.g., all per-unit costs, or all percentage concentrations of the same substance).

Mixing More Than Two Ingredients

The basic rule of alligation is directly applicable only when mixing two ingredients. When more than two ingredients are mixed, the simplest approach is to use the concept of the weighted average directly.

If quantities $Q_1, Q_2, \dots, Q_n$ of $n$ ingredients with values $C_1, C_2, \dots, C_n$ are mixed, the mean value $C_m$ of the mixture is:

If you are given a desired mean value $C_m$ and the values $C_i$ for multiple ingredients, and you need to find the quantities or ratios, you can sometimes pair ingredients (one cheaper than the mean, one dearer than the mean) and apply the rule of alligation pairwise, or use algebraic methods based on the weighted average formula.

Problems Involving Removing and Replacing a Mixture

A specific type of mixture problem involves starting with a vessel full of a pure liquid, removing a certain amount of the liquid, and replacing it with another liquid (usually water). This process is then repeated one or more times. We often need to find the quantity of the original liquid remaining in the vessel after several such operations.

Suppose a vessel initially contains $V$ units of a pure liquid. In each operation, $x$ units of the current mixture are drawn out and replaced with $x$ units of another liquid (e.g., water). This process is repeated $n$ times.

Let's track the quantity of the original liquid.

• Initially (before 1st operation): Quantity of original liquid = $V$. Total quantity = $V$. Concentration of original liquid = $\frac{V}{V} = 1$.

• After 1st operation: $x$ units of milk are removed. Quantity of original liquid removed = $x$. Quantity of original liquid remaining = $V - x$. $x$ units of water are added. Total quantity = $(V-x) + x = V$. Concentration of original liquid = $\frac{V-x}{V} = 1 - \frac{x}{V}$.

• After 2nd operation: $x$ units of the mixture are removed. The concentration of the original liquid in this mixture is $1 - \frac{x}{V}$. Quantity of original liquid removed = $x \times \left(1 - \frac{x}{V}\right)$. Quantity of original liquid remaining = $(V - x) - x \left(1 - \frac{x}{V}\right)$.
  A simpler way is to consider the fraction of original liquid remaining. After the first operation, the fraction is $1 - \frac{x}{V}$. In the second operation, $x$ units of the *mixture* are removed. These $x$ units contain $x \times \left(1 - \frac{x}{V}\right)$ of the original liquid. The remaining quantity of original liquid is $V - x \left(1 - \frac{x}{V}\right)$. The fraction of original liquid remaining in the $V-x$ part of the mixture is still $1 - \frac{x}{V}$. When $x$ units of water are added back, the total volume is $V$. The quantity of original liquid is what was left *before* adding water, $(V-x) \times (1-\frac{x}{V})$.

  Let $V_k$ be the quantity of original liquid after $k$ operations. $V_0 = V$. After 1st op: Quantity removed = $x$. Remaining Original = $V-x$. Quantity water added = $x$. Total volume = $V$. After 2nd op: Quantity of original liquid removed = $x \times (\text{concentration after 1st op}) = x \times \frac{V-x}{V}$. Remaining Original = $(V-x) - x \frac{V-x}{V} = (V-x)\left(1 - \frac{x}{V}\right) = V\left(1 - \frac{x}{V}\right)\left(1 - \frac{x}{V}\right) = V\left(1 - \frac{x}{V}\right)^2$.

Following this pattern, after $n$ such operations, the quantity of the original liquid remaining in the container is given by the formula:

Where:

• $V$ is the initial volume of the pure liquid.
• $x$ is the volume of mixture removed and replaced in each operation.
• $n$ is the total number of times the operation of removing and replacing is performed.

The quantity of water in the container after $n$ operations would be $V - V \left(1 - \frac{x}{V}\right)^n$.

