Algebra for Officer Cadet: Maths Formulas, Tricks, and Practice Questions

Algebra is an important part of Mathematics that covers Highest Common Factors (HCF), Lowest Common Factors (LCF), algebraic fractions, indices, simultaneous equations, quadratic equations, surds, and radical equations. It also includes simplifying algebraic expressions and solving verbal problems using algebraic equations.

For Officer Cadet Entrance Exam preparation, algebra questions help develop logical thinking, mathematical reasoning, and the ability to apply formulas and solve equations accurately. This resource covers essential algebra formulas, solving methods, useful tricks, and practice questions related to algebraic expressions, fractions, indices, simultaneous equations, quadratic equations, surds, and radical equations. Explore the complete Officer Cadet Exam resources and Maths for Officer Cadet for additional practice and study materials.

What is Algebra for Officer Cadet?

Algebra is a branch of mathematics that uses letters, symbols, and numbers to represent quantities and relationships in a general way, so that problems can be solved even when some values aren’t known. Instead of working only with specific numbers, algebra lets you work with unknowns — usually written as letters like x, y, or a, b — and rules for how they combine and relate to each other.

The Basic Idea

In arithmetic, you might say: “3 + 5 = 8.” In algebra, you generalize this idea: “a + b = b + a” — a statement that’s true no matter what numbers $a$ and $b$ actually are. This shift from specific numbers to general symbols is what makes algebra so powerful: it lets you describe patterns, solve unknown quantities, and build formulas that work in every case, not just one.

Algebra for Officer Cadet: Syllabus

This Mathematics for Officer Cadet Entrance Exam Preparation resource covers the following main topics:
1. HCF, LCM & Algebraic Fractions
2. Indices and Equations Involving Indices
3. Verbal Problems
4. Surds and Radical Equations

1. HCF, LCM & Algebraic Fractions

1.1 HCF (Highest Common Factor)

HCF means Highest Common Factor. When we find the factors of two or more numbers, we often discover that some factors are shared by all of them. These shared factors are called common factors. The greatest factor among these common factors is called the HCF.

Example:
Find the HCF of the numbers 12 and 18.
👉 Factors of 12: 1, 2, 3, 4, 6, 12
👉 Factors of 18: 1, 2, 3, 6, 9, 18
The common factors are 1, 2, 3, and 6. The greatest of these is 6. Therefore, the HCF of 12 and 18 is 6. Thus, 6 is the largest number that divides both 12 and 18 exactly.
The HCF is also known as the Greatest Common Factor (GCF) or Greatest Common Divisor (GCD).

1.2 LCM (Lowest Common Multiple)

LCM means Lowest Common Multiple. When we find the multiples of two or more numbers, we often discover that some multiples are common to all of them. These shared multiples are called common multiples. The smallest multiple among these common multiples is called the LCM.

Example:
Find the LCM of the numbers 12 and 18.
👉 Multiples of 12: 12, 24, 36, 48, 60, 72, …
👉 Multiples of 18: 18, 36, 54, 72, 90, …
The common multiples are 36, 72, … The smallest of these is 36. Therefore, the LCM of 12 and 18 is 36.
Thus, 36 is the smallest number that is exactly divisible by both 12 and 18.
The LCM is also known as the Lowest Common Multiple or the Least Common Multiple.

1.3 HCF and LCM: Formulas and Relationships

Product Relationship:
HCF × LCM = Product of two expressions
HCF(a, b) × LCM(a, b) = a × b
This is always true for exactly two numbers or expressions.
Example:
Find the HCF and LCM of 12 and 18, then verify the product relationship.
HCF(12, 18) = 6, LCM(12, 18) = 36
6 × 36 = 216 = 12 × 18
Conclusion: The relationship holds true.
Note: This relationship holds only for exactly two numbers or expressions. For three or more, HCF × LCM ≠ Product.
Application: If HCF(a, b) = 6, LCM(a, b) = 36, and a = 12, find b.
Using the product relationship: HCF × LCM = a × b
Substitute the known values: 6 × 36 = 12 × b
Simplify: 216 = 12b
Solve for b: b = 216 ÷ 12
Therefore, b = 18

1.4 Methods to Find HCF and LCM

Factorization Method
👉 Factorize each expression completely (common factor, grouping, splitting the middle term, or standard identities).
👉 HCF = product of common factors, each raised to the lowest power in which it occurs.
👉 LCM = product of all factors (common + uncommon), each raised to the highest power in which it occurs.

