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ICSE • Class 6 • Mathematics • Ch 9
Estimated Time: 45-60 mins
Study Progress: In Progress

Unitary Method

The Unitary Method is a fundamental arithmetical technique used to solve problems involving proportional quantities by first determining the numerical value of a single unit (one item, one hour, one kilogram, or one litre) through division, and subsequently calculating the value of the required number of units through multiplication. The term 'Unitary' is derived from the word 'Unit', which mathematically signifies 'one' or a single entity. At its heart, the unitary method transforms complex comparative problems into an intuitive, two-step operational bridge: first contract from many items to one single item by dividing, and then expand from that single item to the target number of items by multiplying. This universal procedure serves as the foundational bridge between basic elementary arithmetic and the advanced mathematical fields of direct variation, ratio and proportion, commercial arithmetic, physics rate problems, and financial algebra. In the CISCE and ICSE Class 6 Mathematics curriculum, the study of the Unitary Method primarily centres on Direct Variation. Two related physical quantities are said to be in direct variation (or direct proportion) when an increase in the first quantity causes a strictly proportional increase in the second quantity, and conversely, a decrease in the first quantity causes a strictly proportional decrease in the second quantity, provided the basic rate remains constant throughout the interaction. Under direct variation, the quotient of the two quantities remains invariant and equal to a fixed positive constant k. For instance, if the number of identical pens purchased increases from 5 to 10, the total purchase cost doubles correspondingly. If a vehicle travels at a uniform speed, doubling the travel time doubles the total distance covered. Because the rate per unit remains unchanging, calculating the value of exactly one unit establishes the definitive benchmark upon which any future quantity can be reliably evaluated. Unitary method problems in Class 6 are systematically classified into two primary archetypes based on the target unknown: Type A Problems (Finding Total Value or Cost for a Given Quantity): In this category, the problem statement provides the total cost or value of a known number of items n, and asks the student to determine the total cost of a different number of items m. The standard algebraic formulation proceeds as follows: First, calculate the cost of 1 item by dividing the total given cost by n. Second, calculate the required cost by multiplying the cost of 1 item by the required number m. That is, Required Cost equals (Given Cost divided by n) multiplied by m. Whether m is greater than n (scaling upward) or m is less than n (scaling downward), the exact same sequential logic applies without modification. Type B Problems (Finding the Number of Articles for a Given Amount of Money or Budget): In this inverse perspective, the problem statement provides the total cost of n items, but specifies a fixed available monetary budget B, asking how many articles can be purchased with that sum. Here, students can either determine the cost of 1 article through division and subsequently divide the total budget B by that unit cost, or alternatively, calculate how many articles one single rupee can buy and multiply that fractional rate by B. Both computational routes lead to identical results, reinforcing the deep symmetry of mathematical reasoning. Beyond commercial purchasing of consumer goods, the Unitary Method serves as the foundational computational engine for uniform motion involving speed, distance, and time. Uniform speed is defined precisely as the distance traversed in one unit of time (such as 1 hour, 1 minute, or 1 second). Therefore, determining the distance a car travels in 1 hour by dividing total distance by total elapsed time is nothing other than finding the unitary rate of motion. Once this speed is established, multiplying it by any target duration yields the distance traversed, while dividing any target distance by this unit speed yields the required travel time. A critical area of emphasis in ICSE board examinations is unit homogenisation and the mastery of commercial counting measures. Often, word problems blend metric and commercial units deliberately to test conceptual attentiveness. Students must be thoroughly conversant with standard commercial conversion constants: 1 Dozen equals 12 individual units; 1 Score equals 20 units; and 1 Gross equals 12 Dozen, which equals 144 units. Furthermore, metric conversions between kilograms and grams (1 kg equals 1,000 g), litres and millilitres (1 L equals 1,000 mL), metres and centimetres (1 m equals 100 cm), and hours and minutes (1 hour equals 60 minutes) must be executed before performing division. Attempting to compute a unitary rate using mismatched units (such as dividing rupees by kilograms to calculate the price per gram without conversion) leads to catastrophic numerical errors. Finally, the ICSE syllabus introduces students to the conceptual boundary between Direct Variation and Inverse Variation. While Direct Variation follows the rule 'More items require More money, and Fewer items require Less money' (solved by dividing first, then multiplying), certain scenarios such as men working together or time taken at varying speeds follow Inverse Variation ('More workers take Less time, and Fewer workers take More time'). In Class 6, students learn to carefully inspect whether quantities increase together or move in opposite directions before blindly applying an operational rule. By cultivating this habit of reflective problem analysis, students develop robust mathematical maturity, structured algorithmic thinking, and flawless accuracy in board examinations.

Have You Ever Wondered?

