The Dairy Farm

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):

$$12 \times 35$$

| $\times$ | 30 kg | 5 kg |
|---|---|---|
| 10 | 300 | 50 |
| 2 | 60 | 10 |

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

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

Working using grid method:

| $\times$ | 30 | 2 |
|---|---|---|
| 8 | 240 | 16 |

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

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

Working using grid method:

| $\times$ | 60 | 9 |
|---|---|---|
| 40 | 2400 | 360 |
| 5 | 300 | 45 |

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

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

Breaking up: $78 = 70 + 8$

| $\times$ | 70 | 8 |
|---|---|---|
| 4 | 280 | 32 |

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

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

Breaking up: $83 = 80 + 3$

| $\times$ | 80 | 3 |
|---|---|---|
| 9 | 720 | 27 |

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

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

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

| $\times$ | 60 | 7 |
|---|---|---|
| 20 | 1200 | 140 |
| 8 | 480 | 56 |

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

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

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

| $\times$ | 50 | 3 |
|---|---|---|
| 30 | 1500 | 90 |
| 7 | 350 | 21 |

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

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

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: $548 = 500 + 40 + 8$

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

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

Working:
- $507 \times 20 = 10{,}140$
- $507 \times 3 = 1{,}521$

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)
- $239 \times 400 = 95{,}600$

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

Working:
- $101 \times 2 = 202$
- $101 \times 20 = 2{,}020$

$$202 + 2{,}020 = \mathbf{2{,}222}$$

Working:
$$\begin{array}{r} 807 \\ \times\phantom{00} 5 \\ \hline 4035 \end{array}$$

$5 \times 7 = 35$, write 5 carry 3; $5 \times 0 = 0 + 3 = 3$; $5 \times 8 = 40$

Answer: $807 \times 5 = \mathbf{4{,}035}$

Working:
- $143 \times 8 = 1{,}144$
- $143 \times 20 = 2{,}860$

$$1{,}144 + 2{,}860 = \mathbf{4{,}004}$$

Working:
- $450 \times 8 = 3{,}600$
- $450 \times 30 = 13{,}500$

$$3{,}600 + 13{,}500 = \mathbf{17{,}100}$$

Working:
- $584 \times 3 = 1{,}752$
- $584 \times 20 = 11{,}680$

$$1{,}752 + 11{,}680 = \mathbf{13{,}432}$$

Working:
- $302 \times 3 = 906$
- $302 \times 10 = 3{,}020$

$$906 + 3{,}020 = \mathbf{3{,}926}$$

Working:
- $604 \times 4 = 2{,}416$
- $604 \times 50 = 30{,}200$

$$2{,}416 + 30{,}200 = \mathbf{32{,}616}$$

Working:
- $112 \times 3 = 336$
- $112 \times 20 = 2{,}240$

$$336 + 2{,}240 = \mathbf{2{,}576}$$

Concept: Use properties of multiplication — distributive, commutative — to check equivalence without calculating.

(i) $12 \times 17 = 204$
- $11 \times 18 = 198$ ❌ (Not the same — different numbers, not related by halving/doubling)
- $6 \times 34$: $6 = 12 \div 2$ and $34 = 17 \times 2$, so $6 \times 34 = 12 \times 17$ ✅ Same

(ii) $26 \times 11$
- $26 \times 10 + 26 \times 1 = 26 \times (10 + 1) = 26 \times 11$ ✅ Same (Distributive property)
- $20 \times 11 + 6 \times 11 = (20 + 6) \times 11 = 26 \times 11$ ✅ Same (Distributive property)

(iii) $18 \times 4$
- $9 \times 8$: $9 = 18 \div 2$ and $8 = 4 \times 2$, so $9 \times 8 = 18 \times 4$ ✅ Same
- $20 \times 4 - 8 = 80 - 8 = 72$ and $18 \times 4 = 72$ ✅ Same

(iv) $55 \times 9$
- $50 \times 9 + 5 \times 9 = (50 + 5) \times 9 = 55 \times 9$ ✅ Same
- $54 \times 10 = 540$ but $55 \times 9 = 495$ ❌ Not the same
- $55 \times 10 - 55 = 550 - 55 = 495 = 55 \times 9$ ✅ Same

