Chapter 3 Whole numbers - Praadis Education

Chapter 3

Whole numbers

Exercise 3.1

Page No 45:

Question 1:

Write the next three whole numbers after 30999.

ANSWER:

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

Question 2:

Write the three whole numbers occurring just before 10001.

ANSWER:

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.

Question 3:

How many whole numbers are there between 1032 and 1209?

ANSWER:

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

= 177 − 1

= 176

Question 4:

Which is the smallest whole number?

ANSWER:

0 (zero) is the smallest whole number.

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

Question 5:

Write the successor of:

(i) 2540801

(ii) 9999

(iii) 50904

(iv) 61639

(v) 687890

(vi) 5386700

(vii) 6475999

(viii) 9999999

ANSWER:

(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

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(vii) Successor of 6475999 = 6475999 + 1 = 6476000

(viii) Successor of 9999999 = 9999999 + 1 = 10000000

Question 6:

Write the predecessor of:

(i) 97

(ii) 10000

(iii) 36900

(iv) 7684320

(v) 1566391

(vi) 2456800

(vii) 100000

(viii) 1000000

ANSWER:

(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

Question 7:

Write down three consecutive whole numbers just preceding 7510001.

ANSWER:

The three consecutive whole numbers just preceding 7510001 are as

follows:

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7510001 − 1 = 7510000

7510000 − 1 = 7509999

7509999 − 1 = 7509998

∴ The three consecutive numbers just preceding 7510001 are 7510000,

7509999 and 7509998.

Question 8:

Write (T) for true and (F) for false against each of the following

statements:

(i) Zero is the smallest natural number.

(ii) Zero is the smallest whole number.

(iii) Every whole number is a natural number.

(iv) Every natural number is a whole number.

(v) 1 is the smallest whole number.

(vi) The natural number 1 has no predecessor.

(vii) The whole number 1 has no predecessor.

(viii) The whole number 0 has no predecessor.

(ix) The predecessor of a two-digit number is never a single-digit

number.

(x) The successor of a two-digit number is always a two-digit number.

(xi) 500 is the predecessor of 499.

(xii) 7000 is the successor of 6999.

ANSWER:

(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.

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(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.

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Exercise 3.2

Page No 48:

Question 1:

Fill in the blanks to make each of the following a true statement:

(i) 458 + 639 = 639 + ......

(ii) 864 + 2006 = 2006 + ......

(iii) 1946 + ...... = 984 + 1946

(iv) 8063 + 0 = ......

(v) 53501 + (574 + 799) = 574 + (53501 + ......)

ANSWER:

(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)

Question 2:

Add the following numbers and check by revershing the order of the

addends:

(i) 16509 + 114

(ii) 2359 + 548

(iii) 19753 + 2867

ANSWER:

(i) 16509 + 114 = 16623

By reversing the order of the addends, we get:

114 + 16509 = 16623

∴ 16509 + 114 = 114 + 16509

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(ii) 2359 + 548 = 2907


548 + 2359 = 2907

∴ 2359 + 548 = 548 + 2359

(iii) 19753 + 2867 = 22620


2867 + 19753 = 22620

∴ 19753 + 2867 = 2867 + 19753

Question 3:

Find the sum: (1546 + 498) + 3589.

Also, find the sum: 1546 + (498 + 3589).

Are the two sums equal?

State the property satisfied.

ANSWER:

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.

Question 4:

Determine each of the sums given below using suitable rearrangement.

(i) 953 + 707 + 647

(ii) 1983 + 647 + 217 + 353

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(iii) 15409 + 278 + 691 + 422

(iv) 3259 + 10001 + 2641 + 9999

(v) 1 + 2 + 3 + 4 + 96 + 97 + 98 + 99

(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48

ANSWER:

(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

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(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

Question 5:

Find the sum by short method:

(i) 6784 + 9999

(ii) 10578 + 99999

ANSWER:

(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

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Question 6:

For any whole numbers a, b, c, is it true that (a + b) + c = a + (c + b)?

Give reasons.

ANSWER:

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.

Question 7:

Complete each one of the following magic squares by supplying the

missing numbers:

(i)

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9 2

5

8

(ii)

16 2

10

4

(iii)

2 15 16

9 12

7 10

14 17

(iv)

18 17 4

14 11

9 10

19 16

ANSWER:

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)

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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

Question 8:

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Write (T) for true and (F) for false for each of the following statements:

(i) The sum of two odd numbers is an odd number.

(ii) The sum of two even numbers is an even number.

(iii) The sum of an even number and an odd number is an odd number.

ANSWER:

(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.