Imagine you walk into a stationery store to buy notebooks for the new school term. The shopkeeper tells you that a pack of 6 premium notebooks costs 210 rupees. You only need 4 notebooks for your subjects, while your friend wants to buy 9 notebooks. How can both of you instantly determine the exact amount to pay without guessing or arguing? Similarly, suppose your family car travels 180 kilometres on 12 litres of petrol. How far can you drive on a full tank of 35 litres, and how many litres are needed for a 300-kilometre road trip? The secret key that effortlessly unlocks both of these real-life riddles is one of the most powerful, practical tools in all of arithmetic: The Unitary Method. By reducing any collection of articles, distance, or work down to the fundamental value of a single unit first, you gain the mathematical superpower to scale up or down to any required quantity with complete precision.

Why This Chapter Matters

The Unitary Method is without question one of the most practically pervasive and intellectually enduring concepts encountered in secondary school mathematics. While abstract algebraic theorems may feel distant from day-to-day existence, the unitary method is utilized actively by every individual on a daily basis. Whenever you compare prices in a supermarket to identify which cereal box offers the best value per gram, calculate how many litres of fuel your vehicle requires for a holiday road trip, estimate baking ingredients when doubling a dessert recipe for family guests, or convert currencies while planning international travel, you are executing the unitary method. It provides an indispensable mental framework for evaluating cost-efficiency and making informed consumer decisions. In the academic continuum of STEM education, the unitary method constitutes the direct stepping stone to several higher branches of science and mathematics. In physics, concepts such as speed (distance per unit time), density (mass per unit volume), electric current (charge per unit second), and pressure (force per unit area) are all unit rates established via the unitary principle. In chemistry, stoichiometry, molar mass calculations, and chemical reaction yields rely completely on unitary scaling of atoms and moles. In economics and commerce, simple interest, compound interest, profit and loss percentages, taxation rates, foreign currency exchange, and production labour efficiency all demand flawless fluency in unitary reasoning. Mastering this chapter in Class 6 ensures that a student builds rock-solid numerical confidence, avoiding common traps and laying an unbreakable foundation for secondary school excellence.

Before You Begin (Prerequisites)

  • Fluency in standard long division and multi-digit multiplication with natural numbers and whole numbers.
  • Basic operational understanding of proper, improper, and mixed fractions, including fraction simplification.
  • Familiarity with decimal numbers, decimal place values, and performing basic division and multiplication with decimals.
  • Knowledge of standard metric measurement conversion factors: 1 kg = 1,000 g, 1 L = 1,000 mL, 1 km = 1,000 m, 1 m = 100 cm.
  • Basic understanding of time unit conversions: 1 hour = 60 minutes, 1 minute = 60 seconds.

What You Will Learn (Core Objectives)

  • Articulate the precise mathematical meaning, etymology, and two-step operational architecture of the Unitary Method.
  • Distinguish Direct Variation from Inverse Variation by identifying whether related physical quantities scale in identical or opposite directions.
  • Execute Type A problems with complete algorithmic precision: calculate the unit value by division and find the total cost of any target quantity by multiplication.
  • Solve Type B problems fluently: calculate the maximum number of articles purchasable within a given monetary budget or constraint.
  • Apply the Unitary Method to uniform motion scenarios to determine vehicle speed, distance covered in arbitrary times, and time taken for specified distances.
  • Seamlessly integrate commercial counting units (1 Dozen = 12, 1 Score = 20, 1 Gross = 144) and metric conversion factors into multi-step word problems.
  • Handle fractional and decimal unit values accurately without premature rounding off that could distort the final answer.
  • Employ a systematic 4-step examination protocol: State Given Information, Find Unit Value, Multiply for Target Quantity, and Conduct Reasonableness Sanity Check.

Chapter Roadmap & Progression

1 Module 1: Concept, Etymology & Foun...
2 Module 2: Direct Variation — Concep...
3 Module 3: Type A Problems — Finding...
4 Module 4: Type B Problems — Finding...
5 Module 5: Uniform Motion, Speed, Di...
6 Module 6: Multi-Unit Measures & Con...
7 Module 7: Applied Commercial Mathem...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Concept, Etymology & Foundation of the Unitary Method

In everyday life and commercial arithmetic, we constantly encounter situations where the total value, cost, weight, or duration of a group of items is known, and we are tasked with finding the corresponding value for a different number of items. The mathematical method designed specifically to solve such proportional problems with complete accuracy is known as the Unitary Method.

Etymology and Definition

The word 'Unitary' originates from the root word 'Unit', which in mathematics means 'one' or a single individual entity.

Formal Definition: The Unitary Method is an arithmetical procedure in which we first determine the value of a single unit (one item, one hour, one kilogram, one litre) from the given value of multiple units through division, and then calculate the value of the required number of units through multiplication.

The Universal Two-Step Operational Pipeline

The unitary method acts as a two-phase operational bridge. It breaks down an otherwise daunting problem into two simple, sequential elementary calculations:

PHASE 1 : DIVISION

The Unit Reduction Phase

Transition from Many Items down to 1 Single Item. Whenever we move from a larger number of articles to a single article, the value decreases proportionately. Therefore, we DIVIDE:

Value of 1 Unit = (Total Given Value) ÷ (Given Number of Items n)
PHASE 2 : MULTIPLICATION

The Scaling Expansion Phase

Transition from 1 Single Item up to Target Items (m). Whenever we move from a single article to a larger number of articles, the value increases proportionately. Therefore, we MULTIPLY:

Value of m Units = (Value of 1 Unit) × (Required Number of Items m)

Why Subtraction Fails: Proportional vs Additive Thinking

A common beginner intuition when comparing quantities is to use addition or subtraction. For example, if 4 pencils cost ₹24 and someone asks for the cost of 7 pencils, an untrained student might think: "7 is 3 more pencils, so should I add 3 to ₹24?"