(v) $101 \times 42$
- $100 \times 42 + 100 = 4{,}200 + 100 = 4{,}300$ but $101 \times 42 = 4{,}242$ ❌ Not the same
- $100 \times 42 + 42 = (100 + 1) \times 42 = 101 \times 42$ ✅ Same

(vi) $247 \times 8$
- $250 \times 8 - 24 = 2{,}000 - 24 = 1{,}976$ and $247 \times 8 = 1{,}976$ ✅ Same (since $250 - 3 = 247$, so $250 \times 8 - 3 \times 8 = 250 \times 8 - 24$)
- $247 \times 10 - 247 = 2{,}470 - 247 = 2{,}223$ but $247 \times 8 = 1{,}976$ ❌ Not the same (this would equal $247 \times 9$)

(vii) $1001 \times 5$
- $1{,}000 \times 6 = 6{,}000$ but $1001 \times 5 = 5{,}005$ ❌ Not the same
- $1{,}000 \times 5 + 5 = 5{,}000 + 5 = 5{,}005 = 1001 \times 5$ ✅ Same

(viii) $1999 \times 2$
- $2{,}000 \times 2 - 4 = 4{,}000 - 4 = 3{,}996$ and $1999 \times 2 = 3{,}998$ ❌ Not the same
- $2{,}000 \times 2 - 2 = 4{,}000 - 2 = 3{,}998 = 1999 \times 2$ ✅ Same

Strategy: $25 = 100 \div 4$, so $16 \times 25 = 16 \times 100 \div 4 = 1{,}600 \div 4$

$$16 \times 25 = 4 \times 100 = \mathbf{400}$$

Alternatively: $16 \times 25 = 4 \times 4 \times 25 = 4 \times 100 = \mathbf{400}$

Strategy: $125 = 1000 \div 8$ and $12 = 8 + 4$

Alternatively: $8 \times 125 = 1{,}000$

$$12 \times 125 = (8 + 4) \times 125 = 1{,}000 + 500 = \mathbf{1{,}500}$$

Strategy: $250 = 1000 \div 4$

$$24 \times 250 = 24 \times \frac{1000}{4} = 6 \times 1{,}000 = \mathbf{6{,}000}$$

Strategy: $4 \times 25 = 100$

$$36 \times 25 = 9 \times 4 \times 25 = 9 \times 100 = \mathbf{900}$$

Strategy: $4 \times 75 = 300$

$$28 \times 75 = 7 \times 4 \times 75 = 7 \times 300 = \mathbf{2{,}100}$$

$$300 \times 15 = 3 \times 100 \times 15 = 3 \times 1{,}500 = \mathbf{4{,}500}$$

Or: $300 \times 15 = 300 \times 10 + 300 \times 5 = 3{,}000 + 1{,}500 = \mathbf{4{,}500}$

Strategy: $199 = 200 - 1$

$$199 \times 63 = (200 - 1) \times 63 = (200 \times 63) - (1 \times 63)$$
$$= 12{,}600 - 63 = \mathbf{12{,}537}$$

Strategy: $128 = 4 \times 32$ and $35 \times 4 = 140$

$$128 \times 35 = 4 \times 32 \times 35 = 4 \times 1{,}120 = \mathbf{4{,}480}$$

Or: $128 \times 35 = 128 \times 30 + 128 \times 5 = 3{,}840 + 640 = \mathbf{4{,}480}$

Sample examples using easy strategies:

1. $8 \times 125 = 8 \times \frac{1000}{8} = \mathbf{1{,}000}$

2. $4 \times 250 = 4 \times \frac{1000}{4} = \mathbf{1{,}000}$

3. $99 \times 7 = (100 - 1) \times 7 = 700 - 7 = \mathbf{693}$

4. $25 \times 44 = 25 \times 4 \times 11 = 100 \times 11 = \mathbf{1{,}100}$

5. $50 \times 46 = \frac{100 \times 46}{2} = \frac{4{,}600}{2} = \mathbf{2{,}300}$