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Exercise 3.3

Page No 49:

Question 1:

Perform the following subtractions. Check your results by the

corresponding additions.

(i) 6237 − 694

(ii) 21205 − 10899

(iii) 100000 − 78987

(iv) 1010101 − 656565

ANSWER:

(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

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Question 2:

Replace each * by the correct digit in each of the following:

(i)

9 1 7 9 1 7

- * 5 * ⇒ - 3 5 9

5 * 8 5 5 3

(ii)

6 1 7 2 6 1 7 2

- * * 6 9 ⇒ - 3 2 6 9

2 9 * * 2 9 0 3

(iii)

5 0 0 1 0 0 3 5 0 0 1 0 0 3

- * * 6 9 87 ⇒ - 1 5 6 9 8 7

4 8 4 * * * * 4 8 4 5 0 1 6

(iv)

1 0 0 0 0 0 0 1 0 0 0 0 0 0

- * * * * 1 ⇒ - 2 9 5 7 1

* 7 0 4 2 * 9 7 0 4 2 9

ANSWER:

(i) 917 − *5* = 5*8

9 1 7 9 1 7

- * 5 * ⇒ - 3 5 9

5 * 8 5 5 3

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⇒ 917 − 359 = 558

(ii) 6172 − **69 = 29**

6 1 7 2 6 1 7 2

- * * 6 9 ⇒ - 3 2 6 9

2 9 * * 2 9 0 3

⇒ 6172 − 3269 = 2903

(iii) 5001003 − **6987 = 484****

5 0 0 1 0 0 3 5 0 0 1 0 0 3

- * * 6 9 87 ⇒ - 1 5 6 9 8 7

4 8 4 * * * * 4 8 4 5 0 1 6

⇒ 5001003 − 155987 = 4845016

(iv) 1000000 − ****1 = *7042*

1 0 0 0 0 0 0 1 0 0 0 0 0 0

- * * * * 1 ⇒ - 2 9 5 7 1

* 7 0 4 2 * 9 7 0 4 2 9

⇒ 1000000 − 29571 = 970429

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Question 3:

Find the difference:

(i) 463 − 9

(ii) 5632 − 99

(iii) 8640 − 999

(iv) 13006 − 9999

ANSWER:

(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

Question 4:

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Find the difference between the smallest number of 7 digits and the

largest number of 4 digits.

ANSWER:

Smallest seven-digit number = 1000000

Largest four-digit number = 9999

∴ Their difference = 1000000 − 9999

=1000000 − 10000 + 1

=1000001 − 10000

=990001

Question 5:

Ravi opened his account in a bank by depositing Rs 136000. Next day

he withdrew Rs 73129 from it. How much money was left in his

account?

ANSWER:

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.

Question 6:

Mrs Saxena withdrew Rs 100000 from her bank account. She purchased

a TV set for Rs 38750, a refrigerator for Rs 23890 and jewellery worth

Rs 35560. How much money was left with her?

ANSWER:

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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.

Question 7:

The population of a town was 110500. In one year it increased by 3608

due to new births. However, 8973 persons died or left the town during

the year. What was the population at the end of the year?

ANSWER:

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.

Question 8:

Find the whole number n when:

(i) n + 4 = 9

(ii) n + 35 = 101

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(iii) n − 18 = 39

(iv) n − 20568 = 21403

ANSWER:

(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

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Exercise 3.4

Page No 53:

Question 1: Fill in the blanks to make each of the following a true

statement:

(i) 246 × 1 = ......

(ii) 1369 × 0 = .......

(iii) 593 × 188 = 188 × .......

(iv) 286 × 753 = ...... × 286

(v) 38 × (91 × 37) = ...... × (38 × 37)

(vi) 13 × 100 × ...... = 1300000

(vii) 59 × 66 + 59 × 34 = 59 × (...... + ......)

(viii) 68 × 95 = 68 × 100 − 68 × .......

ANSWER:

(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

Question 2: State the property used in each of the following statements:

(i) 19 × 17 = 17 × 19

(ii) (16 × 32) is a whole number

(iii) (29 × 36) × 18 = 29 × (36 × 18)

(iv) 1480 × 1 = 1480

(v) 1732 × 0 = 0

(vi) 72 × 98 + 72 × 2 = 72 × (98 + 2)

(vii) 63 × 126 − 63 × 26 = 63 × (126 − 26)

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ANSWER:

(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

Question 3: Find the value of each of the following using various

properties:

(i) 647 × 13 + 647 × 7

(ii) 8759 × 94 + 8759 × 6

(iii) 7459 × 999 + 7459

(iv) 9870 × 561 − 9870 × 461

(v) 569 × 17 + 569 × 13 + 569 × 70

(vi) 16825 × 16825 − 16825 × 6825

ANSWER:

(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


(iii) 7459 × 999 + 7459

= 7459× (999 + 1)

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= 7459 × 1000


(iv) 9870 × 561 − 9870 × 461

= 9870 × (561 − 461)

= 9870 × 100


(v) 569 × 17 + 569 × 13 + 569 × 70

= 569 × (17+ 13+ 70)

= 569 × 100


(vi) 16825 × 16825 − 16825 × 6825

= 16825 × (16825 − 6825)

= 16825 × 10000


Question 4: Determine each of the following products by suitable

rearrangements:

(i) 2 × 1658 × 50

(ii) 4 × 927 × 25

(iii) 625 × 20 × 8 × 50

(iv) 574 × 625 × 16

(v) 250 × 60 × 50 × 8

(vi) 8 × 125 × 40 × 25

ANSWER:

(i) 2 × 1658 × 50

= (2 × 50) × 1658

= 100 × 1658

= 165800

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(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

Question 5: Find each of the following products, using distributive laws:

(i) 740 × 105

(ii) 245 × 1008

(iii) 947 × 96

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(iv) 996 × 367

(v) 472 × 1097

(vi) 580 × 64

(vii) 439 × 997

(viii) 1553 × 198

ANSWER:

(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)



= 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)



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= 367000 × 1468

= 365532

(v) 472 × 1097

= 472 × ( 1000 + 97)



= 472000 + 45784

= 517784

(vi) 580 × 64

= 580 × (60 + 4)



= 34800 + 2320

= 37120

(vii) 439 × 997

= 439 × (1000 − 3)



= 439000 − 1317

= 437683

(viii) 1553 × 198

= 1553 × (200 − 2)



= 310600 − 3106

= 307494

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Question 6: Find each of the following products, using distributive

laws:

(i) 3576 × 9

(ii) 847 × 99

(iii) 2437 × 999

ANSWER:

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

Question 7: Find the products:

(i)

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4 5 6

× 6 7

3 2 0 6 Multiplication by 7

2 7 4 8 × Multiplication by 40

3 0 6 8 6

(ii)

3 7 0 9

× 8 9

3 3 3 8 1 Multiplication by 9

2 9 6 7 2 × Multiplication by 80

3 3 0 1 0 1

(iii)

4 6 1 7

× 2 3 4



9 2 3 4 × × Multiplication by 200

1 0 8 0 3 0 7 8

(iv)

1 5 2 0 8

× 5 4 2



7 6 0 4 0 × × × Multiplication by 500

8 2 4 2 7 3 6

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ANSWER:

(i)

4 5 6

× 6 7

3 2 0 6 Multiplication by 7

2 7 4 8 × Multiplication by 40

3 0 6 8 6

458 × 67 = 30686

(ii)

3 7 0 9

× 8 9



3 3 0 1 0 1

3709 × 89 = 330101

(iii)

4 6 1 7

× 2 3 4



9 2 3 4 × × Multiplication by 200

1 0 8 0 3 0 7 8

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4617 × 234 = 1080378

(iv)

1 5 2 0 8

× 5 4 2



7 6 0 4 0 × × × Multiplication by 500

8 2 4 2 7 3 6

15208 × 542 = 8242736

Question 8: Find the product of the largest 3-digit number and the

largest 5-digit number.

ANSWER:

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

Question 9: A car moves at a uniform speed of 75 km per hour. How

much distance will it cover in 98 hours?

ANSWER:

Uniform speed of a car = 75 km/h

Distance = speed × time

= 75 × 98

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=75 × (100 − 2) (Using distributive law)

=75 × 100 − 75 × 2

=7500 − 150

= 7350 km

∴ The distance covered in 98 h is 7350 km.

Question 10: A dealer purchased 139 VCRs. If the cost of each set is Rs

24350, find the cost of all the sets together.

ANSWER:

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.

Question 11: A housing society constructed 197 houses. If the cost of

construction for each house is Rs 450000, what is the total cost for all

the houses?

ANSWER:

Cost of construction of 1 house = Rs 450000

Cost of construction of 197 such houses = 197 × 450000

= 450000 × (200 − 3)

= 450000 × 200 − 450000 × 3

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[Using distributive property of multiplication over subtraction]

= 90000000 − 1350000

= 88650000

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

Question 12: 50 chairs and 30 blackboards were purchased for a school.

If each chair costs Rs 1065 and each blackboard costs Rs 1645, find the

total amount of the bill.

