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**Q.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)**

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

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

**Ans.**** (i) **246 × 1 = 246

(By multiplicative property of 1)**(ii)** 1369 × 0 = 0

(By multiplicative property of 0)**(iii)** 593 × 188 = 188 × 593

(By commutative law of multiplicaton)

**(iv)** 286 × 753 = 753 × 286

(By commutative law of multiplication)

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

(By associative law of multiplication)

**(vi)** 13 × 100 × 1000 = 1300000

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

(By distributive law of multiplication)

**(viii) **68 × 95 = 68 × 100 – 68 × 5**Q.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)**

**Ans. ****(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**Q.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**

**Ans.** Using the law of distribution over addition

and subtraction**(i)** 647 × 13 + 647 × 7

= 647 × (13 + 7) = 647 × 20

= 12940**(ii)** 8759 × 94 + 8759 × 6

= 8759 × (94 + 6) = 8759 × 100

= 875900**(iii)** 7459 × 999 + 7459

= 7459 × 999 + 7459 × 1

= 7459 × (999 + 1) = 7459 × 1000

= 7459000**(iv)** 9870 × 561 – 9870 × 461

= 9870 × (561 – 461) = 9870 × 100

= 987000

**(v) **569 × 17 + 569 × 13 + 569 × 70

= 569 × (17 + 13 + 70) = 569 × 100

= 56900

**(vi)** 16825 × 16825 – 16825 × 6825

= 16825 × (16825 – 6825)

= 16825 × 10000 = 168250000

Q.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**

**Ans. ****(i)** 2 × 1658 × 50 = 1658 × (2 × 50)

(Associative law of multiplication)

1658 × 100 = 165800**(ii)** 4 × 927 × 25 = 927 × (4 × 25)

(Associative law of multiplication)

= 927 × 100 = 92700

**(iii)** 625 × 20 × 8 × 50

(By associative law of multiplication)

(625 × 8) × (20 × 50)

= 5000 × 1000 = 5000000

**(iv)** 574 × (625 × 16)

= 574 × 10000 = 5740000

**(v)** 250 × 60 × 50 × 8

= (250 × 8) × (60 × 50)

(By associative law)

= 2000 × 3000 = 6000000

**(vi)** 8 × 125 × 40 × 25

= (8 × 125) × (40 × 25)

= 1000 × 1000 = 1000000

Q.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**Ans.**

Using distributive law of multiplication over addition or subtraction,

**(i) **740 × 105 = 740 × (100 + 5)

= 740 × 100 + 740 × 5

= 74000 + 3700 = 77700

**(ii)** 245 × 1008 = 245 × (1000 + 8)

= 245 × 1000 + 245 × 8

= 245000 + 1960 = 246960**(iii)** 947 × 96 = 947 × (100 – 4)

= 947 × 100 – 947 × 4

= 94700 – 3788 = 90912

**(iv)** 996 × 367 = 367 × (1000 – 4)

= 367 × 1000 – 367 × 4

= 367000 – 1468 = 365532

**(v)** 472 × 1097 = 472 × (1100 – 3)

= 472 × 1100 – 472 × 3

= 519200 – 1416 = 517784**(vi) **580 × 64 = 580 × (60 + 4)

= 580 × 60 + 580 × 4

= 34800 + 2320 = 37120**(vii)** 439 × 997 = 437 × (1000 – 3)

= 439 × 1000 – 439 × 3

= 439000 – 1317 = 437683**(viii)** 1553 × 198 = 1553 × (200 – 2)

= 1553 × 200 – 1553 × 2

= 310600 – 3106 = 307494**Q.6. Find each of the following products, using distributive laws :(i) 3576 × 9(ii) 847 × 99(iii) 2437 × 999Ans. **

= 3576 × 10 – 3576 × 1

= 35760 – 3576 = 32184

= 847 × 100 – 847 × 1

= 84700 – 847 = 83853

= 2437 × 1000 – 2437 × 1

= 2437000 – 2437 = 2434563

Q.7. Find the products :

(i)

(ii)

(iii)

(iv)

Q.8. Find the product of the largest 3-digit number and the largest 5-digit number.

Largest 5-digit number = 99999

Required product = 99999 × 999

= 99999 × (1000 – 1)

= 99999 × 1000 – 99999 × 1

= 99999000 – 99999

= 9,98,99,001

Q.9. A car moves at a uniform speed of 75 km per hour. How much distance will it cover in 98 hours ?

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 × 98

= 75 × (100 – 2) = 75 × 100 – 75 × 2

= 7500 – 150 = 7350 km

Cost of 1 set of VCR = Rs. 24350

Cost of 139 sets of VCR

= Rs. 24350 × 139 = Rs. 3384650

Ans.

Cost of 1 house = Rs. 450000

Cost of 197 houses = Rs. 450000 × 197

= Rs. 450000 × (200 – 3)

= Rs. (450000 × 200 – 450000 × 3)

= Rs. (90000000 – 1350000) = Rs. 88650000

Q.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.

Cost of each chair = Rs. 1065

∴ Cost of 50 chairs = Rs. 1065 × 50

= Rs. 53250

Cost of each blackboard = Rs. 1645

∴ Cost of 30 blackboards = Rs. 1645 × 30

= Rs. 49350

Total cost of 50 chairs and 30 blackboards

= Rs. 53250 + 49350 = Rs. 102600**Q.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.Ans.** Number of students in 1 section = 45

∴ Number of students in 6 sections

= 45 × 6 = 270

Monthly charges of 1 student = Rs. 1650

∴ Total monthly incomes from the class VI

= Rs. 270 × 1650

= Rs. (300 – 30) × 1650

= Rs. (1650 × 300 – 1650 × 30)

= Rs. (495000 – 49500) = Rs. 445500**Q.14. The product of two whole numbers is zero. What do you conclude ?**

**Ans.** Since the product of two whole numbers is zero

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

zero.**Q.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 × a = a ⇒ a = ?**

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

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

**(iii)** a × a = a **⇒** a = **1 as 1 × 1 = 1**

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