*(Students may write their own valid examples using doubling/halving, rounding, or factor strategies.)*

Given: $17 \times 23 = 391$ (reference fact)

Concept: $17 \times 24 = 17 \times 23 + 17 \times 1 = 17 \times 23 + 17$

$$17 \times 24 = 391 + 17 = \mathbf{408}$$

*(To find $17 \times 24$, we add 17 to $17 \times 23$, i.e., we add 17, not 23.)*

Concept: $17 \times 22 = 17 \times 23 - 17$

$$17 \times 22 = 391 - 17 = \mathbf{374}$$

Concept: $16 \times 23 = 17 \times 23 - 1 \times 23 = 17 \times 23 - 23$

$$16 \times 23 = 391 - 23 = \mathbf{368}$$

Given: $8 \times 9 = 72$ (reference fact)

Concept: $18 \times 9 = (10 + 8) \times 9 = 10 \times 9 + 8 \times 9 = 90 + 72$

$$18 \times 9 = 90 + 72 = \mathbf{162}$$

Concept: $28 \times 9 = (20 + 8) \times 9 = 20 \times 9 + 8 \times 9 = 180 + 72$

$$28 \times 9 = 180 + 72 = \mathbf{252}$$

Concept: $108 \times 9 = (100 + 8) \times 9 = 100 \times 9 + 8 \times 9 = 900 + 72$

$$108 \times 9 = 900 + 72 = \mathbf{972}$$

Concept: $18 \times 23 = 17 \times 23 + 1 \times 23 = 391 + 23$

$$18 \times 23 = 391 + 23 = \mathbf{414}$$

Concept: Round each number to the nearest ten or hundred and multiply.

$25 \times 31$: $\approx 25 \times 30 = 750$ → matches 750

$132 \times 19$: $\approx 130 \times 20 = 2{,}600$ → matches 2,600

$101 \times 11$: $\approx 100 \times 10 = 1{,}000$ → matches 1,000

$248 \times 49$: $\approx 250 \times 50 = 12{,}500$ → matches 12,500

$12 \times 25$: $\approx 10 \times 30 = 300$ → matches 300

Final Matching:
$$25 \times 31 \rightarrow 750$$
$$132 \times 19 \rightarrow 2{,}600$$
$$101 \times 11 \rightarrow 1{,}000$$
$$248 \times 49 \rightarrow 12{,}500$$
$$12 \times 25 \rightarrow 300$$

Minister 1 (Start: 5, double each day for 7 days):

| Day | Coins |
|---|---|
| Start | 5 |
| Day 1 | $5 \times 2 = 10$ |
| Day 2 | $10 \times 2 = 20$ |
| Day 3 | $20 \times 2 = 40$ |
| Day 4 | $40 \times 2 = 80$ |
| Day 5 | $80 \times 2 = 160$ |
| Day 6 | $160 \times 2 = 320$ |
| Day 7 | $320 \times 2 = 640$ |

Total after 7 days = $5 \times 2^7 = 5 \times 128 = \mathbf{640}$ coins

---

Minister 2 (Start: 3, triple each day for 7 days):

| Day | Coins |
|---|---|
| Start | 3 |
| Day 1 | $3 \times 3 = 9$ |
| Day 2 | $9 \times 3 = 27$ |
| Day 3 | $27 \times 3 = 81$ |
| Day 4 | $81 \times 3 = 243$ |
| Day 5 | $243 \times 3 = 729$ |
| Day 6 | $729 \times 3 = 2{,}187$ |
| Day 7 | $2{,}187 \times 3 = 6{,}561$ |

Total after 7 days = $3 \times 3^7 = 3^8 = \mathbf{6{,}561}$ coins

---

Minister 3 (Start: 1, multiply by 5 each day for 7 days):

| Day | Coins |
|---|---|
| Start | 1 |
| Day 1 | $1 \times 5 = 5$ |
| Day 2 | $5 \times 5 = 25$ |
| Day 3 | $25 \times 5 = 125$ |
| Day 4 | $125 \times 5 = 625$ |
| Day 5 | $625 \times 5 = 3{,}125$ |
| Day 6 | $3{,}125 \times 5 = 15{,}625$ |
| Day 7 | $15{,}625 \times 5 = 78{,}125$ |

Total after 7 days = $5^7 = \mathbf{78{,}125}$ coins

---

Comparison:
- Minister 1: 640 coins
- Minister 2: 6,561 coins
- Minister 3: 78,125 coins

Answer: $\mathbf{Minister\ 3}$ received the most gold coins (78,125). Even though he started with only 1 coin, multiplying by 5 each day grew the fastest!

