4.1 Arithmetic Operations, Percentages, Ratio & Proportion

Fundamental Arithmetic & Order of Operations (BODMAS)

In the Pakistan Air Force (PAF) Airman Academic Test, speed and accuracy in basic computation are vital. Numerical reasoning questions frequently combine multiple arithmetic operations, fractions, decimals, and prime factorizations.

The BODMAS / PEMDAS Rule

When an arithmetic expression contains multiple operations, you must follow the standard order of operations: BODMAS (Brackets, Orders/Exponents, Division, Multiplication, Addition, Subtraction) or PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction). Note that Division and Multiplication hold equal priority and are evaluated from left to right, as do Addition and Subtraction.

Worked Example: Multi-Operation Evaluation

Problem: Evaluate the expression: $48 \div 6 \times (3 + 5) - 2^3 + 14 \div 2$.

Step-by-Step Solution:

1. Brackets: Simplify $(3 + 5) = 8$. Expression becomes: $48 \div 6 \times 8 - 2^3 + 14 \div 2$.
2. Orders/Exponents: Calculate $2^3 = 8$. Expression becomes: $48 \div 6 \times 8 - 8 + 14 \div 2$.
3. Division & Multiplication (Left to Right):
   • First division: $48 \div 6 = 8$.
   • Next multiplication: $8 \times 8 = 64$.
   • Next division: $14 \div 2 = 7$.
   • Expression becomes: $64 - 8 + 7$.
4. Addition & Subtraction (Left to Right):
   • $64 - 8 = 56$.
   • $56 + 7 = 63$.

Final Result: $63$.

Fractions, Decimals, LCM and HCF

Highest Common Factor (HCF) and Least Common Multiple (LCM)

• HCF (GCD): The greatest integer that divides two or more numbers without leaving a remainder.
• LCM: The smallest positive integer that is divisible by two or more numbers.
• Fundamental Property: For any two positive integers $a$ and $b$: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

Fractional LCM and HCF Formulas

$\text{HCF of Fractions} = \frac{\text{HCF of Numerators}}{\text{LCM of Denominators}}$ $\text{LCM of Fractions} = \frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$

Percentages, Ratios & Proportional Reasoning

Percentages and ratios form the core of commercial mathematics and quantitative reasoning in PAF Airman recruitment tests.

Percentages & Conversion Benchmarks

A percentage represents a fraction out of $100$. Converting rapidly between fractions, decimals, and percentages saves critical seconds during the exam.

Core Percentage Formulas

1. Percentage of a Quantity: $\text{Value} = \left(\frac{P}{100}\right) \times N$
2. Percentage Change: $\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$
3. Successive Percentage Formula: If a quantity changes by $+a%$ and then by $+b%$, the net percentage change is: $\text{Net Change} = \left(a + b + \frac{a \cdot b}{100}\right)\%$

Worked Example: Successive Discount

Problem: A flight suit is marked at PKR 12,000. It is offered at a successive discount of $20%$ followed by an additional $10%$. What is the final selling price?

Solution:

• Method 1 (Successive Formula): $a = -20$, $b = -10$. $\text{Net Discount} = -20 - 10 + \frac{(-20)(-10)}{100} = -30 + 2 = -28\%$ Total discount is $28%$. $\text{Final Price} = 12,000 \times (1 - 0.28) = 12,000 \times 0.72 = \text{PKR } 8,640$

Ratio & Proportion

A ratio compares two quantities of the same unit ($a:b = \frac{a}{b}$). A proportion states that two ratios are equal ($a:b = c:d$, meaning $\frac{a}{b} = \frac{c}{d} \implies a \cdot d = b \cdot c$).

Direct vs. Inverse Proportion

• Direct Proportion ($y \propto x$): As $x$ increases, $y$ increases proportionally ($y = kx \implies \frac{y_1}{x_1} = \frac{y_2}{x_2}$).
• Inverse Proportion ($y \propto \frac{1}{x}$): As $x$ increases, $y$ decreases proportionally ($y \cdot x = k \implies x_1 \cdot y_1 = x_2 \cdot y_2$).

Worked Example: Compound Ratio / Unitary Method

Problem: If 15 Air Force technicians can service 30 aircraft engines in 6 days, how many days will it take 10 technicians to service 40 aircraft engines?

Solution:

• Let $D$ be the required number of days.
• Technicians ($T$) and Days ($D$) are inversely proportional (fewer technicians take more days).
• Engines ($E$) and Days ($D$) are directly proportional (more engines require more days). $\frac{D_2}{D_1} = \left(\frac{T_1}{T_2}\right) \times \left(\frac{E_2}{E_1}\right)$ $\frac{D}{6} = \left(\frac{15}{10}\right) \times \left(\frac{40}{30}\right)$ $\frac{D}{6} = \frac{3}{2} \times \frac{4}{3} = \frac{12}{6} = 2 \implies D = 6 \times 2 = 12 \text{ days}$