The Dairy Farm — NCERT Solutions

Given: Butter packets arranged in different row × column arrangements.

Concept: Commutative Property of Multiplication — changing the order of factors does not change the product.

Observation from the arrangements:

For example:

• $3 \times 4 = 12$ (3 rows, 4 in each row)
• $4 \times 3 = 12$ (4 rows, 3 in each row)

Similarly:

• $2 \times 6 = 12$ and $6 \times 2 = 12$
• $1 \times 12 = 12$ and $12 \times 1 = 12$

Pattern observed: The number of groups and the group size are interchanged in each case, but the total number of butter packets remains the same.

Conclusion: For any two numbers $a$ and $b$: $$a \times b = b \times a$$

This is called the Commutative Property of Multiplication and it is true for the product of any two numbers.

Given: $4 \times 10$

Concept: Multiplying by 10 shifts each digit one place to the left (adds one zero).

$$4 \times 10 = \mathbf{40}$$

Given: $20 \times 10$

Concept: Multiplying by 10 adds one zero to the number.

$$20 \times 10 = \mathbf{200}$$

Given: $10 \times 40$

Concept: By commutative property, $10 \times 40 = 40 \times 10$. Multiplying by 10 adds one zero.

$$10 \times 40 = \mathbf{400}$$

Given: $10 \times 10$

$$10 \times 10 = \mathbf{100}$$ ✓

Given: $20 \times 50$

Working:
$$20 \times 50 = 2 \times 5 \times 10 \times 10 = 10 \times 100 = \mathbf{1{,}000}$$

Given: $80 \times 10$

$$80 \times 10 = \mathbf{800}$$

Given: $3 \times 100$

Concept: Multiplying by 100 adds two zeros.

$$3 \times 100 = 100 \times 3 = \mathbf{300}$$ ✓

Given: $8 \times 100$

$$8 \times 100 = 100 \times 8 = \mathbf{800}$$

Given: $10 \times 100$

$$10 \times 100 = 100 \times 10 = \mathbf{1{,}000}$$

Concept: To multiply numbers ending in zeros, multiply the non-zero parts and then attach the total number of zeros.

(i) $60 \times 50$:
$$6 \times 5 = 30, \text{ attach } 2 \text{ zeros} \Rightarrow 60 \times 50 = \mathbf{3{,}000}$$
Place value table: Th = 3, H = 0, T = 0, O = 0

(ii) $220 \times 20$:
$$22 \times 2 = 44, \text{ attach } 2 \text{ zeros} \Rightarrow 220 \times 20 = \mathbf{4{,}400}$$
Place value table: Th = 4, H = 4, T = 0, O = 0

(iii) $11 \times 300$:
$$11 \times 3 = 33, \text{ attach } 2 \text{ zeros} \Rightarrow 11 \times 300 = \mathbf{3{,}300}$$
Place value table: Th = 3, H = 3, T = 0, O = 0

(iv) $80 \times 90$:
$$8 \times 9 = 72, \text{ attach } 2 \text{ zeros} \Rightarrow 80 \times 90 = \mathbf{7{,}200}$$
Place value table: Th = 7, H = 2, T = 0, O = 0

(v) $10 \times 63$:
$$10 \times 63 = \mathbf{630}$$
Place value table: Th = 0, H = 6, T = 3, O = 0

(vi) $40 \times 12$:
$$4 \times 12 = 48, \text{ attach } 1 \text{ zero} \Rightarrow 40 \times 12 = \mathbf{480}$$
Place value table: Th = 0, H = 4, T = 8, O = 0

Pattern observed: When we multiply numbers with zeros at the end, we multiply the non-zero digits and then place as many zeros at the end as there are in both numbers combined. Each digit shifts left by the number of zeros being multiplied.

Concept: Multiply the non-zero digits, then attach the combined number of zeros.

