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NCERT Solutions

The Dairy Farm

CBSE · Class 5 · Mathematics

NCERT Solutions for The Dairy Farm — CBSE Class 5 Mathematics.

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Order of Numbers in Multiplication

1Daljeet Kaur has arranged butter packets in different ways (rows × columns). Find the number of butter packets in each case and describe the pattern you notice.Show solution
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:
- 3×4=123 \times 4 = 12 (3 rows, 4 in each row)
- 4×3=124 \times 3 = 12 (4 rows, 3 in each row)

Similarly:
- 2×6=122 \times 6 = 12 and 6×2=126 \times 2 = 12
- 1×12=121 \times 12 = 12 and 12×1=1212 \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 aa and bb: a×b=b×aa \times b = b \times a

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

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Patterns in Multiplication by 10s and 100s — Exercise 1

1a4×10=_4 \times 10 = \_Show solution
Given: 4×104 \times 10

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

4×10=404 \times 10 = \mathbf{40}

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1b20×10=_20 \times 10 = \_Show solution
Given: 20×1020 \times 10

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

20×10=20020 \times 10 = \mathbf{200}

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1c10×40=_10 \times 40 = \_Show solution
Given: 10×4010 \times 40

Concept: By commutative property, 10×40=40×1010 \times 40 = 40 \times 10. Multiplying by 10 adds one zero.

10×40=40010 \times 40 = \mathbf{400}

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1d10×10=10010 \times 10 = 100 (given/verify)Show solution
Given: 10×1010 \times 10

10×10=10010 \times 10 = \mathbf{100} ✓

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1e20×50=_20 \times 50 = \_Show solution
Given: 20×5020 \times 50

Working:
20×50=2×5×10×10=10×100=1,00020 \times 50 = 2 \times 5 \times 10 \times 10 = 10 \times 100 = \mathbf{1{,}000}

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1f80×10=_80 \times 10 = \_Show solution
Given: 80×1080 \times 10

80×10=80080 \times 10 = \mathbf{800}

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1g3×100=100×3=3003 \times 100 = 100 \times 3 = 300 (given/verify)Show solution
Given: 3×1003 \times 100

Concept: Multiplying by 100 adds two zeros.

3×100=100×3=3003 \times 100 = 100 \times 3 = \mathbf{300} ✓

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1h8×100=_=_8 \times 100 = \_ = \_Show solution
Given: 8×1008 \times 100

8×100=100×8=8008 \times 100 = 100 \times 8 = \mathbf{800}

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1i10×100=_=_10 \times 100 = \_ = \_Show solution
Given: 10×10010 \times 100

10×100=100×10=1,00010 \times 100 = 100 \times 10 = \mathbf{1{,}000}

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Patterns in Multiplication by 10s and 100s — Exercise 2 (Fill in the Table)

2Find answers to the following and fill in the place value table. Describe the pattern.
(i) 60×5060 \times 50
(ii) 220×20220 \times 20
(iii) 11×30011 \times 300
(iv) 80×9080 \times 90
(v) 10×6310 \times 63
(vi) 40×1240 \times 12Show solution
Concept: To multiply numbers ending in zeros, multiply the non-zero parts and then attach the total number of zeros.

**(i) 60×5060 \times 50:**
6×5=30, attach 2 zeros⇒60×50=3,0006 \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) 220×20220 \times 20:**
22×2=44, attach 2 zeros⇒220×20=4,40022 \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) 11×30011 \times 300:**
11×3=33, attach 2 zeros⇒11×300=3,30011 \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) 80×9080 \times 90:**
8×9=72, attach 2 zeros⇒80×90=7,2008 \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) 10×6310 \times 63:**
10×63=63010 \times 63 = \mathbf{630}
Place value table: Th = 0, H = 6, T = 3, O = 0

**(vi) 40×1240 \times 12:**
4×12=48, attach 1 zero⇒40×12=4804 \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.

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3Fill in the place value table for multiplication by 1,000:
(i) 3×5,0003 \times 5{,}000
(ii) 8×3,0008 \times 3{,}000
(iii) 5×7,0005 \times 7{,}000
(iv) 20×10020 \times 100
(v) 40×50040 \times 500
(vi) 60×30060 \times 300
(vii) 600×30600 \times 30
(viii) 80×90080 \times 900
(ix) 70×60070 \times 600
(x) 5×7,0005 \times 7{,}000 (repeated)Show solution
Concept: Multiply the non-zero digits, then attach the combined number of zeros.

