1.1 Real Numbers, Ratios, Rates & Percentages

Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.1 Real Numbers, Ratios, Rates & Percentages

Key Takeaways Explain This Topic AI

Master mental math shortcuts and fraction-decimal-percent equivalencies to eliminate scratchwork delays under ACET time constraints.

Apply the unit ratio method for partitive proportions and distinguish clearly between part-to-part and part-to-whole relationships.

Solve combined work and average speed problems using harmonic mean concepts rather than simple arithmetic averages.

Differentiate direct, inverse, and joint variation setups by establishing explicit constants of proportionality.

Last updated: July 2026

Real Numbers, Ratios, Rates & Percentages

The Mathematics subtest of the Ateneo College Entrance Test (ACET) demands extraordinary mental agility. Because calculators are strictly prohibited, your success depends on rapid number sense, quick identification of structural patterns, and instant execution of arithmetic shortcuts. The ACET tests real numbers not merely through straightforward computations, but through multi-step word problems involving rates, proportions, and percentage changes engineered to trigger common psychological traps.

1. Real Number Structure & Order of Operations

The real number system R \mathbb{R} R consists of rational numbers Q \mathbb{Q} Q (integers, terminating decimals, and repeating decimals) and irrational numbers I \mathbb{I} I (non-repeating, non-terminating values such as 2 , 3 , π , \sqrt{2}, \sqrt{3}, \pi, 2 ​ , 3 ​ , π , and e e e ).

Order of Operations (PEMDAS / BODMAS)

When evaluating complex numerical expressions, strictly adhere to the established hierarchy:

Parentheses / Grouping: Evaluate operations inside brackets () , braces {} or fraction bars first.

Exponents / Radicals: Compute powers and roots from left to right.

Multiplication & Division: Perform these with equal priority strictly from left to right.

Addition & Subtraction: Perform these with equal priority strictly from left to right.

ACET Speed Tactic (Left-to-Right Priority): In the expression 24 ÷ 4 × 2 24 \div 4 \times 2 24 ÷ 4 × 2 , many test-takers improperly multiply 4 × 2 = 8 4 \times 2 = 8 4 × 2 = 8 first to get 24 ÷ 8 = 3 24 \div 8 = 3 24 ÷ 8 = 3 . The correct execution is left-to-right: ( 24 ÷ 4 ) × 2 = 6 × 2 = 12 (24 \div 4) \times 2 = 6 \times 2 = 12 ( 24 ÷ 4 ) × 2 = 6 × 2 = 12 .

Essential Divisibility Rules for Fast Prime Factoring

Quickly testing for divisibility allows you to reduce fractions and simplify radical terms instantaneously without long division:

Divisor Divisibility Condition Example ACET Application

3 Sum of digits is divisible by 3 4,251 → 4 + 2 + 5 + 1 = 12 \rightarrow 4+2+5+1 = 12 → 4 + 2 + 5 + 1 = 12 (Divisible) Fast fraction simplification

4 Last two digits form a number divisible by 4 7,316 → 16 \rightarrow 16 → 16 (Divisible) Quick reduction of denominators

6 Number is even AND divisible by 3 8,532 → \rightarrow → Even and 8 + 5 + 3 + 2 = 18 8+5+3+2=18 8 + 5 + 3 + 2 = 18 Factoring large composite values

8 Last three digits form a number divisible by 8 19,120 → 120 ÷ 8 = 15 \rightarrow 120 \div 8 = 15 → 120 ÷ 8 = 15 Reducing higher-order ratios

9 Sum of digits is divisible by 9 6,858 → 6 + 8 + 5 + 8 = 27 \rightarrow 6+8+5+8=27 → 6 + 8 + 5 + 8 = 27 (Divisible) Simplifying radical radicands

11 Alternating sum of digits is divisible by 11 9,185 → ( 9 + 8 ) − ( 1 + 5 ) = 11 \rightarrow (9+8)-(1+5) = 11 → ( 9 + 8 ) − ( 1 + 5 ) = 11 Factoring special polynomials

2. High-Frequency Mental Math Conversions

Memorize these fraction, decimal, and percentage equivalents. Converting complex percentage problems into simple fractional multiplication is often the single fastest way to arrive at an answer. 1 8 = 0.125 = 12.5 % 3 8 = 0.375 = 37.5 % 5 8 = 0.625 = 62.5 % 7 8 = 0.875 = 87.5 % 1 6 ≈ 0.1667 = 16.67 % 5 6 ≈ 0.8333 = 83.33 % 1 12 ≈ 0.0833 = 8.33 % 1 16 = 0.0625 = 6.25 % 1 7 ≈ 0.142857 2 7 ≈ 0.285714 1 9 = 0. 1 ˉ = 11. 1 ˉ % 1 11 = 0. 09 ‾ = 9. 09 ‾ % \begin{aligned}

