Rs Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 3 Whole Numbers are provided here with simple step-by-step explanations. These solutions for Whole Numbers are extremely popular among Class 6 students for Maths Whole Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2020 2021 Book of Class 6 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggrawal 2020 2021 Solutions. All Rs Aggrawal 2020 2021 Solutions for class Class 6 Maths are prepared by experts and are 100% accurate.

#### Page No 45:

The next three whole numbers after 30999 are 31000, 31001, and 31002.

#### Page No 45:

Three whole numbers occurring just before 10001 are as follows:

10001 − 1 = 10000
10000 − 1 = 9999
9999 − 1 = 9998

∴ The three whole numbers just before 10001 are 10000, 9999 and 9998.

#### Page No 45:

Number of whole numbers between 1032 and 1209 = (1209 − 1032) − 1
= 177 − 1
= 176

#### Page No 45:

0 (zero) is the smallest whole number.

All the natural numbers along with 0 are called whole numbers.

#### Page No 45:

(i) Successor of 2540801 = 2540801 + 1 = 2540802
(ii) Successor of 9999 = 9999 + 1 = 10000
(iii) Successor of 50904 = 50904 + 1 = 50905
(iv) Successor of 61639 = 61639 + 1 = 61640
(v) Successor of 687890 = 687890 + 1 = 687891
(vi) Successor of 5386700 = 5386700 + 1 = 5386701
(vii) Successor of 6475999 = 6475999 + 1 = 6476000
(viii) Successor of 9999999 = 9999999 + 1 = 10000000

#### Page No 46:

(i) Predecessor of 97 = 97 − 1 = 96
(ii) Predecessor of 10000 = 10000 − 1 = 9999
(iii) Predecessor of 36900 = 36900 − 1 = 36899
(iv) Predecessor of 7684320 = 7684320 − 1 = 7684319
(v) Predecessor of 1566391 = 1566391 − 1 = 1566390
(vi) Predecessor of 2456800 = 2456800 − 1 = 2456799
(vii) Predecessor of 100000 = 100000 − 1 = 99999
(viii) Predecessor of 1000000 = 1000000 − 1 = 999999

#### Page No 46:

The three consecutive whole numbers just preceding 7510001 are as follows:

7510001 − 1 = 7510000
7510000 − 1 = 7509999
7509999 − 1 = 7509998

∴ The three consecutive numbers just preceding 7510001 are 7510000, 7509999 and 7509998.

#### Page No 46:

(i) False. 0 is not a natural number.1 is the smallest natural number.
(ii) True.
(iii) False. 0 is a whole number but not a natural number.
(iv) True. Natural numbers include 1,2,3 ..., which are whole numbers.
(v) False. 0 is the smallest whole number.
(vi) True. The predecessor of 1 is 1 − 1 = 0, which is not a natural number.
(vii) False. The predecessor of 1 is 1 − 1 = 0, which is a whole number.
(viii) True. The predecessor of 0 is 0 − 1 = −1, which is not a whole number.
(ix) False. The predecessor of a two-digit number can be a single digit number. For example, the predecessor of 10 is 10 − 1, i.e., 9.
(x) False. The successor of a two-digit number is not always a two-digit number. For example, the successor of 99 is 99 + 1, i.e., 100.
(xi) False. The predecessor of 499 is 499 − 1, i.e., 498.
(xii) True. The successor of 6999 is 6999 + 1, i.e., 7000.

#### Page No 48:

(i) 458 + 639 = 639 + 458
(ii) 864 + 2006 = 2006 + 864
​(iii) 1946 + 984 = 984 + 1946
(iv) 8063 + 0 = 8063
(v) 53501 + (574 + 799) = 574 + (53501 + 799)

#### Page No 48:

(i) 16509 + 114 = 16623
By reversing the order of the addends, we get:
114 + 16509 = 16623
∴ 16509 + 114 = 114 + 16509

(ii) 2359 + 548 = 2907
By reversing the order of the addends, we get:
548 + 2359 = 2907
∴ 2359 + 548 = 548 + 2359

(iii) 19753 + 2867 = 22620
By reversing the order of the addends, we get:
2867 + 19753 = 22620
∴ 19753 + 2867 = 2867 + 19753

#### Page No 48:

We have:

(1546 + 498) + 3589 = 2044 + 3589 = 5633

Also, 1546 + (498 + 3589) = 1546 + 4087 = 5633

Yes, the two sums are equal.