Examples

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Numerical Problems

Applied Mathematics often involves solving quantitative problems from various real-world scenarios. This section focuses on revisiting and applying fundamental concepts from arithmetic and algebra to solve numerical problems related to percentages, profit and loss, and simple and compound interest. These topics are essential for developing numerical aptitude and are widely applicable in personal finance, business, and other quantitative fields.

Percentages

The term Percentage means "out of one hundred". It is a way of expressing a number as a fraction of 100. Percentages are used to represent parts of a whole, proportions, or relative changes.

A percentage is denoted by the symbol '%'. If we have $x\%$, it means $\frac{x}{100}$.

Example: $25\% = \frac{25}{100} = 0.25$. $150\% = \frac{150}{100} = 1.5$.

Conversion between Percentage, Fraction, and Decimal:

• Percentage to Fraction/Decimal: Divide the percentage by 100.

  $p\% = \frac{p}{100}$ (Fraction)

  $\text{p}\% = \frac{p}{100}$ (Decimal)

• Fraction/Decimal to Percentage: Multiply the fraction or decimal by 100 and add the '%' symbol.

  $\text{Fraction } F \implies (F \times 100)\%$

  $\text{Decimal } D \implies (D \times 100)\%$

Percentage Increase or Decrease:

To find the percentage change in a value, we compare the amount of change to the original value and express it as a percentage.

Percentage Increase:

If a value increases from an Original Value to a New Value, the Increase is (New Value - Original Value).

Percentage Decrease:

If a value decreases from an Original Value to a New Value, the Decrease is (Original Value - New Value).

Finding a Value After Percentage Change:

If a value undergoes a percentage increase or decrease, the new value can be directly calculated.

Value after $x\%$ Increase:

New Value = Original Value + Increase Amount

Increase Amount = Original Value $\times \frac{x}{100}$

Value after $x\%$ Decrease:

New Value = Original Value - Decrease Amount

Decrease Amount = Original Value $\times \frac{x}{100}$

These formulas are useful for calculating prices after discounts, taxes, or growth/decay over one period.

Profit and Loss

Profit and Loss are terms used in business and commerce to describe the financial outcome of buying and selling goods or services. They are calculated based on the relationship between the cost incurred to acquire or produce an item (Cost Price) and the price at which it is sold (Selling Price).

• Cost Price (CP): The amount for which an article is bought. This may include overhead expenses like transportation, labour, etc.

• Selling Price (SP): The amount for which an article is sold.

Profit or Gain:

A profit is made when the Selling Price (SP) is greater than the Cost Price (CP).

Profit is often expressed as a percentage of the Cost Price.

Loss:

A loss is incurred when the Selling Price (SP) is less than the Cost Price (CP).

Loss is often expressed as a percentage of the Cost Price.

Important Note: Unless explicitly stated otherwise, Profit or Loss percentage is always calculated with respect to the Cost Price (CP).

Relating Selling Price and Cost Price:

We can express the Selling Price directly in terms of the Cost Price and the Profit or Loss percentage.

If there is a Profit of $P\%$ ($P$ is the percentage value, e.g., 15):

If there is a Loss of $L\%$ ($L$ is the percentage value):

Simple and Compound Interest

Interest is the cost of borrowing money or the return on investment. We covered the basic concepts (Principal, Rate, Time, Amount) and the calculation methods (Simple vs. Compound Interest) in detail in Chapter 11 on Financial Mathematics. Here, we'll recall the key formulas for numerical problems.

Let $P$ be the Principal amount.

Let $R$ be the Nominal Annual Rate of Interest (as a percentage, e.g., 10 for 10%).

Let $T$ be the Time period in years.

Let $n$ be the number of times interest is compounded per year.

Let $A$ be the Total Amount (Principal + Interest).

Simple Interest (SI):

Interest is calculated only on the original principal.

Compound Interest (CI):

Interest is calculated on the principal plus accumulated interest.

If interest is compounded annually ($n=1$):

If interest is compounded $n$ times a year:

• Rate per compounding period (as percentage) = $R/n$.
• Total number of compounding periods = $T \times n$.

