Unitary Method
Pointwise AISSEE Class 6 Mathematics guide for Unitary Method (Topic 3). Master single unit calculation, direct vs inverse variation, work-time rate formulas, and army garrison provision problems.
Overview & Fundamental Concept
Pointwise AISSEE Class 6 Mathematics guide for Unitary Method (Topic 3). Master single unit calculation, direct vs inverse variation, work-time rate formulas, and army garrison provision problems.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 3 | Concept of unitary method, direct variation, inverse variation, finding the value of a single unit and then multiple units in real-world scenarios. | Master unit rate extraction, direct and inverse proportionality, multi-step work problems, and military garrison food provision scenarios for AISSEE Class VI. |
1. The Unitary Principle: Finding Value of a Single Unit
- Definition of Unitary Method: The mathematical procedure where we first find the value of a single unit (1 unit) by division, and then find the value of the required number of units by multiplication.
- Two Core Steps:
• Step 1 (Down to One): $\text{Value of 1 unit} = \frac{\text{Total Value}}{\text{Given Number of Units}}$ (Operation: Division).
• Step 2 (Up to Many): $\text{Value of Required Units} = \text{Value of 1 unit} \times \text{Required Units}$ (Operation: Multiplication). - Homogeneity of Units: Before calculating, ensure all quantities are in uniform units (e.g., if cost is given for 2 dozen eggs, convert 2 dozen into $2 \times 12 = 24$ eggs before finding the unit cost).
| Problem Phase | Direction | Mathematical Operation | Example (8 pens cost ₹96) |
|---|---|---|---|
| Phase 1: Unit Rate | Many $\to$ One (Single unit) | Division ($\div$) | 1 pen = ₹96 ÷ 8 = ₹12 |
| Phase 2: Required Total | One $\to$ Many (Target units) | Multiplication ($\times$) | 15 pens = ₹12 × 15 = ₹180 |
Cadet Mnemonic • D-M Unitary Rhythm
Remember the operation sequence:
- Divide down to One
- Multiply up to Many
2. Direct Variation vs. Inverse Variation
- Direct Variation (Proportion): When an increase in one quantity results in a proportional increase in the other quantity (or a decrease results in a proportional decrease). The ratio $\frac{x}{y} = k$ remains constant.
- Direct Variation Examples: Articles purchased vs Total cost; Distance traveled vs Time (at constant speed); Hours worked vs Wages earned.
- Inverse (Indirect) Variation: When an increase in one quantity results in a proportional decrease in the other quantity (or vice versa). The product $x \times y = k$ remains constant.
- Inverse Variation Examples: Number of workers vs Days to finish a task; Speed of vehicle vs Time taken for a fixed distance; Number of cadets in a mess vs Days the food provisions will last.
| Characteristic | Direct Variation | Inverse Variation |
|---|---|---|
| Core Relationship | More brings More; Less brings Less | More takes Less; Less takes More |
| Constant Mathematical Form | $\frac{x}{y} = \text{constant } (k)$ | $x \times y = \text{constant } (k)$ |
| Finding 1 unit | Divide: $\frac{\text{Total}}{\text{Units}}$ | Multiply: $\text{Workers} \times \text{Days}$ |
| Classic AISSEE Example | 12 kg sugar costs ₹480 | 15 men dig a trench in 6 days |
AISSEE Trap Alert • The Inverse Workers Trap
Never use direct proportion for workers and time! If 4 men build a wall in 8 days, 1 man takes $4 \times 8 = 32\text{ days}$ (NOT $8 \div 4 = 2\text{ days}$)! More workers mean FEWER days.
3. Advanced AISSEE Applications: Military Garrison & Work Rates
- Military Garrison / Hostel Food Model: When provisions exist for $M_1$ men for $D_1$ days, the total food capacity is measured in Man-Days: $\text{Total Man-Days} = M_1 \times D_1$.
- Reinforcement or Departure: If after $d$ days, reinforcements arrive or men leave:
• Remaining food $= M_1 \times (D_1 - d)$ man-days.
• If $m$ men join, new days $= \frac{M_1 \times (D_1 - d)}{M_1 + m}$.
• If $m$ men leave, new days $= \frac{M_1 \times (D_1 - d)}{M_1 - m}$. - Time and Work Fundamentals:
• If person $A$ completes a work in $n$ days, work done in 1 day $= \frac{1}{n}$.
• If $A$ takes $a$ days and $B$ takes $b$ days, working together their 1-day work is $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$.
• Total time taken together $= \frac{a \times b}{a + b}\text{ days}$.
