Detailed ExplanationLET Elementary · MathematicsReal content

LET Elementary Mathematics — Fractions, Decimals, Percent, Ratio and ProportionDetailed Explanation

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

Detailed explanations for LET Elementary Mathematics — Fractions, Decimals, Percent, Ratio and Proportion. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) frames Fractions, Decimals, Percent, Ratio and Proportion questions, and explain the underlying reasoning that gets you to the right answer every time.

Exam context

The Licensure Examination for Professional Teachers — Elementary is conducted by Professional Regulation Commission (PRC) and is scheduled for Bi-annual. The Mathematics subtest is marked as "Core" in the official pattern, and Fractions, Decimals, Percent, Ratio and Proportion appears in position 2nd of 7 in the LET Elementary Mathematics review rotation. Passing mark: Weighted average of 75% with no grade below 50%. Recent LET Elementary 2026 papers have drawn roughly a meaningful share of questions from this subject.

Fractions, Decimals, Percent, Ratio and Proportion - Detailed Explanation

This chapter covers the most frequently tested numerical concepts in the LET Mathematics component. Fractions, decimals, percent, ratio, and proportion are not isolated topics — they are four ways of expressing the same fundamental idea: a part compared to a whole. For BEEd graduates preparing for the LET, mastery of these concepts is non-negotiable, because they underpin virtually every quantitative word problem on the Board Examination. In Philippine classrooms, these concepts are taught across Grades 1 to 6 under the K-12 Basic Education Curriculum (BEC), making pedagogical fluency in these areas a professional requirement under RA 7836 (Philippine Teachers Professionalization Act). This detailed explanation walks you through every subtopic with step-by-step worked examples drawn from Filipino daily life — jeepney fares, market prices, classroom statistics, and DepEd allowances — so you can answer LET items with both speed and accuracy.

Concepts

Operations on Fractions

A fraction represents a part of a whole, written as numerator/denominator. The numerator tells how many parts you have; the denominator tells the total number of equal parts the whole is divided into. Mastering the four fundamental operations on fractions is the gateway to solving almost every LET word problem involving parts of quantities. ADDITION AND SUBTRACTION: You can only add or subtract fractions that share the same denominator (called the Least Common Denominator or LCD). Steps: (1) Find the LCD of all denominators. (2) Convert each fraction to an equivalent fraction using the LCD. (3) Add or subtract the numerators. (4) Keep the LCD as the denominator. (5) Simplify the result to lowest terms. MULTIPLICATION: Multiply numerator by numerator and denominator by denominator. Before multiplying, cancel any common factors between any numerator and any denominator (cross-cancellation) to keep numbers manageable. Always simplify the final answer. DIVISION: Multiply the first fraction by the RECIPROCAL (flip) of the second fraction. The phrase 'Keep, Change, Flip' (KCF) is a helpful mnemonic: Keep the first fraction, Change the division sign to multiplication, Flip the second fraction. MIXED NUMBERS: Convert to improper fractions before multiplying or dividing. To convert: multiply the whole number by the denominator, add the numerator, and place the result over the denominator. Example: 3 2/5 = (3×5+2)/5 = 17/5. LOWEST TERMS: A fraction is in lowest terms when the Greatest Common Factor (GCF) of the numerator and denominator is 1. Always express final answers in lowest terms.

Examples

This problem requires addition of unlike fractions and subtraction from the whole. The LCD ensures we compare equal-sized parts. The 'whole' is represented as 12/12 when the LCD is 12.

Scenario

A Grade 4 pupil spent 1/3 of his ₱120 baon on snacks and 1/4 on a notebook. What fraction of his baon was left?

Solution

Step 1: Find what fraction was spent. 1/3 + 1/4 — LCD of 3 and 4 is 12. = 4/12 + 3/12 = 7/12 Step 2: Subtract from 1 (the whole baon). 1 - 7/12 = 12/12 - 7/12 = 5/12 Answer: 5/12 of his baon was left.

Adding mixed numbers by first converting to improper fractions avoids errors when the sum of the fractional parts exceeds 1. Converting back to a mixed number gives a practical, readable answer.

Scenario

A dressmaker needs 2 1/3 meters of cloth for a school uniform top and 1 1/2 meters for a skirt. How many meters does she need in all?

Solution

Step 1: Convert to improper fractions. 2 1/3 = 7/3 1 1/2 = 3/2 Step 2: Find LCD of 3 and 2, which is 6. 7/3 = 14/6 3/2 = 9/6 Step 3: Add. 14/6 + 9/6 = 23/6 Step 4: Convert back to mixed number. 23/6 = 3 5/6 meters Answer: 3 5/6 meters of cloth.

Dividing by a fraction answers 'how many groups of this size fit into this amount?' Cross-cancellation before multiplying simplifies arithmetic significantly.

Scenario

A science teacher divides 3/4 kg of baking soda equally into 3/8-kg portions for an experiment. How many portions can she make?

Solution

Step 1: Divide 3/4 by 3/8 (KCF method). 3/4 ÷ 3/8 = 3/4 × 8/3 Step 2: Cross-cancel: 3 in numerator and 3 in denominator cancel; 8 in numerator and 4 in denominator — 8÷4=2. = 1/1 × 2/1 = 2 Answer: She can make 2 portions.

Applications

Computing portions of a budget or allowance in school financial planning
Measuring cloth or ingredients for Home Economics (TLE) classroom activities
Calculating what fraction of pupils attended, passed, or failed in a class record
Solving map-scale problems where distances are expressed as fractions
Determining how to share materials among groups in cooperative learning

Misconceptions

Adding denominators when adding fractions (e.g., 1/2 + 1/3 ≠ 2/5). Denominators are NEVER added.
Forgetting to find the LCD and simply adding numerators and denominators separately.
Not converting mixed numbers to improper fractions before dividing, leading to wrong answers.
Thinking that dividing by a fraction makes the result smaller (dividing by a fraction less than 1 gives a LARGER result).
Forgetting to simplify the final answer to lowest terms.