Difference between CI and SI:

For the same principal ($P$), rate ($R\%$ p.a.), and time ($T$ years), the compound interest is generally higher than simple interest (except for $T=1$). The difference arises because compound interest earns interest on previous interest.

Let's derive the difference for 2 and 3 years (compounded annually):

Difference (CI - SI) for 2 years:

Difference (CI - SI) for 3 years:

These difference formulas are helpful for quickly finding the difference between CI and SI for 2 or 3 years given P and R.

Examples

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Boats and Streams (upstream and downstream)

The concept of Boats and Streams deals with problems involving motion in water, where the speed of the water current affects the effective speed of a boat or a swimmer. These problems are applications of relative speed principles. The key idea is to understand how the speed of the stream either assists or opposes the motion of the boat or swimmer.

Definitions

To analyze problems involving boats and streams, we use the following terms:

• Speed of the boat or swimmer in still water ($v_b$): This is the speed at which the boat would travel or the swimmer would swim if there were no water current. It is the inherent speed of the boat or swimmer relative to the water itself. The units are typically km/hr or m/sec.

• Speed of the stream or current ($v_s$): This is the speed at which the water is flowing. It represents the speed of the water relative to the ground. The units are typically the same as $v_b$ (km/hr or m/sec).

• Downstream: Refers to the direction in which the boat or swimmer moves along with the flow of the stream. The motion is in the same direction as the current.

• Upstream: Refers to the direction in which the boat or swimmer moves against the flow of the stream. The motion is in the opposite direction to the current.

Effective Speeds

When a boat or swimmer is in a flowing stream, their speed relative to the ground is affected by the speed of the stream. We consider the effective speed in the downstream and upstream directions.

1. Speed Downstream ($v_d$):

When the boat or swimmer moves downstream, the speed of the stream adds to their speed in still water. The water current helps them move faster relative to the ground.

2. Speed Upstream ($v_u$):

When the boat or swimmer moves upstream, the speed of the stream opposes their motion. The water current pushes against them, slowing them down relative to the ground.

For upstream motion to be possible (i.e., for the boat/swimmer to make progress against the current), the speed of the boat or swimmer in still water ($v_b$) must be greater than the speed of the stream ($v_s$). If $v_b \leq v_s$, the boat or swimmer will be carried downstream by the current, even when attempting to move upstream.

Finding Speed in Still Water and Speed of Stream

If we are given the effective speeds downstream ($v_d$) and upstream ($v_u$), we can use the two equations above to find the speed of the boat/swimmer in still water ($v_b$) and the speed of the stream ($v_s$). We have a system of two linear equations:

Equation (i): $v_d = v_b + v_s$

Equation (ii): $v_u = v_b - v_s$

To find $v_b$: Add Equation (i) and Equation (ii).

The speed of the boat/swimmer in still water is the average of the downstream and upstream speeds.

To find $v_s$: Subtract Equation (ii) from Equation (i).

The speed of the stream is half the difference between the downstream and upstream speeds.

Relating Distance, Speed, and Time

Problems in this topic often involve calculating distances, speeds, or times using the fundamental relationship:

This relationship can be rearranged to find speed (Speed = Distance / Time) or time (Time = Distance / Speed).

When applying this relationship in boats and streams problems, always use the effective speed for the direction of travel:

• For Downstream Motion: If distance traveled downstream is $D_d$ and time taken is $T_d$, then $D_d = v_d \times T_d = (v_b + v_s) \times T_d$.

• For Upstream Motion: If distance traveled upstream is $D_u$ and time taken is $T_u$, then $D_u = v_u \times T_u = (v_b - v_s) \times T_u$.

Examples

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Pipes and Cisterns

Pipes and Cisterns is a common topic in numerical problems, which is essentially an application of the concepts used in "Time and Work". Instead of individuals working on a task, the "workers" are pipes (or faucets, taps, etc.), and the "task" is filling or emptying a container, typically a tank or cistern.

