EVALUATE THE NUMERICAL EXPRESSION USING PROPERTIES

Problem 1 :

Find the sum by suitable rearrangement:

713 + 200 + 87

Solution :

= 713 + 200 + 87

By observing the expression, we notice that 713 and 87 are complements.

= 713 + 87 + 200

= 800 + 200

= 1000

Problem 2 :

Simplify

126 × 45 + 126 × 55

by using suitable property.

Solution :

= 126 × 45 + 126 × 55

Factoring 126, we get

= 126 x (45 + 55)

Here 45 and 55 are complements.

= 126 x (100)

= 12600

Problem 3 :

Siddharth purchased 10 boxes of chocolates to be distributed among the students. Each package has 10 packets, each of which contains 10 chocolates. Then how many chocolates did he order?

Solution :

10 boxes, each box will contain 10 packets and each packet will contain 10 chocolates.

Total number of chocolates = 10 x 10 x 10

= 1000

So, number of chocolates he order is 1000.

Problem 4 :

Shelly got 49 marks in Math, 39 marks in English, and 51 in Science. John got 62 marks in Math, 36 in English and 54 in Science. What are their total marks?

Solution :

Shelly got the following marks :

Math = 49, English = 39, Science = 51

John got the following marks :

Math = 62, English = 36, Science = 54

Total marks by Shelly = 49 + 39 + 51

Here 49 and 51 are complements 

= 49 + 51 + 39

= 100 + 39

= 139

Total marks by John = 62 + 36 + 54

Here 36 and 54 are complements.

= 62 + 90

= 152

Problem 5 :

Simplify using distributive property:

23 × 99

Solution :

= 23 × 99

Instead of using direct multiplication, we can write the number for distributive property.

= 23 x (100 - 1)

= 2300 - 23

= 2277

Problem 6 :

Simplify by suitable rearrangement:

i) 643 + 346 + 357

ii) 5 × 241 × 20

Solution :

i) 643 + 346 + 357

Here 643 and 357 are complements.

= 643 + 357 + 346 

= 1000 + 346

= 1346

ii) 5 × 241 × 20

While multiplying 5 and 20, we will get 100.

= (5 x 20) × 241

= 100 x 241

= 24100

Problem 7 :

Simplify the following using the distributive property:

i) 234 × 256 – 234 × 56

ii) 54 × 1001

Solution :

i) 234 × 256 – 234 × 56

Factoring 234 out,

= 234 x (256 - 56)

= 234 x 200

= 234 x 2 x 100

= 468 x 100

= 46800

ii) 54 × 1001

= 54 x (1000 + 1)

Distributing 54, we get

= 54000 + 54

= 54054

Problem 8 :

The cost of a chair is ₹7635 and the cost of a table is ₹12365. Find the total cost of 12 chairs and 12 tables.

Solution :

Cost of 1 chair = ₹7635

Cost of a table = ₹12365

Cost of 12 chairs and tables,

= 12(7635) + 12(12365))

= 12(7635 + 12365)

= 12(20000)

= 240000

So, cost of 12 chairs and tables is ₹240000

Problem 9 :

Find the sum by suitable rearrangement

47+953+6437 

Solution :

= 47 + 953 + 6437 

Here 47 and 953 are complements of 1000.

= 1000 + 6437

= 7437

Problem 10 :

i) 

2698 × (100 + 9) = 2698 × 100 + 2698 × 9

is true by _________property.

ii)  Use the same property to solve the expression:

51887 × 88 + 51887 × 12

Solution :

i)

2698 × (100 + 9) = 2698 × 100 + 2698 × 9

Using distributive property, they have distributed 2698.

ii) 51887 × 88 + 51887 × 12

Factoring 51887, we get

= 51887 x (88 + 12)

= 51887 x 100

= 5188700

Problem 11 :

Identify the property demonstrated by the following statements:

a) 12 + 13 = 13 + 12

b) 10 × (46 + 24) = 10 × 46 + 10 × 24

c) 100 + 0 = 0 + 100 = 100

Solution :

a) Comparing with a + b = b + a

12 + 13 = 13 + 12 is under commutative property.

b) Comparing with a x (b + c) = a x b + a x c

10 × (46 + 24) = 10 × 46 + 10 × 24 is under distributive property. 

c) 100 + 0 = 0 + 100 = 100

identity property.