The associative property of addition is satisfied.

#### Page No 48:

(i) 953 + 707 + 647
953 + (707 + 647)                                   (Using associative property of addition)
= 953 + 1354
= 2307

(ii) 1983 + 647 + 217 + 353
(1983 + 647)  + (217 +353)                    (Using associative property of addition)
= 2630 + 570
=  3200

(iii) 15409 + 278 + 691 + 422
(15409 + 278) + (691 + 422)                     (Using associative property of addition)
= 15687 + 1113
= 16800

(iv) 3259 + 10001 + 2641 + 9999
(3259 + 10001) + (2641 +  9999)             (Using associative property of addition)
= 13260 + 12640
= 25900

(v)1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(1 + 2 + 3 + 4) + (96 + 97 + 98 + 99)       (Using associative property of addition)
= (10) + (390)
=  400

(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
(2 + 3 + 4 + 5) + (45 + 46 + 47 + 48)                 (Using associative property of addition)
= 14 + 186
= 200

#### Page No 48:

(i)  6784 + 9999
=  6784 + (10000 − 1)
=  (6784 + 10000) − 1                              (Using associative property of addition)
= 16784 − 1
= 16783

(ii) 10578 + 99999
= 10578 + (100000 − 1)
= (10578 + 100000) − 1                         (Using associative property of addition)
= 110578 − 1
= 110577

#### Page No 48:

For any whole numbers a, b and c, we have:
(a + b) + c = a + b + c​)

Let a = 2, b = 3 and c = 4 [we can take any values for a, b and c]

LHS = (a + b​) + c
(2 + 3) + 4
= 5 + 4
= 9

RHS = a + (c + b)
= a + (b + c)       [∵ Whole numbers follow the commutative law]
= 2 + (3 + 4)
= 2 + 7
= 9

∴ This shows that associativity (in addition) is one of the properties of whole numbers.

#### Page No 48:

In a magic square, the  sum of each row is equal to the sum of each column and the sum of each main diagonal.  By using this concept, we have:
(i)

 4 9 2 3 5 7 8 1 6

(ii)
 16 2 12 6 10 14 8 18 4

(iii)
 2 15 16 5 9 12 11 6 13 8 7 10 14 3 4 17

(iv)
 7 18 17 4 8 13 14 11 12 9 10 15 19 6 5 16

#### Page No 48:

(i)  F (false). The sum of two odd numbers may not be an odd number. Example: 3 + 5 = 8, which is an even number.

(ii) T (true). The sum of two even numbers is an even number. Example: 2 + 4 = 6, which is an even number.

(iii) T (true). The sum of an even and an odd number is an odd number. Example: 5 + 4 = 9, which is an odd number.

#### Page No 49:

(i) Subtraction: 6237 − 694 = 5543
Addition: 5543 + 694 = 6237

(ii) Subtraction: 21205 − 10899 = 10306
Addition: 10306 + 10899 = 21205

(iii) Subtraction: 100000 − 78987 = 21013
Addition: 21013 + 78987 = 100000

(iv) Subtraction: 1010101 − 656565 = 353536
Addition: 353536 + 656565 = 1010101

#### Page No 49:

(i)   917 − *5* =  5*8 ⇒ 917 − 359 =  558

(ii) 6172 − **69 =  29** ⇒ 6172 − 3269 = 2903

(iii) 5001003 − **6987 =  484**** ⇒ 5001003 − 155987 = 4845016

(iv)  1000000 − ****1 = *7042* ⇒ 1000000 − 29571 = 970429

#### Page No 49:

(i) 463 − 9
= 463 − 10 + 1
= 464 − 10
= 454

(ii) 5632 − 99
= 5632 − 100 + 1
= 5633 − 100
=  5533

(iii) 8640 − 999
= 8640 − 1000 + 1
= 8641 − 1000
= 7641

(iv) 13006 − 9999
= 13006 − 10000 + 1
= 13007 − 10000
= 3007

#### Page No 50:

Smallest seven-digit number = 1000000
Largest four-digit number = 9999
∴ Their difference = 1000000 − 9999
=1000000 − 10000 + 1
=1000001 − 10000
=990001