The core idea remains the same: determine the rate at which each pipe performs its task and then combine these rates to find the rate when multiple pipes work together.

Basic Concepts

In Pipes and Cisterns problems:

• Total Work: The total work is usually considered as filling or emptying the entire cistern. This can be represented as 1 unit of work or, more abstractly, as the total capacity of the cistern.

• Rate of Work (Efficiency): The rate of a pipe is the amount of the cistern that can be filled or emptied by that pipe in one unit of time (e.g., per hour, per minute, per second). If a pipe completes the total work (fills or empties the entire cistern) in a certain amount of time, its rate is the reciprocal of that time.

1. Filling Pipes (Inlet Pipes):

These are pipes that add liquid to the cistern, hence filling it. If a filling pipe can fill the entire cistern in $T_{\text{fill}}$ units of time, its rate of filling is $\frac{1}{T_{\text{fill}}}$ of the cistern per unit of time. The work done by a filling pipe is considered positive.

2. Emptying Pipes (Outlet Pipes or Leaks):

These are pipes or leaks that remove liquid from the cistern, hence emptying it. If an emptying pipe can empty the entire cistern in $T_{\text{empty}}$ units of time, its rate of emptying is $\frac{1}{T_{\text{empty}}}$ of the cistern per unit of time. The work done by an emptying pipe is considered negative.

It is assumed that the rate of flow of liquid through a pipe is constant unless stated otherwise.

Pipes Working Together

When multiple pipes are operating simultaneously, their individual rates of filling or emptying are combined to find the net effect on the cistern. The combined rate is the algebraic sum of their individual rates.

Let there be $k$ pipes, $P_1, P_2, \dots, P_k$, connected to a cistern. Let their individual rates be $r_1, r_2, \dots, r_k$ respectively, where $r_i = \frac{1}{T_i}$ if $P_i$ can fill or empty the cistern in $T_i$ time units. $r_i$ is positive for filling pipes and negative for emptying pipes.

The Combined Rate ($r_{\text{combined}}$) when all pipes work together is:

If $r_{\text{combined}}$ is positive, the cistern will fill. If $r_{\text{combined}}$ is negative, the cistern will empty (assuming it initially contained liquid).

The time taken to fill or empty the entire cistern (1 unit of work) is the reciprocal of the combined rate:

Specific Cases:

• Two Filling Pipes: If pipe A fills in $T_A$ hours and pipe B fills in $T_B$ hours.

  Rate of A = $\frac{1}{T_A}$, Rate of B = $\frac{1}{T_B}$.

  Combined Rate = $\frac{1}{T_A} + \frac{1}{T_B} = \frac{T_B + T_A}{T_A T_B}$.

  Time taken together to fill = $\frac{1}{\text{Combined Rate}} = \frac{T_A T_B}{T_A + T_B}$.

• One Filling Pipe and One Emptying Pipe: If pipe A fills in $T_A$ hours and pipe B empties in $T_B$ hours. For the tank to fill, the filling rate must be greater than the emptying rate, which implies $T_A < T_B$.

  Rate of A = $\frac{1}{T_A}$, Rate of B = $-\frac{1}{T_B}$.

  Combined Rate = $\frac{1}{T_A} + \left(-\frac{1}{T_B}\right) = \frac{1}{T_A} - \frac{1}{T_B} = \frac{T_B - T_A}{T_A T_B}$.

  Time taken together to fill = $\frac{1}{\text{Combined Rate}} = \frac{T_A T_B}{T_B - T_A}$ (assuming $T_A < T_B$).

  If $T_A > T_B$, the emptying pipe is faster, the tank will empty, and the time to empty would be $\frac{T_A T_B}{T_A - T_B}$ (using the magnitude of the negative combined rate).

