## RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3A
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3B
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3C
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E
- RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F

**Question 1.**

**Solution:**

(i) 246 x 1 = 246

(By multiplicative property of 1)

(ii) 1369 x 0 = 0

(By multiplicative property of 0)

(iii) 593 x 188 = 188 x 593

(By commutative law of multiplication)

(iv) 286 x 753 = 753 x 286

(By commutative law of multiplication)

(v) 38 x (91 x 37) = 91 x (38 x 37)

(By associative law of multiplication)

(vi) 13 x 100 x 1000 = 1300000

(vii) 59 x 66 + 59 x 34 = 59 x (66 + 34)

(By distributive law of multiplication)

(vii) 68 x 95 = 68 x 100 – 68 x 5

**Question 2.**

**Solution:**

(i) Commutative law of multiplication

(ii) Closure property

(iii) Associative law of multiplication

(iv) Multiplicative property of 1

(v) Multiplicative property of 0

(vi) Distributive law of multiplication over addition in whole numbers

(vii) Distributive law of multiplication over subtraction in whole numbers

**Question 3.**

**Solution:**

Using the law of distribution over addition and subtraction

(i) 647 x 13 + 647 x 7

= 647 x (13 + 7)

= 647 x 20

= 12940

(ii) 8759 x 94 + 8759 x 6

= 8759 x (94 + 6)

= 8759 x 100

= 875900

(iii) 7459 x 999 + 7459

= 7459 x 999 + 7459 x 1

= 7459 x (999 + 1)

= 7459 x 1000

= 7459000

(iv) 9870 x 561 – 9870 x 461

= 9870 x (561 – 461)

= 9870 x 100

= 987000

(v) 569 x 17 + 569 x 13 + 569 x 70

= 569 x (17 + 13 + 70)

= 569 x 100

= 56900

(vi) 16825 x 16825 – 16825 x 6825

= 16825 x (16825 – 6825)

= 16825 x 10000

= 168250000

**Question 4.**

**Solution:**

(i) 2 x 1658 x 50 = 1658 x (2 x 50) (Associative law of multiplication)

= 1658 x 100

= 165800

(ii) 4 x 927 x 25 = 927 x (4 x 25) (Associative law of multiplication)

= 927 x 100

= 92700

(iii) 625 x 20 x 8 x 50

(By associative law of multiplication)

(625 x 8) x (20 x 50)

= 5000 x 1000

= 5000000

(iv) 574 x (625 x 16)

= 574 x 10000

= 5740000

(v) 250 x 60 x 50 x 8

= (250 x 8) x (60 x 50)

(By associative law)

= 2000 x 3000

= 6000000

(vi) 8 x 125 x 40 x 25

= (8 x 125) x (40 x 25)

= 1000 x 1000

= 1000000

**Question 5.**

**Solution:**

Using distributive law of multiplication over addition or subtraction,

(i) 740 x 105

= 740 x (100 + 5)

= 740 x 100 + 740 x 5

= 74000 + 3700

= 77700

(ii) 245 x 1008

= 245 x (1000 + 8)

= 245 x 1000 + 245 x 8

= 245000 + 1960

= 246960

(iii) 947 x 96

= 947 x (100 – 4)

= 947 x 100 – 947 x 4

= 94700 – 3788

= 90912

(iv) 996 x 367

= 367 x (1000 – 4)

= 367 x 1000 – 367 x 4

= 367000 – 1468

= 365532

(v) 472 x 1097

= 472 x (1100 – 3)

= 472 x 1100 – 472 x 3

= 519200 – 1416

= 517784

(vi) 580 x 64

= 580 x (60 + 4)

= 580 x 60 + 580 x 4

= 34800 + 2320

= 37120

(vii) 439 x 997

= 437 x (1000 – 3)

= 439 x 1000 – 439 x 3

= 439000 – 1317

= 437683

(viii) 1553 x 198

= 1553 x (200 – 2)

= 1553 x 200 – 1553 x 2

= 310600 – 3106

= 307494

**Question 6.**

**Solution:**

(i) 3576 x 9 = 3576 x (10 – 1)

= 3576 x 10 – 3576 x 1

= 35760 – 3576

= 32184

(ii) 847 x 99 = 84 x (100 – 1)

= 847 x 100 – 847 x 1

= 84700 – 847

= 83853

(iii) 2437 x 999 = 2437 x (1000 – 1)

= 2437 x 1000 – 2437 x 1

= 2437000 – 2437

= 2434563

**Question 7.**

**Solution:**

**Question 8.**

**Solution:**

Largest 3-digit number = 999

Largest 5-digit number = 99999

Required product = 99999 x 999

= 99999 x (1000 – 1)

= 99999 x 1000-99999 x 1

= 99999000 – 99999

= 9,98,99,001

**Question 9.**

**Solution:**

Speed of car = 75 km per hour

In 1 hour, distance covered by a car = 75 km

.’. In 98 hours, distance will be covered = 75 x 98

= 75 x (100 – 2)

= 75 x 100 – 75 x 2

= 7500 – 150

= 7350 km

**Question 10.**

**Solution:**

Cost of 1 set of VCR = Rs. 24350

Cost of 139 sets of VCR = Rs. 24350 x 139

= Rs. 3384650

**Question 11.**

**Solution:**

Cost of 1 house = Rs. 450000

Cost of 197 houses = Rs. 450000 x 197

= Rs. 450000 x (200 – 3)

= Rs. (450000 x 200 – 450000 x 3)

= Rs. (90000000 – 1350000)

= Rs. 88650000

**Question 12.**

**Solution:**

Cost of each chair = Rs. 1065

Cost of 50 chairs = Rs. 1065 x 50

= Rs. 53250

Cost of each blackboard = Rs. 1645

Cost of 30 blackboards = Rs. 1645 x 30

= Rs. 49350

Total cost of 50 chairs and 30 blackboards = Rs. 53250 + 49350

= Rs. 102600

**Question 13.**

**Solution:**

Number of students in 1 section = 45

Number of students in 6 sections = 45 x 6 = 270

Monthly charges of 1 student = Rs. 1650

Total monthly incomes from the class VI = Rs. 270 x 1650

= Rs. (300 – 30) x 1650

= Rs. (1650 x 300 – 1650 x 30)

= Rs. (495000 – 49500)

= Rs. 445500

**Question 14.**

**Solution:**

Since the product of two whole numbers is zero

.’. From multiplicative property of zero, we conclude that one of the whole numbers is zero.

**Question 15.**

**Solution:**

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

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

(iii) a x a = a => a = 1 as 1 x 1 = 1

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D are helpful to complete your math homework.

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