Model Textbook of Mathematics Grade 5 (FBISE / NBF)
Class 5 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 5 (FBISE / NBF)

Mastery Guide: The Unitary Method, Direct Proportion, Rates & Financial Applications

📖 Chapter 6: Unitary Method 📅 Updated: Sep 09, 2026
Teacher Pedagogical Roadmap Grade 5 Mathematics • FBISE / SNC Aligned • Chapter 6

Instructional Blueprint: Unit 6 Unitary Method

Target Learning Outcomes
  • Understand the meaning of "Unitary" (finding the value of 1 unit first).
  • Calculate the value of many items when the value of 1 item is given (Multiplication).
  • Calculate the value of 1 item when the value of many items is given (Division).
  • Master the full two-step Unitary Method: Given Many → Find 1 → Find Required Many.
  • Solve multi-step real-world problems involving shopping, weights, distances, bus capacities, and costs.
Pacing & Lesson Sequence (60 Min)
  • 00–15m: Hook & Intuition: The Vegetable Vendor Story & "Unit = 1".
  • 15–30m: Type 1 (Multiply: $1 \to \text{Many}$) & Type 2 (Divide: $\text{Many} \to 1$).
  • 30–45m: The 2-Step Master Strategy ($\text{Many} \to 1 \to \text{Target}$).
  • 45–60m: Exercise 1 & Exercise 2 Step-by-Step Problem Solving.
Child-Centric Visual Aids & Cues
  • The Stepping Stone Metaphor: "You cannot jump across the river in one leap; step on the stone of $1$ unit first!"
  • Memory Rule: "To go to ONE → DIVIDE. To go to MANY → MULTIPLY!"

💡 Study Cues & Key Inquiries

1. What Does "Unitary" Mean?

The word unitary comes from unit, which means ONE ($1$). The method is called unitary because we always find the value of $1$ single thing first!

2. The Golden Rule of Operations

Many → 1: Use Division ($\div$).
1 → Many: Use Multiplication ($\times$).

3. The 2-Step Magic Bridge

When given $5\text{ items}$ and asked for $13\text{ items}$: First divide by $5$ to get the cost of $1\text{ item}$, then multiply by $13$!

🛒 Real-Life Challenge: The Vegetable Market Story

Imagine you go to the market with your mother. The vegetable vendor says: "5 kilograms of fresh vegetables cost Rs. 400."
Your mother wants to buy 13 kilograms. How much should she pay?

The Child's Strategy: You cannot jump directly from $5\text{ kg}$ to $13\text{ kg}$ easily. But look how simple it is if you find $1\text{ kg}$ first:
Step 1 (Find 1 kg): $\text{Price of } 1\text{ kg} = \text{Rs. } 400 \div 5 = \mathbf{\text{Rs. } 80}$
Step 2 (Find 13 kg): $\text{Price of } 13\text{ kg} = \text{Rs. } 80 \times 13 = \mathbf{\text{Rs. } 1,040}$
This clever method is called the Unitary Method!

1. Type 1: Calculating the Value of Many Objects from One Object

When you know the price or weight of one ($1$) item, finding the total for many items is super easy: simply MULTIPLY ($\times$)!

Formula for Type 1: $$\text{Value of Many Items} = \text{Value of } 1 \text{ Item} \times \text{Number of Required Items}$$
🔬 Science Lab Example: Magnifying Glasses

For a science experiment, students were divided into $8\text{ groups}$ and each group was given $1\text{ magnifying glass}$. If the price of $1\text{ magnifying glass}$ is $\text{Rs. } 245$, find the price of $8\text{ glasses}$.

$$\text{Price of } 1\text{ glass} = \text{Rs. } 245$$ $$\text{Price of } 8\text{ glasses} = 245 \times 8 = \mathbf{\text{Rs. } 1,960}$$

✏️ Stationery Example: Pencils

If the price of a pencil is $\text{Rs. } 10$, what will be the price of $15\text{ pencils}$?

$$\text{Price of } 1\text{ pencil} = \text{Rs. } 10$$ $$\text{Price of } 15\text{ pencils} = 10 \times 15 = \mathbf{\text{Rs. } 150}$$

2. Type 2: Calculating the Value of One Object from Many Objects

When you know the total cost or quantity of many items together, finding the cost of just one ($1$) single item requires sharing or distributing equally: simply DIVIDE ($\div$)!