Examples

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Races and Games

The topic of Races and Games in numerical problems deals with comparing the performance of participants in competitive scenarios. These problems often involve concepts of speed, distance, and time, or comparing the abilities of players based on points or scores. They require understanding relative speeds, interpreting scenarios where one participant gives a "start" to another, or comparing performances over different distances or formats.

Races

Problems related to races usually involve participants running, swimming, or rowing over a fixed distance. Comparing how participants finish relative to each other (who wins, by what margin) or how they perform when given a head start are common themes.

1. Giving a Start:

When one participant "gives a start" to another, it means the slower participant is given an advantage before the race begins.

• Start in Distance: If participant A gives participant B a start of $x$ meters in a race of $L$ meters, it implies that B starts $x$ meters ahead of the starting line. Effectively, A has to cover the full distance $L$ meters, while B only has to cover $(L - x)$ meters from their starting position to the finish line. If they start at the same time, the time taken by A to cover $L$ meters is equal to the time taken by B to cover $(L - x)$ meters.

• Start in Time: If participant A gives participant B a start of $t$ seconds, it means B starts running $t$ seconds before A. If they both cover the full distance $L$, A will take $t$ seconds less than B to complete the race. If B takes $T_B$ seconds, A takes $T_B - t$ seconds.

2. Winning a Race:

Describing how one participant wins a race relative to another provides information about their speeds.

• Winning by a Distance: If participant A beats participant B by $x$ meters in a race of $L$ meters (assuming they start at the same time from the same starting line), it means when A finishes the race (covers $L$ meters), B is $x$ meters behind the finish line. In the same amount of time that A covers $L$ meters, B covers $(L - x)$ meters.

• Winning by a Time: If participant A beats participant B by $t$ seconds in a race of $L$ meters (assuming they start at the same time from the same starting line), it means A finishes the race $t$ seconds earlier than B. If A takes $T_A$ seconds to cover $L$ meters, B takes $(T_A + t)$ seconds to cover the same $L$ meters.

3. Comparing Speeds:

The relationship between distance, speed, and time ($D = S \times T$) is fundamental. When comparing participants in a race:

• For the same distance (D): If participants A and B cover the same distance $D$, and their speeds are $S_A, S_B$ and times taken are $T_A, T_B$, then $D = S_A \times T_A = S_B \times T_B$.

  This implies that the ratio of their speeds is the inverse of the ratio of their times:

  $\frac{S_A}{S_B} = \frac{T_B}{T_A}$

  ... (i)

• For the same time (T): If participants A and B run for the same amount of time $T$, and their speeds are $S_A, S_B$ and distances covered are $D_A, D_B$, then $D_A = S_A \times T$ and $D_B = S_B \times T$.

  This implies that the ratio of their speeds is the same as the ratio of the distances covered:

  $\frac{S_A}{S_B} = \frac{D_A}{D_B}$

  ... (ii)

Often, problems combine these ideas, requiring you to find the ratio of speeds between multiple participants based on their performance in one race and then apply that ratio to a different scenario or distance.

Games of Skill

Problems categorized as "Games" often involve comparing the scoring abilities of players in a game played to a certain number of points.

If player A can give player B $x$ points in a game of $P$ points, it means that in a game where A scores $P$ points, B only manages to score $(P - x)$ points. The ratio of their scoring abilities (points scored in the same duration or 'game') is $P : (P - x)$.

If A can beat B by $t$ seconds in a game, this relates to the time taken to score points, which can be converted to a scoring rate.

Examples

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Numerical Inequalities

Numerical inequalities are fundamental mathematical statements that compare two expressions using inequality symbols. Unlike equations which state that two expressions are equal, inequalities state that one quantity is greater than ($>$), less than ($<$), greater than or equal to ($\ge$), or less than or equal to ($\le$) another quantity. Understanding and solving inequalities are crucial for describing ranges of values, solving problems involving constraints, and working with various mathematical concepts in Applied Mathematics.