ANSWER:

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

Question 13: There are six sections of Class VI in a school and there are

45 students in each section. If the monthly charges from each student be

Rs 1650, find the total monthly collection from Class VI.

ANSWER:

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

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Question 14: The product of two whole numbers is zero. What do you

conclude?

ANSWER:

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.

Question 15: Fill in the blanks:

(i) Sum of two odd numbers is an ...... number.

(ii) Product of two odd numbers is an ...... number.

(iii) a ≠ 0a ≠ 0 and a × a = a ⇒⇒ a = ?

ANSWER:

(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) a ≠ 0 and a × a = a

Given: a × a = a

⇒ a = 𝑎

𝑎 = 1, a ≠ 0

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Exercise 3.5

Page No 56:

Question 1:

Divide and check your answer by the corresponding multiplication in

each of the following:

(i) 1936 ÷ 36

(ii) 19881 ÷ 47

(iii) 257796 ÷ 341

(iv) 612846 ÷ 582

(v) 34419 ÷ 149

(vi) 39039 ÷ 1001

ANSWER:

(i)

53

36 1936

– 180

136

– 180

28

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

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423

47 19881

– 188

108

– 94

141

– 141

0

Dividend = 19881, Divisor = 47 , Quotient = 423, Remainder = 0

Check: Divisor ×Quotient + Remainder= 47 × 423 + 0

= 19881

=Dividend


Verified.

(iii)

756

341 257796

– 2387

1909

– 1705

2046

– 2046

0

Dividend = 257796 , Divisor = 341 , Quotient = 756 , Remainder = 0

Check : Divisor × Quotient + Remainder = 341 × 756 + 0

= 257796

= Dividend

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Verified.

(iv) 612846 ÷ 582

1053

582 612846

– 582

3084

– 2910

1746

– 1746

0

Dividend = 612846 , Divisor = 582, Quotient = 1053 , Remainder = 0

Check : Divisor × Quotient + Remainder= 582 × 1053 + 0

= 612846

=Dividend


Verified.

(v) 34419 ÷ 149

231

149 34419

– 298

461

– 447

149

– 149

0

Dividend = 34419, Divisor = 149 , Quotient = 231, Remainder = 0


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= 34419

=Dividend


Verified.

(vi) 39039 ÷ 1001

1053

1001 39039

– 3003

9009

– 9009

0

Dividend = 39039 , Divisor = 1001 , Quotient = 39 , Remainder = 0


= 39039

=Dividend


Verified.

Page No 56:

Question 2: Divide, and find out the quotient and remainder. Check your

answer.

(i) 6971 ÷ 47

(ii) 4178 ÷ 35

(iii) 36195 ÷ 153

(iv) 93575 ÷ 400

(v) 23025 ÷ 1000

(vi) 16135 ÷ 875

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ANSWER:

(i) 6971 ÷ 47

148

47 6971

– 47

227

– 188

391

– 376

15

Quotient = 148 and Remainder = 15


= 6971

= Dividend

∴ Dividend = Divisor × Quotient + Remainder

Verified.

(ii) 4178 ÷ 35

119

35 4178

– 35

67

– 35

328

– 315

13

Dividend = 119 and Remainder = 13

Check: Divisor × Quotient + remainder = 35 × 119 + 13

= 4178

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= Dividend

∴ Dividend= Divisor × Quotient + Remainder

Verified.

(iii) 36195 ÷ 153

236

153 36195

– 306

559

– 459

1005

– 918

87



= 36195

= Dividend

∴ Dividend= Divisor × Quotient +Remainder

Verified.

(iv) 93575 ÷ 400

233

400 93575

– 800

1357

– 1200

1575

– 1200

375



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= 93575

= Dividend

∴ Dividend= Divisor × Quotient + Remainder

Verified.

(v) 23025 ÷ 1000

1053

1000 23025

– 2000

3025

– 3000

25

Quotient = 23 and remainder = 25

Check: Divisor × Quotient + Remainder =1000 × 23 + 25

= 23025

= Dividend


Verified.

(vi) 16135 ÷ 875

18

875 16135

– 875

7385

– 7000

385


Check: Divisor × Quotient + Remainder =875 × 18 + 385

= 16135

= Dividend

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Verified.

Page No 56:

Question 3: Find the value of

(i) 65007 ÷ 1

(ii) 0 ÷ 879

(iii) 981 + 5720 ÷ 10

(iv) 1507 − (625 ÷ 25)

(v) 32277 ÷ (648 − 39)

(vi) (1573 ÷ 1573) − (1573 ÷ 1573)

ANSWER:

(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

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(vi) (1573 ÷ 1573) − (1573 ÷ 1573) (Following BODMAS

property)

= 1 − 1

= 0

Page No 56:

Question 4: Find a whole number n such that n ÷ n = n.