#### Page No 50:

Money deposited by Ravi = Rs 1,36,000
Money withdrawn by Ravi= Rs 73,129
Money left in his account  =  money deposited − money withdrawn
= Rs (136000 − 73129)
= Rs 62871

∴ Rs 62,871 is left in Ravi's account.

#### Page No 50:

Money withdrawn by Mrs Saxena  = Rs 1,00,000
Cost of the TV set = Rs 38,750
Cost of the refrigerator = Rs 23,890
Cost of the jewellery = Rs 35,560
Total money spent = Rs (38750 + 23890 + 35560) = Rs 98200

Now, money left = money withdrawn − money spent
= Rs (100000 − 98200)
= Rs 1800

∴ Rs 1,800 is left with Mrs Saxena.

#### Page No 50:

Population of the town = 110500
Increased population = 110500 + 3608 = 114108
Number of persons who died or left the town = 8973
Population at the end of the year = 114108 − 8973 = 105135

∴ The population at the end of the year will be 105135.

#### Page No 50:

(i) n + 4 = 9
⇒ n = 9 − 4 = 5

(ii) n + 35 = 101
⇒ n = 101 − 35 = 66

(iii) n - 18 = 39
⇒ n =  18 + 39 = 57

(iv) n  20568 = 21403
⇒ n  = 21403 + 20568 = 41971

#### Page No 53:

(i) 246 × 1 = 246
(ii) 1369 × 0 = 0
(iii) 593 × 188  = 188 × 593
(iv) 286 × 753 = 753 × 286
(v) 38 × (91 × 37) = 91 × (38 × 37)
(vi) 13 × 100 × 1000 = 1300000
(vii) 59 × 66 + 59  ×  34 = 59 × ( 66 + 34)
(viii) 68 × 95 = 68 × 100 − 68 × 5

#### Page No 53:

(i) Commutative law in multiplication
(ii) Closure property
(iii) Associativity of multiplication
(iv) Multiplicative identity
(v) Property of zero
(vi) Distributive law of multiplication over addition
(vii) Distributive law of multiplication over subtraction

#### Page No 53:

(i) 647 × 13 + 647 × 7
=  647 × (13 + 7)
= 647 ×  20
= 12940                                 (By using distributive property)

(ii)  8759 × 94 + 8759 × 6
= 8759 × (94 + 6)
= 8759 ×  100
= 875900                              (By using distributive property)

(iii) 7459 × 999 + 7459
= 7459× (999 + 1)
= 7459 × 1000
= 7459000                         (By using distributive property)

(iv) 9870 × 561 − 9870 × 461
= 9870 × (561 − 461)
= 9870 × 100
= 987000                           (By using distributive property)

(v)  569 × 17 + 569 × 13 + 569 × 70
= 569 × (17+ 13+ 70)
= 569  × 100
= 56900                            (By using distributive property)

(vi) 16825 × 16825 − 16825 × 6825
= 16825 × (16825 − 6825)
=  16825 × 10000
= 168250000                        (By using distributive property)

#### Page No 53:

(i) 2 × 1658 × 50
= (2 × 50) × 1658
= 100 × 1658
= 165800

(ii) 4 × 927 × 25
= (4 × 25) × 927
= 100 × 927
= 92700

(iii) 625 × 20 × 8 × 50
= (20  × 50) ×  8 × 625
= 1000 ×  8 × 625
= 8000 × 625
= 5000000