Formula for Type 2: $$\text{Value of } 1 \text{ Item} = \frac{\text{Total Value of All Given Items}}{\text{Total Number of Given Items}}$$
🪑 Furniture Example: Chairs

The price of $20\text{ chairs}$ is $\text{Rs. } 7,240$. How can we find the price of $1\text{ such chair}$?

$$\text{Price of } 1\text{ chair} = 7,240 \div 20 = \mathbf{\text{Rs. } 362}$$

🌳 Orchard Example: Mango Trees

In a mango orchard, there are $576\text{ trees}$ in $32\text{ identical rows}$. How many trees will there be in $1\text{ row}$?

$$\text{Number of trees in } 1\text{ row} = 576 \div 32 = \mathbf{18\text{ trees}}$$

🌟 Summary of the Fundamental Unitary Rules:

  • Rule 1: When the value of one item is known, the value of many items of the same kind can be found by MULTIPLICATION.
  • Rule 2: When the value of many items is known, then the price of one item can be found by DIVISION.

3. Type 3: The Complete Two-Step Unitary Method

In most real-life exam questions, you are given the value of some items ($A$) and asked to find the value of different items ($B$). You use the two-step bridge:

Given Value of Many ($n$)
Step 1: Divide ($\div n$)
Value of ONE Unit ($1$)
Step 2: Multiply ($\times k$)
Required Value of Target ($k$)
💡 Textbook Example: Tube Lights

Father bought $12\text{ tube lights}$ for our home which cost $\text{Rs. } 6,900$ altogether. What will be the price of $7\text{ such tube lights}$?

Step 1: Price of $1\text{ tube light} = 6,900 \div 12 = \text{Rs. } 575$
Step 2: Price of $7\text{ tube lights} = 575 \times 7 = \mathbf{\text{Rs. } 4,025}$

📖 Textbook Example: Notebook Pages

There are $2,940\text{ pages}$ in $30\text{ notebooks}$. How many pages will be there in $14\text{ such notebooks}$?

Step 1: Pages in $1\text{ notebook} = 2,940 \div 30 = 98\text{ pages}$
Step 2: Pages in $14\text{ notebooks} = 98 \times 14 = \mathbf{1,372\text{ pages}}$

📝 100% Solved Textbook Exercises • Grade 5 Mathematics Chapter 6

Exercise 1 • Single-Step Operations (Page 150)
Q1. The price of $1\text{ geometry box}$ is $\text{Rs. } 76$. Find the price of $26\text{ geometry boxes}$.

Given:
Price of $1\text{ geometry box} = \text{Rs. } 76$
Number of geometry boxes $= 26$

Working:
$$\text{Price of } 26\text{ boxes} = 76 \times 26$$ $$\begin{array}{r@{\quad}l} & 76 \\ \times & 26 \\ \hline & 456 \quad (76 \times 6) \\ + & 1520 \quad (76 \times 20) \\ \hline & \mathbf{1,976} \end{array}$$

Answer: The price of $26\text{ geometry boxes}$ is $\text{Rs. } 1,976$.

Q2. If the price of one ice cream is $\text{Rs. } 55$, find the price of $17\text{ ice creams}$.

Given:
Price of $1\text{ ice cream} = \text{Rs. } 55$
Number of ice creams $= 17$

Working:
$$\text{Price of } 17\text{ ice creams} = 55 \times 17$$ $$\begin{array}{r@{\quad}l} & 55 \\ \times & 17 \\ \hline & 385 \quad (55 \times 7) \\ + & 550 \quad (55 \times 10) \\ \hline & \mathbf{935} \end{array}$$

Answer: The price of $17\text{ ice creams}$ is $\text{Rs. } 935$.

Q3. If the price of a popcorn pack is $\text{Rs. } 37$, find out the price of $71\text{ popcorn packs}$.

Given:
Price of $1\text{ pack} = \text{Rs. } 37$
Number of packs $= 71$

Working:
$$\text{Total price} = 37 \times 71$$ $$\begin{array}{r@{\quad}l} & 37 \\ \times & 71 \\ \hline & 37 \quad (37 \times 1) \\ + & 2590 \quad (37 \times 70) \\ \hline & \mathbf{2,627} \end{array}$$

Answer: The price of $71\text{ popcorn packs}$ is $\text{Rs. } 2,627$.

Q4. A bag weighs $21\text{ kilograms}$. What will be the weight of $15\text{ bags}$?