ANSWER:

Given: 𝑛 ÷ 𝑛 = 𝑛

⇒ 𝑛𝑛𝑛𝑛 = 𝑛

⇒ 𝑛 = 𝑛2

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

∴ The given whole number must be 1.

Page No 56:

Question 5: The product of two numbers is 504347. If one of the numbers

is 317, find the other.

1591

317 504347

– 317

1873

– 1585

2884

– 2853

317

– 317

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0

ANSWER:

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

1591

317 504347

– 317

1873

– 1585

2884

– 2853

317

– 317

0

y= 1591

∴ The other number is 1591.

Page No 56:

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Question 6: On dividing 59761 by a certain number, the quotient is 189

and the remainder is 37. Find the divisor.

ANSWER:

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

316

189 59724

– 567

302

– 189

1134

– 1134

0

Hence, divisor =316

Page No 56:

Question 7: On dividing 55390 by 299, the remainder is 75. Find the

quotient using the division algorethm.

ANSWER:

Here, Dividend = 55390, Divisor = 299 and Remainder = 75

We have to find the quotient.

Now, Dividend = Divisor × Quotient + Remainder

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⇒ 55390 = 299 × Quotient + 75

⇒ 55390 − 75 = 299 × Quotient

⇒ 55315 = 299 × Quotient

⇒ Quotient = 55315 ÷ 299

185

299 55315

– 299

2541

– 2392

1495

– 1495

0

Hence, quotient =185

Page No 56:

Question 8: What least number must be subtracted from 13601 to get a

number exactly divisible by 87?

ANSWER:

First, we will divide 13601 by 87.

156

87 13601

– 87

490

– 435

551

– 522

29

Remainder = 29

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So, 29 must be subtracted from 13601 to get a number exactly divisible

by 87.

i.e., 13601 − 29 = 13572

Now, we have:

156

87 13572

– 87

487

– 435

522

– 522

29

∴ 29 must be subtracted from 13601 to make it divisible by 87.

Page No 56:

Question 9: What least number must be added to 1056 to get a number

exactly divisible by 23?

ANSWER:

First, we will divide 1056 by 23.

45

23 1056

– 92

136

– 115

21

Required number = 23 − 21 = 2

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So, 2 must be added to 1056 to make it exactly divisible by 23.

i.e., 1056 + 2 = 1058

Now, we have:

46

23 1058

– 92

138

– 138

0

∴ 1058 is exactly divisible by 23.

Page No 56:

Question 10: Find the largest 4-digit number divisible by 16.

ANSWER: We have to find the largest four-digit number divisible by 16.

The largest four-digit number = 9999

Therefore, dividend =9999

Divisor =16

62

16 9999

– 96

39

– 32

79

– 64

15

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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:

Question 11: Divide the largest 5-digit number by 653. Check your

answer by the division algorithm.

ANSWER:

Largest five-digit number =99999

153

653 99999

– 653

3469

– 3265

2049

– 1959

90

Dividend = 99999, Divisor = 653, Quotient = 153 and Remainder = 90

Check: Divisor ×Quotient + Remainder

= 653 × 153 + 90

= 99909 + 90

= 99999

= Dividend

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∴ Dividend = Divisor × Quotient + Remainder

Verified.

Page No 56:

Question 12: Find the least 6-digit number exactly divisible by 83.

ANSWER:

Least six-digit number = 100000

Here, dividend = 100000 and divisor = 83

1204

83 100000

– 83

170

– 166

40

– 0

400

– 332

68

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:

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Question 13: 1 dozen bananas cost Rs 29. How many dozens can be

purchased for Rs 1392?

ANSWER:

Cost of 1 dozen bananas = Rs 29

Number of dozens purchased for Rs 1392 = 1392 ÷ 29

48

29 1392

– 116

232

– 232

0

Hence, 48 dozen of bananas can be purchased with Rs. 1392.

Page No 56:

Question 14: 19625 trees have been equally planted in 157 rows. Find the

number of trees in each row.

ANSWER:

Number of trees planted in 157 rows = 19625

Trees planted in 1 row = 19625 ÷ 157

123

157 19625

– 157

392

– 314

785

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– 785

0

∴ 125 trees are planted in each row.

Page No 56:

Question 15: The population of a town is 517530. If one out of every 15

is reported to be literate, find how many literate persons are there in the

town.