Given: $16 \times 44 = 704$

1) $8 \times 88$:
$8 = 16 \div 2$ (halved) and $88 = 44 \times 2$ (doubled) → Product stays the same.
$$8 \times 88 = \mathbf{704}$$

2) $8 \times 22$:
$8 = 16 \div 2$ (halved) and $22 = 44 \div 2$ (halved) → Product is halved twice = $\frac{1}{4}$ of original.
$$8 \times 22 = 704 \div 4 = \mathbf{176}$$

3) $16 \times 22$:
$16$ stays same, $22 = 44 \div 2$ (halved) → Product is halved.
$$16 \times 22 = 704 \div 2 = \mathbf{352}$$

4) $32 \times 44$:
$32 = 16 \times 2$ (doubled), $44$ stays same → Product is doubled.
$$32 \times 44 = 704 \times 2 = \mathbf{1{,}408}$$

Given: $12 \times 32 = 384$

1) $6 \times 16$:
$6 = 12 \div 2$ (halved) and $16 = 32 \div 2$ (halved) → Product = $384 \div 4$
$$6 \times 16 = \mathbf{96}$$

2) $24 \times 16$:
$24 = 12 \times 2$ (doubled) and $16 = 32 \div 2$ (halved) → Product stays same.
$$24 \times 16 = \mathbf{384}$$

3) $24 \times 64$:
$24 = 12 \times 2$ (doubled) and $64 = 32 \times 2$ (doubled) → Product = $384 \times 4$
$$24 \times 64 = \mathbf{1{,}536}$$

4) $12 \times 16$:
$12$ stays same, $16 = 32 \div 2$ (halved) → Product is halved.
$$12 \times 16 = 384 \div 2 = \mathbf{192}$$

Given: 18 books bought; 3 books cost ₹150; ₹20 left after purchase

Step 1: Number of sets of 3 books:
$$18 \div 3 = 6 \text{ sets}$$

Step 2: Total cost of books:
$$6 \times 150 = ₹900$$

Step 3: Money at the beginning:
$$900 + 20 = ₹\mathbf{920}$$

Answer: Mala had $₹\mathbf{920}$ at the beginning.

(a) Total collection:

Ticket sales:
$$57 \times 115$$
$$= 57 \times 100 + 57 \times 15$$
$$= 5{,}700 + 855 = ₹6{,}555$$

Participation fees:
$$3 \times 1{,}599 = 3 \times 1{,}600 - 3 = 4{,}800 - 3 = ₹4{,}797$$

Total collection:
$$6{,}555 + 4{,}797 = ₹\mathbf{11{,}352}$$

(b) Total expenses:
$$1{,}750 + 1{,}129 = ₹\mathbf{2{,}879}$$

Answer:
(a) Total collection = $₹\mathbf{11{,}352}$
(b) Total expenses = $₹\mathbf{2{,}879}$

(a) Total rows:

Rows in front + Ananya's row + rows behind:
$$12 + 1 + 17 = \mathbf{30} \text{ rows}$$

(b) Students in Ananya's row:

Students to right + Ananya + students to left:
$$18 + 1 + 22 = \mathbf{41} \text{ students}$$

(c) Total students:
$$30 \times 41$$
$$= 30 \times 40 + 30 \times 1$$
$$= 1{,}200 + 30$$
$$= \mathbf{1{,}230} \text{ students}$$

Answer:
(a) 30 rows
(b) 41 students in Ananya's row
(c) Total = $\mathbf{1{,}230}$ students