**(i) 3×5,0003 \times 5{,}000:**
3×5=15, attach 3 zeros⇒15,0003 \times 5 = 15, \text{ attach 3 zeros} \Rightarrow \mathbf{15{,}000}
TTh = 1, Th = 5, H = 0, T = 0, O = 0

**(ii) 8×3,0008 \times 3{,}000:**
8×3=24, attach 3 zeros⇒24,0008 \times 3 = 24, \text{ attach 3 zeros} \Rightarrow \mathbf{24{,}000}
TTh = 2, Th = 4, H = 0, T = 0, O = 0

**(iii) 5×7,0005 \times 7{,}000:**
5×7=35, attach 3 zeros⇒35,0005 \times 7 = 35, \text{ attach 3 zeros} \Rightarrow \mathbf{35{,}000}
TTh = 3, Th = 5, H = 0, T = 0, O = 0

**(iv) 20×10020 \times 100:**
2×1=2, attach 3 zeros⇒2,0002 \times 1 = 2, \text{ attach 3 zeros} \Rightarrow \mathbf{2{,}000}
TTh = 0, Th = 2, H = 0, T = 0, O = 0

**(v) 40×50040 \times 500:**
4×5=20, attach 3 zeros⇒20,0004 \times 5 = 20, \text{ attach 3 zeros} \Rightarrow \mathbf{20{,}000}
TTh = 2, Th = 0, H = 0, T = 0, O = 0

**(vi) 60×30060 \times 300:**
6×3=18, attach 3 zeros⇒18,0006 \times 3 = 18, \text{ attach 3 zeros} \Rightarrow \mathbf{18{,}000}
TTh = 1, Th = 8, H = 0, T = 0, O = 0

**(vii) 600×30600 \times 30:**
6×3=18, attach 3 zeros⇒18,0006 \times 3 = 18, \text{ attach 3 zeros} \Rightarrow \mathbf{18{,}000}
TTh = 1, Th = 8, H = 0, T = 0, O = 0

**(viii) 80×90080 \times 900:**
8×9=72, attach 3 zeros⇒72,0008 \times 9 = 72, \text{ attach 3 zeros} \Rightarrow \mathbf{72{,}000}
TTh = 7, Th = 2, H = 0, T = 0, O = 0

**(ix) 70×60070 \times 600:**
7×6=42, attach 3 zeros⇒42,0007 \times 6 = 42, \text{ attach 3 zeros} \Rightarrow \mathbf{42{,}000}
TTh = 4, Th = 2, H = 0, T = 0, O = 0

**(x) 5×7,0005 \times 7{,}000:**
5×7=35, attach 3 zeros⇒35,0005 \times 7 = 35, \text{ attach 3 zeros} \Rightarrow \mathbf{35{,}000}
TTh = 3, Th = 5, H = 0, T = 0, O = 0

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Let Us Solve (Word Problems — Multiplication)

1A school has an auditorium with 35 rows, with 42 seats in each row. How many people can sit in this auditorium?Show solution
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:
35×4235 \times 42
=35×(40+2)= 35 \times (40 + 2)
=(35×40)+(35×2)= (35 \times 40) + (35 \times 2)
=1,400+70= 1{,}400 + 70
=1,470= \mathbf{1{,}470}

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

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2Priya jogs 4 kilometres every day. How many kilometres will she jog in 31 days?Show solution
Given: Distance jogged per day = 4 km; Number of days = 31

To find: Total distance in 31 days

Working:
31×4=(30+1)×4=120+4=12431 \times 4 = (30 + 1) \times 4 = 120 + 4 = \mathbf{124}