Inequality Symbols

The basic symbols used to denote inequalities are essential vocabulary:

Inequalities involving $>$ or $<$ are called strict inequalities, while those involving $\ge$ or $\le$ are called non-strict inequalities.

Properties of Inequalities

Manipulating inequalities requires understanding their properties, which dictate how the inequality sign behaves under various operations. Let $a, b, c$ be real numbers.

Addition Property

Adding the same real number $c$ to both sides of an inequality does not change the direction of the inequality sign. This is because adding a constant shifts both sides by the same amount, maintaining their relative order.

If $a > b$, then $a + c > b + c$.

If $a < b$, then $a + c < b + c$.

Similarly, if $a \ge b$, then $a + c \ge b + c$, and if $a \le b$, then $a + c \le b + c$.

Example: If $x - 5 > 2$, adding 5 to both sides gives $(x - 5) + 5 > 2 + 5$, which simplifies to $x > 7$. The inequality direction remains the same.

Subtraction Property

Subtracting the same real number $c$ from both sides of an inequality does not change the direction of the inequality sign. This is equivalent to adding $-c$.

If $a > b$, then $a - c > b - c$.

If $a < b$, then $a - c < b - c$.

This property also holds for $\ge$ and $\le$.

Example: If $y + 3 \le 8$, subtracting 3 from both sides gives $(y + 3) - 3 \le 8 - 3$, which simplifies to $y \le 5$. The inequality direction remains the same.

Multiplication and Division Properties

These properties are crucial and depend entirely on the sign of the number you are multiplying or dividing by.

Case 1: Multiplying or dividing by a positive number ($c > 0$).
If you multiply or divide both sides of an inequality by the same positive real number $c$, the direction of the inequality sign remains the same. This is because scaling both sides by a positive factor maintains their relative proportion.

If $a > b$ and $c > 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$.

If $a < b$ and $c > 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$.

This property also holds for $\ge$ and $\le$.

Example: If $2x < 10$, dividing by 2 (a positive number) gives $\frac{2x}{2} < \frac{10}{2}$, which simplifies to $x < 5$. The inequality direction remains the same.

Case 2: Multiplying or dividing by a negative number ($c < 0$).
If you multiply or divide both sides of an inequality by the same negative real number $c$, the direction of the inequality sign must be reversed. This is because multiplying or dividing by a negative number reflects the numbers across zero on the number line, thus reversing their order.

If $a > b$ and $c < 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$.

If $a < b$ and $c < 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$.

This property also holds for $\ge$ and $\le$ (they become $\le$ and $\ge$ respectively).

Example: If $-3x \ge 9$, dividing by -3 (a negative number) requires reversing the inequality sign: $\frac{-3x}{-3} \le \frac{9}{-3}$, which simplifies to $x \le -3$. The inequality direction is reversed.

Case 3: Multiplying or dividing by zero ($c = 0$).
Multiplication by zero on both sides results in $0 > 0$, $0 < 0$, $0 \ge 0$, or $0 \le 0$, which are either false or don't help in finding the value of a variable. Division by zero is undefined. Therefore, these operations are not used when solving inequalities involving variables.

Reciprocal Property

Taking the reciprocal ($\frac{1}{x}$) of both sides of an inequality requires careful consideration of the signs of the expressions involved.

If $a$ and $b$ are both positive or both negative (i.e., they have the same sign and are non-zero):
If $a > b$, then $\frac{1}{a} < \frac{1}{b}$. The direction of the inequality is reversed.

Examples:
If $x > 2$ (where $x$ is positive), then $\frac{1}{x} < \frac{1}{2}$.
If $y < -1$ (where $y$ is negative), then $\frac{1}{y} > \frac{1}{-1}$, i.e., $\frac{1}{y} > -1$.

If $a$ and $b$ have opposite signs (one positive, one negative):
If $a > b$, then $\frac{1}{a} > \frac{1}{b}$. The direction of the inequality remains the same. (This is because $a > 0$ and $b < 0$, so $\frac{1}{a} > 0$ and $\frac{1}{b} < 0$. Any positive number is greater than any negative number).