This is incorrect because physical articles in commerce scale multiplicatively, not additively. Each individual pencil possesses an inherent commercial value of ₹24 ÷ 4 = ₹6. Therefore, 7 pencils cost 7 × ₹6 = ₹42 (not ₹24 + 3 = ₹27). The unitary method systematically protects students from this additive trap by forcing an explicit intermediate calculation: the valuation of exactly one single item.

Key Pedagogical Takeaway:

Whenever you are given the value of many and need to find the value of many, you cannot leap directly across in one step. You must always pass through the unitary base (1 unit) as your arithmetical bridge!

Module 2: Direct Variation — Concept, Characteristics & Identification

Before blindly applying division and multiplication to any word problem, an ICSE student must examine the mathematical relationship connecting the two variables. The standard Unitary Method taught in Class 6 operates under the mathematical law of Direct Variation (also known as Direct Proportion).

Mathematical Law of Direct Variation

Two quantities x and y are said to vary directly if an increase in x causes a proportional increase in y, or a decrease in x causes a proportional decrease in y, such that their ratio (quotient) remains constant:

x / y = k     (where k is a positive constant known as the Constant of Variation or Unit Rate)

Equivalently: y = k × x. If quantity x is doubled, y doubles. If x is halved, y is halved.

Classic Real-World Manifestations of Direct Variation

Recognizing direct variation in word problems is intuitive once you consider the physical nature of the quantities involved:

Physical Domain Quantity 1 (x) Quantity 2 (y) Meaning of Unit Rate (k)
Commerce / Retail Number of Articles Total Cost (₹) Price per article (₹/unit)
Automobile / Travel Fuel Consumed (L) Distance Travelled (km) Fuel Mileage (km/L)
Uniform Motion Time Elapsed (hours) Distance Covered (km) Speed (km/h)
Labour / Employment Days Worked Total Wages Earned (₹) Daily Wage rate (₹/day)
Grocery / Weights Mass of Groceries (kg) Total Price (₹) Cost per kilogram (₹/kg)

The Diagnostic Test for Direct Variation

Before writing down your mathematical steps in an examination, always perform the Sanity Direction Check:

  • Ask yourself: "If I buy MORE items, will the total bill be MORE or LESS?" If MORE, it is Direct Variation.
  • Ask yourself: "If a car travels for MORE hours at constant speed, will it cover MORE or LESS distance?" If MORE, it is Direct Variation.
  • The Direct Rule: In all direct variation scenarios, find 1 unit by division, then find required units by multiplication.
Examiner Alert (Inverse Variation Contrast):

Be cautious when questions involve number of workers and time taken to complete a job. If 5 workers build a wall in 12 days, will 10 workers take MORE days? No! More workers finish faster (LESS days). That is Inverse Variation, where 1 worker takes 5 × 12 = 60 days (multiplication first!). Always confirm the direction before dividing.

Module 3: Type A Problems — Finding Total Cost or Value for a Given Quantity

The most common problem type in Class 6 examinations is Type A: Given the cost or value of a certain number of articles, determine the total cost or value of a different number of articles.

Standard Algorithmic Template for Type A

Given: Cost of n articles = ₹C
To Find: Cost of m articles

Statement 1: Cost of n articles = ₹C
Statement 2: Cost of 1 article = ₹(C / n)         [Division]
Statement 3: Cost of m articles = ₹(C / n) × m    [Multiplication]

Case 1: Scaling Up (Required Quantity m is Greater than Given Quantity n)

When the target quantity is larger than the given quantity, the unit value is multiplied by a larger integer.

Pedagogical Example 1:

If 8 identical geometry boxes cost ₹360, what is the cost of 15 such geometry boxes?

• Cost of 8 geometry boxes = ₹360
• Cost of 1 geometry box = ₹360 ÷ 8 = ₹45
• Cost of 15 geometry boxes = 15 × ₹45 = ₹675
Final Answer: The cost of 15 geometry boxes is ₹675.

Case 2: Scaling Down (Required Quantity m is Smaller than Given Quantity n)

Students often mistakenly believe the unitary method is only for buying more. It works with identical elegance when reducing from a bulk purchase to a smaller quantity:

Pedagogical Example 2:

A wholesale pack of 25 files costs ₹875. A student only requires 6 files. What amount must the student pay?

• Cost of 25 files = ₹875
• Cost of 1 file = ₹875 ÷ 25 = ₹35
• Cost of 6 files = 6 × ₹35 = ₹210
Final Answer: The student must pay ₹210.