Related Concepts

Least Common Multiple (LCM) for finding LCD
Greatest Common Factor (GCF) for simplifying fractions
Equivalent fractions
Decimals and percent (converting from fractions)
Ratio (a fraction comparing two quantities)

Common Exam Questions

Example

A teacher spent 2/5 of his salary on food, 1/4 on rent, and 1/10 on utilities. What fraction is left? LCD=20: 8/20+5/20+2/20=15/20=3/4 spent. Left: 1-3/4=1/4.

Approach

Add all fractional parts spent, then subtract from 1 (the whole). Use LCD for unlike denominators.

Question Type

Finding the remaining fraction after spending

Example

3 1/2 ÷ 1 3/4 = 7/2 ÷ 7/4 = 7/2 × 4/7 = 28/14 = 2.

Approach

Convert to improper fractions, perform the operation, convert back to mixed number and simplify.

Question Type

Operations on mixed numbers in context

Example

A class of 40 pupils — 3/5 are girls. How many are girls? 3/5 × 40 = 24 girls.

Approach

Identify the fraction and the whole, then multiply.

Question Type

Word problems using 'of' to mean multiply

Key Points To Remember

Addition and subtraction require a common denominator (LCD); multiplication and division do not.
To divide fractions, multiply by the reciprocal of the divisor (Keep, Change, Flip).
Convert mixed numbers to improper fractions before multiplying or dividing.
Cross-cancel common factors before multiplying to simplify computation.
Always reduce the final answer to lowest terms using the GCF.
The word 'OF' in a problem means MULTIPLY (e.g., 3/4 of 20 = 3/4 × 20 = 15).
To find the LCD, use prime factorization or the Venn diagram method for two or more denominators.

Operations on Decimals

Decimals are another way of writing fractions whose denominators are powers of 10 (10, 100, 1000, etc.). The decimal point separates the whole-number part from the fractional part. Tenths, hundredths, thousandths are the place values immediately to the right of the decimal point. ADDITION AND SUBTRACTION: Align the decimal points vertically, filling in zeros as placeholders where needed. Then add or subtract exactly as with whole numbers, and bring the decimal point straight down into the answer. MULTIPLICATION: (1) Ignore the decimal points and multiply the numbers as if they were whole numbers. (2) Count the total number of decimal places in BOTH factors combined. (3) Place the decimal point in the product so that it has exactly that many decimal places, counting from the right. DIVISION: (1) If the divisor has a decimal, move its decimal point to the right until it becomes a whole number. (2) Move the dividend's decimal point the SAME number of places to the right. (3) Divide as with whole numbers. (4) Place the decimal point directly above its new position in the dividend. ROUNDING: To round a decimal, look at the digit immediately to the RIGHT of the desired place. If it is 5 or more, round up; if it is 4 or less, round down (keep the digit as is).

Examples

Aligning the decimal points ensures each digit is in the correct place value column. The zero in 128.50 is a placeholder and does not change the value.

Scenario

A school canteen sold snacks worth ₱45.75, ₱128.50, and ₱9.05 in three breaks. What was the total sales?

Solution

Step 1: Align decimal points. 45.75 128.50 + 9.05 ——————— 183.30 Answer: ₱183.30 total sales.

Moving the decimal of both numbers the same number of places preserves the value of the quotient. This is equivalent to multiplying both by 100.

Scenario

A piece of ribbon 0.6 m long is cut into pieces 0.15 m each for a classroom bulletin board. How many pieces can be cut?

Solution

Step 1: Divide 0.6 ÷ 0.15. Move both decimals 2 places right: 60 ÷ 15 = 4. Answer: 4 pieces.

Writing 85.5 as 85.50 (adding a placeholder zero) makes alignment easier. Division of decimals by a whole number follows the same process as whole-number division.

Scenario

A pupil's grades in three subjects were 85.5, 91.25, and 88.75. What is the average grade?

Solution

Step 1: Add all grades. 85.5 + 91.25 + 88.75 = 85.50 + 91.25 + 88.75 = 265.50 Step 2: Divide by 3. 265.50 ÷ 3 = 88.50 Answer: Average grade is 88.50.

Applications

Computing grades and grade averages in the class record
Calculating total expenses and change in school fund transactions
Measuring lengths, weights, and volumes in Science and TLE experiments
Reading and interpreting data in tables with decimal values
Computing salaries, deductions, and net pay in professional contexts

Misconceptions

Not aligning decimal points when adding or subtracting (treating the decimal as irrelevant).
Placing the decimal point incorrectly in multiplication (common error: matching decimal places of only one factor).
Moving only the divisor's decimal point but forgetting to move the dividend's point by the same amount.
Thinking 0.9 > 0.90 or that more decimal digits always means a larger number (0.09 < 0.9).

Related Concepts

Place value system
Fractions with denominators of 10, 100, 1000
Percent (decimal × 100)
Rounding and estimation
Money computation in Philippine pesos

Common Exam Questions

Example

Cost of 3.5 kg of rice at ₱54.50 per kg: 3.5 × 54.50 = 190.75. (3.5 has 1 decimal place, 54.50 has 2, so product has 3: 190.750 = ₱190.75)

Approach

Multiply ignoring decimal points, count total decimal places, place in product.

Question Type

Multiplication of decimals (finding total cost)

Example

₱337.50 ÷ 4.5 kg = 3375 ÷ 45 = ₱75 per kg.

Approach

Convert divisor to whole number by moving decimal, apply same move to dividend.

Question Type

Division of decimals (finding unit rate or number of pieces)

Key Points To Remember

ALWAYS align decimal points when adding or subtracting — this is the most common source of errors.
For multiplication: decimal places in answer = sum of decimal places in both factors.
For division: make the divisor a whole number by moving the decimal point right, and move the dividend's point the same number of places.
Adding zeros after the last decimal digit does not change the value (0.5 = 0.50 = 0.500).
Multiplying a decimal by 10, 100, or 1000 moves the decimal point 1, 2, or 3 places to the RIGHT.
Dividing a decimal by 10, 100, or 1000 moves the decimal point 1, 2, or 3 places to the LEFT.

