Set the Formula, get the Formula is the chapter 12 maths of class 6. So, the class 6 last chapter is Set the Formula, get the Formula. Here, you can see the solution of the chapter Set the Formula, get the Formula. In addition to, this chapter is difficult than other chapter. So, we discuss about the Set the Formula, get the Formula in this article. Here, you get the all solution of Set the Formula, get the Formula. you read the article and get your solution of chapter 12. Here, you get the information about the all chapter solution of class 6 maths.

## Class 6 maths chapter 12

- The geometrical figures below are made up of lines of equal length.

a) Make the fourth diagram and find the number of lines.

b) Which mathematical formula or rule is satisfied by the number of lines,

explain with logic.

c) Find out the number of lines required to make the 1st 100 diagrams

A) The required number of lines in 4th figure = 21

B) In figure 1, the number of lines = 6 = 5×1+1

In figure 2, the number of lines =11 = 10+1= 5×2+1

So, In figure 3, the number of lines =16= 15+1=5×3+1

In figure 4, the number of lines =21= 20+1= 5×4+1

So, in the same way in the figure,

N number lines= 5×n+1= 5n +1

The required mathematical formula is = 5n +1

C) From “b” we get,

The algebraic formula is= 5n + 1

So, The number of lines required for

100 figure = 5 × 100 × 1 = 501

Now, the total number in 1st 100 figure

= 6+11+16+21+……….+501

Here, the 1st number = 6

The last number = 501

Number of terms = 100

So, the required total number of lines

= ( 1st number + last number) × Number of terms } ÷2

= (6 +501)×100/2

Now, = 507 × 50

= 25350

## Read more

CHAPTER 1 : The Story of Numbers

CHAPTER 2 : The Story of Two Dimensional Objects

Now, CHAPTER 3 : Information Investigation and Analysis

CHAPTER 4 : Trees of Prime Factors

CHAPTER 5 : Measurement of Length

Now, CHAPTER 6 : The World of Integers

CHAPTER 7 : The Game of Fractions part 1

The Game of Fractions : part 2

CHAPTER 8: World of Unknown expressions

CHAPTER 9: Linear Equation

Now, CHAPTER 10 : Story of Three Dimensional Objects

CHAPTER 11 : Unitary Method, Percentages and Ratio

Chapter 12 : Set the Formula, get the Formula

## Set the Formula, get the Formula

- Anowara Begum saves Tk 500 in the first month from her salary and in the

following month she saves Tk 100 more than the previous month.

a) Express, with explanation, the account of savings with a mathematical

formula or rule.

b) How much does she save on the 30th month?

c) What are her total savings in the first 3 years?

A) Solution:

Anowara Begum saves Tk 500 the first month from her salary and in the following month she saves Tk. 100 more than the previous month.

Now, How much

1″ month savings

= 500

= 100 × 1 + 400

2nd month savings

= 500+100

= 100 × 2 + 400

So, 3rd month savings

= 600 +100

= 100 × 3 + 400

4th month savings

= 700 +100

= 100 × 4 + 400

So, required the n month = 100 × n + 400

The required mathematical formula is 100n + 400

B) Solution :

Now, we get from a

Algebraic formula is = 100n + 400

So, Nth month savings = 100n + 400

And, 30th month savings = 100 × 30 +400

= 3000 + 400 = 3400

C) Solution

We know that,

1 year = 12 month

So, 3 year = 12 × 3 = 36 month

Nth month savings = 100n + 400

36th month savings = 100 × 36 + 400

= 3600 + 400 = 4000

Now, the total savings in 3 years or 36 months

= ( 500 + 600 + 700 + 800 +………………. + 4000)

The first months = 500

The last month = 600

Number of terms = 36

So, the required total

= ( 1st month + last month) × Number of terms } /2

= 500 + 4000 × 36/2

Now, = 4500 × 18

= 81000

## Set the Formula, get the Formula

- Aurobindu Chakma bought 3 yearly savings certificates with 3 monthly

interests, for 5 lac taka from his pension money. The rate of interest is 8% per

annum.

a) Find a Mathematical formula or rule, with explanation, to find the interest.

b) How much interest will he get in the first installment i.e. that is after the

first 3 months, use your formula to find that.

A) Solution

Let the principal is = p and rate of interest = r%

Interest of Tk 100 in 1 year is = r tk

Interest of Tk 1 in 1 year is = r/100

So, Interest of Tk P in 1 year is = r × P/100

Interest of Tk p in n year is = p×n×r /100

The formula of interest = pnr/100

B) Solution :

The principal p = 500000 tk and rate of interest r = 8%

Here, 12 month = 1 year

So, 1 month = 1/12

3 month =3/12 = ¼ month

So, time = ¼ month

## class 6 Math

Interest after 3 month I = 500000 × ¼ × 8/100

= 10000 tk

After 3 month he will get tk 10,000 tk.