(i) $3 \times 5{,}000$:
$$3 \times 5 = 15, \text{ attach 3 zeros} \Rightarrow \mathbf{15{,}000}$$
TTh = 1, Th = 5, H = 0, T = 0, O = 0

(ii) $8 \times 3{,}000$:
$$8 \times 3 = 24, \text{ attach 3 zeros} \Rightarrow \mathbf{24{,}000}$$
TTh = 2, Th = 4, H = 0, T = 0, O = 0

(iii) $5 \times 7{,}000$:
$$5 \times 7 = 35, \text{ attach 3 zeros} \Rightarrow \mathbf{35{,}000}$$
TTh = 3, Th = 5, H = 0, T = 0, O = 0

(iv) $20 \times 100$:
$$2 \times 1 = 2, \text{ attach 3 zeros} \Rightarrow \mathbf{2{,}000}$$
TTh = 0, Th = 2, H = 0, T = 0, O = 0

(v) $40 \times 500$:
$$4 \times 5 = 20, \text{ attach 3 zeros} \Rightarrow \mathbf{20{,}000}$$
TTh = 2, Th = 0, H = 0, T = 0, O = 0

(vi) $60 \times 300$:
$$6 \times 3 = 18, \text{ attach 3 zeros} \Rightarrow \mathbf{18{,}000}$$
TTh = 1, Th = 8, H = 0, T = 0, O = 0

(vii) $600 \times 30$:
$$6 \times 3 = 18, \text{ attach 3 zeros} \Rightarrow \mathbf{18{,}000}$$
TTh = 1, Th = 8, H = 0, T = 0, O = 0

(viii) $80 \times 900$:
$$8 \times 9 = 72, \text{ attach 3 zeros} \Rightarrow \mathbf{72{,}000}$$
TTh = 7, Th = 2, H = 0, T = 0, O = 0

(ix) $70 \times 600$:
$$7 \times 6 = 42, \text{ attach 3 zeros} \Rightarrow \mathbf{42{,}000}$$
TTh = 4, Th = 2, H = 0, T = 0, O = 0

(x) $5 \times 7{,}000$:
$$5 \times 7 = 35, \text{ attach 3 zeros} \Rightarrow \mathbf{35{,}000}$$
TTh = 3, Th = 5, H = 0, T = 0, O = 0

Given: Number of rows = 35; Seats in each row = 42

To find: Total number of seats

Formula: Total seats = Number of rows × Seats per row

Working:
$$35 \times 42$$
$$= 35 \times (40 + 2)$$
$$= (35 \times 40) + (35 \times 2)$$
$$= 1{,}400 + 70$$
$$= \mathbf{1{,}470}$$

Answer: $\mathbf{1{,}470}$ people can sit in the auditorium.

Given: Distance jogged per day = 4 km; Number of days = 31

To find: Total distance in 31 days

Working:
$$31 \times 4 = (30 + 1) \times 4 = 120 + 4 = \mathbf{124}$$

Answer: Priya will jog $\mathbf{124}$ kilometres in 31 days.

Given: Number of boxes = 36; Books in each box = 48

To find: Total number of books

Working:
$$36 \times 48$$
$$= 36 \times (40 + 8)$$
$$= (36 \times 40) + (36 \times 8)$$
$$= 1{,}440 + 288$$
$$= \mathbf{1{,}728}$$

Answer: The school received $\mathbf{1{,}728}$ books in total.

Given: Cloth for 4 kurtas = 16 m

Step 1: Cloth for 1 kurta:
$$16 \div 4 = 4 \text{ m}$$

Step 2: Cloth for 8 kurtas:
$$8 \times 4 = \mathbf{32} \text{ m}$$

Alternate approach: 8 kurtas = 2 × 4 kurtas, so cloth = $2 \times 16 = 32$ m

Answer: Priya would need $\mathbf{32}$ metres of cloth to make 8 kurtas.

Given: Number of cows = 29; Milk per cow per day = 5 litres

Working:
$$29 \times 5 = (30 - 1) \times 5 = 150 - 5 = \mathbf{145}$$

Answer: The cows produce $\mathbf{145}$ litres of milk in total each day.

Given: Number of cows = 297; Fodder per cow per day = 18 kg

Working:
$$297 \times 18$$
$$= (300 - 3) \times 18$$
$$= (300 \times 18) - (3 \times 18)$$
$$= 5{,}400 - 54$$
$$= \mathbf{5{,}346}$$

Answer: $\mathbf{5{,}346}$ kg of fodder is needed every day.

Given: Waste in 1 month = 35 kg; Number of months in a year = 12

To find: Waste in 12 months

Working (using grid/Nida's method):

$$12 \times 35$$

$$300 + 60 = 360$$
$$50 + 10 = 60$$
$$360 + 60 = \mathbf{420}$$

Answer: The family will produce $\mathbf{420}$ kg of kitchen waste in a year.