Converting Among Fractions, Decimals, and Percent

Percent means 'per hundred' — it is a ratio with a denominator of 100. The ability to move fluently among fractions, decimals, and percents is essential because LET problems often mix the three forms in a single item. CONVERSION RULES: 1. FRACTION to DECIMAL: Divide the numerator by the denominator. Example: 3/8 = 3 ÷ 8 = 0.375 2. DECIMAL to PERCENT: Multiply by 100 (move the decimal point 2 places to the RIGHT, then add the % sign). Example: 0.375 × 100 = 37.5% 3. PERCENT to DECIMAL: Divide by 100 (move the decimal point 2 places to the LEFT, remove the % sign). Example: 37.5% ÷ 100 = 0.375 4. PERCENT to FRACTION: Write the percent value over 100, then simplify. Example: 75% = 75/100 = 3/4 5. FRACTION to PERCENT: Convert to decimal first (divide), then multiply by 100. Example: 3/4 = 0.75 = 75% 6. DECIMAL to FRACTION: Write the decimal as a fraction over the appropriate power of 10, then simplify. Example: 0.6 = 6/10 = 3/5; 0.125 = 125/1000 = 1/8 BENCHMARK FRACTIONS: Memorize the most common equivalents to save time on the exam.

Examples

Converting to a single form (decimals) allows direct comparison. This technique also works for problems asking which value is largest or smallest.

Scenario

Arrange from least to greatest: 3/5, 0.58, 61%, 11/20.

Solution

Step 1: Convert all to decimals. 3/5 = 0.60 0.58 = 0.58 61% = 0.61 11/20 = 0.55 Step 2: Order. 0.55 < 0.58 < 0.60 < 0.61 Answer: 11/20, 0.58, 3/5, 61%.

When the percent has a decimal part, multiplying by a power of 10 creates a whole-number fraction before simplifying. This is a common LET technique.

Scenario

Express 37.5% as a fraction in lowest terms.

Solution

Step 1: Write over 100. 37.5% = 37.5/100 Step 2: Eliminate the decimal by multiplying numerator and denominator by 10. = 375/1000 Step 3: Simplify using GCF. GCF(375, 1000) = 125 375 ÷ 125 = 3 1000 ÷ 125 = 8 Answer: 3/8.

Applications

Reading quiz scores expressed as percent and converting to fraction for grading purposes
Converting nutritional information on food labels (often in grams and percent) for Health/Science
Interpreting statistical data in DepEd reports (graduation rates, dropout rates expressed as percent)
Converting map scale ratios for Geography/Sibika
Understanding VAT and discount rates in TLE Home Management lessons

Misconceptions

Moving the decimal point to the RIGHT when converting percent to decimal (should move LEFT).
Forgetting to simplify when converting percent to fraction (leaving 25/100 instead of 1/4).
Thinking 1/3 = 0.33 exactly — it is a repeating decimal 0.333... or 33⅓%.
Confusing 'per cent' (out of 100) with 'per thousand' or other ratios.

Related Concepts

Fractions and equivalent fractions
Place value and powers of 10
Ratio (percent is a special ratio with denominator 100)
GCF for simplifying converted fractions
Ordering and comparing numbers

Common Exam Questions

Example

Which is greatest: 2/3, 0.67, 65%? Convert: 2/3≈0.667, 0.67, 65%=0.65. Greatest: 0.67.

Approach

Convert all values to decimals, order them, then re-label with original forms.

Question Type

Ordering mixed forms from least to greatest or greatest to least

Example

12.5% = 12.5/100 = 125/1000 = 1/8.

Approach

Write over 100, multiply to clear any decimal in the percent, then simplify with GCF.

Question Type

Expressing a percent as a fraction in lowest terms

Key Points To Remember

Percent to decimal: DIVIDE by 100 (move decimal 2 LEFT). Decimal to percent: MULTIPLY by 100 (move decimal 2 RIGHT).
Fraction to decimal: divide numerator by denominator.
Percent to fraction: place over 100 and simplify using GCF.
Benchmark equivalents to memorize: 1/2=0.5=50%; 1/4=0.25=25%; 3/4=0.75=75%; 1/3=0.333=33⅓%; 1/5=0.2=20%; 1/8=0.125=12.5%; 1/10=0.1=10%.
When comparing mixed forms (fractions, decimals, percents), convert everything to DECIMALS first.
A percent greater than 100% means more than the whole (e.g., 150% = 1.5 = 3/2).

The Three Percentage Cases

Every percentage word problem on the LET can be classified into one of three cases. All three are based on one master formula: Rate × Base = Part Where: - RATE (R) = the percent (expressed as a decimal in calculations) - BASE (B) = the whole or total amount (the number that follows 'of' or 'of what') - PART (P) = the portion or percentage of the base THREE CASES: CASE 1 — FIND THE PART: P = R × B 'What is 30% of 250?' P = 0.30 × 250 = 75 CASE 2 — FIND THE RATE: R = P ÷ B (then convert to percent) '36 is what percent of 150?' R = 36 ÷ 150 = 0.24 = 24% CASE 3 — FIND THE BASE: B = P ÷ R '18 is 25% of what number?' B = 18 ÷ 0.25 = 72 IDENTIFYING THE BASE: The BASE always follows the word 'OF' or 'OF WHAT NUMBER.' It is the reference quantity — the original or total. The part is the result of applying the rate to the base. PERCENT INCREASE OR DECREASE: Percent Change = (Change ÷ Original) × 100 For increase: Change = New - Original For decrease: Change = Original - New CRITICAL: The ORIGINAL amount is ALWAYS the base, never the new amount.