- You are told to donate 100 kg rice. But you cannot donate the whole amount at

a time. On the 1st day you can donate half of 100 kg, i.e. 50 kg; on 2nd day can

donate half of 50 kg, i.e., 25 kg. In this way, every day you must donate half

the remaining rice. How many days will you take to donate the entire amount

of rice in this way?

[N.B. you cannot donate less than 1 kg in any way]

Solution :

Here, total rice = 100 kg

1st day donated = 100/2 = 50

2nd day donated = 50/2= 25

Now, 3rd day donated = 25/2= 12.5

4th day donated = 12.5 / 2 = 6.25

5th day donated = 6.25/2 = 3.125

Now, 6th day donated = 3 125 /2 = 1.5625

7th day donated = 1.5625/2 = 0.78126

The 7 day cannot be accepted because donation

Here, 7th day can not be expected because donation cannot be less than 1 kg in any way. So, he can donate for 6 days

It will take 6 days for him to donate the entire amount of rice in this way.

## Set the Formula, get the Formula

5) solution

A) From the figure we get,

The number of tiles at the bottom row = 20

So, The number of tiles at the 2nd row from the bottom = (20 – 1) = 19

The number of tiles at the 3rd row from the bottom = ( 18 – 1) = 17

The number of tiles at all the rows from the bottom,

S = 20 + 19 + 18 +17 +……. +10

That is, the total number of tiles is the sum of the numbers from 10 to 20

There are total = ( 20 -10) +1 = 11 numbers from 10 to 20

So, number of total tiles

= sum of total tiles in 1st and Last row /2 × number of total rows

= ( 20 + 10) ×11 /2

Now, = 30 ×11 /2

= 15 × 11

=165

b) From” a” number of total tiles = 165

Area of 1 tile

= 12 square inches

= 1 square feet ( 12 inches = 1 feet)

So, area of 165 tiles = ( 1×165) square feet = 165 square feet

The cost for 1 square feet of tiles is 75 tk

So, The cost for 165 square feet of tiles is 75 ×165 = 12375 tk

## Chapter 12 maths class 6

6)

a) How many bricks will be there at the topmost step?

b) Express the process of brick arrangements using a Mathematical formula or

rule, with logical explanations.

c) How many bricks has he arranged?

A) Solution

Number of bricks in each row in the first step or lowest step = 30

Number of bricks in each row in the 2nd step is = 28

= ( 30 – 2)

= 30 – ( 2-1)× 2

Number of bricks in each row in the 3rd step = 26

=(30-4)

=30-(3-1)×2

Number of bricks in each row in 4th step is = 24

= (30-6)

= 30-(4-1)×2

Similarly, the number of bricks in each row in the 15th step

=30-(15-1)×2

=30-28

=2

Since there are two rows per step,

So number of bricks at the top most step is = 2×2=4

b) Number of bricks in each row first to last step = 30

Number of bricks in each row in the 2nd step is = 28

= 30 – ( 2 -1) × 2

Number of bricks in each row in the 15th step

= 30 – ( 15 -1) ×2

= 30 – 28 = 2

Number of bricks in each row in the nth step is

= 30- (n -1) ×2

= 30 – (2n – 2)

So, = 30 – 2n + 2

= 32 -2n

Since, there are 2 rows per step

So, Number of bricks in each row in n step is = 2×( 32 -2n)

= 64 – 4n

So, the mathematical formula two process of bricks arrangements is 64 – 4n, where n is the number of steps.

## Class 6 maths

c)

Number of bricks in two rows of the lowest steps = 2×30=60

So, Number of bricks in two rows of 2nd step is = 2× ( 30 -2) = 56

Number of bricks in two rows of 3nd step is = 2 × ( 28 – 2) = 52

So, Number of bricks in two rows of 4nd step is = 2 × ( 26 -2)= 48

Number of bricks in two rows of 15 step is = 2 × 2 =4

So, total number of bricks in all step is, S = 60 + 56 + 52 + 48+……… +4

Number of bricks 15th steps = 1st number + last number /2 × number of steps

= 60 + 4 /2 × 15 = 480

The Total number of bricks = 480

## Chapter 12 class 6 maths

7) solution

B) Diagram 1 number of tiles 1

Diagram 2 number of tiles 5

So, Diagram 3 number of tiles 9

Diagram 4 number of tiles 13

So, Diagram 5 number of tiles 17

Diagram 6 number of tiles 21

Diagram 7 number of tiles 25

So, Diagram 8 number of tiles 29

Diagram 9 number of tiles 33

Diagram 10 number of tiles 37

c) In diagram 1, the Number of tiles = 1

diagram 2, the Number of tiles = 5 = 4+1= (2-1)×4 +1

diagram 3, the Number of tiles = 9= 2 × 4 +1 = ( 3-1) × 4 +1

So, Diagram 4 number of tiles 13 = 3 × 4 +1 = ( 4 -1) × 4 +1

Diagram n number of tiles (n -1) × 4 +1 = 4n – 4 + 1 = 4n – 3

So, the required formula to find the number of tiles is 4n – 3,

Where, n is the number of Diagrams.

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