ANSWER:

Population of the town = 517530

(115)115 of the population is reported to be literate, i.e., (115)115 ×

517530 = 517530 ÷÷ 15

34502

15 517530

– 45

67

– 60

75

– 75

030

– 30

0

∴ There are 34502 illiterate persons in the given town.

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Page No 56:

Question 16: The cost price of 23 colour television sets is Rs 570055.

Determine the cost price of each TV set if each costs the same.

ANSWER:

Cost price of 23 colour TV sets = Rs 5,70,055

Cost price of 1 TV set = Rs 570055 ÷ 23

24785

23 57005

– 46

110

– 92

180

– 184

115

– 115

0

∴ The cost price of one TV set is Rs 24,785.

© PRAADIS

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EXERCISE – 3.6

Page No 56:

Question 1: The smallest whole number is

(a) 1

(b) 0

(c) 2

(d) none of these

ANSWER:

(b) 0

The smallest whole number is 0.

Page No 56:

Question 2:

The least number of 4 digits which is exactly divisible by 9 is

(a) 1018

(b) 1026

(c) 1009

(d) 1008

ANSWER:

(d) 1008

(a)

© PRAADIS

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113

36 1018

– 9

11

– 19

28

– 27

1

Hence, 1018 is not exactly divisible by 9.

(b)

114

9 1026

– 9

12

– 9

28

– 27

1

Hence, 1026 is exactly divisible by 9.

(c)

112

9 1009

– 9

10

– 9

19

– 18

1


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(d)

112

9 1008

– 9

10

– 9

18

– 18

0


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

exactly divisible by 9.

Page No 57:

Question 3: The largest number of 6 digits which is exactly divisible by

16 is

(a) 999980

(b) 999982

(c) 999984

(d) 999964

ANSWER:

(c) 999984

(a)

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62498

16 999980

– 96

39

– 32

79

– 64

158

– 144

140

– 128

12


(b)

62498

16 999982

– 96

39

– 32

79

– 64

158

– 144

142

– 128

14


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(c)

62499

16 999984

– 96

39

– 32

79

– 64

158

– 144

144

– 144

0


(d)

62497

16 999964

– 96

39

– 32

79

– 64

156

– 144

124

– 112

12


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

Page No 57:

Question 4: What least number should be subtracted from 10004 to get

a number exactly divisible by 12?

(a) 4

(b) 6

(c) 8

(d) 20

ANSWER:

(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.

833

12 10004

– 96

40

– 36

44

– 36

8

Remainder = 8

So, 8 should be subtracted from 10004 to get the number exactly

divisible by 12.

i.e., 10004 − 8 = 9996

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833

12 9996

– 96

39

– 36

36

– 36

0


Page No 57:

Question 5: What least number should be added to 10056 to get a number

exactly divisible by 23?

(a) 5

(b) 18

(c) 13

(d) 10

ANSWER:

(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

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437

23 10056

– 92

85

– 69

166

– 161

5

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

438

23 10074

– 92

87

– 69

184

– 184

0


Page No 57:

Question 6: What whole number is nearest to 457 which is divisible by

11?

(a) 450

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(b) 451

(c) 460

(d) 462

ANSWER:

(d) 462

(a)

4

11 450

– 44

10

Hence, 450 is not divisible by 11.

(b)

4

11 450

– 44

11

– 11

0

Hence, 451 is divisible by 11.

(c)

41

11 460

– 44

20

– 11

9

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Hence, 460 is not divisible by 11.

(d)

42

11 462

– 44

22

– 22

0

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:

Question 7: How many whole numbers are there between 1018 and 1203?

(a) 185

(b) 186

(c) 184

(d) none of these

ANSWER:

(c) 184

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

= 185 − 1

= 184

Page No 57:

Question 8: A number when divided by 46 gives 11 as quotient and 15 as

remainder. The number is

(a) 491

(b) 521

(c) 701

(d) 679

ANSWER:

(b) 521

Divisor = 46

Quotient = 11

Remainder = 15


= 46 × 11 + 15

= 506 + 15

= 521

Page No 57:

Question 9: In a division sum, we have dividend = 199, quotient = 16

and remainder = 7. The divisor is

(a) 11

(b) 23

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(c) 12

(d) none of these

ANSWER:

(c) 12

Dividend = 199

Quotient = 16

Remainder = 7

According to the division algorithm, we have:


⇒ 199 = divisor × 16 + 7

⇒ 199 − 7 = divisor × 16

⇒ Divisor = 192 ÷ 16

12

16 192

– 16

32

– 32

0

Page No 57:

Question 10: 7589 −? = 3434

(a) 11023

(b) 4245

(c) 4155

(d) none of these

ANSWER:

(a) 11023

7589 − ? = 3434

⇒ 7589 − x = 3434

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⇒ x = 7589 + 3434

⇒ x = 11023

Page No 57:

Question 11: 587 × 99 = ?

