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TET Syllabus - Mathematics Study Materials in Number System (Part III)

 

           Teachers Eligibility Test - Paper 2

 

Class – VI Mathematics

 

                                       NUMBER SYSTEM

 

            Use of Large Numbers in Daily Life Situations

 

Example: 1

    In an exhibition, the number of tickets sold on the first, second, third and fourth days are 1, 10,000, 75,060, 25,700 and 30,606 respectively. Find the total number of tickets sold on all the 4 days.

Solution:

Number of tickets sold on the first day      = 1, 10,000

Number of tickets sold on the second day = 75,060

Number of tickets sold on the third day     = 25,700

Number of tickets sold on the fourth day   = 30,606

Adding all the above,

            The total number of tickets sold on all the 4 days = 2,41,366.

 

Example: 2

       In a year, a whole-sale paper firm sold 6, 25,600 notebooks out of 7, 50,000 notebooks. Find the number of notebooks left unsold.

Solution:

Number of notebooks in the store = 7, 50,000

Number of notebooks sold = 6, 25,600

Subtracting the above,

Number of notebooks unsold = 1, 24,400.

Example: 3

In a mobile store, the number of mobiles sold during a month is 1250. Assuming that the same numbers of mobiles are sold every month, find the number of mobiles sold in 2 years.

Solution:

Number of mobiles sold in 1 month = 1250

1 year = 12 months

2 years = 2 × 12 = 24 months

Number of mobiles sold in 12 months =1250 × 12

                                                               =15,000

Number of mobiles sold in 24 months = 1250 × 24

                                                             = 30,000

Number of mobiles sold in 2 years = 30,000.

 

Example: 4

           If ` 10,00,000 was distributed in a Government scheme to 500 women in the Self Help Groups, then find the amount given to each woman.

Solution:

Amount to be given to 500 women = 10,00,000

Amount given to each woman = 10,00,000 ÷ 500

                                                 = 2000

Each woman in the Self Help Group was given 2000.

 

                            Order of Operations

 

BIDMAS

Valli and her four friends went to a butter milk shop. Each had a cup of butter milk and paid 30, assuming that the cost of one cup of butter milk to be 6. But the shop keeper told that the cost of butter milk had increased by 2. Then, Valli decided to give 2 more and paid 32. But the shop keeper claimed that she had to pay 40. Who is correct?

Valli calculated as = (5x6)+2 =30 + 2= 32

 

Shop keeper calculated as, = 5x (6+2)

                                           =5x8

                                           =40

The amount 40 claimed by the Shop keeper is correct. This confusion can be avoided by using the brackets in the correct places like 5 × (6 + 2).

The rule of order of operations is called as BIDMAS.

Expansion of BIDMAS

B

Bracket ( )

I

Indices (you will learn it later)

D

Division ÷ or /

M

Multiplication ×

A

Addition +

S

Subtraction −

 

 

          

 

 

 

 

 

Example: 5

Simplify: 24 + 2 × 8 ÷ 2 + 1

Solution:

Given, 24 + 2 × 8 ÷ 2 – 1                      (first we solve the division ÷)

                                     =24 + 2 × 4 − 1 (second we solve the multiplication)

                                     =24 + 8 – 1        (we solve the addition)

                                     =32 – 1               (then subtract)

                                      =31

Hence, 24 + 2 × 8 ÷ 2 + 1 = 31

                                               

Example: 6

Simplify: 20 + [8 × 2 + {  − 10 ÷ 5}]

Solution:

Given, 20 + [8 × 2 + { − 10 ÷ 5}]

                               =20 + [8 × 2 + {18 − 10 ÷ 5}]    (bar completed first)

                               =20 + [8 × 2 + {18 − 2}]            completed second)

                               =20 + [8 × 2 + 16]                      ({ } completed third)

                               =20 + [16 + 16]                           (× completed fourth)

                               =20 + 32                          ([ ] operation completed fifth)

                               =52                                  (+ completed last)

 

Hence, 20 + [8 × 2 + {  − 10 ÷ 5}] = 52

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