Exam Trick • Combined Work Shortcut Formula
For two individuals taking $A$ and $B$ days respectively, the combined time is $\frac{\text{Product}}{\text{Sum}} = \frac{A \times B}{A + B}$. Example: If A takes 6 days and B takes 12 days, together they take $\frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\text{ days}$!
Visual Concept Blueprint • Vector Architecture
Speed Shortcuts & Mental Calculation Strategies
Master these high-yield time-saving heuristics curated by top AISSEE scorers to eliminate calculation bottlenecks and answer questions in under 45 seconds.
Shortcut Technique 1: Product Constancy for Inverse Variation
In any inverse variation problem, equate the products directly: $M_1 \times D_1 = M_2 \times D_2$. Solve for the unknown in one step without setting up fractions.
Shortcut Technique 2: Unit Conversion Pre-Check
Ensure quantities on both sides match. If rate is per dozen, convert target dozens to units or vice versa before multiplying.
Shortcut Technique 3: Arrow Analysis for Compound Unitary
Write given values in columns. Up arrow for direct proportion (flip fraction), down arrow for inverse proportion (keep fraction as is).
Interactive 3D Revision Flashcards Deck
Click any card to flip and reveal the key exam rule or topper secret.
REVISION FLASHCARD
What is the core rule & definition for Unitary Method?
CADET MEMORY KEY
Unit Cost (Direct): \text{Unit Value} = \frac{\text{Total Value}}{\text{Given Quantity}}
REVISION FLASHCARD
Cadet Shortcut: How to solve Unitary Method in under 30 seconds?
CADET MEMORY KEY
Fast MCQ Elimination: Check the last digit (unit digit) and estimate the order of magnitude to eliminate 2 distractors in under 10 seconds.
REVISION FLASHCARD
Exam Trap Alert: What pitfall do most candidates fall into?
CADET MEMORY KEY
Direct Variation: If quantity increases, total cost increases (Divide then Multiply).
REVISION FLASHCARD
Rapid Memory Anchor for AISSEE Hall Recall:
CADET MEMORY KEY
Inverse Variation: If workers increase, days taken decrease (Multiply then Divide).
REVISION FLASHCARD
If 8 men can reap a field in 15 days, how many men can reap the same field in 10 days?
CADET MEMORY KEY
Inverse variation: M1 × D1 = M2 × D2. 8 × 15 = M2 × 10 => 120 = 10 × M2 => M2 = 12 men.
Common Traps & Examination Pitfalls
Avoid the Speed-Guess Trap: Under AISSEE pressure, candidates frequently misread units (e.g., mixing metres and centimetres, or per hour and per second) and confuse complementary ($90^\circ$) with supplementary ($180^\circ$) angles. Read every stem carefully and check units before selecting your OMR option.
Key Identities & Formulas
Direct Variation Equation
$$\frac{x_1}{y_1} = \frac{x_2}{y_2} \implies y_2 = y_1 \times \frac{x_2}{x_1}$$
Cost, distance, production scaling.
Inverse Variation Equation
$$M_1 \times D_1 = M_2 \times D_2 \implies D_2 = \frac{M_1 \times D_1}{M_2}$$
Work, speed-time, and military food supply.
Combined Work Formula
$$\text{Time Together} = \frac{A \times B}{A + B}$$
Total days for two workers working concurrently.
Step-by-Step Solved Examination Models
Solved Example 1
Model Problem
If 15 pens cost ₹180, what is the cost of 8 such pens?
Detailed Solution:
Cost of 15 pens = ₹180. Cost of 1 pen = 180 / 15 = ₹12. Cost of 8 pens = 8 × 12 = ₹96.
Solved Example 2
Model Problem
12 workers can build a wall in 8 days. In how many days can 16 workers build the same wall?
Detailed Solution:
This is an inverse variation: More workers take fewer days. Total worker-days = 12 × 8 = 96. Days for 16 workers = 96 / 16 = 6 days.
Solved Example 3
Model Problem
A military garrison of 300 soldiers has food provisions for 45 days. After 15 days, a reinforcement of 150 soldiers arrives. How many days will the remaining provisions last?