(iv) 574 × 625 × 16
= 574 × (625 × 16)
=  574 × 10000
= 5740000

(v)  250 × 60 × 50 × 8
= (250 × 8) × (60 × 50)
=  2000  × 3000
=  6000000

(vi)  8 × 125 × 40 × 25
=  (8 × 125) × (40 × 25)
= 1000 × 1000
= 1000000

#### Page No 53:

(i)  740 × 105
= 740 × (100 + 5)
= 740 × 100 + 740 × 5                    (Using distributive law of multiplication over addition)
= 74000 + 3700
= 77700

(ii) 245 × 1008
= 245 × (1000 + 8)
= 245 × 1000 + 245 × 8                  (Using distributive law of multiplication over addition)
= 245000 + 1960
= 246960

(iii) 947 × 96
= 947 × ( 100 − 4)
=  947 × 100 − 947 × 4                    (Using distributive law of multiplication over subtraction)
= 94700 − 3788
= 90912

(iv)  996 × 367
=  367 × (1000 − 4)
=   367 × 1000 − 367 × 4             (Using distributive law of multiplication over subtraction)
= 367000 × 1468
= 365532

(v) 472 × 1097
= 472 × ( 1000 + 97)
= 472 × 1000 + 472 × 97                 (Using distributive law of multiplication over addition)
= 472000 + 45784
= 517784

(vi)  580 × 64
=  580 × (60 + 4)
=  580 × 60 + 580 × 4                        (Using distributive law of multiplication over addition)
= 34800 + 2320
= 37120

(vii) 439 × 997
= 439 × (1000 − 3)
= 439 × 1000 − 439 × 3                   (Using distributive law of multiplication over subtraction)
= 439000 − 1317
= 437683

(viii) 1553 × 198
= 1553 × (200 − 2)
= 1553 × 200 − 1553 × 2                 (Using distributive law of multiplication over subtraction)
= 310600 − 3106
= 307494

#### Page No 53:

Distributive property of multiplication over addition states that a (b + c) = ab + ac
Distributive property of multiplication over subtraction  states that a (b c) = ab - ac
(i) 3576  ​×  9
= 3576 × (10 − 1)
= 3576 ​× 10 − 3576 × 1
= 35760 − 3576
= 32184

(ii) 847 ×  99
= 847 × (100 − 1)
= 847 × 100 − 847 × 1
=  84700 − 847
= 83853

(iii) 2437 × 999
= 2437 × (1000 − 1)
= 2437 × 1000 − 2437 × 1
=  2437000 − 2437
= 2434563

#### Page No 54:

(i) 458 × 67 = 30686

(ii) 3709 × 89 = 330101

(iii) 4617 × 234 = 1080378

(iv) 15208 × 542 = 8242736

#### Page No 54:

Largest three-digit number = 999
Largest five-digit number = 99999
∴ Product of the two numbers = 999 × 99999
= 999 × (100000 − 1)                  (Using distributive law)
= 99900000 − 999
= 99899001

#### Page No 54:

Uniform speed of a car = 75 km/h

Distance = speed × time
= 75 × 98
=75 × (100 − 2)                     (Using distributive law)
=75 × 100 − 75 × 2
=7500 − 150
= 7350 km

∴ The distance covered in 98 h is 7350 km.

#### Page No 54:

Cost of 1 VCR set = Rs 24350
Cost of 139 VCR sets = 139 × 24350
=24350 × (140 − 1)                 (Using distributive property)
=24350 × 140 − 24350
=3409000 − 24350
= Rs. 3384650

∴ The cost of all the VCR sets is Rs 33,84,650.

#### Page No 54:

Cost of construction of 1 house = Rs 450000
Cost of construction of 197 such houses = 197 × 450000
= 450000 × (200 − 3)
= 450000 × 200 − 450000 × 3               [Using distributive property of multiplication over subtraction]
= 90000000 − 1350000
= 88650000

∴ The total cost of construction of 197 houses is Rs 8,86,50,000.