\frac{1}{8} &= 0.125 = 12.5\% & \frac{3}{8} &= 0.375 = 37.5\% & \frac{5}{8} &= 0.625 = 62.5\% & \frac{7}{8} &= 0.875 = 87.5\%

\\ \frac{1}{6} &\approx 0.1667 = 16.67\% & \frac{5}{6} &\approx 0.8333 = 83.33\% & \frac{1}{12} &\approx 0.0833 = 8.33\% & \frac{1}{16} &= 0.0625 = 6.25\%

\\ \frac{1}{7} &\approx 0.142857 & \frac{2}{7} &\approx 0.285714 & \frac{1}{9} &= 0.\bar{1} = 11.\bar{1}\% & \frac{1}{11} &= 0.\overline{09} = 9.\overline{09}\%

\end{aligned} 8 1 ​ 6 1 ​ 7 1 ​ ​ = 0.125 = 12.5% ≈ 0.1667 = 16.67% ≈ 0.142857 ​ 8 3 ​ 6 5 ​ 7 2 ​ ​ = 0.375 = 37.5% ≈ 0.8333 = 83.33% ≈ 0.285714 ​ 8 5 ​ 12 1 ​ 9 1 ​ ​ = 0.625 = 62.5% ≈ 0.0833 = 8.33% = 0. 1 ˉ = 11. 1 ˉ % ​ 8 7 ​ 16 1 ​ 11 1 ​ ​ = 0.875 = 87.5% = 0.0625 = 6.25% = 0. 09 = 9. 09 % ​

3. Ratios, Proportions & Partitive Partitioning

A ratio compares two quantities a a a and b b b as a : b a:b a : b or a b \frac{a}{b} b a ​ . A proportion states that two ratios are equivalent: a b = c d ⟹ a d = b c \frac{a}{b} = \frac{c}{d} \implies ad = bc b a ​ = d c ​ ⟹ a d = b c .

Part-to-Part vs. Part-to-Whole Relationships

If a mixture contains red and blue marbles in the ratio 3 : 7 3:7 3 : 7 :

The part-to-part ratio of red to blue is 3 : 7 3:7 3 : 7 .

The total number of units is 3 + 7 = 10 3 + 7 = 10 3 + 7 = 10 .

The part-to-whole ratio for red marbles is 3 10 = 30 % \frac{3}{10} = 30\% 10 3 ​ = 30% , and for blue marbles is 7 10 = 70 % \frac{7}{10} = 70\% 10 7 ​ = 70% .

The Multiplier ( x x x ) Method for Partitive Proportions

When a total quantity T T T is divided into a multi-part ratio a : b : c a : b : c a : b : c :

Express the parts as a x ax a x , b x bx b x , and c x cx c x .

Set up the sum: a x + b x + c x = T ⟹ x ( a + b + c ) = T ax + bx + cx = T \implies x(a + b + c) = T a x + b x + c x = T ⟹ x ( a + b + c ) = T .

Solve for x = T a + b + c x = \frac{T}{a+b+c} x = a + b + c T ​ and multiply by the required ratio coefficient.

• 4. Rates: Motion, Combined Work & Pipes
• Rate of Motion & The Average Speed Trap

Distance formula: d = r ⋅ t ⟹ r = d t d = r \cdot t \implies r = \frac{d}{t} d = r ⋅ t ⟹ r = t d ​ .

CRITICAL ACET TRAP (Average Speed): If a car travels from Town A to Town B at speed v 1 v_1 v 1 ​ and returns along the exact same distance at speed v 2 v_2 v 2 ​ , the average speed is NOT the arithmetic mean v 1 + v 2 2 \frac{v_1 + v_2}{2} 2 v 1 ​ + v 2 ​ ​ . Because time spent on each leg varies, the overall average speed is the harmonic mean : v avg = Total Distance Total Time = 2 d d v 1 + d v 2 = 2 v 1 v 2 v 1 + v 2 v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2 d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2 v_1 v_2}{v_1 + v_2} v avg ​ = Total Time Total Distance ​ = v 1 ​ d ​ + v 2 ​ d ​ 2 d ​ = v 1 ​ + v 2 ​ 2 v 1 ​ v 2 ​ ​