Crucial Strategy: Handling Fractional Unit Values

In many board questions, the cost of 1 article is not a whole number. When (C ÷ n) results in a non-terminating or recurring decimal, never round off prematurely! Instead, keep the unit cost as an unreduced or reduced common fraction:

Example: Suppose 14 notebooks cost ₹245. Find the cost of 18 notebooks.

Cost of 1 notebook = ₹(245 / 14). Simplifying by dividing both numerator and denominator by 7 gives: ₹(35 / 2).
Do not write 17.5 yet if you prefer working with fractions: Cost of 18 notebooks = 18 × (35 / 2) = 9 × 35 = ₹315.
Leaving it as a fraction allows common factors to cancel out effortlessly with the required number!

Module 4: Type B Problems — Finding the Number of Articles for a Given Amount of Money

In practical shopping and commercial transactions, we often face budget constraints where we do not ask "How much will m items cost?", but rather: "Given a fixed budget of ₹B, how many articles can I purchase?" This is categorized as a Type B Problem.

The Dual Mathematical Approaches for Type B

There are two mathematically valid pathways to structure Type B problems. Both yield identical answers, but Method 1 is the standard approach favored in ICSE board marking schemes:

Method 1: Cost of 1 Article First (Standard)

1. Find the cost of 1 article = (Given Cost C) ÷ n.
2. Divide the total available money B by the cost of 1 article:

Number of Articles = Total Budget B ÷ (Cost of 1 Article)
Method 2: Inverted Unitary Rate (Unit Rupee)

1. Number of articles bought for ₹C = n.
2. Number of articles bought for ₹1 = n / C.
3. Articles bought for ₹B = (n / C) × B.

Number of Articles = (n ÷ C) × B

Step-by-Step Pedagogical Demonstration

Pedagogical Example:

A stationery store sells a pack of 18 gel pens for ₹270. How many such gel pens can a student purchase with ₹405?

• Cost of 18 gel pens = ₹270
• Therefore, cost of 1 gel pen = ₹270 ÷ 18 = ₹15
• Total budget available = ₹405
• Number of pens purchasable for ₹405 = Total Budget ÷ Cost of 1 Pen = 405 ÷ 15 = 27
Final Answer: The student can purchase 27 gel pens.
The Logical Statement Rule:

In Indian ICSE examinations, teachers insist that the quantity to be calculated must always be placed on the right-hand side of the statement. In Type A, you seek Cost, so Cost goes on the right: "Cost of 1 item = ₹...". In Type B, you seek Quantity, so Quantity goes on the right: "For ₹270, number of pens = 18". Following this rule guarantees maximum presentation marks!

Module 5: Uniform Motion, Speed, Distance & Time via Unitary Method

One of the most elegant applications of the Unitary Method in Class 6 Mathematics connects arithmetic with introductory physics: Uniform Motion. When an object travels at a constant speed, the distance covered varies directly with the elapsed time.

Speed as a Fundamental Unit Rate

What is Speed? Mathematically, Speed is defined as the distance covered in ONE unit of time (such as 1 hour, 1 minute, or 1 second).

Speed = (Distance Covered in 1 Hour) = Total Distance ÷ Total Time in Hours

Because speed is simply the unitary rate of distance per hour, all motion problems can be solved using the exact two-step unitary method without memorizing complex physics formulas!

The Two Core Motion Question Archetypes

Question Type 1: Finding Distance for Target Time

• Step 1: Distance covered in 1 hour = (Total Distance) ÷ (Given Hours)
• Step 2: Distance covered in t hours = (Distance in 1 hour) × t

Example: A train covers 240 km in 4 hours.
Distance in 1 hour = 240 ÷ 4 = 60 km.
Distance in 7 hours = 60 × 7 = 420 km.

Question Type 2: Finding Time for Target Distance

• Step 1: Time taken to cover 1 km = (Total Hours) ÷ (Total Distance)
• Or Step 1: Find Speed (km/h), then Time = (Target Distance) ÷ Speed

Example: Same train (Speed = 60 km/h).
How long to cover 540 km?
Time = 540 ÷ 60 = 9 hours.

Handling Unit Conversions: Hours to Minutes

A classic examiner trap in ICSE Class 6 involves mixing hours and minutes. If a question states that a car travels 90 km in 2 hours, and asks for the distance covered in 40 minutes, you must convert the unitary rate to minutes:

• Distance covered in 2 hours (120 minutes) = 90 km
• Distance covered in 1 minute = 90 ÷ 120 = 9/12 = 3/4 km (or 0.75 km)
• Distance covered in 40 minutes = 40 × (3/4) = 10 × 3 = 30 km.
Always ensure your time units are fully homogenised before multiplying!

Module 6: Multi-Unit Measures & Conversions (Dozen, Gross, Score, Metric System)

In commercial word problems, articles are frequently bought and sold in specialized collective counting measures such as dozens, scores, and grosses. Simultaneously, commodities like foodgrains, liquids, and textiles are measured in the metric system (kilograms, grams, litres, metres). Homogenising units before applying the unitary method is an absolute mathematical requirement.