#### Page No 54:

Cost of a chair = Rs 1065
Cost of a blackboard = Rs 1645
Cost of 50 chairs = 50 × 1065 = Rs 53250
Cost of 30 blackboards = 30 × 1645 = Rs 49350
∴ Total amount of the bill = cost of 50 chairs + cost of 30 blackboards
= Rs (53250 + 49350)
= Rs 1,02,600

#### Page No 54:

Number of student in 1 section = 45
Number of students in 6 sections = 45 × 6 = 270
Monthly charges from 1 student = Rs 1650
∴ Total monthly collection from class VI = Rs 1650 × 270 = Rs 4,45,500

#### Page No 54:

If the product of two whole numbers is zero, then one of them is definitely zero.
Example: 0 × 2 = 0 and 0 × 15 = 0

If the product of whole numbers is zero, then both of them may be zero.
i.e., 0 × 0 = 0

Now, 2 × 5 = 10. Here, the product will be non-zero because the numbers to be multiplied are not equal to zero.

#### Page No 54:

(i) Sum of two odd numbers is an even number. Example: 3 + 5 = 8, which is an even number.
(ii) Product of two odd numbers is an odd number. Example: 5 × 7 = 35, which is an odd number.
(iii)  ≠ 0 and a × a = a
Given: a × a = a
⇒ a = $\frac{a}{a}=1$
, ≠ 0

#### Page No 56:

(i) Dividend = 1936, Divisor = 36 , Quotient = 53 , Remainder = 28
Check: Divisor × Quotient + Remainder  =  36 × 53 + 28
= 1936
=Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.
(ii) 19881 ​÷ 47 Dividend = 19881, Divisor = 47 , Quotient = 423, Remainder = 0
Check: Divisor ×Quotient + Remainder= 47 × 423 + 0
= 19881
=Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iii) Dividend = 257796 , Divisor = 341 , Quotient = 756 , Remainder = 0
Check : Divisor × Quotient + Remainder = 341 × 756 + 0
= 257796
= Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iv) 612846 ​÷ 582 Dividend = 612846 , Divisor = 582, Quotient = 1053 , Remainder = 0
Check :  Divisor × Quotient + Remainder= 582 × 1053 + 0
= 612846
=Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(v) 34419 ​÷ 149 Dividend = 34419, Divisor = 149 , Quotient = 231, Remainder = 0
Check : Divisor × Quotient + Remainder  = 149 × 231 + 0
= 34419
=Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(vi) 39039 ​÷ 1001 Dividend = 39039 , Divisor = 1001 , Quotient = 39 , Remainder = 0
Check : Divisor × Quotient + Remainder = 1001 × 39 + 0
= 39039
=Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

#### Page No 56:

(i)  6971 ​÷ 47 Quotient = 148 and Remainder = 15
Check: Divisor × Quotient + Remainder = 47 × 148 + 15
= 6971
= Dividend
∴ Dividend = Divisor × Quotient + Remainder
Verified.

(ii)  4178 ​÷ 35 Dividend = 119 and Remainder = 13
Check: Divisor × Quotient + remainder = 35 ×  119 + 13
= 4178
= Dividend

∴ Dividend= Divisor × Quotient + Remainder
Verified.
(iii) 36195 ​÷ 153 Quotient = 236 and Remainder = 87
Check: Divisor × Quotient + Remainder =  153 × 236 + 87
= 36195
= Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(iv) 93575 ​÷ 400 Quotient = 233 and Remainder = 375
Check: Divisor × Quotient + Remainder =  400 ×  233 + 375
= 93575
= Dividend
∴ Dividend= Divisor × Quotient + Remainder
Verified.

(v)  23025 ​÷ 1000 Quotient = 23 and remainder = 25
Check: Divisor × Quotient + Remainder =1000  × 23 + 25
= 23025
= Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(vi) 16135 ​÷ 875 Quotient = 18 and Remainder = 385
Check: Divisor × Quotient + Remainder =875  ×  18 + 385
= 16135
= Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.

#### Page No 56:

(i) 65007 ​÷ 1 = 65007

(ii) 0 ​÷ 879  = 0

(iii) 981 + 5720 ​÷ 10
= 981 + (5720 ​÷ 10)                               (Following DMAS property)
= 981 + 572
= 1553

(iv) 1507 − (625 ÷ 25)                             (Following BODMAS property)
= ​​1507 − 25
= 1482

(v) 32277 ÷ (648 − 39)                                 (Following BODMAS property)
= ​32277 ÷ (609)
=  53

(vi)  (1573 ÷ 1573) − (1573 ÷ 1573)            (Following BODMAS property)
= 1 − 1
= 0

#### Page No 56:

Given:  n ÷ = n
⇒  $\frac{n}{n}$= n
​⇒  n = n2

i.e., the whole number n is equal to n2.