(a) 57213

(b) 58513

(c) 58113

(d) 56413

ANSWER:

(c) 58113

587 × 99

= 587 × (100 − 1)

= 587 × 100 − 587 × 1 [Using distributive property of

multiplication over subtraction]

= 58700 − 587

= 58113

Page No 57:

Question 12: 4 × 538 × 25 = ?

(a) 32280

(b) 26900

(c) 53800

(d) 10760

ANSWER:

(c) 53800

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4 × 538 × 25

= (4 × 25) × 538

= 100 × 538

= 53800

Page No 57:

Question 13: 24679 × 92 + 24679 × 8 = ?

(a) 493580

(b) 1233950

(c) 2467900

(d) none of these

ANSWER:

(c) 2467900

By using the distributive property, we have:

24679 × 92 + 24679 × 8

= 24679 × (92 + 8)

= 24679 × 100

= 2467900

Page No 57:

Question 14: 1625 × 1625 − 1625 × 625 = ?

(a) 1625000

(b) 162500

(c) 325000

(d) 812500

ANSWER:

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(a) 1625000


1625 × 1625 − 1625 × 625

= 1625 × (1625 − 625)

=1625 × 1000

= 1625000

Page No 57:

Question 15: 1568 × 185 − 1568 × 85 = ?

(a) 7840

(b) 15680

(c) 156800

(d) none of these

ANSWER:

(c) 156800


1568 × 185 − 1568 × 85

= 1568 × (185 − 85)

= 1568 × 100

= 156800

Page No 57:

Question 16: (888 + 777 + 555) = (111 × ?)

(a) 120

(b) 280

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(c) 20

(d) 140

ANSWER:

(c) 20

(888 + 777 + 555) = (111 × ?)

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

= 111 × (20) = 2220

Page No 57:

Question 17: The sum of two odd numbers is

(a) an odd number

(b) an even number

(c) a prime number

(d) a multiple of 3

ANSWER:

(b) an even number

The sum of two odd numbers is an even number.

Example: 5 + 3 = 8

Page No 57:

Question 18: The product of two odd numbers is

(a) an odd number

(b) an even number

(c) a prime number

(d) none of these

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ANSWER:

(a) an odd number

The product of two odd numbers is an odd number.

Example: 5 × 3 = 15

Page No 57:

Question 19: If a is a whole number such that a + a = a, then a = ?

(a) 1

(b) 2

(c) 3

(d) none of these

ANSWER:

(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:

Question 20: The predecessor of 10000 is

(a) 10001

(b) 9999

(c) none of these

ANSWER:

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(b) 9999

Predecessor of 10000 = 10000 − 1 = 9999

Page No 57:

Question 21: The successor of 1001 is

(a) 1000

(b) 1002

(c) none of these

ANSWER:

(b) 1002

Successor of 1001 = 1001 + 1 = 1002

Page No 57:

Question 22: The smallest even whole number is

(a) 0

(b) 2

(c) none of these

ANSWER:

(b) 2

The smallest even whole number is 2. Zero (0) is neither an even

number nor an odd number.

Page No 59:

Question 1: How many whole numbers are there between 1064 and

1201?

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ANSWER:

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

= 137 − 1

= 136

Page No 59:

Question 2: Fill in the blanks.

1000000

−****1

*7042*

ANSWER:

1000000

− ****1

*7042*

Then, we have:

1000000

− 29571

970429

Page No 59:

Question 3: Use distributive law to find the value of

1063 × 128 − 1063 × 28.

ANSWER:

Using distributive law, we have:

1063 × 128 − 1063 × 28

= 1063 × (128 − 28)

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= 1063 × 100

= 106300

Page No 59:

Question 4: Find the product of the largest 5-digit number and the

largest 3-digit number using distributive law.

ANSWER:

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:

Question 5: Divide 53968 by 267 and check the result by the division

algorithm.

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ANSWER:

202

11 53968

– 534

568

– 534

34

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:

Question 6: Find the largest 6-digit number divisible by 16.

ANSWER:

Largest six-digit number = 999999

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62499

16 999999

– 96

39

– 32

72

– 64

159

– 144

159

– 144

15

Remainder = 15

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

62499

16 999984

– 96

39

– 32

79

– 64

158

– 144

144

– 144

0

∴ 999984 is divisible by 16.

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Page No 59:

Question 7: The cost price of 23 TV sets is Rs 570055. Find the cost of

each such set.