Example: If $x > -4$, where we know $x$ can be positive or negative. If we consider $x > 0$ and $x < 0$ separately, it becomes complex. The reciprocal property for opposite signs applies when comparing a positive and a negative number directly, e.g., $2 > -3 \implies \frac{1}{2} > \frac{1}{-3}$.

Important Note: When dealing with inequalities involving variables where the sign is unknown (e.g., $\frac{1}{x} < 5$), you cannot simply take reciprocals or multiply by the variable without considering cases for the variable being positive, negative, or zero (which is excluded). It's often safer to move all terms to one side and find a common denominator or multiply by the square of the denominator (which is always positive, provided the denominator is non-zero).

Solving Linear Inequalities

Solving a linear inequality involves finding the range (or set) of real numbers that make the inequality true. This process uses the properties discussed above to isolate the variable on one side of the inequality sign. The steps are very similar to solving linear equations, with the crucial difference being the potential reversal of the inequality sign when multiplying or dividing by a negative number.

The general approach is to:

1. Simplify both sides of the inequality (remove brackets, combine like terms).
2. Move all terms containing the variable to one side and constant terms to the other using addition or subtraction properties.
3. Isolate the variable by multiplying or dividing by its coefficient. Remember to reverse the inequality sign if you multiply or divide by a negative number.
4. Express the solution in terms of the variable, interval notation, or represent it on a number line.

Answer:

We need to solve the inequality $2x - 3 < 7$.

Add 3 to both sides to isolate the term with $x$:

$2x - 3 + 3 < 7 + 3$

[Using Addition Property]

$2x < 10$

... (i)

Divide both sides by 2. Since 2 is positive, the inequality direction remains unchanged:

$\frac{2x}{2} < \frac{10}{2}$

[Using Multiplication/Division Property ($c>0$)]

$x < 5$

... (ii)

The solution set is $\{x \in \mathbb{R} \mid x < 5\}$. In interval notation, this is $(-\infty, 5)$.

Representation on a number line: An open circle at 5 indicates that 5 is not included. An arrow extends to the left from 5, covering all numbers less than 5.

Answer:

We need to solve the inequality $-4x + 5 \le 17$.

Subtract 5 from both sides:

$-4x + 5 - 5 \le 17 - 5$

[Using Subtraction Property]

$-4x \le 12$

... (i)

Divide both sides by -4. Since -4 is negative, we must reverse the direction of the inequality:

$\frac{-4x}{-4} \ge \frac{12}{-4}$

[Dividing by negative number reverses inequality]

$x \ge -3$

... (ii)

The solution set is $\{x \in \mathbb{R} \mid x \ge -3\}$. In interval notation, this is $[-3, \infty)$.

Representation on a number line: A closed circle (or solid dot) at -3 indicates that -3 is included. An arrow extends to the right from -3, covering all numbers greater than or equal to -3.

Answer:

Answer:

Compound Inequalities

A compound inequality consists of two or more inequalities joined by the words "and" or "or".

"And" inequalities: These represent the intersection of the solution sets of the individual inequalities. A common form is $a < x < b$, which means $a < x$ AND $x < b$. To solve, you can often work with all three parts simultaneously, applying operations to all parts to isolate the variable in the middle. Remember that the order of the numbers in the inequality ($a < b$) must be maintained.

"Or" inequalities: These represent the union of the solution sets of the individual inequalities. To solve, you solve each inequality separately and then combine their solution sets.

Answer:

We need to solve the compound inequality $-5 \le 2x + 1 < 7$. This means we need to find values of $x$ that satisfy both $-5 \le 2x + 1$ AND $2x + 1 < 7$ simultaneously.

We can solve this by performing the same operations on all three parts of the inequality to isolate $x$ in the middle.