Essential Commercial Counting Constants

1 DOZEN
12 Units
Eggs, Bananas, Pens
1 SCORE
20 Units
Apples, Mangoes, Bangles
1 GROSS
144 Units
12 Dozens (Buttons, Pencils)

The Metric System Conversion Matrix

Measurement Dimension Higher Unit Lower Unit Conversion Relationship
Mass / Weight 1 Kilogram (kg) Gram (g) 1 kg = 1,000 g
Volume / Capacity 1 Litre (L) Millilitre (mL) 1 L = 1,000 mL
Length / Distance 1 Kilometre (km) Metre (m) 1 km = 1,000 m
Length / Distance 1 Metre (m) Centimetre (cm) 1 m = 100 cm

Worked Pedagogical Example with Unit Homogenisation

Pedagogical Example:

The cost of 2.5 kg of premium basmati rice is ₹225. What is the cost of 800 grams of this rice?

• Step 1: Homogenise units into grams: 2.5 kg = 2.5 × 1,000 g = 2,500 g
• Cost of 2,500 g of rice = ₹225
• Step 2: Cost of 1 g of rice = ₹225 ÷ 2,500 = ₹(225 / 2,500) = ₹0.09 (or ₹9/100)
• Step 3: Cost of 800 g of rice = 800 × ₹0.09 = ₹72
Final Answer: The cost of 800 grams of rice is ₹72.

Module 7: Applied Commercial Mathematics & 4-Step Exam Strategy

In advanced CISCE Class 6 exam problems, the unitary method is deployed in multi-step scenarios such as calculating worker wages, comparing consumer pack sizes to find the 'Best Buy', and budgeting fuel expenses. To achieve full marks, students must follow a disciplined, four-step problem-solving strategy.

Consumer Arithmetic: Finding the 'Best Buy' via Unit Rate

Supermarkets often offer products in multiple package sizes with different price tags. The unitary method allows us to find the price per single unit for each option to determine which provides the greatest consumer value:

Best Buy Example:

Brand A sells 500 g of washing powder for ₹140. Brand B sells 750 g of washing powder for ₹195. Which brand is more economical?

• Brand A: Cost of 1 g = 140 ÷ 500 = ₹0.28 per gram (or ₹28 per 100 g)
• Brand B: Cost of 1 g = 195 ÷ 750 = ₹0.26 per gram (or ₹26 per 100 g)
• Comparison: Since ₹0.26 is less than ₹0.28, Brand B costs less per unit.
Final Conclusion: Brand B is the more economical purchase (the Best Buy).

The 4-Step ICSE Board Examination Solution Protocol

STEP 1: Identify & Align Given Information

Read the question carefully. Identify the given pair and the required unknown. Place the required variable on the right-hand side of your written statement.

STEP 2: Execute Unit Homogenisation

Ensure both quantities are expressed in identical units (convert kg to g, hours to minutes, dozens to units) BEFORE writing your division statement.

STEP 3: Divide to Find Single Unit Value

Calculate the value of exactly 1 unit. If the quotient is a non-terminating fraction, keep it in lowest fractional form to prevent premature rounding errors.

STEP 4: Multiply & Perform Sanity Check

Multiply the unit rate by target units. Write the final numerical value accompanied by appropriate physical units (₹, kg, km, hours), and verify that more items cost more!

Key Formulas, Identities & Theorems

Unit Value Equation (Phase 1: Division)
Value of 1 Unit = (Total Given Value) / (Given Number of Units n)
Target Expansion Equation (Phase 2: Multiplication)
Required Value = (Value of 1 Unit) × (Required Number of Units m)
Budget Purchasing Formula (Type B Quantity Calculation)
Number of Articles = (Total Available Budget B) / (Cost of 1 Article)
Direct Variation Invariant (Constant of Proportionality)
x₁ / y₁ = x₂ / y₂ = k (Constant Unit Rate)
Uniform Speed & Distance Formula
Distance in t hours = [Total Distance / Total Hours] × t = (Speed) × t
Motion Time Duration Formula
Time Required = (Target Distance) / [Distance Covered in 1 Unit Time (Speed)]
Commercial Multi-Unit Conversion Constants
1 Dozen = 12 units | 1 Score = 20 units | 1 Gross = 12 Dozens = 144 units
Consumer Best Buy Metric (Unit Rate Comparison)
Unit Price = (Total Package Price) / (Total Package Quantity)

Conceptual Solved Examples & Case Studies

Example 1
If 15 identical gel pens cost ₹225, find the cost of 8 such gel pens.
Step-by-Step Solution:

Step 1: Write down the given statement. Cost of 15 gel pens = ₹225

Step 2: Find the cost of 1 gel pen by division. Cost of 1 gel pen = Total Cost ÷ 15 = ₹225 / 15 = ₹15

Step 3: Find the cost of the required 8 gel pens by multiplication. Cost of 8 gel pens = 8 × Cost of 1 gel pen = 8 × ₹15 = ₹120

Final Answer: The cost of 8 gel pens is ₹120.