Examples

The base (B) is 800 (total pupils). The rate (R) is 65% = 0.65. Multiply to find the part.

Scenario

Case 1: A school has 800 pupils. If 65% attended a program, how many pupils were present?

Solution

P = R × B P = 0.65 × 800 P = 520 pupils Answer: 520 pupils were present.

The base (B) is 60 (total items). The part (P) is 48. Divide part by base, then convert to percent.

Scenario

Case 2: Out of 60 test items, a pupil got 48 correct. What is his percentage score?

Solution

R = P ÷ B R = 48 ÷ 60 R = 0.80 R = 80% Answer: His percentage score is 80%.

The part (P) is the increase amount ₱3,750. The rate (R) is 15% = 0.15. The base (B) — the original salary — is found by dividing.

Scenario

Case 3: A teacher received a 15% salary increase amounting to ₱3,750. What was her original salary?

Solution

B = P ÷ R B = 3,750 ÷ 0.15 B = ₱25,000 Answer: Her original salary was ₱25,000.

The ORIGINAL price (₱360) is the base. Always divide the change by the original, not the new amount. Dividing by 450 would give the wrong answer.

Scenario

Percent Increase: A textbook that originally cost ₱360 now costs ₱450. What is the percent increase?

Solution

Step 1: Find the change. Change = 450 - 360 = ₱90 Step 2: Divide by the original (base). 90 ÷ 360 = 0.25 Step 3: Convert to percent. 0.25 × 100 = 25% Answer: 25% increase.

Applications

Computing quarterly test scores and MPS (Mean Percentage Score) in DepEd class monitoring
Calculating discounts during school supplies shopping
Determining VAT on purchases for school projects
Computing salary increases, deductions, and net pay for teachers
Interpreting DepEd enrollment and dropout rate statistics

Misconceptions

Using the NEW value instead of the ORIGINAL as the base when computing percent change.
Forgetting to convert percent to decimal before multiplying (using 30 instead of 0.30).
Adding percent change rates directly without considering different bases (see successive percents).
Confusing 'part' and 'base' when the part is larger than 100% of the original.
When finding the base, dividing the base by the part instead of the part by the rate.

Related Concepts

Fractions and decimals (rate as a decimal)
Successive percents (applying percent changes sequentially)
Simple interest (a Case 1 application: I = P × r × t)
VAT and discount in everyday transactions
Ratio (percent as a ratio per 100)

Common Exam Questions

Example

72 pupils passed, representing 90% of the class. How many pupils are in the class? B = 72 ÷ 0.90 = 80 pupils.

Approach

Recognize that the given number is the PART and the percent is the RATE. Use B = P ÷ R. Convert percent to decimal first.

Question Type

Find the base (most commonly tricky LET case)

Example

Enrollment dropped from 250 to 200. Percent decrease = (50 ÷ 250) × 100 = 20%.

Approach

Compute the change (new minus original or original minus new). Divide by the ORIGINAL. Multiply by 100.

Question Type

Percent change (increase or decrease)

Example

A ₱600 school bag is 25% off. Sale price = 0.75 × 600 = ₱450.

Approach

Discount = rate × original price. Sale price = original - discount. Shortcut: sale price = (1 - rate) × original.

Question Type

Discount and sale price

Key Points To Remember

Formula: Rate × Base = Part. Rearrange for each case: P=R×B; R=P÷B; B=P÷R.
The BASE is the whole or original amount — it follows the word 'OF' or 'OF WHAT.'
Convert the rate to a decimal before calculating (divide percent by 100).
Percent change always uses the ORIGINAL value as the base, not the new value.
A shortcut for discounts: if the discount is 20%, the customer pays 80% (multiply by 0.80).
A shortcut for VAT: if VAT is 12%, total = price × 1.12.
When the PART is larger than the BASE, the rate is greater than 100%.

Ratio and Proportion

A RATIO is a comparison of two or more quantities of the same kind, expressed in the form a:b, a/b, or 'a to b.' Ratios must always compare quantities in the same unit. If units differ, convert first. RATIO IN LOWEST TERMS: Divide all terms by their GCF. Example: 12:18 → GCF is 6 → 2:3. A PROPORTION is a statement that two ratios are equal: a/b = c/d, also written a:b = c:d. The cross-multiplication property states that if a/b = c/d, then a × d = b × c (the product of the means equals the product of the extremes). THREE TYPES OF PROPORTION: 1. DIRECT PROPORTION: As one quantity increases, the other increases at the same rate (and vice versa). Relationship: y/x = constant, or x₁/y₁ = x₂/y₂. Example: More pupils → more chairs needed. Method: Set ratios in the SAME ORDER and cross-multiply. 2. INVERSE (INDIRECT) PROPORTION: As one quantity increases, the other decreases proportionally. Relationship: x × y = constant, or x₁ × y₁ = x₂ × y₂. Example: More workers → fewer days to finish a job. Method: The products of corresponding values are EQUAL. 3. PARTITIVE PROPORTION: A whole quantity is divided into parts according to a given ratio. Method: Find the total number of ratio parts, compute one part's value, then multiply each ratio term by that value. SCALE DRAWINGS AND MAPS: The scale (e.g., 1 cm : 50,000 cm) is a ratio of map distance to actual distance. Cross-multiply to find actual distance from map distance, or vice versa.

Examples

As servings increase, cups needed increase — direct proportion. Keep the same quantity on the same side: servings on one side, cups on the other. Cross-multiplying gives a simple equation.

Scenario

Direct Proportion: A recipe for 6 servings of champorado needs 1½ cups of chocolate. How many cups are needed for 10 servings?

Solution

Step 1: Set up the proportion (servings and cups in same order). 6/1.5 = 10/x Step 2: Cross-multiply. 6x = 1.5 × 10 = 15 Step 3: Solve for x. x = 15 ÷ 6 = 2.5 cups Answer: 2.5 cups of chocolate.