Subtract 1 from all three parts:

$-5 - 1 \le 2x + 1 - 1 < 7 - 1$

[Using Subtraction Property]

$-6 \le 2x < 6$

... (i)

Divide all three parts by 2. Since 2 is positive, the inequality directions remain unchanged:

$\frac{-6}{2} \le \frac{2x}{2} < \frac{6}{2}$

[Using Multiplication/Division Property ($c>0$)]

$-3 \le x < 3$

... (ii)

The solution set is $\{x \in \mathbb{R} \mid -3 \le x < 3\}$. In interval notation, this is $[-3, 3)$.

Representation on a number line: A closed circle (or solid dot) at -3 indicates that -3 is included. An open circle at 3 indicates that 3 is not included. A solid line segment connects the two circles, showing that all numbers between -3 and 3 are included.

Answer:

We need to solve the compound inequality $3x - 2 < 4$ or $4x + 5 > 17$. This means we need to find values of $x$ that satisfy either the first inequality or the second inequality (or both).

Solve the first inequality:

$3x - 2 < 4$

Add 2 to both sides:

$3x < 6$

Divide by 3 (positive):

$x < 2$

... (i)

The solution set for the first inequality is $(-\infty, 2)$.

Solve the second inequality:

$4x + 5 > 17$

Subtract 5 from both sides:

$4x > 12$

Divide by 4 (positive):

$x > 3$

... (ii)

The solution set for the second inequality is $(3, \infty)$.

The solution to the compound inequality "$x < 2$ or $x > 3$" is the union of the two individual solution sets.

Solution Set: $\{x \in \mathbb{R} \mid x < 2 \text{ or } x > 3\}$.

In interval notation, this is $(-\infty, 2) \cup (3, \infty)$.

Representation on a number line: An open circle at 2 with an arrow to the left, AND an open circle at 3 with an arrow to the right. The solution consists of the two separate shaded regions.

Intervals

The solution sets of inequalities are typically expressed as intervals on the real number line. Interval notation is a concise way to represent these sets.

Inequality	Interval Notation	Description	Number Line Representation (example)
$a < x < b$	$(a, b)$	Open interval: All real numbers $x$ such that $x$ is strictly between $a$ and $b$. Endpoints $a$ and $b$ are not included.
$a \le x \le b$	$[a, b]$	Closed interval: All real numbers $x$ such that $x$ is between $a$ and $b$, including $a$ and $b$. Endpoints $a$ and $b$ are included.
$a \le x < b$	$[a, b)$	Half-open/half-closed interval: All real numbers $x$ such that $x$ is between $a$ and $b$, including $a$ but not $b$.
$a < x \le b$	$(a, b]$	Half-open/half-closed interval: All real numbers $x$ such that $x$ is between $a$ and $b$, including $b$ but not $a$.
$x > a$	$(a, \infty)$	Unbounded interval: All real numbers $x$ strictly greater than $a$. The interval extends infinitely to the right. $a$ is not included.
$x \ge a$	$[a, \infty)$	Unbounded interval: All real numbers $x$ greater than or equal to $a$. The interval extends infinitely to the right. $a$ is included.
$x < a$	$(-\infty, a)$	Unbounded interval: All real numbers $x$ strictly less than $a$. The interval extends infinitely to the left. $a$ is not included.
$x \le a$	$(-\infty, a]$	Unbounded interval: All real numbers $x$ less than or equal to $a$. The interval extends infinitely to the left. $a$ is included.
All real numbers	$(-\infty, \infty)$ or $\mathbb{R}$	The set of all real numbers. The entire number line.

Parentheses $(\ )$ are always used with $\infty$ or $-\infty$ because they are symbols representing unboundedness, not actual numbers that can be included as endpoints. Parentheses are also used for endpoints in strict inequalities ($>$ or $<$). Square brackets $[\ ]$ are used for finite endpoints that are included in the interval (in non-strict inequalities $\ge$ or $\le$).