5.3 Essential Mathematics: Arithmetic, Percentages, Ratios, and Profit & Loss

5.3 Essential Mathematics: Arithmetic, Percentages, Ratios, and Profit & Loss

Core Principle: Mathematics in the PMA Academic preliminary test evaluates applied quantitative reasoning across core secondary and intermediate arithmetic principles. Candidates should develop reliable mental calculations while following the live instructions on timing and permitted aids. Understanding fraction-decimal shortcuts, ratio balancing, work-rate identities, and commercial formulas eliminates guesswork and secures high quantitative scores.

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Arithmetic Foundations and Mental Math

At the AS&RC computer lab, candidates are provided with a pencil and scratch sheet. Because spending more than 45 to 60 seconds on a single arithmetic problem causes severe time shortages at the end of the module, mastering rapid mental calculation shortcuts is indispensable.

The BODMAS / PEMDAS Priority Hierarchy

Multi-operator arithmetic expressions must follow the rigid order of mathematical operations:

1. Brackets (Parentheses: resolve innermost parentheses first, then curly brackets, then square brackets).
2. Orders (Powers, exponents, and square roots).
3. Division and Multiplication (evaluated from left to right at equal priority).
4. Addition and Subtraction (evaluated from left to right at equal priority).

Worked Example: Multi-Operator Expression

Evaluate the expression: $36 - 6 \times (9 - 5) \div 3 + 4^2 \div 8$

• Step 1 (Brackets): $(9 - 5) = 4 \implies 36 - 6 \times 4 \div 3 + 4^2 \div 8$
• Step 2 (Orders / Powers): $4^2 = 16 \implies 36 - 6 \times 4 \div 3 + 16 \div 8$
• Step 3 (Division & Multiplication from left to right):
  • $6 \times 4 = 24$, then $24 \div 3 = 8$
  • $16 \div 8 = 2$
  • Expression becomes: $36 - 8 + 2$
• Step 4 (Addition & Subtraction from left to right):
  • $36 - 8 = 28$, then $28 + 2 = 30$.

High-Speed Mental Arithmetic Shortcuts

• Multiplying by 5: Halve the number, then append a zero (multiply by 10).
  Example: $68 \times 5 = (68 \div 2) \times 10 = 34 \times 10 = 340$.
• Multiplying by 25: Divide the number by 4, then multiply by 100.
  Example: $84 \times 25 = (84 \div 4) \times 100 = 21 \times 100 = 2{,}100$.
• Dividing by 5: Double the number, then shift the decimal point one place to the left (divide by 10).
  Example: $365 \div 5 = (365 \times 2) \div 10 = 730 \div 10 = 73$.
• Dividing by 25: Multiply the number by 4, then divide by 100.
  Example: $450 \div 25 = (450 \times 4) \div 100 = 1{,}800 \div 100 = 18$.

High-Yield Fraction-Decimal-Percentage Equivalence Table

Memorizing these standard conversions saves 20 to 30 seconds per question on the terminal screen:

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Percentages and Consecutive Percentage Changes

A percentage is a dimensionless ratio expressed as a fraction of 100:

$$\text{Value} = \frac{\text{Percentage}}{100} \times \text{Base Quantity}$$

The Reversible Percentage Shortcut

A useful algebraic identity for rapid computation is:

$$x\% \text{ of } y = y\% \text{ of } x$$

Example: Finding $16\% \text{ of } 50$ mentally is identical to computing $50\% \text{ of } 16 = 8$. Example: Finding $64\% \text{ of } 25$ is identical to computing $25\% \text{ of } 64 = \frac{64}{4} = 16$.

Percentage Increase and Decrease

$$\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$$

• If the result is positive, it represents a percentage increase.
• If negative, it represents a percentage decrease.

Consecutive Percentage Changes Formula

When a quantity undergoes two successive percentage alterations of $x\%$ and $y\%$ (where a positive sign denotes an increase and a negative sign denotes a decrease), the net overall percentage change is given by:

$$\text{Net Percentage Change} = x + y + \frac{x \cdot y}{100}$$

Worked Example: Successive Markup and Markdown

A field equipment dealer marks up prices by $25\%$, but later offers a clearance discount of $20\%$ on the marked price. What is the net percentage change relative to the original cost?

• Let $x = +25$ and $y = -20$.
• Apply the formula:

$$\text{Net Change} = 25 + (-20) + \frac{(25)(-20)}{100} = 5 - \frac{500}{100} = 5 - 5 = 0\%$$

• The net change is exactly $0\%$. (The final selling price equals the original cost price).