Combined Work Rate Model

If Worker A completes a job in t A t_A t A ​ hours and Worker B completes the same job in t B t_B t B ​ hours, their individual rates are R A = 1 t A R_A = \frac{1}{t_A} R A ​ = t A ​ 1 ​ and R B = 1 t B R_B = \frac{1}{t_B} R B ​ = t B ​ 1 ​ jobs/hour. Working together, their combined rate R combined R_{\text{combined}} R combined ​ is: R combined = 1 t A + 1 t B = t A + t B t A t B ⟹ T together = t A t B t A + t B R_{\text{combined}} = \frac{1}{t_A} + \frac{1}{t_B} = \frac{t_A + t_B}{t_A t_B} \implies T_{\text{together}} = \frac{t_A t_B}{t_A + t_B} R combined ​ = t A ​ 1 ​ + t B ​ 1 ​ = t A ​ t B ​ t A ​ + t B ​ ​ ⟹ T together ​ = t A ​ + t B ​ t A ​ t B ​ ​

If Pipe A fills a tank in F F F hours and Pipe B drains the tank in D D D hours (where D > F D > F D > F ), the net filling rate is: R net = 1 F − 1 D = D − F F D R_{\text{net}} = \frac{1}{F} - \frac{1}{D} = \frac{D - F}{F D} R net ​ = F 1 ​ − D 1 ​ = F D D − F ​

5. Variations & Proportional Dependencies

Direct Variation: y ∝ x ⟹ y = k x ⟹ y 1 x 1 = y 2 x 2 y \propto x \implies y = k x \implies \frac{y_1}{x_1} = \frac{y_2}{x_2} y ∝ x ⟹ y = k x ⟹ x 1 ​ y 1 ​ ​ = x 2 ​ y 2 ​ ​

Inverse Variation: y ∝ 1 x ⟹ y = k x ⟹ x 1 y 1 = x 2 y 2 y \propto \frac{1}{x} \implies y = \frac{k}{x} \implies x_1 y_1 = x_2 y_2 y ∝ x 1 ​ ⟹ y = x k ​ ⟹ x 1 ​ y 1 ​ = x 2 ​ y 2 ​

Joint & Combined Variation: y ∝ x z w ⟹ y = k x z w ⟹ w 1 y 1 x 1 z 1 = w 2 y 2 x 2 z 2 y \propto \frac{x z}{w} \implies y = \frac{k x z}{w} \implies \frac{w_1 y_1}{x_1 z_1} = \frac{w_2 y_2}{x_2 z_2} y ∝ w x z ​ ⟹ y = w k x z ​ ⟹ x 1 ​ z 1 ​ w 1 ​ y 1 ​ ​ = x 2 ​ z 2 ​ w 2 ​ y 2 ​ ​

• 6. Percentages, Percent Change & Successive Discounts
• Basic Percent Translation Formula

Translate English statements directly into algebraic equations:

• "Is" → = \rightarrow = →=
• "Of" → × \rightarrow \times → ×

"What number / percent" → x \rightarrow x → x Percent Change = New Value − Original Value Original Value × 100 % \text{Percent Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\% Percent Change = Original Value New Value − Original Value ​ × 100%

The Successive Discount / Increase Trap

Applying a discount of d 1 % d_1\% d 1 ​ % followed by another discount of d 2 % d_2\% d 2 ​ % is NOT equal to a single discount of ( d 1 + d 2 ) % (d_1 + d_2)\% ( d 1 ​ + d 2 ​ ) % .

Final Price = P 0 × ( 1 − d 1 ) × ( 1 − d 2 ) = P_0 \times (1 - d_1) \times (1 - d_2) = P 0 ​ × ( 1 − d 1 ​ ) × ( 1 − d 2 ​ )

Single Equivalent Discount Rate = 1 − ( 1 − d 1 ) ( 1 − d 2 ) = 1 - (1 - d_1)(1 - d_2) = 1 − ( 1 − d 1 ​ ) ( 1 − d 2 ​ )

• Worked Step-by-Step ACET Exam Problems
• Problem 1: Combined Work with Variable Timelines

Worker A can build a wall in 12 hours. Worker B can build the same wall in 15 hours. Worker A starts working alone for 3 hours, after which Worker B joins to complete the rest. How many total hours were required to complete the wall?

Step 1: Determine individual work rates.

Rate of A: R A = 1 12 R_A = \frac{1}{12} R A ​ = 12 1 ​ wall/hour.

Rate of B: R B = 1 15 R_B = \frac{1}{15} R B ​ = 15 1 ​ wall/hour.