Detailed Solution:
1. For 300 soldiers, after 15 days, remaining food would last for: $45 - 15 = 30\text{ days}$.
2. Remaining food in soldier-days: $300 \times 30 = 9,000\text{ soldier-days}$.
3. New total strength of garrison: $300 + 150 = 450\text{ soldiers}$.
4. Number of days the remaining food will last: $\frac{9,000}{450} = 20\text{ days}$. Correct Option is (A).
2. Remaining food in soldier-days: $300 \times 30 = 9,000\text{ soldier-days}$.
3. New total strength of garrison: $300 + 150 = 450\text{ soldiers}$.
4. Number of days the remaining food will last: $\frac{9,000}{450} = 20\text{ days}$. Correct Option is (A).
AISSEE High-Yield Practice Drills (10 Questions)
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Question 1 of 10
Exam Standard
A car travels 165 km in 3 hours. How far will it travel in 5 hours at the same speed?
Correct Answer: Option (B)
Distance in 1 hour = 165 / 3 = 55 km. Distance in 5 hours = 55 × 5 = 275 km.
Question 2 of 10
Exam Standard
If 25 metres of cloth costs ₹1,500, how many metres of cloth can be purchased for ₹960?
Correct Answer: Option (C)
Cost of 1 metre = 1,500 / 25 = ₹60. Metres for ₹960 = 960 / 60 = 16 metres.
Question 3 of 10
Exam Standard
A garrison of 120 soldiers has food provisions for 30 days. If 30 more soldiers join, how long will the food last?
Correct Answer: Option (B)
Total soldier-days = 120 × 30 = 3,600. New total soldiers = 120 + 30 = 150. Days = 3,600 / 150 = 24 days.
Question 4 of 10
Exam Standard
If 8 men can reap a field in 6 days, how many men are required to reap it in 4 days?
Correct Answer: Option (B)
Total man-days = 8 × 6 = 48. Men required for 4 days = 48 / 4 = 12 men.
Question 5 of 10
Exam Standard
A train covers 240 km in 4 hours. How much time will it take to cover 360 km at the same speed?
Correct Answer: Option (B)
Speed = 240 / 4 = 60 km/h. Time for 360 km = 360 / 60 = 6 hours.
Question 6 of 10
Exam Standard
If 12 notebooks cost ₹216, what will be the cost of 20 notebooks?
Correct Answer: Option (B)
Cost of 1 notebook = 216 / 12 = ₹18. Cost of 20 notebooks = 20 × 18 = ₹360.
Question 7 of 10
Exam Standard
A tap fills a 600-litre tank in 40 minutes. How many litres does it fill in 25 minutes?
Correct Answer: Option (B)
Rate = 600 / 40 = 15 litres/min. In 25 minutes = 25 × 15 = 375 litres.
Question 8 of 10
Exam Standard
6 pipes can fill a tank in 80 minutes. How long will 8 pipes take to fill the same tank?
Correct Answer: Option (B)
Inverse variation: 6 × 80 = 480 pipe-minutes. Time for 8 pipes = 480 / 8 = 60 minutes.
Question 9 of 10
Exam Standard
If the cost of 5 kg of sugar is ₹220, find the cost of 7.5 kg of sugar.
Correct Answer: Option (B)
Cost of 1 kg = 220 / 5 = ₹44. Cost of 7.5 kg = 7.5 × 44 = ₹330.
Question 10 of 10
Exam Standard
15 men complete a job in 20 days. How many more men must be employed to complete it in 12 days?
Correct Answer: Option (B)
Total man-days = 15 × 20 = 300. Men required for 12 days = 300 / 12 = 25 men. More men needed = 25 − 15 = 10 men.
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Last-Minute Cadet Revision Matrix
Rapid recall checkpoints to review 24 hours before stepping into the examination hall.
| Concept Checkpoint | High-Yield Exam Rule & Memory Anchor |
|---|---|
| Single Unit Rule | In direct variation: Divide to find 1 unit, Multiply to find required units. |
| Worker-Day Product | Total work = Workers × Days. 1 worker takes (Workers × Days) days to finish the work. |
| Garrison Remainder Rule | First subtract the days elapsed before calculating remaining food for new total strength. |
| Combined Work | Combined time for two persons is always strictly less than the time taken by the faster person alone. |
Frequently Asked Questions
How many questions from Unitary Method appear in the AISSEE Class VI exam?
Typically, 2 to 3 questions directly test Unitary Method in the 50-question Mathematics paper, contributing 6 to 9 marks towards the 150-mark section total.
Is there any negative marking for incorrect answers in AISSEE Class VI?
No. The National Testing Agency (NTA) does not apply negative marking in the AISSEE Class VI entrance examination. Candidates should attempt every single question.
What is the best way to revise Unitary Method before the exam?
Review the high-yield formulas, practice the 10 interactive drill MCQs on this page, and memorize the revision checkpoints matrix.