Working using grid method:

$$240 + 16 = \mathbf{256}$$

Verification using standard method:
$$32 \times 8 = (30 + 2) \times 8 = 240 + 16 = \mathbf{256}$$

Working using grid method:

$$2400 + 360 + 300 + 45$$
$$= 2760 + 345$$
$$= \mathbf{3{,}105}$$

Verification:

• $5 \times 69 = 345$
• $40 \times 69 = 2{,}760$
• Total $= 345 + 2{,}760 = \mathbf{3{,}105}$

Breaking up: $78 = 70 + 8$

$$280 + 32 = \mathbf{312}$$

Answer: $78 \times 4 = \mathbf{312}$

Breaking up: $83 = 80 + 3$

$$720 + 27 = \mathbf{747}$$

Answer: $83 \times 9 = \mathbf{747}$

Breaking up: $67 = 60 + 7$; $28 = 20 + 8$

$$1200 + 140 + 480 + 56$$
$$= 1340 + 536$$
$$= \mathbf{1{,}876}$$

Answer: $67 \times 28 = \mathbf{1{,}876}$

Breaking up: $53 = 50 + 3$; $37 = 30 + 7$

$$1500 + 90 + 350 + 21$$
$$= 1590 + 371$$
$$= \mathbf{1{,}961}$$

Answer: $53 \times 37 = \mathbf{1{,}961}$

Working:
$$94 \times 5 = (90 + 4) \times 5$$
$$= (90 \times 5) + (4 \times 5)$$
$$= 450 + 20$$
$$= \mathbf{470}$$

Working:
$$49 \times 6 = (40 + 9) \times 6$$
$$= (40 \times 6) + (9 \times 6)$$
$$= 240 + 54$$
$$= \mathbf{294}$$

Working:
$$37 \times 53 = 37 \times (50 + 3)$$
$$= (37 \times 50) + (37 \times 3)$$
$$= 1{,}850 + 111$$
$$= \mathbf{1{,}961}$$

Working:
$$28 \times 79 = 28 \times (80 - 1)$$
$$= (28 \times 80) - (28 \times 1)$$
$$= 2{,}240 - 28$$
$$= \mathbf{2{,}212}$$

Alternatively:
$$28 \times 79 = (20 + 8) \times 79$$
$$= (20 \times 79) + (8 \times 79)$$
$$= 1{,}580 + 632 = \mathbf{2{,}212}$$

Working:
$$\begin{array}{r} 86 \\ \times\phantom{0} 3 \\ \hline 258 \end{array}$$

$3 \times 6 = 18$, write 8 carry 1; $3 \times 8 = 24 + 1 = 25$

Answer: $86 \times 3 = \mathbf{258}$

Working:
$$\begin{array}{r} 72 \\ \times\phantom{0} 7 \\ \hline 504 \end{array}$$

$7 \times 2 = 14$, write 4 carry 1; $7 \times 7 = 49 + 1 = 50$

Answer: $72 \times 7 = \mathbf{504}$

Working:
$$\begin{array}{r} 94 \\ \times 36 \\ \hline 564 \quad (94 \times 6)\\ 2820 \quad (94 \times 30)\\ \hline 3384 \end{array}$$

• $94 \times 6 = 564$
• $94 \times 30 = 2{,}820$
• $564 + 2{,}820 = \mathbf{3{,}384}$

Answer: $94 \times 36 = \mathbf{3{,}384}$

Working:
$$\begin{array}{r} 66 \\ \times 22 \\ \hline 132 \quad (66 \times 2)\\ 1320 \quad (66 \times 20)\\ \hline 1452 \end{array}$$

• $66 \times 2 = 132$
• $66 \times 20 = 1{,}320$
• $132 + 1{,}320 = \mathbf{1{,}452}$

Answer: $66 \times 22 = \mathbf{1{,}452}$

Given: 8 rows, 12 seats per row

Step 1: Total seats:
$$8 \times 12 = 96 \text{ seats}$$

Step 2: Half the seats filled:
$$96 \div 2 = 48 \text{ people}$$

Step 3: 3 more rows get filled:
$$3 \times 12 = 36 \text{ more people}$$

Step 4: Total people:
$$48 + 36 = \mathbf{84} \text{ people}$$

Answer: 48 people are initially watching. After 3 more rows fill up, there are $\mathbf{84}$ people in total.