ANSWER:

Cost price of 23 TV sets = Rs 5,70,055

Cost price of 1 TV set = 570055 ÷ 23

24785

23 570055

– 46

110

– 92

180

– 161

195

– 184

115

– 115

0

∴ The cost of one TV set is Rs 24,785.

Page No 59:

Question 8: What least number must be subtracted from 13801 to get a

number exactly divisible by 87?

ANSWER:

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

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158

87 13801

– 87

510

– 435

751

– 696

55

Remainder = 55

The number 55 should be subtracted from 13801 to get a number

divisible by 87.

i.e., 13801 − 55 = 13746

158

87 13746

– 87

504

– 435

696

– 696

0

∴ 13746 is divisible by 87.

Page No 59:

Question 9: The value of (89 × 76 + 89 × 24) is

(a) 890

(b) 8900

(c) 89000

(d) 10420

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ANSWER:

(b) 8900

(89 × 76 + 89 × 24)

= 89 × (76 + 24) [Using distributive property of multiplication over

addition]

= 89 × 100

= 8900

Page No 59:

Question 10: On dividing a number by 53 we get 8 as quotient and 5 as

remainder. The number is

(a) 419

(b) 423

(c) 429

(d) none of these

ANSWER:

(c) 429

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

Now, Dividend = Quotient × Divisor +Remainder

= 8 × 53 + 5

= 429

Page No 59:

Question 11: The whole number which has no predecessor is

(a) 1

(b) 0

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(c) 2

(d) none of these

ANSWER:

(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:

Question 12: 67 + 33 = 33 + 67 is an example of

(a) closure property

(b) associative property

(c) commutative property

(d) distributive property

ANSWER:

(c) Commutative property

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

Page No 59:

Question 13: Additive inverse of 36 is

(a) 136136

(b) 0

(c) −36

(d) none of these

ANSWER:

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(c) -36

The additive inverse of 36 is −36.

i.e., 36 + (−36) = 0

Page No 59:

Question 14: Which of the following is not zero?

(a) 0 × 0

(b) 0202

(c) (8 − 8)28 - 82

(d) 2 + 0

ANSWER:

(d) 2+0

(a) 0 × 0 = 0

(b) 0/2 = 0

(c) (8−8)2=028-82=02 =0

(d) 2 + 0 = 2

Page No 59:

Question 15: The predecessor of the smallest 3-digit number is

(a) 999

(b) 100

(c) 101

(d) 99

ANSWER:

(d) 99

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Smallest three-digit number = 100

∴ Predecessor of 100 = 100 − 1 = 99

Page No 59:

Question 16: The number of whole numbers between the smallest

whole number and the greatest 2-digit number is

(a) 88

(b) 98

(c) 99

(d) 101

ANSWER:

(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:

Question 17: Fill in the blanks.

(i) The smallest natural number is ...... .

(ii) The smallest whole number is ...... .

(iii) Division by ...... is not defined.

(iv) ...... is a whole number which is not a natural number.

(v)...... is a whole number which is not a natural number.

ANSWER:

(i) The smallest natural number is 1.

(ii) The smallest whole number is 0.

(iii) Division by 0 is not defined.

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(iv) 0 is a whole number which is not a natural number.

(v) 1 is the multiplicative identity for whole numbers.

Page No 60:

Question 18: Write 'T' for true and 'F' for false in each of the

following:

(i) 0 is the smallest natural number.

(ii) Every natural number is a whole number.

(iii) Every whole number is a natural number.

(iv) 1 has no predecessor in whole numbers.

ANSWER:

(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.

Page No 60:

Question 19: Match the following columns on whole numbers:

column A column B

(a) 137 + 63 = 63 + 137 (i) Associativity of multiplication

(b) (16 × 25) is a number (ii) Commutativity of multiplication

(c) 365 × 18 = 18 × 365 (iii) Distributive law of multiplication

over addition

(d) (86 × 14) × 25 = 86 × (14 × 25) (iv) Commutativity of addition

(e) 23 × (80 + 5) = (23 × 80) + (23

× 5) (v) Closure property for multiplication

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ANSWER:

Column A Column B

(a) 137 + 63 = 63 + 137 (iv) Commutativity of addition

(b) (16 × 25) is a number (v) Closure property for

multiplication

(c) 365 × 18 = 18 × 365 (ii) Commutativity of multiplication

(d) (86 × 14) × 25 = 86 × (14 × 25) (i) Associativity of multiplication

(e) 23 × (80 + 5) = (23 × 80) + (23

× 5)

(iii) Distributive law of

multiplication over addition

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Chapter 3 Whole numbers - Praadis Education

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