#### Exercise 4.3

*Q1. Fill in the blanks to make each of the following a true statement:*

Solution: (i) 785 x 0 = 0

(ii) 4567 x 1 = 4567 (Multiplicative identity)

(iii) 475 x 129 = 129 x 475 (Commutativity)

(iv) 1243 x 8975 = 8975 x 1243 (Commutativity)

(v) 10 x 100 x 10 = 10000

(vi) 27 x 18 = 27 x 9 + 27 x 4 + 27 x 5

(vii) 12 x 45 = 12 x 50 – 12 x 5

(viii) 78 x 89 = 78 x 100 – 78 x 16 + 78 x 5

(ix) 66 x 85 = 66 x 90 – 66 x 4 – 66

(x) 49 x 66 + 49 x 34 = 49 x (66 + 34)

*Q2. Determine each of the following products by suitable rearrangements: *

Solution: (i) 2 x 1497 x 50

= (2 x 50) x 1497 = 100 x 1497 = 149700

(ii) 4 x 358 x 25

= (4 x 25) x 358 = 100 x 358 = 35800

(iii) 495 x 625 x 16

= (625 x 16) x 495 = 10000 x 495 = 4950000

(iv) 625 x 20 x 8 x 50

= (625 x 8) x (20 x 50) = 5000 x 1000 = 5000000

*Q3. Using distributivity of multiplication over addition of whole numbers, find each of the following products: *

Solution: (i) 736 x 103 = 736 x (100 + 3)

{Using distributivity of multiplication over addition of whole numbers}

= (736 x 100) + (736 x 3)

= 73600 + 2208 = 75808

(ii) 258 x 1008 = 258 x (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 x 1000) + (258 x 8)

= 258000 + 2064 = 260064

(iii) 258 x 1008 = 258 x (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 x 1000) + (258 x 8)

= 258000 + 2064 = 260064

*Q4. Find each of the following products:*

Solution: (i) 736 x 93

Since, 93 = (100 – 7)

Therefore, 736 x (100 – 7)

= (736 x 100) – (736 x 7)

(Using distributivity of multiplication over subtraction of whole numbers)

= 73600 – 5152 = 68448

(ii) 816 x 745

Since, 745 = (750 – 5)

Therefore, 816 x (750 – 5)

= (816 x 750) – (816 x 5)

(Using distributivity of multiplication over subtraction of whole numbers)

= 612000 – 4080 = 607920

(iii) 2032 x 613

Since, 613 = (600 +13)

Therefore, 2032 x (600 + 13)

= (2032 x 600) + ( 2032 x 13)

= 1219200 + 26416 = 1245616

*Q5. Find the values of each of the following using properties:*

Solution: (i) 493 x 8 + 493 x 2

= 493 x (8 + 2)

(Using distributivity of multiplication over addition of whole numbers)

= 493 x 10 = 4930

(ii) 24579 x 93 + 7 x 24579

= 24579 x (93 + 7)

(Using distributivity of multiplication over addition of whole numbers)

= 24579 x 100 = 2457900

(iii) 1568 x 184 – 1568 x 84

= 1568 x (184 – 84)

(Using distributivity of multiplication over subtraction of whole numbers)

= 1568 x 100 = 156800

(iv) 15625 x 15625 – 15625 x 5625

= 15625 x (15625 – 5625)

(Using distributivity of multiplication over subtraction of whole numbers)

= 15625 x 10000 = 156250000

*Q6. Determine the product of:*

*(i) the greatest number of four digits and the smallest number of three digits.*

* (ii) the greatest number of five digits and the greatest number of three digits.*

Solution: (i) The largest four-digit number = 9999

The smallest three – digit number = 100

Therefore, Product of the smallest three-digit number and the largest four-digit number = 9999 x 100 = 999900

(ii) The largest five – digit number = 9999

The largest number of three digits = 999

Therefore, Product of the largest three-digit number and the largest five-digit number

= 9999 x 999

= 9999 x (1000 — 1)

= (9999 x 1000) — (9999 x 1)

= 9999000 – 9999

= 9989001

*Q7. In each of the following, fill in the blanks, so that the statement is true:*

Solution: (i) (500 + 7) (300 – 1)

= 507 x 299

= 299 x 507 (Commutativity)

(ii) 888 + 777 + 555

= 111 (8 + 7 + 5)

= 111 x 20 (Distributivity)

(iii) 75 x 425

= (70 + 5) x 425

= (70 + 5) (340 + 85)

(iv) 89 x (100 – 2)

= 89 x 98

= 98 x 89

= 98 x (100 – 11) (Commutativity)

(v) (15 + 5) (15 – 5)

= 20 x 10

= 200

= 225 – 25

(vi) 9 x (10000 + 974)

= 98766

*Q8. A dealer purchased 125 color television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.*

Solution: Cost of 1 color television set = Rs 19820

Therefore, Cost of 125 color television sets = Rs (19820 x 125)

= Rs 19820 x (100 + 25)

= Rs (19820 x 100) + (19820 x 25)

= Rs 1982000 + 495500

= Rs 2477500

*Q9. The annual fee charged from a student of class 6 ^{th} in a school is Rs 8880. If there are, in all, 235 students in class 6^{th}, find the total collection.*

Solution: Fees charged from 1 student = Rs 8880

Therefore, Fees charged from 235 students = Rs 8880 x 235

= 2086800

Thus, the total collection from class VI students is Rs 2086800.

*Q10. A group housing society constructed 350 flats. If the cost of construction for each flat is Rs 993570, what is the total cost of construction of all the flats.*

Solution: Cost of construction of 1 flat = Rs 993,570

Total number of flats constructed = 350

Total cost of construction of 350 flats = Rs (993,570 x 350)

= Rs 347,749,500

*Q11. The product of two whole numbers is zero. What do you conclude?*

Solution: If the product of two whole numbers is zero, then it means that either one of them is zero or both of them are zero.

*Q12. What are the whole numbers which when multiplied with itself gives the same number?*

Solution: There are two numbers which when multiplied with themselves give the same numbers.

(i) 0 x 0 = 0

(ii) 1 x 1 = 1

*Q13. In a large housing complex, there are 15 small buildings and 22 large building. Each of the large buildings has 10 floors with 2 apartments on each floor. Each of the small buildings has 12 floors with 3 apartments on each floor. How many apartments are there in all.*

Solution: Number of large buildings = 22

Number of small buildings = 15

Number of floors in 1 large building = 10

Number of apartments on 1 floor = 2

Therefore, Total apartments in 1 large building = 10 x 2 = 20

Similarly,

Total apartments in 1 small building = 12 x 3 = 36

Therefore, Total apartments in the entire housing complex = (22 x 20) + (15 x 36)

= 440 + 540

= 980