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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