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

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3A school has received 36 boxes of books with 48 books in each box. How many total books did the school receive?Show solution
Given: Number of boxes = 36; Books in each box = 48

To find: Total number of books

Working:
36×4836 \times 48
=36×(40+8)= 36 \times (40 + 8)
=(36×40)+(36×8)= (36 \times 40) + (36 \times 8)
=1,440+288= 1{,}440 + 288
=1,728= \mathbf{1{,}728}

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

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4Priya uses 16 metres of cloth to make 4 kurtas. How much cloth would she need to make 8 kurtas?Show solution
Given: Cloth for 4 kurtas = 16 m

Step 1: Cloth for 1 kurta:
16÷4=4 m16 \div 4 = 4 \text{ m}

Step 2: Cloth for 8 kurtas:
8×4=32 m8 \times 4 = \mathbf{32} \text{ m}

Alternate approach: 8 kurtas = 2 × 4 kurtas, so cloth = 2×16=322 \times 16 = 32 m

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

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5Gollappa has 29 cows on his farm. Each cow produces 5 litres of milk per day. How many litres of milk do the cows produce in total each day?Show solution
Given: Number of cows = 29; Milk per cow per day = 5 litres

Working:
29×5=(30−1)×5=150−5=14529 \times 5 = (30 - 1) \times 5 = 150 - 5 = \mathbf{145}

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

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6Maska Cow Farm has 297 cows. Each cow requires 18 kg of fodder per day. How much total fodder is needed to feed 297 cows every day?Show solution
Given: Number of cows = 297; Fodder per cow per day = 18 kg

Working:
297×18297 \times 18
=(300−3)×18= (300 - 3) \times 18
=(300×18)−(3×18)= (300 \times 18) - (3 \times 18)
=5,400−54= 5{,}400 - 54
=5,346= \mathbf{5{,}346}

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

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Waste and Composting

1A family of 4 produces around 35 kg of kitchen waste in a month. How much waste will the family produce in a year?Show solution
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):

12×3512 \times 35

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

300+60=360300 + 60 = 360
50+10=6050 + 10 = 60
360+60=420360 + 60 = \mathbf{420}

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

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Let Us Multiply — Practice Problems

aFind 32×832 \times 8 using the grid (area) method.Show solution
Working using grid method:

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

240+16=256240 + 16 = \mathbf{256}

Verification using standard method:
32×8=(30+2)×8=240+16=25632 \times 8 = (30 + 2) \times 8 = 240 + 16 = \mathbf{256}

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bFind 69×4569 \times 45 using the grid (area) method.Show solution
Working using grid method:

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

2400+360+300+452400 + 360 + 300 + 45
=2760+345= 2760 + 345
=3,105= \mathbf{3{,}105}

Verification:
- 5×69=3455 \times 69 = 345
- 40×69=2,76040 \times 69 = 2{,}760
- Total =345+2,760=3,105= 345 + 2{,}760 = \mathbf{3{,}105}

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Let Us Do — Exercise 1 (Grid Method)

1aSolve 78×478 \times 4 using the grid method (like Nida).Show solution
Breaking up: 78=70+878 = 70 + 8

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

280+32=312280 + 32 = \mathbf{312}

Answer: 78×4=31278 \times 4 = \mathbf{312}

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1bSolve 83×983 \times 9 using the grid method (like Nida).Show solution
Breaking up: 83=80+383 = 80 + 3

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

720+27=747720 + 27 = \mathbf{747}

Answer: 83×9=74783 \times 9 = \mathbf{747}

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1cSolve 67×2867 \times 28 using the grid method (like Nida).Show solution
Breaking up: 67=60+767 = 60 + 7; 28=20+828 = 20 + 8

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

1200+140+480+561200 + 140 + 480 + 56
=1340+536= 1340 + 536
=1,876= \mathbf{1{,}876}

Answer: 67×28=1,87667 \times 28 = \mathbf{1{,}876}

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1dSolve 53×3753 \times 37 using the grid method (like Nida).Show solution
Breaking up: 53=50+353 = 50 + 3; 37=30+737 = 30 + 7

