Exercise 3A
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.
Page No 45:
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.
Page No 45:
Question 3:
How many whole numbers are there between 1032 and 1209?
ANSWER:
Number of whole numbers between 1032 and 1209 = (1209 − 1032) − 1
= 177 − 1
= 176
Page No 45:
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.
Page No 45:
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
(vii) Successor of 6475999 = 6475999 + 1 = 6476000
(viii) Successor of 9999999 = 9999999 + 1 = 10000000
Page No 46:
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
Page No 46:
Question 7:
Write down three consecutive whole numbers just preceding 7510001.
ANSWER:
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:
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.
(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:
Exercise 3B
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)
Page No 48:
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
(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:
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.
Page No 48:
Question 4:
Determine each of the sums given below using suitable rearrangement.
(i) 953 + 707 + 647
(ii) 1983 + 647 + 217 + 353
(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
(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:
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
Page No 48:
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.
Page No 48:
Question 7:
Complete each one of the following magic squares by supplying the missing numbers:
(i)
| 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)
| 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:
Question 8:
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.
Page No 49
Exercise 3C
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
Page No 49:
Question 2:
Replace each * by the correct digit in each of the following:
(i) 
(ii) 
(iii) 
(iv) 
ANSWER:
(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:
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
Page No 50:
Question 4:
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
Page No 50:
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.
Page No 50:
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:
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:
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.
Page No 50:
Question 8:
Find the whole number n when:
(i) n + 4 = 9
(ii) n + 35 = 101
(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
Page No 53:
Exercise 3D
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
Page No 53:
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)
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
Page No 53:
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
= 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:
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
(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:
Question 5:
Find each of the following products, using distributive laws:
(i) 740 × 105
(ii) 245 × 1008
(iii) 947 × 96
(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)
= 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:
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
Page No 54:
Question 7:
Find the products:
(i) 
(ii) 
(iii) 
(iv)
ANSWER:
(i)
458 × 67 = 30686
(ii)
3709 × 89 = 330101
(iii)
4617 × 234 = 1080378
(iv)
15208 × 542 = 8242736
Page No 54:
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
Page No 54:
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
=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:
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.
Page No 54:
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 [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:
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
Page No 54:
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
Page No 54:
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.
Page No 54:
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 anda × a = a
Given: a × a = a
⇒ a = aa=1aa=1, a ≠ 0
Page No 56:
Exercise 3E
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)
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:
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
ANSWER:
(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:
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
(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.
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
y= 1591
∴ The other number is 1591.
Page No 56:
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
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
⇒ 55390 = 299 × Quotient + 75
⇒ 55390 − 75 = 299 × Quotient
⇒ 55315 = 299 × Quotient
⇒ Quotient = 55315 ÷ 299
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.
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:
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.
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:
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
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
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:
Question 12:
Find the least 6-digit number exactly divisible by 83.
ANSWER:
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:
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
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
∴ 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
∴ There are 34502 illiterate persons in the given town.
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
∴ The cost price of one TV set is Rs 24,785.
Page No 56:
Exercise 3F
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)
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:
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)
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:
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.
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:
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
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:
Question 6:
What whole number is nearest to 457 which is divisible by 11?
(a) 450
(b) 451
(c) 460
(d) 462
ANSWER:
(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:
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
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
Dividend = divisor × quotient + remainder
= 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
(c) 12
(d) none of these
ANSWER:
(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:
Question 10:
7589 − ? = 3434
(a) 11023
(b) 4245
(c) 4155
(d) none of these
ANSWER:
(a) 11023
7589 − ? = 3434
⇒ 7589 − x = 3434
⇒ 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
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:
(a) 1625000
By using the distributive property, we have:
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
By using the distributive property, we have:
1568 × 185 − 1568 × 85
= 1568 × (185 − 85)
= 1568 × 100
= 156800
Page No 57:
Question 16:
(888 + 777 + 555) = (111 × ?)
(a) 120
(b) 280
(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
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:
(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:
Exercise 3G
Question 1:
How many whole numbers are there between 1064 and 1201?
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* 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)
= 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.
ANSWER:
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
Remainder = 15
Largest six-digit number divisible by 16 = 999999 − 15 = 999984
∴ 999984 is divisible by 16.
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
∴ 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
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:
Question 9:
The value of (89 × 76 + 89 × 24) is
(a) 890
(b) 8900
(c) 89000
(d) 10420
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
(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:
(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
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.
(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 |
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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