4.1 Fundamental Arithmetic & Percentage Applications

4.1 Fundamental Arithmetic & Percentage Applications

Arithmetic and number properties form the fundamental bedrock of the Quantitative Reasoning section in the Higher Education Aptitude Test (HAT). Whether you are taking HAT 1 (Engineering/STEM), HAT 2 (Management Sciences), HAT 3 (Social Sciences/Law), HAT 4 (Biological/Medical Sciences), or HAT-General, computational speed, numerical fluency, and conceptual mastery of number systems are critical. On the paper-based HEC HAT administered by the Education Testing Council (ETC), calculators are strictly prohibited. Consequently, candidates must rely on mental mathematical agility, structural number properties, and shortcut identities to solve quantitative problems efficiently.

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Classification of the Real Number System

All numerical quantities evaluated on the HEC HAT belong to the Real Number System (denoted by $\mathbb{R}$). Understanding the hierarchical subset relations among real numbers is essential for questions involving variables, constraints, and set definitions.

                     REAL NUMBERS (R)
                       /        \
          Rational (Q)            Irrational (I)
          /          \             (e.g., √2, π, e)
    Integers (Z)   Fractions/Decimals
      /      \
 Negative  Whole (W)
             /     \
           Zero   Natural (N)

Subset Hierarchy

1. Natural Numbers ($\mathbb{N}$): The set of positive counting numbers: $\mathbb{N} = {1, 2, 3, 4, 5, \dots}$.
2. Whole Numbers ($\mathbb{W}$): Natural numbers including zero: $\mathbb{W} = {0, 1, 2, 3, 4, \dots}$.
3. Integers ($\mathbb{Z}$): All positive numbers, negative whole numbers, and zero: $\mathbb{Z} = {\dots, -3, -2, -1, 0, 1, 2, 3, \dots}$.
4. Rational Numbers ($\mathbb{Q}$): Any number that can be expressed as a quotient $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$. Rational numbers either terminate (e.g., $0.75 = \frac{3}{4}$) or recur infinitely (e.g., $0.333\dots = \frac{1}{3}$).
5. Irrational Numbers ($\mathbb{I}$ or $\mathbb{R} \setminus \mathbb{Q}$): Numbers that cannot be expressed as a ratio of two integers. Their decimal expansions are non-terminating and non-recurring (e.g., $\sqrt{2} \approx 1.4142$, $\sqrt{3}$, $\pi \approx 3.14159$, $e \approx 2.71828$).

Prime and Composite Numbers

• Prime Numbers: Integers strictly greater than $1$ that have exactly two distinct positive divisors: $1$ and the number itself. The smallest prime number is $2$, which is also the only even prime number. The primes below $30$ are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.
• Composite Numbers: Integers strictly greater than $1$ that have more than two positive divisors (e.g., $4, 6, 8, 9, 10, 12$).
• Important Rules: The number $1$ is neither prime nor composite because it has only one positive divisor ($1$). The number $0$ is neither positive nor negative, but it is an even integer.
• Co-Prime (Relatively Prime) Numbers: Two positive integers $a$ and $b$ are co-prime if their Greatest Common Divisor is $1$ (i.e., $\gcd(a,b) = 1$). For example, $8$ and $15$ are co-prime, even though neither number is individually prime.

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Divisibility Rules (Integers 2 through 11)

Divisibility shortcuts allow test takers to factor large numbers rapidly without performing tedious long division on paper.

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Prime Factorization, LCM & GCD

The Fundamental Theorem of Arithmetic

Every composite integer greater than $1$ can be uniquely expressed as a product of prime factors, up to the order of the factors.

$\text{Example: } 360 = 2^3 \times 3^2 \times 5^1$

Greatest Common Divisor (GCD) and Least Common Multiple (LCM)

• GCD (or HCF): The largest positive integer that divides two or more integers without a remainder. Obtained by taking the lowest power of common prime factors.
• LCM: The smallest positive integer that is divisible by two or more integers. Obtained by taking the highest power of all prime factors.

Fundamental LCM-GCD Identity

For any two positive integers $a$ and $b$:

$\gcd(a,b) \times \text{lcm}(a,b) = a \times b$

Worked Example: Find the GCD and LCM of $48$ and $72$, and verify the product identity.

1. Prime Factorization: $48 = 2^4 \times 3^1$, and $72 = 2^3 \times 3^2$.
2. $\gcd(48, 72) = 2^{\min(4,3)} \times 3^{\min(1,2)} = 2^3 \times 3^1 = 8 \times 3 = 24$.
3. $\text{lcm}(48, 72) = 2^{\max(4,3)} \times 3^{\max(1,2)} = 2^4 \times 3^2 = 16 \times 9 = 144$.
4. Verification: $\gcd \times \text{lcm} = 24 \times 144 = 3,456$. Product $a \times b = 48 \times 72 = 3,456$. Identical!

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Converting Recurring Decimals to Vulgar Fractions

Recurring decimals represent rational numbers. On the HEC HAT, questions frequently test conversion of repeating decimals into reduced fractions $\frac{P}{Q}$.