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

1500+90+350+211500 + 90 + 350 + 21
=1590+371= 1590 + 371
=1,961= \mathbf{1{,}961}

Answer: 53×37=1,96153 \times 37 = \mathbf{1{,}961}

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Let Us Do — Exercise 2 (Kanti's Method — Partial Products Written Vertically)

2aSolve 94×594 \times 5 like Kanti (expanded/partial products method).Show solution
Working:
94×5=(90+4)×594 \times 5 = (90 + 4) \times 5
=(90×5)+(4×5)= (90 \times 5) + (4 \times 5)
=450+20= 450 + 20
=470= \mathbf{470}

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2bSolve 49×649 \times 6 like Kanti.Show solution
Working:
49×6=(40+9)×649 \times 6 = (40 + 9) \times 6
=(40×6)+(9×6)= (40 \times 6) + (9 \times 6)
=240+54= 240 + 54
=294= \mathbf{294}

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2cSolve 37×5337 \times 53 like Kanti.Show solution
Working:
37×53=37×(50+3)37 \times 53 = 37 \times (50 + 3)
=(37×50)+(37×3)= (37 \times 50) + (37 \times 3)
=1,850+111= 1{,}850 + 111
=1,961= \mathbf{1{,}961}

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2dSolve 28×7928 \times 79 like Kanti.Show solution
Working:
28×79=28×(80−1)28 \times 79 = 28 \times (80 - 1)
=(28×80)−(28×1)= (28 \times 80) - (28 \times 1)
=2,240−28= 2{,}240 - 28
=2,212= \mathbf{2{,}212}

Alternatively:
28×79=(20+8)×7928 \times 79 = (20 + 8) \times 79
=(20×79)+(8×79)= (20 \times 79) + (8 \times 79)
=1,580+632=2,212= 1{,}580 + 632 = \mathbf{2{,}212}

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Let Us Do — Exercise 3 (John's Method — Standard Algorithm)

3aSolve 86×386 \times 3 like John (standard algorithm).Show solution
Working:
86×03258\begin{array}{r} 86 \\ \times\phantom{0} 3 \\ \hline 258 \end{array}

3×6=183 \times 6 = 18, write 8 carry 1; 3×8=24+1=253 \times 8 = 24 + 1 = 25

Answer: 86×3=25886 \times 3 = \mathbf{258}

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3bSolve 72×772 \times 7 like John.Show solution
Working:
72×07504\begin{array}{r} 72 \\ \times\phantom{0} 7 \\ \hline 504 \end{array}

7×2=147 \times 2 = 14, write 4 carry 1; 7×7=49+1=507 \times 7 = 49 + 1 = 50

Answer: 72×7=50472 \times 7 = \mathbf{504}

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3cSolve 94×3694 \times 36 like John.Show solution
Working:
94×36564(94×6)2820(94×30)3384\begin{array}{r} 94 \\ \times 36 \\ \hline 564 \quad (94 \times 6)\\ 2820 \quad (94 \times 30)\\ \hline 3384 \end{array}

- 94×6=56494 \times 6 = 564
- 94×30=2,82094 \times 30 = 2{,}820
- 564+2,820=3,384564 + 2{,}820 = \mathbf{3{,}384}

Answer: 94×36=3,38494 \times 36 = \mathbf{3{,}384}

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3dSolve 66×2266 \times 22 like John.Show solution
Working:
66×22132(66×2)1320(66×20)1452\begin{array}{r} 66 \\ \times 22 \\ \hline 132 \quad (66 \times 2)\\ 1320 \quad (66 \times 20)\\ \hline 1452 \end{array}

- 66×2=13266 \times 2 = 132
- 66×20=1,32066 \times 20 = 1{,}320
- 132+1,320=1,452132 + 1{,}320 = \mathbf{1{,}452}

Answer: 66×22=1,45266 \times 22 = \mathbf{1{,}452}

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Let Us Do — Exercise 4 (Word Problems)

4aA movie theater has 8 rows of seats, and each row has 12 seats. If half the seats are filled, how many people are watching the movie? If 3 more rows get filled, how many total people will be there?Show solution
Given: 8 rows, 12 seats per row