∴ The given whole number must be 1.

#### Page No 56:

Let x and y be the two numbers.

Product of the two numbers = x × y = 504347

If x = 317, we have:

317 × y = 504347
⇒ y = 504347 ÷ 317 y=  1591

∴ The other number is 1591.

#### Page No 56:

Dividend = 59761, quotient = 189, remainder = 37 and divisor = ?

Dividend = divisor × quotient + remainder
⇒ 59761 = divisor × 189 + 37
⇒ 59761 − 37 = divisor × 189
⇒ 59724 = divisor × 189
⇒ Divisor = 59724 ​÷ 189 Hence, divisor =316

#### Page No 56:

Here, Dividend = 55390, Divisor = 299 and Remainder = 75
We have to find the quotient.
Now, Dividend = Divisor × Quotient + Remainder
⇒ 55390 = 299 × Quotient + 75
⇒ 55390 − 75 = 299 × Quotient
⇒ 55315 = 299 × Quotient
⇒ Quotient = 55315 ​÷ 299 Hence, quotient =185

#### Page No 56:

First, we will divide 13601 by 87. Remainder = 29
So, 29 must be subtracted from 13601 to get a number exactly divisible by 87.
i.e., 13601 − 29 = 13572

Now, we have: ∴ 29 must be subtracted from 13601 to make it divisible by 87.

#### Page No 56:

First, we will divide 1056 by 23. Required number = 23 − 21 = 2
So, 2 must be added to 1056 to make it exactly divisible by 23.
i.e., 1056 + 2 = 1058

Now, we have: ∴ 1058 is exactly divisible by 23.

#### Page No 56:

We have to find the largest four digit number divisible by 16 .
The largest four-digit number = 9999
Therefore, dividend =9999
Divisor =16 Here, we get remainder =15
Therefore, 15 must be subtracted from 9999 to get the largest four digit number that is divisible by 16.
i.e., 9999 − 15 = 9984

Thus, 9984 is the largest four-digit number that is divisible by 16.

#### Page No 56:

Largest five-digit number =99999 Dividend = 99999, Divisor = 653, Quotient = 153 and Remainder = 90
Check: Divisor ×Quotient + Remainder
= 653 × 153 + 90
= 99909 + 90
= 99999
= Dividend

∴ Dividend = Divisor × Quotient + Remainder
Verified.

#### Page No 56:

Least six-digit number = 100000
Here, dividend = 100000 and divisor = 83 In order to find the least 6-digit number exactly divisible by 83, we have to add 83 − 68 = 15 to the dividend.
I.e., 100000 + 15 = 100015

So, 100015 is the least six-digit number exactly divisible by 83.

#### Page No 56:

Cost of 1 dozen bananas = Rs 29
Number of dozens purchased for Rs 1392 = 1392 ÷ 29 Hence, 48 dozen of bananas can be purchased with Rs. 1392.

#### Page No 56:

Number of trees planted in 157 rows = 19625
Trees planted in 1 row = 19625 ÷ 157 ∴ 125 trees are planted in each row.

#### Page No 56:

Population of the town = 517530
$\left(\frac{1}{15}\right)$ of the population is reported to be literate, i.e., $\left(\frac{1}{15}\right)$ × 517530 = 517530 $÷$ 15 ∴ There are 34502 illiterate persons in the given town.

#### Page No 56:

Cost price of 23 colour TV sets = Rs 5,70,055
Cost price of 1 TV set  = Rs 570055 ÷ 23 ∴ The cost price of one TV set is Rs 24,785.

#### Page No 56:

(b) 0

The smallest whole number is 0.

#### Page No 56:

(d) 1008

(a) Hence, 1018 is not exactly divisible by 9.

(b) Hence, 1026 is exactly divisible by 9.
(c) Hence, 1009 is not exactly divisible by 9.