Given:
Weight of $1\text{ bag} = 21\text{ kg}$
Number of bags $= 15$

Working:
$$\text{Total weight} = 21 \times 15$$ $$\begin{array}{r@{\quad}l} & 21 \\ \times & 15 \\ \hline & 105 \quad (21 \times 5) \\ + & 210 \quad (21 \times 10) \\ \hline & \mathbf{315} \end{array}$$

Answer: The total weight of $15\text{ bags}$ is $315\text{ kilograms}$.

Q5. The capacity of $15\text{ identical buses}$ is $555\text{ passengers}$. What will be the capacity of $1\text{ bus}$?

Given:
Capacity of $15\text{ buses} = 555\text{ passengers}$
Number of buses $= 15$

Working:
$$\text{Capacity of } 1\text{ bus} = 555 \div 15$$ $$15 \times 30 = 450,\quad 555 - 450 = 105,\quad 15 \times 7 = 105 \implies 30 + 7 = \mathbf{37}$$

Answer: The capacity of $1\text{ bus}$ is $37\text{ passengers}$.

Q6. The price of $26\text{ kilograms of sugar}$ is $\text{Rs. } 1,170$. What will be the price of $1\text{ kilogram of sugar}$?

Given:
Price of $26\text{ kg sugar} = \text{Rs. } 1,170$
Quantity $= 26\text{ kg}$

Working:
$$\text{Price of } 1\text{ kg sugar} = 1,170 \div 26$$ $$26 \times 40 = 1,040,\quad 1,170 - 1,040 = 130,\quad 26 \times 5 = 130 \implies 40 + 5 = \mathbf{45}$$

Answer: The price of $1\text{ kilogram of sugar}$ is $\text{Rs. } 45$.

Q7. If the rent of a house for $8\text{ months}$ is $\text{Rs. } 145,680$, what will be the rent of the house for $1\text{ month}$?

Given:
Rent for $8\text{ months} = \text{Rs. } 145,680$
Number of months $= 8$

Working:
$$\text{Rent for } 1\text{ month} = 145,680 \div 8$$ $$145,680 \div 8 = \mathbf{18,210}$$

Answer: The monthly rent of the house is $\text{Rs. } 18,210$.

Q8. The cost of $35\text{ registers}$ is $\text{Rs. } 5,075$. Find the cost of $1\text{ register}$.

Given:
Cost of $35\text{ registers} = \text{Rs. } 5,075$
Number of registers $= 35$

Working:
$$\text{Cost of } 1\text{ register} = 5,075 \div 35$$ $$35 \times 100 = 3,500,\quad 5,075 - 3,500 = 1,575,\quad 35 \times 40 = 1,400,\quad 175 \div 35 = 5 \implies 100 + 40 + 5 = \mathbf{145}$$

Answer: The cost of $1\text{ register}$ is $\text{Rs. } 145$.

Exercise 2 • Two-Step Unitary Method Problems (Page 152)
Q1. There are $4,900\text{ markers}$ in $50\text{ boxes}$. How many markers will be there in $75\text{ such boxes}$?

Step 1 (Find markers in $1\text{ box}$):
$$\text{Markers in } 1\text{ box} = 4,900 \div 50 = \mathbf{98\text{ markers}}$$

Step 2 (Find markers in $75\text{ boxes}$):
$$\text{Markers in } 75\text{ boxes} = 98 \times 75$$ $$\begin{array}{r@{\quad}l} & 98 \\ \times & 75 \\ \hline & 490 \quad (98 \times 5) \\ + & 6860 \quad (98 \times 70) \\ \hline & \mathbf{7,350} \end{array}$$

Answer: There will be $7,350\text{ markers}$ in $75\text{ boxes}$.

Q2. The price of $10\text{ mobile phones}$ is $\text{Rs. } 348,290$. What will be the price of $29\text{ mobile phones}$ of the same model?

Step 1 (Find price of $1\text{ mobile phone}$):
$$\text{Price of } 1\text{ phone} = 348,290 \div 10 = \mathbf{\text{Rs. } 34,829}$$

Step 2 (Find price of $29\text{ mobile phones}$):
$$\text{Price of } 29\text{ phones} = 34,829 \times 29$$ $$\begin{array}{r@{\quad}l} & 34,829 \\ \times & 29 \\ \hline & 313,461 \quad (34,829 \times 9) \\ + & 696,580 \quad (34,829 \times 20) \\ \hline & \mathbf{1,010,041} \end{array}$$

Answer: The price of $29\text{ mobile phones}$ is $\text{Rs. } 1,010,041$.