Given: Twenty-four 4s, eighteen 6s, 234 runs by running, 23 extras

Step 1: Runs from 4s:
$$24 \times 4 = 96 \text{ runs}$$

Step 2: Runs from 6s:
$$18 \times 6 = 108 \text{ runs}$$

Step 3: Total runs:
$$96 + 108 + 234 + 23$$
$$= 204 + 234 + 23$$
$$= 438 + 23$$
$$= \mathbf{461} \text{ runs}$$

Answer: Runs from 4s = 96, Runs from 6s = 108. Total runs scored by the Indian team = $\mathbf{461}$ runs.

Given: 15 bulbs at ₹25 each; 12 tube lights at ₹34 each

Step 1: Cost of bulbs:
$$15 \times 25 = ₹375$$

Step 2: Cost of tube lights:
$$12 \times 34 = 12 \times 30 + 12 \times 4 = 360 + 48 = ₹408$$

Step 3: Total amount:
$$375 + 408 = ₹\mathbf{783}$$

Answer: Anjali should give $₹\mathbf{783}$ to the shopkeeper.

Given: 28 bags; Cost per bag = ₹350

Working:
$$28 \times 350 = 28 \times 35 \times 10$$
$$28 \times 35 = 28 \times 30 + 28 \times 5 = 840 + 140 = 980$$
$$980 \times 10 = \mathbf{9{,}800}$$

Answer: The shopkeeper earned $₹\mathbf{9{,}800}$ by selling rice bags.

Given: 86 shelves; 162 books per shelf

Working:
$$86 \times 162$$
$$= 86 \times (100 + 60 + 2)$$
$$= (86 \times 100) + (86 \times 60) + (86 \times 2)$$
$$= 8{,}600 + 5{,}160 + 172$$
$$= 13{,}760 + 172$$
$$= \mathbf{13{,}932}$$

Answer: The school library has $\mathbf{13{,}932}$ books.

Breaking up: $548 = 500 + 40 + 8$

$$548 \times 6 = (500 \times 6) + (40 \times 6) + (8 \times 6)$$
$$= 3{,}000 + 240 + 48$$
$$= \mathbf{3{,}288}$$

$$682 \times 3 = (600 + 80 + 2) \times 3$$
$$= 1{,}800 + 240 + 6$$
$$= \mathbf{2{,}046}$$

$$324 \times 18 = 324 \times (10 + 8)$$
$$= (324 \times 10) + (324 \times 8)$$
$$= 3{,}240 + 2{,}592$$
$$= \mathbf{5{,}832}$$

Working for $324 \times 8$:
$(300 \times 8) + (20 \times 8) + (4 \times 8) = 2{,}400 + 160 + 32 = 2{,}592$

$$507 \times 23 = 507 \times (20 + 3)$$
$$= (507 \times 20) + (507 \times 3)$$
$$= 10{,}140 + 1{,}521$$
$$= \mathbf{11{,}661}$$

Working:

• $507 \times 20 = 10{,}140$
• $507 \times 3 = 1{,}521$

$$190 \times 65 = 190 \times (60 + 5)$$
$$= (190 \times 60) + (190 \times 5)$$
$$= 11{,}400 + 950$$
$$= \mathbf{12{,}350}$$

Working:
$$123 \times 84$$

• $123 \times 4 = 492$
• $123 \times 80 = 9{,}840$

$$492 + 9{,}840 = \mathbf{10{,}332}$$

Working:

• $368 \times 2 = 736$
• $368 \times 30 = 11{,}040$

$$736 + 11{,}040 = \mathbf{11{,}776}$$

Working:

• $159 \times 4 = 636$
• $159 \times 20 = 3{,}180$
• $159 \times 300 = 47{,}700$

$$636 + 3{,}180 + 47{,}700 = \mathbf{51{,}516}$$

Working:

• $239 \times 1 = 239$
• $239 \times 0 = 0$ (tens place)
• $239 \times 400 = 95{,}600$

$$239 + 0 + 95{,}600 = \mathbf{95{,}839}$$