(d) Hence, 1008 is exactly divisible by 9.

(b) and (d) are exactly divisible by 9, but (d) is the least number which is exactly divisible by 9.

#### Page No 57:

(c) 999984

(a) Hence, 999980 is not exactly divisible by 16.
(b) Hence, 999982 is not exactly divisible by 16.
(c) Hence, 999984 is exactly divisible by 16.
(d) Hence, 999964 is not exactly divisible by 16.

The largest six-digit number which is exactly divisible by 16 is 999984.

#### Page No 57:

(c) 8

Here we have to tell what least number should be subtracted from 10004 to get a number exactly divisible by 12
So, we will first divide 10004 by 12. Remainder = 8
So, 8 should be subtracted from 10004 to get the number exactly divisible by 12.
i.e., 10004 − 8 = 9996 Hence, 9996 is exactly divisible by 12.

#### Page No 57:

(a) 18

Here , we have to tell that what least number must be added to 10056 to get a number exactly divisible by 23
So, first we will divide 10056 by 23 Remainder = 5
Required number = 23 − 5 = 18

So, 18 must be added to 10056 to get a number exactly divisible by 23.
i.e., 10056 + 18 = 10074 Hence, 10074 is exactly divisible by 23 .

#### Page No 57:

(d) 462

(a) Hence, 450 is not divisible by 11.
(b) Hence, 451 is divisible by 11.
(c) Hence, 460 is not divisible by 11.
(d) Hence, 462 is divisible by 11.

Here, the numbers given in options (b) and (d) are divisible by 11. However, we want a whole number nearest to 457 which is divisible by 11.
So, 462 is whole number nearest to 457 and divisible by 11.

#### Page No 57:

(c) 184

Number of whole numbers = (1203 − 1018) − 1
= 185 − 1
=  184

#### Page No 57:

(b) 521

Divisor = 46
Quotient = 11
Remainder = 15
Dividend = divisor × quotient + remainder
= 46 × 11 + 15
= 506 + 15
= 521

#### Page No 57:

(c) 12

Dividend = 199
Quotient = 16
Remainder = 7
According to the division algorithm, we have:
Dividend = divisor × quotient + remainder
⇒ 199 = divisor × 16 + 7
⇒ 199 − 7 = divisor × 16
⇒ Divisor = 192 ÷ 16 #### Page No 57:

(a) 11023

7589 − ? = 3434
⇒ 7589 − x = 3434
x = 7589 + 3434
x = 11023

#### Page No 57:

(c) 58113

587 × 99
= 587 × (100 − 1)
= 587 × 100 − 587 × 1         [Using distributive property of multiplication over subtraction]
= 58700 − 587
= 58113

(c) 53800

4 × 538 × 25
= (4 × 25) × 538
=  100 × 538
= 53800

#### Page No 57:

(c) 2467900

By using the distributive property, we have:
24679 × 92 + 24679 × 8
= 24679 ×  (92 + 8)
= 24679 × 100
= 2467900

#### Page No 57:

(a) 1625000

By using the distributive property, we have:

1625 × 1625 − 1625 × 625
= 1625 × (1625 − 625)
=1625 × 1000
= 1625000

#### Page No 57:

(c) 156800

By using the distributive property, we have:
1568 × 185 − 1568 × 85
= 1568 × (185 − 85)
= 1568 × 100
= 156800

#### Page No 57:

(c) 20

(888 + 777 + 555) = (111 × ?)
⇒ (888 + 777 + 555) = 111 × (8 + 7 + 5)          [By taking 111 common]
= ​111 × (20) = 2220

#### Page No 57:

(b) an even number

The sum of two odd numbers is an even number.

Example: 5 + 3 = 8

#### Page No 57:

(a) an odd number

The product of two odd numbers is an odd number.

Example: 5 × 3 = 15

#### Page No 57:

(d) none of these

Given: a is a whole number such that a + a = a.