Worked Example: Symmetrical Increase and Decrease Trap

A military technician's salary is increased by $20\%$ and subsequently reduced by $20\%$. What is the net effect?

• Intuitive error: Candidates assume there is no change ($0\%$).
• Correct algebraic solution: $x = +20, y = -20$.

$$\text{Net Change} = 20 - 20 + \frac{(20)(-20)}{100} = 0 - \frac{400}{100} = -4\%$$

• The salary suffers a net decrease of 4% because the $20\%$ decrease operated on a larger, marked-up base.

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Ratios, Proportions, and Continued Ratios

A ratio ($a : b = \frac{a}{b}$) compares two quantities of the same unit. A proportion ($a : b :: c : d$) equates two ratios ($a \times d = b \times c$, where product of extremes equals product of means).

Direct vs. Inverse Proportion

1. Direct Proportion: When an increase in one variable causes a proportional increase in the other ($\frac{y_1}{x_1} = \frac{y_2}{x_2}$).
   • Example: Fuel consumed vs. distance traversed by a military convoy.
2. Inverse Proportion: When an increase in one variable causes a proportional decrease in the other ($x_1 \cdot y_1 = x_2 \cdot y_2$).
   • Example: Number of soldiers deployed vs. days required to construct a defensive berm.

Dividing a Total Quantity into a Given Ratio

To divide a total quantity $T$ into the ratio $a : b : c$:

$$\text{Sum of parts} = S = a + b + c$$ $$\text{Share of } a = \frac{a}{S} \times T, \quad \text{Share of } b = \frac{b}{S} \times T, \quad \text{Share of } c = \frac{c}{S} \times T$$

Worked Example: Divide Rs. 72,000 among three military units in the ratio $2 : 3 : 7$.

• Sum of parts: $2 + 3 + 7 = 12$.
• Unit 1: $\frac{2}{12} \times 72{,}000 = Rs.\ 12{,}000$.
• Unit 2: $\frac{3}{12} \times 72{,}000 = Rs.\ 18{,}000$.
• Unit 3: $\frac{7}{12} \times 72{,}000 = Rs.\ 42{,}000$.

Continued Ratio Alignment ($A : B$ and $B : C$ to $A : B : C$)

When two separate ratios share a common intermediate variable with different numerical values, equalize the middle term by finding the Least Common Multiple (LCM).

Problem: If $A : B = 3 : 4$ and $B : C = 6 : 7$, find the unified ratio $A : B : C$.

• Middle term $B$ has values $4$ and $6$. The LCM of $4$ and $6$ is $12$.
• Multiply first ratio by $3$: $A : B = (3 \times 3) : (4 \times 3) = 9 : 12$.
• Multiply second ratio by $2$: $B : C = (6 \times 2) : (7 \times 2) = 12 : 14$.
• Resulting unified ratio: $A : B : C = 9 : 12 : 14$.

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Unitary Method: Man-Days, Work Rates & Cistern Problems

Work and labor problems represent a core question type on the AS&RC computer test. Because individual working capacities combine additively per unit of time, problems are solved using daily or hourly unit rates.

The Master Compound Man-Days Identity

When evaluating scenarios involving different numbers of men ($M$), days ($D$), working hours per day ($H$), and total work completed ($W$):

$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$

Worked Example: Man-Days Formula

If 20 military engineers can excavate a 400-meter anti-tank ditch in 6 days working 8 hours a day, how many days will 30 engineers require to excavate a 600-meter ditch working 8 hours a day?

• Assign variables: $M_1 = 20, D_1 = 6, H_1 = 8, W_1 = 400$.
• Target variables: $M_2 = 30, D_2 = ?, H_2 = 8, W_2 = 600$.
• Apply the identity:

$$\frac{20 \times 6 \times 8}{400} = \frac{30 \times D_2 \times 8}{600}$$ $$\frac{960}{400} = \frac{240 \times D_2}{600}$$ $$2.4 = 0.4 \times D_2 \implies D_2 = \frac{2.4}{0.4} = 6\text{ days}$$

• The 30 engineers will take exactly 6 days.