Step 1: Total seats:
8×12=96 seats8 \times 12 = 96 \text{ seats}

Step 2: Half the seats filled:
96÷2=48 people96 \div 2 = 48 \text{ people}

Step 3: 3 more rows get filled:
3×12=36 more people3 \times 12 = 36 \text{ more people}

Step 4: Total people:
48+36=84 people48 + 36 = \mathbf{84} \text{ people}

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

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4bIn a test match between India and West Indies, the Indian team hit twenty-four 4s and eighteen 6s across the two innings. How many runs were scored in 4s and 6s each? 234 runs were made by running between the wickets. If 23 runs were extras, how many runs were scored by the Indian team in the two innings?Show solution
Given: Twenty-four 4s, eighteen 6s, 234 runs by running, 23 extras

Step 1: Runs from 4s:
24×4=96 runs24 \times 4 = 96 \text{ runs}

Step 2: Runs from 6s:
18×6=108 runs18 \times 6 = 108 \text{ runs}

Step 3: Total runs:
96+108+234+2396 + 108 + 234 + 23
=204+234+23= 204 + 234 + 23
=438+23= 438 + 23
=461 runs= \mathbf{461} \text{ runs}

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

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4cAnjali buys 15 bulbs and 12 tube lights from Sudha Electricals. Each bulb costs ₹25 and each tube light costs ₹34. How much money should Anjali give to the shopkeeper?Show solution
Given: 15 bulbs at ₹25 each; 12 tube lights at ₹34 each

Step 1: Cost of bulbs:
15×25=₹37515 \times 25 = ₹375

Step 2: Cost of tube lights:
12×34=12×30+12×4=360+48=₹40812 \times 34 = 12 \times 30 + 12 \times 4 = 360 + 48 = ₹408

Step 3: Total amount:
375+408=₹783375 + 408 = ₹\mathbf{783}

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

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4dA shopkeeper sold 28 bags of rice. Each bag costs ₹350. How much money did he earn by selling rice bags?Show solution
Given: 28 bags; Cost per bag = ₹350

Working:
28×350=28×35×1028 \times 350 = 28 \times 35 \times 10
28×35=28×30+28×5=840+140=98028 \times 35 = 28 \times 30 + 28 \times 5 = 840 + 140 = 980
980×10=9,800980 \times 10 = \mathbf{9{,}800}

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

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4eA school library has 86 shelves and each shelf has 162 books. Find the number of books in the library.Show solution
Given: 86 shelves; 162 books per shelf

Working:
86×16286 \times 162
=86×(100+60+2)= 86 \times (100 + 60 + 2)
=(86×100)+(86×60)+(86×2)= (86 \times 100) + (86 \times 60) + (86 \times 2)
=8,600+5,160+172= 8{,}600 + 5{,}160 + 172
=13,760+172= 13{,}760 + 172
=13,932= \mathbf{13{,}932}

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

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Let Us Solve — Three-digit Multiplication (Nida's Method)

1aSolve 548×6548 \times 6 like Nida (grid/expanded method).Show solution
Breaking up: 548=500+40+8548 = 500 + 40 + 8

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

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1bSolve 682×3682 \times 3 like Nida.Show solution
682×3=(600+80+2)×3682 \times 3 = (600 + 80 + 2) \times 3
=1,800+240+6= 1{,}800 + 240 + 6
=2,046= \mathbf{2{,}046}

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1cSolve 324×18324 \times 18 like Nida.Show solution
324×18=324×(10+8)324 \times 18 = 324 \times (10 + 8)
=(324×10)+(324×8)= (324 \times 10) + (324 \times 8)
=3,240+2,592= 3{,}240 + 2{,}592
=5,832= \mathbf{5{,}832}

**Working for 324×8324 \times 8:**
(300×8)+(20×8)+(4×8)=2,400+160+32=2,592(300 \times 8) + (20 \times 8) + (4 \times 8) = 2{,}400 + 160 + 32 = 2{,}592