Division Method (for HCF)
Useful when factorization is difficult — divide the expression of higher degree by the lower one, then continue dividing the divisor by the remainder until the remainder is zero. The last non-zero divisor is the HCF.

1.5 Useful Identities to Remember

Solved Example:
Find the HCF and LCM of: x² – 9 and x² + x – 6
Factorize:
👉 x² – 9 = (x-3)(x+3)
👉 x² + x – 6 = (x+3)(x-2)
HCF = x + 3)
LCM = (x-3)(x+3)(x-2)

1.6 Simplification of Algebraic Fractions

Golden Rule: Factorize numerator and denominator fully before cancelling. Never cancel terms that are added or subtracted — only cancel common factors.
Steps for combining fractions:
👉 Factorize every denominator.
👉 Find the LCM of denominators.
👉 Convert each fraction to have the LCM as denominator.
👉 Add/subtract numerators, then simplify (factorize again if possible).

Solved Example:

Simplify: 1x − 2 + 1x + 2 − 4x² − 4

x² − 4 = (x − 2)(x + 2), so LCM = (x − 2)(x + 2)

(x + 2) + (x − 2) − 4(x − 2)(x + 2) = 2x − 4(x − 2)(x + 2) = 2(x − 2)(x − 2)(x + 2) = 2x + 2

1.7 Effective Tricks

👉 Always check for a difference of squares or perfect square trinomial first — these appear in 70%+ of questions.
👉 If an expression won’t factorize normally, try splitting the middle term using two numbers whose product = (coefficient of x²) × (constant) and sum = coefficient of x.
👉 For three expressions, find HCF of any two first, then find HCF of that result with the third.

2. Indices and Equations Involving Indices

2.1 Indices (Exponents)

An index (plural: indices) refers to the power to which a number or variable is raised. It is written as aⁿ, where:
👉 a is the base
👉 n is the index or exponent
This means a is multiplied by itself n times. For example,
2³ = 2X2X2 = 8.

2.2 Laws of Indices (Memorize All)

2.3 Equations Involving Indices

These are equations in which the unknown variable appears in the exponent (e.g., $2^{x+1}=32$) or as a fractional or radical power (e.g., $x^{2\mathrm{/}3}=4$).

Solving them typically involves:
👉 Express both sides with the same base, then equate the exponents.

Example: Solve 2^x+1 = 32
or, 2^x+1 = 2⁵
or, x+1 = 5
or, x = 4

Example:
Solve: $\sqrt[3]{x^2} = 4$
x^2/3 = 4
or, x² = 4³ = 64
or, x = ±8

Solve 3²ˣ − 10·3ˣ + 9 = 0
Let y = 3ˣ. Then:
y² − 10y + 9 = 0
(y − 1)(y − 9) = 0
y = 1 or y = 9
3ˣ = 1 ⇒ x = 0
3ˣ = 9 ⇒ x = 2
Therefore, x = 0 or 2

2.3 Effective Tricks to solve Indices Questions

👉 $\text{If bases don’t match directly, break both sides into prime factors first (e.g., } 8 = 2^3, \ 27 = 3^3, \ 81 = 3^4 \text{).}$
👉 $\text{An equation has } a^x + a^{-x} = k \text{ form, substitute } y = a^x \text{ and solve the resulting quadratic } y + \dfrac{1}{y} = k \Rightarrow y^2 – ky + 1 = 0.$
👉 $\text{Watch for equations disguised as indices, like } \sqrt{x} \cdot \sqrt[3]{x} = x^{5/6} \text{ — convert everything to index form first.}$