Fewer workers means MORE days — quantities move in opposite directions (inverse). The constant product (48) is the total 'worker-days' of effort required.

Scenario

Inverse Proportion: 6 workers can repaint a school building in 8 days. How many days would 4 workers need at the same rate?

Solution

Step 1: Use inverse proportion (workers × days = constant). 6 × 8 = 4 × d Step 2: Solve. 48 = 4d d = 48 ÷ 4 = 12 days Answer: 4 workers would take 12 days.

Partitive proportion divides a total into shares proportional to each ratio term. Always verify your answer by summing the shares — they must equal the original total.

Scenario

Partitive Proportion: Divide ₱2,400 among three teachers for classroom materials in the ratio 1:2:3. How much does each teacher receive?

Solution

Step 1: Find total parts. 1 + 2 + 3 = 6 parts Step 2: Value of one part. ₱2,400 ÷ 6 = ₱400 Step 3: Multiply each ratio term by ₱400. Teacher 1: 1 × ₱400 = ₱400 Teacher 2: 2 × ₱400 = ₱800 Teacher 3: 3 × ₱400 = ₱1,200 Check: 400 + 800 + 1,200 = ₱2,400 ✓ Answer: ₱400, ₱800, and ₱1,200.

Map scale is a direct proportion. The scale ratio tells you how many actual units equal one map unit. Multiply the map distance by the scale factor, then convert units.

Scenario

Map Scale: On a map with scale 1:200,000, two barangays are 3.5 cm apart. What is the actual distance in kilometers?

Solution

Step 1: Actual distance in cm. 1 cm on map = 200,000 cm actual 3.5 cm × 200,000 = 700,000 cm Step 2: Convert cm to km. 700,000 cm ÷ 100,000 cm/km = 7 km Answer: The actual distance is 7 km.

Applications

Dividing class seats, materials, or time using partitive proportion
Reading and computing distances on maps in Sibika and Araling Panlipunan
Planning class schedules where more teachers reduce individual load (inverse)
Scaling up or down recipes in TLE/Home Economics (direct proportion)
Computing recommended pupil-teacher ratios per DepEd standards

Misconceptions

Setting up a direct proportion equation with quantities in different order (e.g., putting workers on opposite sides from days in a direct proportion).
Treating an inverse proportion as a direct proportion by setting up equal ratios instead of equal products.
Forgetting to convert units before writing a ratio (e.g., comparing cm and m without conversion).
In partitive proportion, dividing by the number of people instead of the sum of ratio terms.
Not verifying that the shares in partitive proportion add up to the original total.

Related Concepts

Fractions (ratio as a fraction)
Percent (percent as a ratio per 100)
Scale drawing and maps
Rate (a ratio comparing different units)
Direct, inverse, and partitive variation in algebra

Common Exam Questions

Example

If 5 books cost ₱375, how much do 8 books cost? 5/375 = 8/x → 5x = 3000 → x = ₱600.

Approach

Identify the two quantities. If both change in the same direction, set up a/b = c/x (same quantity same side). Cross-multiply and solve.

Question Type

Direct proportion word problem

Example

10 machines produce 500 units in 3 hours. How long for 6 machines? 10×3 = 6×t → t = 5 hours.

Approach

Identify that one quantity increases while the other decreases. Set up x₁ × y₁ = x₂ × y₂. Solve for the unknown.

Question Type

Inverse proportion word problem

Example

Split ₱900 in ratio 2:3:4. Total parts=9. One part=₱100. Shares: ₱200, ₱300, ₱400.

Approach

Sum the ratio terms to find total parts. Divide the whole by total parts to find value of one part. Multiply each term by one part's value.

Question Type

Partitive proportion with three or more parts

Key Points To Remember

Ratios must compare quantities in the SAME UNIT — convert before writing the ratio.
Always simplify a ratio to lowest terms by dividing all terms by their GCF.
Cross-multiplication: in a/b = c/d, the cross products a×d and b×c are EQUAL.
Direct proportion: both quantities move in the SAME direction. Set ratios in same order.
Inverse proportion: quantities move in OPPOSITE directions. Products of pairs are equal.
Partitive proportion: Total parts = sum of all ratio terms. One part = total ÷ sum of ratio terms.
Map scale problems are direct proportion problems solved by cross-multiplication.
A ratio of 50 cm to 2 m must be converted: 50 cm to 200 cm = 1:4.

Rate and Mixture Applications

A RATE is a special ratio that compares quantities of DIFFERENT kinds (unlike ratio, which compares same-kind quantities). Common rates include: kilometers per hour (speed), pesos per kilogram (unit price), pupils per class (density), liters per minute (flow). UNIT RATE: A rate with a denominator of 1. To find the unit rate, divide the numerator by the denominator. Example: If 4.5 kg of bangus costs ₱495, the unit rate is ₱495 ÷ 4.5 = ₱110 per kg. SPEED-DISTANCE-TIME (D = R × T): Distance (D) = Rate (R) × Time (T) Rate = Distance ÷ Time Time = Distance ÷ Rate This is the foundational formula for all motion problems. BEST BUY PROBLEMS: Compare unit rates. The lower unit price is the better buy. MIXTURE PROBLEMS: These involve combining quantities with different concentrations or prices. The key equation is: (quantity₁ × value₁) + (quantity₂ × value₂) = (total quantity × average value) WORK-RATE PROBLEMS: If a person completes a job in 'n' days, his/her work rate per day is 1/n of the job. Combined work rate = sum of individual rates. Time to finish together = 1 ÷ (combined rate).

Examples

Comparing unit prices (price per gram) puts both options on equal footing. The lower unit price always gives more value for money.

Scenario

Best Buy: A 500 g pack of powdered milk costs ₱85, and a 750 g pack costs ₱120. Which is the better buy?