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1dSolve 507×23507 \times 23 like Nida.Show solution
507×23=507×(20+3)507 \times 23 = 507 \times (20 + 3)
=(507×20)+(507×3)= (507 \times 20) + (507 \times 3)
=10,140+1,521= 10{,}140 + 1{,}521
=11,661= \mathbf{11{,}661}

Working:
- 507×20=10,140507 \times 20 = 10{,}140
- 507×3=1,521507 \times 3 = 1{,}521

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1eSolve 190×65190 \times 65 like Nida.Show solution
190×65=190×(60+5)190 \times 65 = 190 \times (60 + 5)
=(190×60)+(190×5)= (190 \times 60) + (190 \times 5)
=11,400+950= 11{,}400 + 950
=12,350= \mathbf{12{,}350}

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Let Us Solve — Three-digit Multiplication (John's Method — Standard Algorithm)

2aSolve 123×84123 \times 84 like John.Show solution
Working:
123×84123 \times 84
- 123×4=492123 \times 4 = 492
- 123×80=9,840123 \times 80 = 9{,}840

492+9,840=10,332492 + 9{,}840 = \mathbf{10{,}332}

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2bSolve 368×32368 \times 32 like John.Show solution
Working:
- 368×2=736368 \times 2 = 736
- 368×30=11,040368 \times 30 = 11{,}040

736+11,040=11,776736 + 11{,}040 = \mathbf{11{,}776}

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2cSolve 159×324159 \times 324 like John.Show solution
Working:
- 159×4=636159 \times 4 = 636
- 159×20=3,180159 \times 20 = 3{,}180
- 159×300=47,700159 \times 300 = 47{,}700

636+3,180+47,700=51,516636 + 3{,}180 + 47{,}700 = \mathbf{51{,}516}

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2dSolve 239×401239 \times 401 like John.Show solution
Working:
- 239×1=239239 \times 1 = 239
- 239×0=0239 \times 0 = 0 (tens place)
- 239×400=95,600239 \times 400 = 95{,}600

239+0+95,600=95,839239 + 0 + 95{,}600 = \mathbf{95{,}839}

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2eSolve 592×5592 \times 5 like John.
2fSolve 101×22101 \times 22 like John.

Let Us Solve — Mili's Father's Method (Standard Long Multiplication)

3aSolve 807×5807 \times 5 using Mili's father's method (standard algorithm).
3bSolve 143×28143 \times 28 using Mili's father's method.
3cSolve 309×9309 \times 9 using Mili's father's method.
3dSolve 450×38450 \times 38 using Mili's father's method.
3eSolve 584×23584 \times 23 using Mili's father's method.
3fSolve 302×13302 \times 13 using Mili's father's method.
3gSolve 604×54604 \times 54 using Mili's father's method.
3hSolve 112×23112 \times 23 using Mili's father's method.
3iSolve 237×19237 \times 19 using Mili's father's method.

Let Us Do — Identify Same-Answer Problems (No Calculation)

1Identify the problems that have the same answer as the one given at the top of each box. Do not calculate.
(i) 12×1712 \times 17: Is 11×1811 \times 18 the same? Is 6×346 \times 34 the same?
(ii) 26×1126 \times 11: Is '26×1026 \times 10 and 26×126 \times 1' the same? Is '20×1120 \times 11 and 6×116 \times 11' the same?
(iii) 18×418 \times 4: Is 9×89 \times 8 the same? Is 20×4−820 \times 4 - 8 the same?
(iv) 55×955 \times 9: Is '50×950 \times 9 and 5×95 \times 9' the same? Is 54×1054 \times 10 the same? Is 55×10−5555 \times 10 - 55 the same?
(v) 101×42101 \times 42: Is '100×42100 \times 42 and 100100' the same? Is '100×42100 \times 42 and 4242' the same?
(vi) 247×8247 \times 8: Is 250×8−24250 \times 8 - 24 the same? Is 247×10−247247 \times 10 - 247 the same?
(vii) 1001×51001 \times 5: Is 1,000×61{,}000 \times 6 the same? Is '1,000×51{,}000 \times 5 and 55' the same?
(viii) 1999×21999 \times 2: Is 2,000×2−42{,}000 \times 2 - 4 the same? Is 2,000×2−22{,}000 \times 2 - 2 the same?