3. Verbal Problems

3.1 Verbal Problems on Simultaneous Equations

Simultaneous Linear Equations — Solving Methods
👉 Substitution Method: Express one variable in terms of the other from one equation, substitute into the second.
👉 Elimination Method: Multiply equations so that coefficients of one variable match, then add/subtract.
👉 Cross-Multiplication Method: For $a_1x + b_1y = c_1 \text{ and } a_2x + b_2y = c_2:$
$\frac{x}{b_1c_2 – b_2c_1} = \frac{y}{c_1a_2 – c_2a_1} = \frac{1}{a_1b_2 – a_2b_1}$

Framework for Word Problems
👉 Identify the unknowns and assign variables (e.g., x, y).
👉 Translate each sentence into an equation.
👉 Solve the system.
👉 Verify the answer makes sense in context (ages, speeds, and quantities must be positive).

Common Word Problem Types:
👉 Age problems – “5 years ago… in 10 years…”
👉 Number problems – digits, sum/difference of numbers
👉 Speed–distance–time – boats/streams, trains
👉 Money/cost problems – mixtures, articles and prices
👉 Work and time

Solved Example:
The sum of the ages of a father and son is 45 years. Five years ago, the father’s age was four times the son’s age. Find their present ages.
Let father’s age = x, son’s age = y.
x + y = 45 … (i)
x − 5 = 4(y − 5) ⇒ x − 4y = −15 … (ii)
Subtract (ii) from (i): 5y = 60
⇒ y = 12, then x = 33.
Father = 33 years, Son = 12 years

Solved Example:
A boat goes 30 km upstream and 44 km downstream in 10 hours. It goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.
Let boat speed = x km/h, stream speed = y km/h.
$\text{Let } u = \dfrac{1}{x-y}, \ v = \dfrac{1}{x+y}$ (upstream/downstream unit times).
30u + 44v = 10, 40u + 55v = 13
Solving simultaneously: u = 1/5, v = 1/11
or, x − y = 5, x + y = 11
or, x = 8 km/h, y = 3 km/h

3.2 Verbal Problems on Quadratic Equations

Quadratic Equations — Key Formulas
For ax² + bx + c = 0:
x = $\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
Nature of Roots (Discriminant D = b² − 4ac):
D > 0: real and unequal roots
D = 0: real and equal roots
D < 0: imaginary (no real) roots
Relation between roots and coefficients:
Sum of roots = − $\frac{b}{a}$, Product of roots = $\frac{c}{a}$

Solved Example:
The product of two consecutive positive integers is 132. Find the integers.
Solution:
Let the integers be x and x + 1.
x(x + 1) = 132 ⇒ x² + x − 132 = 0 ⇒ (x + 12)(x − 11) = 0
x = 11 (rejecting x = −12 since integers must be positive)
The integers are 11 and 12.

Solved Example:
A rectangular field is 4 m longer than it is wide. Its area is 96 m². Find its dimensions.
Solution:
Let width = x, length = x + 4:
x(x + 4) = 96 ⇒ x² + 4x − 96 = 0
⇒ (x + 12)(x − 8) = 0
⇒ x = 8
Width = 8 m, Length = 12 m

Solved Example:
A train travels 360 km at uniform speed. If the speed were 5 km/h more, it would take 1 hour less. Find the speed.
Solution:
Let the original speed = x km/h.
Time taken at speed x = 360/x hours.
Time taken at speed (x + 5) = 360/(x + 5) hours.
Since the faster speed takes 1 hour less:
360/x − 360/(x + 5) = 1
Multiply both sides by x(x + 5):
360(x + 5) − 360x = x(x + 5)
360x + 1800 − 360x = x² + 5x
1800 = x² + 5x
x² + 5x − 1800 = 0
Factories:
(x + 45)(x − 40) = 0
x = −45 or x = 40
Reject x = −45 (speed cannot be negative).
Therefore, the original speed = 40 km/h.