Algebraic Procedure

1. Let $x$ equal the recurring decimal.
2. Multiply $x$ by $10^m$ to shift non-repeating digits past the decimal point.
3. Multiply $x$ by $10^{m+k}$ to shift one full repeating block past the decimal point (where $k$ is the repeating block length).
4. Subtract the equations to eliminate the infinitely repeating decimal tail.
5. Solve for $x$ and reduce to lowest terms.

Worked Example: Convert $x = 0.2\overline{45} = 0.2454545\dots$ into a simplified fraction.

• Let $x = 0.2454545\dots$
• Multiply by $10$: $10x = 2.454545\dots$ (Equation 1)
• Multiply by $1000$: $1000x = 245.454545\dots$ (Equation 2)
• Subtract Eq 1 from Eq 2: $1000x - 10x = 245.4545\dots - 2.4545\dots \Rightarrow 990x = 243$.
• Simplify: $x = \frac{243}{990}$. Dividing numerator and denominator by $9$: $x = \frac{27}{110}$.

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Order of Operations (PEMDAS / BODMAS)

When evaluating compound numerical expressions, strict precedence must be followed:

1. P / B: Parentheses / Brackets — innermost grouping symbols first: (), [], {}.
2. E / O: Exponents / Orders (powers, roots).
3. MD: Multiplication and Division — evaluated left to right in order of appearance.
4. AS: Addition and Subtraction — evaluated left to right in order of appearance.

$\text{Evaluate: } 18 - \left[ 6 + 2 \times \left( 5 - 3^2 \div 3 \right) \right]$

1. Inner exponent: $3^2 = 9 \rightarrow (5 - 9 \div 3)$.
2. Inner division: $9 \div 3 = 3 \rightarrow (5 - 3) = 2$.
3. Multiplication in brackets: $2 \times 2 = 4 \rightarrow [6 + 4] = 10$.
4. Final subtraction: $18 - 10 = 8$.

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Percentages, Profit, Loss & Discount Applications

A percentage expresses a fraction with a denominator of $100$.

Key Percentage Formulas

• Percentage Change: $\% \text{ Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$
• Restoring Original Value: If a quantity increases by $r%$, the percentage reduction required to restore the original value is: $\text{Percentage Reduction} = \left( \frac{r}{100 + r} \right) \times 100\%$

Profit, Loss & Markup Definitions

• Cost Price (CP): The total acquisition cost incurred by the seller.
• Selling Price (SP): The final transaction price paid by the buyer.
• Marked Price (MP): The list price printed on the label before discounts.
• Profit %: $\text{Profit %} = \frac{SP - CP}{CP} \times 100%$
• Loss %: $\text{Loss %} = \frac{CP - SP}{CP} \times 100%$
• Discount %: $\text{Discount %} = \frac{MP - SP}{MP} \times 100%$

Successive Discounts Formula

If an item is subjected to two consecutive discounts of $d_1%$ and $d_2%$, the single net equivalent discount percentage $D_{\text{net}}$ is:

$D_{\text{net}} = d_1 + d_2 - \frac{d_1 \times d_2}{100}$

Worked Example: An article marked at PKR $8,000$ is offered at successive discounts of $20%$ and $10%$. Find the final selling price.

• Net Discount $= 20 + 10 - \frac{20 \times 10}{100} = 30 - 2 = 28%$.
• Final $SP = \text{MP} \times (1 - 0.28) = 8,000 \times 0.72 = \text{PKR } 5,760$.

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Simple vs. Compound Interest

Simple Interest (SI)

Interest is calculated strictly on the initial principal amount $P$ over time $t$ years at interest rate $r%$ per annum:

$I = \frac{P \times r \times t}{100}, \quad A = P + I = P \left(1 + \frac{r \cdot t}{100}\right)$

Compound Interest (CI)

Interest is added to the principal at the end of each compounding period, so interest earns interest:

$A = P \left(1 + \frac{r}{100n}\right)^{n \cdot t}, \quad CI = A - P$

where $n$ is the compounding frequency per year ($n=1$ for annual, $n=2$ for semi-annual, $n=4$ for quarterly).

Two-Year Difference Formula

For $t = 2$ years compounded annually, the difference between Compound Interest and Simple Interest is:

$CI - SI = P \left( \frac{r}{100} \right)^2$

Worked Example: PKR $100,000$ is invested for $2$ years at $10%$ per annum. Calculate SI, CI (annual), and their difference.

• $SI = \frac{100,000 \times 10 \times 2}{100} = \text{PKR } 20,000$.
• $A = 100,000 \times (1 + 0.10)^2 = 100,000 \times 1.21 = \text{PKR } 121,000 \Rightarrow CI = \text{PKR } 21,000$.
• Difference $= CI - SI = 21,000 - 20,000 = \text{PKR } 1,000$.
• Check via shortcut formula: $100,000 \times (10/100)^2 = 100,000 \times 0.01 = \text{PKR } 1,000$. Exact match!