Solution

Unit price of 500 g pack: ₱85 ÷ 500 = ₱0.170 per gram Unit price of 750 g pack: ₱120 ÷ 750 = ₱0.160 per gram Answer: The 750 g pack is the better buy (₱0.16 < ₱0.17 per gram).

The formula D = R × T has three variables. When given two (D and T), solve for the third (R). Always label units in the answer.

Scenario

Speed Problem: A school service travels 120 km in 2.5 hours. What is its average speed?

Solution

R = D ÷ T R = 120 km ÷ 2.5 hours R = 48 km/h Answer: The average speed is 48 km/h.

Work rate problems use the reciprocal relationship between time and rate. Adding rates gives the combined rate. The time to complete the whole job is the reciprocal of the combined rate.

Scenario

Work Rate: Teacher A can arrange 60 school chairs in 30 minutes. Teacher B can do the same in 20 minutes. How long will they take working together?

Solution

Step 1: Find individual rates per minute. A's rate: 1/30 of the job per minute B's rate: 1/20 of the job per minute Step 2: Combined rate. 1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12 Step 3: Time together = 1 ÷ (1/12) = 12 minutes. Answer: Together they take 12 minutes.

Applications

Comparing prices at the school canteen or local market (best buy)
Computing travel time for school field trips using D = R × T
Mixing ingredients of different concentrations in Science or TLE
Estimating how long multiple teachers or students can complete a task
Computing fuel efficiency (km per liter) for school service vehicles

Misconceptions

Averaging two speeds to find average speed for a round trip — use total distance ÷ total time instead.
Forgetting to convert units before applying D = R × T (e.g., mixing km and minutes).
In work-rate problems, adding times instead of rates (adding 30+20=50 minutes is WRONG).
Choosing the bigger package as 'better buy' without computing the unit price.
Confusing rate (different units) with ratio (same units).

Related Concepts

Ratio and proportion
Fractions (work rates as fractions)
Algebra (solving equations with one variable)
Unit conversion (km to m, minutes to hours)
Percent and mixture problems

Common Exam Questions

Example

3 notebooks for ₱75 vs 5 notebooks for ₱115. Unit prices: ₱25 vs ₱23. Five-pack is better buy.

Approach

Compute cost per unit (gram, mL, piece) for each option. The lower value is the better buy.

Question Type

Best buy (comparing unit rates)

Example

How long to travel 180 km at 60 km/h? T = 180 ÷ 60 = 3 hours.

Approach

Identify which of D, R, or T is unknown. Use D=R×T and solve. Watch units (km vs m, hr vs min).

Question Type

Distance-Rate-Time problem

Example

A finishes in 4 hours, B in 6 hours. Combined: 1/4+1/6=5/12 per hour. Time=12/5=2.4 hours=2 hr 24 min.

Approach

Find 1/n rate for each worker. Add the rates. Take the reciprocal for total time.

Question Type

Combined work rate

Key Points To Remember

Rate compares DIFFERENT units; ratio compares SAME units.
Unit rate = value ÷ quantity (denominator becomes 1). Use this for 'best buy' comparisons.
D = R × T: memorize this and its two rearrangements (R = D÷T and T = D÷R).
For best buy, compute unit price for each option; lower unit price = better deal.
Mixture equation: (q₁ × v₁) + (q₂ × v₂) = total quantity × average value.
Work rate per day: if you finish in n days, your rate is 1/n per day.
Average speed for a round trip is NOT the average of the two speeds; use total distance ÷ total time.

Successive Percents and Percent Change

This is one of the most common TRAP topics in the LET. When two or more percent changes are applied one after another, each change acts on the RESULT of the previous step — NOT on the original amount. This means the percents CANNOT simply be added together. SUCCESSIVE DISCOUNTS: If an item is discounted by 10%, then by 20%: New price = Original × (1 - 0.10) × (1 - 0.20) = Original × 0.90 × 0.80 = Original × 0.72 So the overall effect is a 28% discount, NOT 10%+20%=30%. SUCCESSIVE INCREASE THEN DECREASE (or vice versa): A 20% decrease followed by a 20% increase does NOT return to the original. Start: 100 → After 20% decrease: 80 → After 20% increase: 80 × 1.20 = 96 (NOT 100). FORMULA FOR OVERALL EFFECT OF TWO SUCCESSIVE PERCENTS: If the two rates are a% and b% (both increases: use +; discount: use -): Overall percent = a + b + (a × b)/100 For 10% off then 20% off: -10 + (-20) + [(-10)×(-20)]/100 = -30 + 2 = -28% (28% total discount) This formula works for any combination of increases and decreases; use negative values for decreases. SIMPLE INTEREST: I = P × r × t Where: I = Interest, P = Principal, r = annual rate (decimal), t = time in years. Amount = P + I = P(1 + rt) This is a straightforward Case 1 percentage application over time.

Examples

The second 10% is applied to ₱4,250, not ₱5,000. This is why 15% + 10% ≠ 25% overall. Using the formula: -15 + (-10) + [(-15)×(-10)/100] = -25+1.5 = -23.5% confirms the answer.

Scenario

An appliance costing ₱5,000 is discounted 15% for a sale, then an additional 10% for a teachers' festival promo. What is the final price?

Solution

Step 1: Apply first discount. ₱5,000 × (1 - 0.15) = ₱5,000 × 0.85 = ₱4,250 Step 2: Apply second discount on the REDUCED price. ₱4,250 × (1 - 0.10) = ₱4,250 × 0.90 = ₱3,825 Overall discount = (5,000 - 3,825) ÷ 5,000 × 100 = 23.5% Answer: Final price is ₱3,825 (overall 23.5% discount, NOT 25%).

Simple interest uses the ORIGINAL principal each year (not the growing amount). This distinguishes it from compound interest. The rate must be in decimal form.

Scenario

A teacher deposited ₱50,000 in a bank at 6% simple interest per year. How much interest does she earn in 3 years? What is the total amount?