3.3 Effective Tricks

👉 In work-time and speed-distance problems, always define variables as rates (1/time) rather than time directly — equations become linear instead of fractional.
👉 For “sum and product of roots” type questions, you often don’t need to solve the equation at all — just use -b/a and c/a directly.
👉 Always reject negative/impossible solutions in real-world context (age, length, speed cannot be negative).

4. Surds and Radical Equations

4.1 What is a Surd?

A surd is an irrational root that cannot be simplified to remove the radical, e.g., $\sqrt{2}, \ \sqrt[3]{5}$. The order of a surd is the root index (2 for square root, 3 for cube root, etc.).

4.2 Laws of Surds

Rationalization
To remove a surd from the denominator, multiply numerator and denominator by the conjugate.

Example: Rationalize 1√5 − √3

1√5 − √3 × √5 + √3√5 + √3 = √5 + √35 − 3 = √5 + √32

Solving Radical (Surd) Equations
Steps:
👉 Isolate one radical term on one side.
👉 Square both sides to eliminate the radical.
👉 If a radical remains, repeat.
👉 Solve the resulting equation.
👉 Always verify each solution in the original equation — squaring can introduce extraneous (false) roots.

Solved Example: Solve $\sqrt{2x + 3} = 5$
Square both sides: 2x + 3 = 25
⇒ x = 11

Solved Example: Solve √(x + 7) − √(x − 5) = 2

Isolate: √(x + 7) = 2 + √(x − 5)

Square: x + 7 = 4 + 4√(x − 5) + (x − 5)

x + 7 = x − 1 + 4√(x − 5)  ⇒  8 = 4√(x − 5)  ⇒  √(x − 5) = 2

Square again: x − 5 = 4 ⇒ x = 9

Effective Tricks
👉 If both sides of the equation involve surds, isolate the more complicated radical first before squaring.
👉 Always substitute back to check — Officer Cadet exams frequently include an extraneous root as a trap answer choice.
👉 To simplify nested surds like $\sqrt{a + \sqrt{b}}$, try writing it as $\sqrt{x} + \sqrt{y}$ and solve for x, y using x+y = a and 4xy = b (when b is under an inner square root of the form $\sqrt{4xy}$).

Important Tips to Solve Algebra Questions

👉 Don’t skip the identities — nearly every HCF/LCM and surd question relies on recognizing a standard algebraic identity at a glance.
👉 Practice substitution tricks for indices (letting y = a²) — this single technique converts a large share of “hard” index equations into simple quadratics.
👉 Draw a table of variables for every word problem before writing equations; this prevents translation errors, which are the most common mistake in this section.
👉 Always verify radical equation answers — this is the easiest way to avoid losing marks to extraneous roots.
👉 Practice a mix of past-paper questions across all four sub-topics rather than mastering one and neglecting others, since Officer Cadet papers typically draw at least one question from each area.

Practice Questions

Conclusion

Algebra is a high-scoring and logical topic in the mathematics syllabus for Officer Cadet Entrance Exam preparation. Most algebra questions can be solved efficiently once you can correctly identify the type of equation — linear, quadratic, simultaneous, or radical — and apply the matching formula or shortcut. Focus on key identities, practice factoring and substitution techniques, and use the discriminant to quickly determine the nature of roots. With regular practice, algebra questions can become a quick-scoring area in the Mathematics section.

This resource is part of the Officer Cadet Entrance Exam Preparation series, focusing on Maths for Officer Cadet Entrance Exam. For structured lessons, comprehensive practice question banks, and mock tests, explore the Algebra for Officer Cadet section of the Maths for Officer Cadet Entrance Exam course on the Pratima Academy LMS.

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