Example 2
An automated factory machine manufactures 720 steel bolts in 6 hours of continuous operation. At this constant production rate, how many hours will it take to produce 1,080 bolts?
Step-by-Step Solution:

Step 1: Write down the given production statement. In 6 hours, the machine produces = 720 bolts

Step 2: Find the production rate for 1 hour (Unit Rate). Bolts produced in 1 hour = 720 ÷ 6 = 120 bolts per hour

Step 3: Calculate the time required to produce 1,080 bolts. Time required = Total Required Bolts ÷ Production in 1 Hour = 1,080 / 120 = 9 hours

Alternative Inverted Statement Formulation: Time taken to produce 720 bolts = 6 hours Time taken to produce 1 bolt = 6 / 720 hours = 1 / 120 hours Time taken to produce 1,080 bolts = 1,080 × (1 / 120) = 9 hours

Final Answer: It will take 9 hours to produce 1,080 bolts.

Example 3
A dozen exercise books cost ₹192. (a) What is the cost of 7 such exercise books? (b) How many exercise books can be purchased for ₹320?
Step-by-Step Solution:

Unit Homogenisation: Recall that 1 dozen = 12 units. Therefore, Cost of 12 exercise books = ₹192

Find the cost of 1 exercise book: Cost of 1 exercise book = ₹192 ÷ 12 = ₹16


(a) Cost of 7 exercise books: Cost of 7 exercise books = 7 × Cost of 1 book = 7 × ₹16 = ₹112


(b) Number of exercise books purchasable for ₹320: Total money available = ₹320 Number of exercise books = Total Money ÷ Cost of 1 Book = 320 ÷ 16 = 20 books

Final Answer: (a) The cost of 7 exercise books is ₹112. (b) 20 exercise books can be purchased for ₹320.

Example 4
The cost of 3.5 kg of premium sugar is ₹154. Find: (i) the cost of 8 kg of sugar, (ii) the cost of 500 grams of sugar, and (iii) how many kilograms of sugar can be purchased for ₹352.
Step-by-Step Solution:

Step 1: Calculate the cost of 1 kg of sugar (Unit Rate). Cost of 3.5 kg of sugar = ₹154 Cost of 1 kg of sugar = ₹154 ÷ 3.5 = 154 / (35 / 10) = (154 × 10) / 35 = 1540 / 35 = ₹44 per kg


(i) Cost of 8 kg of sugar: Cost of 8 kg = 8 × Cost of 1 kg = 8 × ₹44 = ₹352


(ii) Cost of 500 grams of sugar: Convert grams to kilograms: 500 g = 500 / 1000 kg = 0.5 kg (or 1/2 kg) Cost of 500 g = 0.5 × ₹44 = ₹22 (Alternatively: Cost of 1 g = 44 / 1000 = ₹0.044; Cost of 500 g = 500 × 0.044 = ₹22)


(iii) Kilograms of sugar bought for ₹352: Quantity of sugar = Total Money ÷ Cost of 1 kg = 352 ÷ 44 = 8 kg

Final Answer: (i) Cost of 8 kg of sugar = ₹352 (ii) Cost of 500 g of sugar = ₹22 (iii) Sugar bought for ₹352 = 8 kg

Example 5
A tourist express bus covers a distance of 180 km in 3 hours at a constant uniform speed. Find: (i) the distance covered by the bus in 5 hours, (ii) the distance covered in 40 minutes, and (iii) the time taken by the bus to cover a journey of 450 km.
Step-by-Step Solution:

Step 1: Calculate the speed (Distance covered in 1 unit of time). Distance covered in 3 hours = 180 km Distance covered in 1 hour (Speed) = 180 ÷ 3 = 60 km per hour


(i) Distance covered in 5 hours: Distance = Speed × Time = 60 km/h × 5 hours = 300 km


(ii) Distance covered in 40 minutes: Method A (Converting minutes to hours): 40 minutes = 40 / 60 hours = 2 / 3 hours Distance = 60 × (2 / 3) = 20 × 2 = 40 km

Method B (Unitary rate per minute): Distance in 60 minutes = 60 km Distance in 1 minute = 60 ÷ 60 = 1 km Distance in 40 minutes = 40 × 1 km = 40 km


(iii) Time taken to cover 450 km: Time = Total Distance ÷ Speed = 450 ÷ 60 = 45 / 6 = 15 / 2 = 7.5 hours Converting 0.5 hours to minutes: 0.5 × 60 = 30 minutes Total time = 7 hours 30 minutes

Final Answer: (i) Distance in 5 hours = 300 km (ii) Distance in 40 minutes = 40 km (iii) Time taken for 450 km = 7 hours 30 minutes (or 7.5 hours)

Example 6
A civil construction contractor employs 12 skilled workers who collectively earn ₹10,800 for 1 day of work. (i) What is the daily wage of 1 worker? (ii) What will be the total earnings of 1 worker in a 24-day working month? (iii) If the contractor hires 18 workers for a special project lasting 15 days, calculate the total wage expenditure. (iv) If a rival contractor pays ₹4,800 for 6 days of work to a team of 1 worker, which contractor offers a higher daily wage rate, and by how much?
Step-by-Step Solution:

Step 1: Calculate the daily wage of 1 worker. Total wages of 12 workers for 1 day = ₹10,800 Daily wage of 1 worker = ₹10,800 ÷ 12 = ₹900 per day


(i) Daily wage of 1 worker: = ₹900


(ii) Total earnings of 1 worker in 24 working days: Earnings = Daily Wage × Number of Days = ₹900 × 24 = ₹21,600


(iii) Total wage bill for 18 workers working for 15 days: Wage of 1 worker for 1 day = ₹900 Wage of 18 workers for 1 day = 18 × ₹900 = ₹16,200 Wage of 18 workers for 15 days = 15 × ₹16,200 = ₹2,43,000 (Alternatively: Total man-days = 18 × 15 = 270 man-days; Total wage = 270 × ₹900 = ₹2,43,000)


(iv) Wage comparison with rival contractor: Rival contractor: 1 worker earns ₹4,800 in 6 days Rival daily wage rate = ₹4,800 ÷ 6 = ₹800 per day First contractor's daily wage rate = ₹900 per day

Comparison: ₹900 is greater than ₹800. Difference = ₹900 - ₹800 = ₹100 per day

Final Answer: (i) Daily wage of 1 worker = ₹900 (ii) Earnings of 1 worker in 24 days = ₹21,600 (iii) Total wage bill for 18 workers in 15 days = ₹2,43,000 (iv) The first contractor offers a higher daily wage rate, exceeding the rival contractor by ₹100 per day.

Common Misconceptions & Examiner Traps

Common Misconception

Inverting the operational sequence by multiplying first instead of dividing (e.g. if 6 notebooks cost ₹120, calculating 120 × 6 instead of 120 ÷ 6).

Scientific Reality & Correction

Always follow the strict two-step order in direct variation: Step 1 (Divide) to find the value of 1 single unit, Step 2 (Multiply) to scale up to the required quantity.

Common Misconception

Premature rounding off of fractional or decimal unit values (e.g. rounding ₹14.285 to ₹14.3 when finding the cost of 1 item, leading to accumulated error in the final step).

Scientific Reality & Correction

Retain fractional unit values in simplest fractional form (e.g. ₹100/7) until the final multiplication step, allowing common factors to cancel cleanly.

Common Misconception

Omitting unit homogenisation before calculating the unit rate (e.g. finding the cost of 500 grams when 2 kg cost ₹120 by dividing 120 by 2 and multiplying directly by 500).

Scientific Reality & Correction

Always convert all quantities to the same unit first: 2 kg = 2,000 g, giving a unit cost of 120 ÷ 2000 = ₹0.06 per gram, so 500 g costs 500 × 0.06 = ₹30.

Common Misconception

Confusing Type A (finding total cost) with Type B (finding quantity of articles purchasable for a budget).

Scientific Reality & Correction

Carefully identify what is required: If finding cost, multiply unit cost by quantity. If finding quantity, divide total budget by the cost of 1 item.

Common Misconception

Inverting uniform speed relationships (e.g. confusing distance covered in 1 hour with the time taken to cover 1 kilometre).

Scientific Reality & Correction

Speed = Distance ÷ Time gives km per hour. To find distance, multiply speed by hours. To find time, divide distance by speed.

Common Misconception

Applying direct variation rules to inverse variation problems (e.g. assuming more workers will take more days to complete a job).

Scientific Reality & Correction

Check the variation direction: If more workers take fewer days, it is inverse variation (multiply number of workers by days to find total man-days, then divide by new worker count).

Visual Learning & Conceptual Map

CISCE / ICSE CLASS 6 MATHEMATICS : THE UNITARY METHOD Fundamental Two-Step Mechanism • Direct Variation Architecture • Speed, Work & Commercial Applications ICSE CURRICULUM PHASE 1 : DIVIDE Unit Reduction Step Find the Value of ONE Single Unit (Unitary Base): Value of 1 Unit = (Total Given Value) ÷ (Given Number of Units n) • Many Items ➔ 1 Item (Contracting Operation via Division) • Example: Cost of 8 pens = ₹160 ➔ Cost of 1 pen = 160 ÷ 8 = ₹20 UNIT BASE 1 UNIT PHASE 2 : MULTIPLY Scaling Expansion Step Find the Value of REQUIRED Number of Units (m): Value of m Units = (Value of 1 Unit) × (Required Count m) • 1 Item ➔ Many Items (Expanding Operation via Multiplication) • Example: Cost of 1 pen = ₹20 ➔ Cost of 12 pens = 20 × 12 = ₹240 TYPE A Finding Total Cost / Value for Given Quantity (m) Given: Cost of n articles = ₹C. Target: Cost of m articles. Formula: Required Cost = (C / n) × m ✓ Step 1: Write statement & divide: 1 unit cost = C ÷ n ✓ Step 2: Multiply unit cost by desired number of units m Example: 15 books cost ₹600 ➔ 1 book = ₹40 ➔ 9 books = 40 × 9 = ₹360 TYPE B Finding Number of Items for Given Budget / Amount Given: Cost of n articles = ₹C. Target: Articles bought for ₹B. Formula: Number of Items = Total Budget B ÷ (Cost of 1 Unit) ✓ Step 1: Find cost of 1 article: Unit Cost = C ÷ n ✓ Step 2: Divide available budget B by unit cost: Items = B ÷ Unit Cost Example: 12 pens cost ₹180 ➔ 1 pen = ₹15 ➔ Pens for ₹105 = 105 ÷ 15 = 7 1 Uniform Speed & Motion • Speed = Distance in 1 Unit Time Distance in 1 h = Total Dist ÷ Total Time • Distance in t hours = (Speed) × t • Time for d km = d ÷ (Speed) 2 Commercial Units System • Essential Conversion Constants: • 1 Dozen = 12 Units • 1 Score = 20 • 1 Gross = 12 Dozen = 144 Units • 1 kg = 1,000 g • 1 L = 1,000 mL • 1 h = 60 min 3 Direct Variation Rule • Constant Proportionality Ratio: x / y = k (Constant Unit Rate) • As Quantity Doubles, Total Cost Doubles • Sanity Check: More items ➔ More cost