Q3. $40\text{ books}$ weigh $88\text{ kilograms}$. What is the weight of $6\text{ such books}$?

Step 1 (Find weight of $1\text{ book}$):
$$\text{Weight of } 1\text{ book} = 88 \div 40 = \frac{88}{40} = \frac{22}{10} = \mathbf{2.2\text{ kg}}\quad (= 2\text{ kg } 200\text{ g})$$

Step 2 (Find weight of $6\text{ books}$):
$$\text{Weight of } 6\text{ books} = 2.2 \times 6 = \mathbf{13.2\text{ kg}}\quad (= 13\text{ kg } 200\text{ g})$$

Answer: The weight of $6\text{ books}$ is $13.2\text{ kg}$ (or $13\text{ kg } 200\text{ g}$).

Q4. A train travels $6,136\text{ kilometres}$ in $52\text{ hours}$. What distance will it cover at the same speed in $45\text{ hours}$?

Step 1 (Find distance covered in $1\text{ hour}$ / Speed):
$$\text{Distance in } 1\text{ hour} = 6,136 \div 52$$ $$52 \times 100 = 5,200,\quad 6,136 - 5,200 = 936,\quad 52 \times 18 = 936 \implies \mathbf{118\text{ km/h}}$$

Step 2 (Find distance covered in $45\text{ hours}$):
$$\text{Distance in } 45\text{ hours} = 118 \times 45$$ $$\begin{array}{r@{\quad}l} & 118 \\ \times & 45 \\ \hline & 590 \quad (118 \times 5) \\ + & 4720 \quad (118 \times 40) \\ \hline & \mathbf{5,310} \end{array}$$

Answer: The train will cover $5,310\text{ kilometres}$ in $45\text{ hours}$.

Q5. Amir buys $3\text{ computers}$ for $\text{Rs. } 838,155$. What will be the cost of $52\text{ such computers}$?

Step 1 (Find cost of $1\text{ computer}$):
$$\text{Cost of } 1\text{ computer} = 838,155 \div 3 = \mathbf{\text{Rs. } 279,385}$$

Step 2 (Find cost of $52\text{ computers}$):
$$\text{Cost of } 52\text{ computers} = 279,385 \times 52$$ $$\begin{array}{r@{\quad}l} & 279,385 \\ \times & 52 \\ \hline & 558,770 \quad (279,385 \times 2) \\ + & 13,969,250 \quad (279,385 \times 50) \\ \hline & \mathbf{14,528,020} \end{array}$$

Answer: The cost of $52\text{ computers}$ is $\text{Rs. } 14,528,020$.

Q6. If the fare for $34\text{ kilometres}$ is $\text{Rs. } 1,190$, what will be the fare for $48\text{ kilometres}$?

Step 1 (Find fare for $1\text{ kilometre}$):
$$\text{Fare for } 1\text{ km} = 1,190 \div 34 = \mathbf{\text{Rs. } 35}$$

Step 2 (Find fare for $48\text{ kilometres}$):
$$\text{Fare for } 48\text{ km} = 35 \times 48 = \mathbf{\text{Rs. } 1,680}$$

Answer: The fare for $48\text{ kilometres}$ is $\text{Rs. } 1,680$.

Q7. The price of $19\text{ pairs of shoes}$ is $\text{Rs. } 23,750$. What will be the price of $65\text{ such pairs}$?

Step 1 (Find price of $1\text{ pair of shoes}$):
$$\text{Price of } 1\text{ pair} = 23,750 \div 19 = \mathbf{\text{Rs. } 1,250}$$

Step 2 (Find price of $65\text{ pairs}$):
$$\text{Price of } 65\text{ pairs} = 1,250 \times 65 = \mathbf{\text{Rs. } 81,250}$$

Answer: The price of $65\text{ pairs of shoes}$ is $\text{Rs. } 81,250$.