Let Us Do — Exercise 2 (Easy Ways to Multiply)

2aFind 16×2516 \times 25 using an easy method.
2bFind 12×12512 \times 125 using an easy method.
2cFind 24×25024 \times 250 using an easy method.
2dFind 36×2536 \times 25 using an easy method.
2eFind 28×7528 \times 75 using an easy method.
2fFind 300×15300 \times 15 using an easy method.
2gFind 50×7850 \times 78 using an easy method.
2hFind 199×63199 \times 63 using an easy method.
2iFind 128×35128 \times 35 using an easy method.
3Write 5 other examples for which you can find easy ways of getting products.

Let Us Do — Exercise 4 (Using Known Products)

4aGiven: 17×23=39117 \times 23 = 391. This is given as reference.
4bFind 17×2417 \times 24 given that 17×23=39117 \times 23 = 391.
4cFind 17×2217 \times 22 given that 17×23=39117 \times 23 = 391.
4dFind 16×2316 \times 23 given that 17×23=39117 \times 23 = 391.
4eGiven: 8×9=728 \times 9 = 72. This is given as reference.
4fFind 18×918 \times 9 given that 8×9=728 \times 9 = 72.
4gFind 28×928 \times 9 given that 8×9=728 \times 9 = 72.
4hFind 108×9108 \times 9 given that 8×9=728 \times 9 = 72.
4iFind 18×2318 \times 23 given that 17×23=39117 \times 23 = 391.

Estimate and Match

2Estimate the products on the left and match them to the numbers given on the right:
25×3125 \times 31, 132×19132 \times 19, 101×11101 \times 11, 248×49248 \times 49, 12×2512 \times 25
Options: 2,600 | 12,500 | 300 | 750 | 1,000

The King's Reward

1Three ministers choose different rewards:
- Minister 1: Start with 5 gold coins, double every day for 7 days.
- Minister 2: Start with 3 gold coins, triple every day for 7 days.
- Minister 3: Start with 1 gold coin, multiply by 5 every day for 7 days.
Calculate how many gold coins each minister received after 7 days. Who received the most?

Multiplication Patterns — Exercise 1

1aGiven 16×44=70416 \times 44 = 704. Solve completely and predict:
1) 8×888 \times 88
2) 8×228 \times 22
3) 16×2216 \times 22
4) 32×4432 \times 44
1bGiven 12×32=38412 \times 32 = 384. Find:
1) 6×166 \times 16
2) 24×1624 \times 16
3) 24×6424 \times 64
4) 12×1612 \times 16

Let Us Solve — Final Word Problems

1Mala went to a book exhibition and bought 18 books. The shop was selling 3 books for ₹150. After buying the books, she still had ₹20 left. How much money did Mala have at the beginning?
2A village sports club organises a women's football tournament.
- Sold 57 tickets for ₹115 each.
- 3 teams paid ₹1,599 each as participation fee.
- ₹1,750 paid to rent the football ground.
- ₹1,129 for food and water.
(a) How much money did the club collect in total from ticket sales and team participation fees?
(b) What were the total expenses on renting the ground and food and water?
3Ananya is watching Republic Day celebrations. There are 12 rows of students in front of her and 17 rows behind her. There are 18 students to her right and 22 students to her left.
(a) How many rows of students are there in total?
(b) How many students are there in Ananya's row?
(c) What is the total number of students on the ground?
4aMultiply: 67×7867 \times 78
4bMultiply: 34×5634 \times 56
4cMultiply: 45×26345 \times 263
4dMultiply: 86×54286 \times 542
4eMultiply: 432×107432 \times 107
4fMultiply: 310×120310 \times 120
5If 67×67=448967 \times 67 = 4489, without multiplication find 67×6867 \times 68.
6If 99×100=990099 \times 100 = 9900, without multiplication find 99×9999 \times 99.

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Frequently Asked Questions

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