Chapter Summary & 10 Key Takeaways

Takeaway 1
The Unitary Method is an arithmetical method where the value of a single unit is determined first by division, and the value of the required number of units is subsequently calculated by multiplication.
Takeaway 2
The term 'Unitary' is derived from 'Unit', meaning one single individual entity or measurement.
Takeaway 3
The method operates primarily on the principle of Direct Variation, where two quantities increase or decrease in the same ratio such that x / y = k (a constant unit rate).
Takeaway 4
The universal two-step engine comprises Phase 1: Division (Unit Value = Total Value ÷ Given Units n) and Phase 2: Multiplication (Required Value = Unit Value × Target Units m).
Takeaway 5
In Type A problems, students determine the total cost or value for a specified number of items: Required Cost = (Given Cost / n) × m.
Takeaway 6
In Type B problems, students determine the quantity of articles purchasable within a given budget: Number of Articles = Total Budget ÷ (Cost of 1 Article).
Takeaway 7
In uniform motion problems, Speed is the unitary rate of distance per hour: Distance in t hours = Speed × t, and Time for d km = d ÷ Speed.
Takeaway 8
Commercial count measures must be converted before calculation: 1 Dozen = 12 units, 1 Score = 20 units, and 1 Gross = 12 Dozen = 144 units.
Takeaway 9
Metric units must be completely homogenised before division: 1 kg = 1,000 g, 1 L = 1,000 mL, 1 km = 1,000 m, and 1 hour = 60 minutes.
Takeaway 10
When evaluating consumer options, the 'Best Buy' is determined by calculating the unit price (Price ÷ Quantity); the option with the lower unit price is the more economical purchase.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
If 9 identical notebooks cost ₹216, find the cost of 14 such notebooks.
Reveal Answer & Explanation
Answer: Cost of 1 notebook = 216 ÷ 9 = ₹24. Cost of 14 notebooks = 14 × 24 = ₹336.
First divide 216 by 9 to find the cost of 1 notebook, then multiply by 14.
2
A worker earns ₹3,150 in 7 days. How many days must the worker work to earn ₹5,850 at the same daily rate?
Reveal Answer & Explanation
Answer: Daily wage = 3150 ÷ 7 = ₹450 per day. Number of days required = 5850 ÷ 450 = 13 days.
Find the daily wage (earnings for 1 day) first by division, then divide 5,850 by that daily wage.
3
The cost of 2 dozen bananas is ₹96. Find the cost of 5 bananas.
Reveal Answer & Explanation
Answer: 2 dozen = 24 bananas. Cost of 1 banana = 96 ÷ 24 = ₹4. Cost of 5 bananas = 5 × 4 = ₹20.
2 dozen = 2 × 12 = 24 bananas. Find the cost of 1 banana first.
4
A car travels 195 km on 15 litres of petrol. How far will it travel on 22 litres of petrol?
Reveal Answer & Explanation
Answer: Distance per 1 litre = 195 ÷ 15 = 13 km. Distance on 22 litres = 22 × 13 = 286 km.
Find mileage (distance per 1 litre) by dividing 195 by 15, then multiply by 22.
5
If 4 kg of wheat flour costs ₹144, find the cost of 750 grams of wheat flour.
Reveal Answer & Explanation
Answer: Cost of 1 kg = 144 ÷ 4 = ₹36. 750 g = 750 / 1000 kg = 0.75 kg (or 3/4 kg). Cost of 750 g = (3/4) × 36 = 3 × 9 = ₹27.
Convert 4 kg to 4,000 g, or find cost of 1 kg = ₹36 and multiply by 0.75 kg.
6
A printing machine prints 450 pages in 15 minutes. How many pages can it print in 1 hour at this speed?
Reveal Answer & Explanation
Answer: Pages in 1 minute = 450 ÷ 15 = 30 pages. Pages in 1 hour (60 minutes) = 60 × 30 = 1,800 pages.
1 hour = 60 minutes. Find pages printed in 1 minute first, then multiply by 60.
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