Step 2: Calculate work completed by A in the first 3 hours. W 1 = R A × 3 = 1 12 × 3 = 1 4 of the wall W_1 = R_A \times 3 = \frac{1}{12} \times 3 = \frac{1}{4} \text{ of the wall} W 1 ​ = R A ​ × 3 = 12 1 ​ × 3 = 4 1 ​ of the wall

Step 3: Calculate remaining work. W remaining = 1 − 1 4 = 3 4 of the wall W_{\text{remaining}} = 1 - \frac{1}{4} = \frac{3}{4} \text{ of the wall} W remaining ​ = 1 − 4 1 ​ = 4 3 ​ of the wall

Step 4: Find combined rate of A and B. R combined = 1 12 + 1 15 = 5 + 4 60 = 9 60 = 3 20 wall/hour R_{\text{combined}} = \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} = \frac{3}{20} \text{ wall/hour} R combined ​ = 12 1 ​ + 15 1 ​ = 60 5 + 4 ​ = 60 9 ​ = 20 3 ​ wall/hour

Step 5: Compute time required for remaining work. t together = W remaining R combined = 3 / 4 3 / 20 = 3 4 × 20 3 = 5 hours t_{\text{together}} = \frac{W_{\text{remaining}}}{R_{\text{combined}}} = \frac{3/4}{3/20} = \frac{3}{4} \times \frac{20}{3} = 5 \text{ hours} t together ​ = R combined ​ W remaining ​ ​ = 3/20 3/4 ​ = 4 3 ​ × 3 20 ​ = 5 hours

Step 6: Total time elapsed. Total Time = 3 hours (alone) + 5 hours (together) = 8 hours \text{Total Time} = 3 \text{ hours (alone)} + 5 \text{ hours (together)} = 8 \text{ hours} Total Time = 3 hours (alone) + 5 hours (together) = 8 hours

Problem 2: Multi-Stage Successive Percent Change

An item's wholesale price is increased by 25 % 25\% 25% . During a weekend clearance sale, the new price is discounted by 20 % 20\% 20% . Later, an administrative surcharge of 10 % 10\% 10% is added to the sale price. What is the net overall percentage change relative to the initial wholesale price?

Step 1: Set a convenient initial value. Let original wholesale price P 0 = 100 P_0 = 100 P 0 ​ = 100 .

Step 2: Apply stage 1 ( 25 % 25\% 25% markup). P 1 = 100 × ( 1 + 0.25 ) = 125 P_1 = 100 \times (1 + 0.25) = 125 P 1 ​ = 100 × ( 1 + 0.25 ) = 125

Step 3: Apply stage 2 ( 20 % 20\% 20% discount). P 2 = 125 × ( 1 − 0.20 ) = 125 × 0.80 = 100 P_2 = 125 \times (1 - 0.20) = 125 \times 0.80 = 100 P 2 ​ = 125 × ( 1 − 0.20 ) = 125 × 0.80 = 100

Step 4: Apply stage 3 ( 10 % 10\% 10% surcharge). P 3 = 100 × ( 1 + 0.10 ) = 110 P_3 = 100 \times (1 + 0.10) = 110 P 3 ​ = 100 × ( 1 + 0.10 ) = 110

Step 5: Compute net overall percentage change. Net Change = 110 − 100 100 × 100 % = + 10 % \text{Net Change} = \frac{110 - 100}{100} \times 100\% = +10\% Net Change = 100 110 − 100 ​ × 100% = + 10%

ACET Speed Tactics & Traps Summary

Problem Type Common Student Trap ACET Speed Tactic

Average Speed Averaging rates: v 1 + v 2 2 \frac{v_1+v_2}{2} 2 v 1 ​ + v 2 ​ ​ Use harmonic mean: 2 v 1 v 2 v 1 + v 2 \frac{2v_1v_2}{v_1+v_2} v 1 ​ + v 2 ​ 2 v 1 ​ v 2 ​ ​ when distance is equal

Combined Work Adding times directly: t 1 + t 2 t_1 + t_2 t 1 ​ + t 2 ​ Add rates: R total = 1 t 1 + 1 t 2 R_{\text{total}} = \frac{1}{t_1} + \frac{1}{t_2} R total ​ = t 1 ​ 1 ​ + t 2 ​ 1 ​

Successive Percent Adding percentages: + 20 % − 20 % = 0 % +20\% - 20\% = 0\% + 20% − 20% = 0% Use multiplier method: 1.20 × 0.80 = 0.96 1.20 \times 0.80 = 0.96 1.20 × 0.80 = 0.96 (Net − 4 % -4\% − 4% )

Inverse Variation Setting up direct ratio: x 1 y 1 = x 2 y 2 \frac{x_1}{y_1} = \frac{x_2}{y_2} y 1 ​ x 1 ​ ​ = y 2 ​ x 2 ​ ​ Product is constant: x 1 y 1 = x 2 y 2 x_1 y_1 = x_2 y_2 x 1 ​ y 1 ​ = x 2 ​ y 2 ​