RS Aggarwal 2021-2022 for Class 6 Maths Chapter 3 - Whole Number
Rs Aggarwal Class 6 Math Solution Chapter 3 Whole Number Exercise 3D is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 3 Whole Number
Exercise 3D
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 ×
Solution:
(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
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 )
Solution:
(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
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
Solution:
(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)
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
Solution:
(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
5: Find each of the following products, using distributive laws:
(i) 740 x 105
(ii) 245 x 1008
(iii) 947 x 96
(iv) 996 x 367
(v) 472 x 1097
(vi) 580 × 64
(vii) 439 × 997
(viii) 1553 × 198
Solution:
(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
6: Find each of the following products, using distributive laws:
(i) 3576 × 9
(ii) 847 × 99
(iii) 2437 × 999
Solution: (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
7: Find the products:
8: Find the product of the largest 3-digit number and the largest 5-digit number.
Solution:
Largest 3-digit number = 999
Largest 5-digit number = 99999
Required product = 999 × 99999
=999 × (100000 - 1)
= 999 × 100000 - 999 × 1
=99900000 - 999
=99899001
9: A car moves at a uniform speed of 75 km per hour. How much distance will it cover in 98 hours?
Solution:
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.
10: A dealer purchased 139 VCRS. If the cost of each set is Rs. 24350, find the cost of all the sets together.
Solution:
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.
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?
Solution:
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.
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.
Solution:
Cost of a chair = Rs 1065
Cost of 50 chairs = 50 × 1065 = Rs 53250
Cost of a blackboard = Rs 1645
Cost of 30 blackboards = 30 × 1645 = Rs 49350
∴ Total amount = cost of 50 chairs + cost of 30 blackboards
= Rs (53250 + 49350)
= Rs 1,02,600
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.
Solution:
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
14. The product of two whole numbers is zero. What do you conclude?
Solution:
If the product of two whole numbers is zero, then one of them is definitely zero or both numbers are 0
Example: 0 × 2 = 0 and 0 × 15 = 0 Or, 0 × 0 = 0
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 ≠ 0 and a × a = a ⇒ a = ?
Solution:
(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 = `frac\{a}{a}` = 1, a ≠ 0
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