If a = 1, then 1+ 1 = 2 ≠ 1
If a =2, then 2 + 2 = 4 ≠ 2
If a =3, then 3 + 3 = 6 ≠ 3

#### Page No 57:

(b) 9999

Predecessor of 10000 = 10000 − 1 = 9999

#### Page No 57:

(b) 1002

Successor of 1001 = 1001 + 1 = 1002

#### Page No 57:

(b) 2

The smallest even whole number is 2. Zero (0) is neither an even number nor an odd number.

#### Page No 59:

Number of whole numbers between 1201 and 1064 = ( 1201 − 1064 ) − 1
= 137 − 1
= 136

#### Page No 59:

1000000
−      ****1

*7042*

Then, we have:

1000000
−    29571

970429

#### Page No 59:

Using distributive law, we have:
1063 × 128 − 1063 × 28
= 1063 × (128 − 28)
= 1063 × 100
= 106300

#### Page No 59:

Largest five-digit number = 99999
Largest three-digit number = 999

By using distributive law, we have:

Product = 99999 × 999
= 99999 × (1000 − 1)                                  [By using distributive law]
= 99999 × 1000 −  99999 × 1
= 99999000 − 99999
= 99899001

OR

999 × 99999
= 999 × ( 100000 − 1)                                   [By using distributive law]
= 999 × 100000 − 999 × 1
= 99900000 − 999
= 99899001

#### Page No 59: Dividend = 53968, Divisor = 267, Quotient = 202 and Remainder = 34

Check: Quotient × Divisor + Remainder
= 267  × 202 + 34
=  53934 + 34
= 53968
= Dividend

∴ Dividend = Quotient × Divisor + Remainder

Verified.

#### Page No 59:

Largest six-digit number = 999999 Remainder = 15

Largest six-digit number divisible by 16 = 999999 − 15 = 999984 ∴ 999984 is divisible by 16.

#### Page No 59:

Cost price of 23 TV sets = Rs 5,70,055
Cost price of 1 TV set = 570055 ÷ 23 ∴ The cost of one TV set is Rs 24,785.

#### Page No 59:

We have to find the least number that must be subtracted from 13801 to get a number exactly divisible by 87
So, first we will divide 13801 by 87 Remainder = 55
The number 55 should be subtracted from 13801 to get a number divisible by 87.
i.e., 13801 − 55 = 13746 ∴ 13746 is divisible by 87.

#### Page No 59:

(b) 8900

(89 × 76 + 89 × 24)
= 89 × (76 + 24)       [Using distributive property of multiplication over addition]
= 89 × 100
= 8900

#### Page No 59:

(c) 429

Divisor = 53, Quotient = 8, Remainder = 5 and Dividend = ?

Now, Dividend = Quotient × Divisor +Remainder
= 8 × 53 + 5
= 429

#### Page No 59:

(b) 0

The whole number which has no predecessor is 0.

i.e., 0 − 1 = −1, which is not a whole number.

#### Page No 59:

(c) Commutative property

67 + 33 = 33 + 67 is an example of​ commutative property of addition.

#### Page No 59:

(c) -36
The additive inverse of 36 is −36.

i.e., 36 + (−36) = 0

#### Page No 59:

(d) 2+0

(a) 0 × 0 = 0
(b) 0/2 = 0
(c) $\frac{\left(8-8\right)}{2}=\frac{0}{2}$ =0
(d) 2 + 0 = 2

#### Page No 59:

(d) 99

Smallest three-digit number = 100
∴ Predecessor of 100 = 100 − 1 = 99

#### Page No 59:

(b) 98
Smallest whole number = 0
Greatest two-digit number = 99
Number of whole numbers between 0 and 99 = (99 − 0 ) − 1 = 98

#### Page No 59:

(i) The smallest natural number is 1.
(ii) The smallest whole number is 0.
(iii) Division by 0 is not defined.
(iv) 0 is a whole number which is not a natural number.
(v) 1 is the multiplicative identity for whole numbers.

#### Page No 60:

(i)  F (false). 0 is not a natural number.
​(ii) T (true).
(iii) F (false). 0 is a whole number but not a natural number.
(iv) F (false). 1 − 1 = 0 is a predecessor of 1, which is a whole number.