Solution

Step 1: Compute interest. I = P × r × t I = ₱50,000 × 0.06 × 3 I = ₱9,000 Step 2: Total amount. A = P + I = ₱50,000 + ₱9,000 = ₱59,000 Answer: Interest = ₱9,000; Total = ₱59,000.

Applications

Evaluating sale offers with multiple successive discounts in school supply stores
Computing final salary after successive increases or deductions
Understanding why two equal-percentage raises and cuts do not cancel out
Simple interest calculations for loans or savings accounts of school cooperatives
Analyzing price changes in DepEd procurement using percent

Misconceptions

Adding successive percent rates directly (e.g., assuming 15% + 10% = 25% total discount).
Believing a percent decrease followed by an equal percent increase returns to the original.
Applying the second percent to the original price instead of the already-discounted price.
Confusing simple interest (rate applied to original principal each period) with compound interest.

Related Concepts

Three percentage cases (Rate × Base = Part)
Percent change (using original as base)
Compound interest (extension of successive percents)
Multiplication of decimals
Practical discounts and VAT in commerce

Common Exam Questions

Example

₱2,000 item: 20% off then 10% off. Final: 2000×0.80×0.90=₱1,440. Overall: (560÷2000)×100=28%.

Approach

Multiply original by (1-r₁)(1-r₂). Find overall discount by comparing final and original prices. Do NOT add the two rates.

Question Type

Successive discounts — finding final price or overall discount

Example

Price drops 25%, then rises 25%. Start: 100 → 75 → 93.75. Net: 6.25% below original.

Approach

Calculate step by step. Show that the two equal rates applied in opposite directions do NOT cancel.

Question Type

Percent decrease then increase (or vice versa) — asking if result equals original

Key Points To Remember

Successive percents are MULTIPLIED, not added. Apply each percent to the RESULT of the previous step.
A 20% discount and a 20% increase do NOT cancel out — the net result is a 4% LOSS.
Overall two-successive-percent = a + b + (a×b/100); use negative values for decreases.
Simple interest: I = Prt. The principal (P) is the base, r is the rate, t is time.
For compound discounts, multiply the factors: price × (1-r₁) × (1-r₂).
Always identify what the CURRENT base is before applying the next percent.

Practice Problems

First find what fraction was spent using LCD, then subtract from 1 to get remaining fraction, then multiply by total baon to convert to pesos.

Problem

1. A pupil spent 1/4 of his ₱200 baon on lunch, 1/5 on snacks, and 1/10 on transportation. How much money did he have left?

Solution

Step 1: Find the fraction spent. 1/4 + 1/5 + 1/10 — LCD of 4, 5, 10 is 20. = 5/20 + 4/20 + 2/20 = 11/20 Step 2: Fraction left = 1 - 11/20 = 9/20. Step 3: Amount left = 9/20 × ₱200 = ₱90. Answer: ₱90 was left.

Successive discounts are applied sequentially to the NEW (reduced) price, not the original. The rates cannot be added directly.

Problem

2. A shirt originally priced at ₱750 is on sale at 30% off. An additional 10% birthday discount applies. What is the final price?

Solution

Step 1: Apply 30% discount. ₱750 × 0.70 = ₱525 Step 2: Apply 10% on reduced price. ₱525 × 0.90 = ₱472.50 Answer: Final price is ₱472.50. Note: Overall discount = (750-472.50)/750 × 100 = 37%, NOT 40%.

This chains two Case 1 percentage problems. The result of the first (30 girls) becomes the base of the second. Be attentive to whether the problem expects rounding.

Problem

3. In a class of 50 pupils, 60% are girls. Of the girls, 75% belong to the Math Club. How many girls are in the Math Club?

Solution

Step 1: Find number of girls. 60% of 50 = 0.60 × 50 = 30 girls Step 2: Find girls in Math Club. 75% of 30 = 0.75 × 30 = 22.5 ≈ 22 or 23 If exact: 0.75 × 30 = 22.5 girls. Answer: 22 or 23 girls (depending on rounding context); mathematically 22.5.

Partitive proportion: sum all ratio terms, find value of one part, then multiply each ratio term. Always verify by summing all parts.

Problem

4. Three teachers share a prize in the ratio 3:4:5. If the prize is ₱36,000, how much does the teacher with the largest share receive?

Solution

Step 1: Total ratio parts. 3 + 4 + 5 = 12 parts Step 2: Value of one part. ₱36,000 ÷ 12 = ₱3,000 Step 3: Largest share (ratio 5). 5 × ₱3,000 = ₱15,000 Verification: 9,000 + 12,000 + 15,000 = 36,000 ✓ Answer: ₱15,000.

As the number of typists increases, time decreases — classic inverse proportion. Set up equal products (not equal ratios).

Problem

5. 8 typists can encode a document in 5 hours. How long will 10 typists take at the same rate?

Solution

More typists → less time → Inverse proportion. 8 × 5 = 10 × t 40 = 10t t = 4 hours Answer: 10 typists take 4 hours.

Reduce both options to unit rates (price per pencil). The smaller unit price gives more value.

Problem

6. A school store sells 3 pencils for ₱18 or 5 pencils for ₱28. Which is the better buy?

Solution

Unit price of 3-pencil set: ₱18 ÷ 3 = ₱6.00 per pencil Unit price of 5-pencil set: ₱28 ÷ 5 = ₱5.60 per pencil Answer: The 5-pencil set is the better buy at ₱5.60 per pencil.

Always convert time to the same unit as the rate denominator (hours in km/h). 36 minutes = 36/60 = 0.6 hours.

Problem

7. A teacher drove 156 km in 2 hours and 36 minutes for a school division seminar. What was her average speed in km/h?

Solution

Step 1: Convert time to hours. 2 hr 36 min = 2 + 36/60 = 2 + 0.6 = 2.6 hours Step 2: Apply R = D ÷ T. R = 156 ÷ 2.6 = 60 km/h Answer: Average speed is 60 km/h.