Review Exercise 6 • Comprehensive Assessment (Pages 152–153)
Q1. Fill in the blanks (MCQs):

a) The price of a book is $\text{Rs. } 250$. The price of $5\text{ books}$ will be $\text{Rs. }$ ______.
(i) $50$   (ii) $1,000$   (iii) $1,250$ ✓   (iv) $2,500$
Explanation: $\text{Price} = 250 \times 5 = \mathbf{\text{Rs. } 1,250}$.

b) The price of $11\text{ carpets}$ is $\text{Rs. } 35,805$. The price of one carpet is ______.
(i) $3,855$   (ii) $3,055$   (iii) $2,355$   (iv) $3,255$ ✓
Explanation: $\text{Price of } 1\text{ carpet} = 35,805 \div 11 = \mathbf{\text{Rs. } 3,255}$.

c) The price of a book is $\text{Rs. } 555$. The price of $10\text{ books}$ will be $\text{Rs. }$ ______.
(i) $2,550$   (ii) $5,050$   (iii) $3,250$   (iv) $5,550$ ✓
Explanation: $\text{Price} = 555 \times 10 = \mathbf{\text{Rs. } 5,550}$.

d) The price of $6\text{ oranges}$ is $\text{Rs. } 48$. The price of $72\text{ oranges}$ will be $\text{Rs. }$ ______.
(i) $567$   (ii) $96$   (iii) $112$   (iv) $576$ ✓
Explanation: $1\text{ orange} = 48 \div 6 = \text{Rs. } 8 \implies 72\text{ oranges} = 8 \times 72 = \mathbf{\text{Rs. } 576}$.

e) The price of $3\text{ chairs}$ is $\text{Rs. } 645$. The price of $16\text{ chairs}$ will be $\text{Rs. }$ ______.
(i) $3,040$   (ii) $3,004$   (iii) $3,440$ ✓   (iv) $3,404$
Explanation: $1\text{ chair} = 645 \div 3 = \text{Rs. } 215 \implies 16\text{ chairs} = 215 \times 16 = \mathbf{\text{Rs. } 3,440}$.

Q2. A storybook costs $\text{Rs. } 440$. Find the price of $15\text{ such storybooks}$.

Given:
Cost of $1\text{ storybook} = \text{Rs. } 440$
Number of storybooks $= 15$

Working:
$$\text{Price of } 15\text{ storybooks} = 440 \times 15 = \mathbf{\text{Rs. } 6,600}$$

Answer: The price of $15\text{ storybooks}$ is $\text{Rs. } 6,600$.

Q3. If the price of $16\text{ school bags}$ is $\text{Rs. } 24,000$, then find the price of $1\text{ such bag}$.

Given:
Price of $16\text{ bags} = \text{Rs. } 24,000$
Number of bags $= 16$

Working:
$$\text{Price of } 1\text{ bag} = 24,000 \div 16 = \mathbf{\text{Rs. } 1,500}$$

Answer: The price of $1\text{ school bag}$ is $\text{Rs. } 1,500$.

Q4. If $425\text{ pearls}$ were used to make a necklace:

a) How many pearls will be used to make $25\text{ necklaces}$?
$$\text{Pearls} = 425 \times 25 = \mathbf{10,625\text{ pearls}}$$

b) How many pearls will be used to make $100\text{ necklaces}$?
$$\text{Pearls} = 425 \times 100 = \mathbf{42,500\text{ pearls}}$$

Q5. If the price of $5\text{ washing machines}$ is $\text{Rs. } 92,465$, then find:

First, find the price of $1\text{ washing machine}$:
$$\text{Price of } 1\text{ machine} = 92,465 \div 5 = \mathbf{\text{Rs. } 18,493}$$

a) The price of $10\text{ such washing machines}$:
$$\text{Price} = 18,493 \times 10 = \mathbf{\text{Rs. } 184,930}$$

b) The price of $24\text{ such washing machines}$:
$$\text{Price} = 18,493 \times 24 = \mathbf{\text{Rs. } 443,832}$$

📚 Unit 6 Key Vocabulary & Summary

Unitary Method: A technique of finding the value of a single unit first, then using multiplication to find the value of the required number of units.
Unit: A single object, quantity, or individual measure ($1$).
Value / Price: The cost, weight, capacity, or measure associated with an item.

Pro Tip for Exams: Always write the given quantities clearly with units ($\text{Rs.}$, $\text{kg}$, $\text{km}$, $\text{items}$). State the value of $1\text{ unit}$ clearly before computing the target answer!

Self-Assessment Practice

Test Your Knowledge on Chapter 6: Mastery Guide: The Unitary Method, Direct Proportion, Rates & Financial Applications

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