Combined Work-Rate Problems

If Worker A completes an entire task in $x$ days, A's single-day work rate is $\frac{1}{x}$.
If Worker B completes the same task in $y$ days, B's single-day work rate is $\frac{1}{y}$.
Working simultaneously, their combined single-day rate is:

$$\text{Combined Daily Rate} = \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}$$ $$\text{Total Days to Complete Work} = \frac{xy}{x + y}$$

Worked Example: Sapper Tariq can build a communication shelter in 12 hours, while Sapper Asim can build it in 6 hours. How long will it take them working together?

$$T = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\text{ hours}$$

Pipes and Cisterns

• Inlet Pipe: Fills a tank at rate $+\frac{1}{A}$ per hour.
• Drain / Leak Pipe: Empties a tank at rate $-\frac{1}{B}$ per hour.
• If an inlet fills a water reservoir in 4 hours and a leak empties it in 6 hours, the net filling rate per hour with both open is:

$$\text{Net Rate} = \frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12} \implies \text{Full Tank in } 12\text{ hours}$$

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Commercial Mathematics: Profit, Loss, and Discount

Commercial mathematics tests basic buying and selling equations under timed conditions. Definitions must be strictly maintained:

• Cost Price (CP): The purchase or production cost.
• Selling Price (SP): The final transaction price.
• Marked Price (MP) / List Price: The initial catalog price prior to discount.

Fundamental Profit and Loss Formulas

$$\text{When } \text{SP} > \text{CP} \implies \text{Profit} = \text{SP} - \text{CP}, \quad \mathbf{\text{Profit \%}} = \frac{\text{Profit}}{\mathbf{\text{CP}}} \times 100$$ $$\text{When } \text{CP} > \text{SP} \implies \text{Loss} = \text{CP} - \text{SP}, \quad \mathbf{\text{Loss \%}} = \frac{\text{Loss}}{\mathbf{\text{CP}}} \times 100$$

Direct Formulas for Selling Price:

$$\text{SP} = \text{CP} \times \left(\frac{100 + \text{Profit \%}}{100}\right) \quad \text{or} \quad \text{SP} = \text{CP} \times \left(\frac{100 - \text{Loss \%}}{100}\right)$$

Direct Formulas for Cost Price:

$$\text{CP} = \frac{\text{SP} \times 100}{100 + \text{Profit \%}} \quad \text{or} \quad \text{CP} = \frac{\text{SP} \times 100}{100 - \text{Loss \%}}$$

Marked Price and Commercial Discount

Discount is a reduction granted on the Marked Price (MP):

$$\text{Discount} = \text{MP} - \text{SP}, \quad \mathbf{\text{Discount \%}} = \frac{\text{Discount}}{\mathbf{\text{MP}}} \times 100$$ $$\text{SP} = \text{MP} \times \left(\frac{100 - \text{Discount \%}}{100}\right)$$

Single Equivalent Discount for Successive Discounts

If a supplier offers two consecutive discounts of $d_1\%$ and $d_2\%$:

$$\text{Equivalent Single Discount \%} = d_1 + d_2 - \frac{d_1 \times d_2}{100}$$

Worked Example: What single discount is equivalent to two successive discounts of $20\%$ and $10\%$?

$$\text{Equivalent Discount} = 20 + 10 - \frac{20 \times 10}{100} = 30 - 2 = 28\%$$

(Note: It is not $30\%$).

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Arithmetic Averages and Weighted Means

The arithmetic average (mean) represents the central value of a set of observations:

$$\text{Average} = \frac{\text{Sum of All Observations}}{\text{Total Number of Observations} (n)} \implies \mathbf{\text{Sum}} = \text{Average} \times n$$

Group Addition and Removal Problems

Many AS&RC average problems test the effect of adding or removing an individual from a known group average.

Worked Example: Platoon Commander Inclusion

A section of 10 soldiers has an average weight of $68\text{ kg}$. When their section commander joins them, the group average increases to $69\text{ kg}$. What is the weight of the section commander?

• Method 1 (Total Sum Comparison):
  • Original total weight (10 soldiers) $= 10 \times 68 = 680\text{ kg}$.
  • New total weight (11 individuals) $= 11 \times 69 = 759\text{ kg}$.
  • Commander's weight $= 759 - 680 = 79\text{ kg}$.
• Method 2 (Mental Shift Shortcut):
  • Baseline average was $68\text{ kg}$.
  • The commander brings enough extra weight to elevate all 11 people by $1\text{ kg}$:
  • Commander's weight $= 68 + (11 \times 1) = 68 + 11 = 79\text{ kg}$.