Alternatively: 0.5625 × 16 = 9, and 1 × 16 = 16, confirming 9/16. Knowing that 1/16=0.0625 helps verify.

Problem

8. Express 0.5625 as a fraction in lowest terms.

Solution

Step 1: Write as a fraction over 10,000 (4 decimal places). 0.5625 = 5625/10000 Step 2: Find GCF(5625, 10000). 5625 = 5⁴ × 3² = 625 × 9 10000 = 10⁴ = 2⁴ × 5⁴ GCF = 5⁴ = 625 Step 3: Simplify. 5625 ÷ 625 = 9 10000 ÷ 625 = 16 Answer: 9/16.

First find total interest (final amount minus principal). Then use I=Prt, substituting known values for I, P, and r, and solving for t.

Problem

9. A principal invested ₱80,000 at 8% simple interest per year. After how many years will the investment reach ₱99,200?

Solution

Step 1: Find the interest earned. I = 99,200 - 80,000 = ₱19,200 Step 2: Use I = Prt and solve for t. 19,200 = 80,000 × 0.08 × t 19,200 = 6,400t t = 19,200 ÷ 6,400 = 3 years Answer: 3 years.

Scale 1:500,000 means 1 cm on map = 500,000 cm actual. Multiply map distance by scale factor, then convert to desired units (100 cm = 1 m; 1,000 m = 1 km; so 100,000 cm = 1 km).

Problem

10. On a map with scale 1:500,000, the distance between two cities measures 6.4 cm. What is the actual distance in km?

Solution

Step 1: Actual distance in cm. 6.4 cm × 500,000 = 3,200,000 cm Step 2: Convert to km. 3,200,000 cm ÷ 100,000 cm per km = 32 km Answer: The actual distance is 32 km.

Exam Preparation Tips

MEMORIZE BENCHMARK CONVERSIONS: Know these cold — 1/2=0.5=50%, 1/4=0.25=25%, 3/4=0.75=75%, 1/3=33⅓%, 1/8=0.125=12.5%, 1/5=0.2=20%. These save 30+ seconds per item.
IDENTIFY THE THREE PERCENT CASES QUICKLY: Find Part (×), Find Rate (÷ by base), Find Base (÷ by rate). Spot the BASE by finding the word 'OF' or 'OF WHAT NUMBER' in the problem.
SET UP PROPORTIONS IN THE SAME ORDER: In direct proportion, put the same type of quantity on the same side. Write: (quantity A₁)/(quantity A₂) = (quantity B₁)/(quantity B₂). Cross-multiply to solve.
NEVER ADD SUCCESSIVE PERCENTS: When two percent changes are applied one after another, multiply the decimal factors. Example: 10% off then 20% off = ×0.90 × 0.80 = ×0.72 (28% off, not 30%).
PERCENT CHANGE: ALWAYS DIVIDE BY THE ORIGINAL. The base in percent change is always the original (starting) amount, never the new amount.
CONVERT UNITS BEFORE WRITING RATIOS: Ratios must compare the same unit. If given 50 cm and 2 m, convert to 50 cm and 200 cm first, then write the ratio (1:4).
USE CROSS-CANCELLATION IN FRACTION MULTIPLICATION: Cancel common factors before multiplying to keep numbers small and avoid arithmetic errors.
FOR INVERSE PROPORTION, SET UP EQUAL PRODUCTS: workers₁ × days₁ = workers₂ × days₂. This is the most common error — students set up equal ratios for inverse proportion, which is WRONG.
WORK-RATE PROBLEMS: Add RATES (1/n), not times. Take the reciprocal of the combined rate to get total time.
VERIFY ALL PROPORTION ANSWERS: After solving, substitute back into the original proportion and check that both sides are equal (cross-multiply to verify).
FOR PARTITIVE PROPORTION, ALWAYS VERIFY: The shares must sum to the original total. If they do not, you made an arithmetic error.
ALIGN DECIMAL POINTS FOR ADDITION/SUBTRACTION: Draw a vertical line through all the decimal points as a visual guide to avoid column misalignment.
D = R × T TRIANGLE: Draw a triangle with D on top, R and T on the bottom. Cover the unknown to see the formula: cover D → D=R×T; cover R → R=D/T; cover T → T=D/R.
LET WORD PROBLEM STRATEGY: (1) Read carefully and identify what is given and what is asked. (2) Identify the concept (fraction, percent, ratio). (3) Write the formula or equation. (4) Solve step by step. (5) Check if the answer is reasonable in context.
PRACTICE SPEED: LET multiple-choice items must be solved in about 1 minute each. Practice timed drills focusing on these topics. Use estimation to eliminate obviously wrong choices quickly.

Loading diagram…

In summary

Fractions, decimals, percent, ratio, and proportion are the cornerstones of the LET Mathematics component. As future elementary teachers, you will teach these concepts to Grades 1–6 pupils under the K-12 BEC, and under RA 7836, your professional competence in these areas is a requirement for licensure. The mastery pathway runs through five principles: (1) fluent conversion among the three numerical forms; (2) accurate identification of the three percentage cases; (3) correct distinction among direct, inverse, and partitive proportion; (4) awareness of the successive-percents trap; and (5) consistent use of unit rates to solve best-buy and speed problems. Use the Mermaid diagrams in this chapter as visual anchors — return to them for a quick conceptual refresher before your exam. Practice every worked example actively (cover the solution, solve it yourself, then check). With deliberate, structured practice grounded in Philippine classroom realities — peso prices, class sizes, map scales — you will approach your LET Mathematics questions with confidence, speed, and precision. Makakaasa ka: kaya mo ito!

Previous chapter

Numbers, Number Sense and the Four Operations

Next chapter

Elementary Algebra

Ready to practise for the LET Elementary 2026?

Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target LET Elementary exam date.

Start free See the full LET Elementary guide LET Elementary preparation guide

← Back to Mathematics hub