Regrouping Math Problems

Introduction to Regrouping problems:

Regrouping problems are important problems in math.  These are mainly used in addition and subtraction problems. Regrouping and no regrouping is the main concept when we can add or subtract the numbers. In this topic we have to discuss how we can use the regrouping concepts in addition and subtraction problems.

Brief Explanation about Regrouping Addition Method in Math

Brief Explanation about regrouping addition method in math:

This involves the following steps.  They are,

Now we can add the two digit numbers
First we can add the unit digit values.
Suppose the sum of the unit digit is less than 9 we can keep the number as it is and then add the ten’s digit place values. This method is called as no regrouping.
Suppose the sum of the unit digit is greater than 9 we can keep the unit digit place number as it is and then take the ten’s digit place value as the remainder. This remainder value is added with ten’s digit place values. This method is called as regrouping.
This procedure can be followed when the sum of three digit and four digit numbers.

Brief Explanation about Regrouping Subtraction Method in Math

This involves the following steps.  They are,

Now we can subtract the two digit numbers
First we can subtract the unit digit values.
Suppose the unit digit place value of the largest number is greater than the another unit digit place value then we can directly subtract the unit digit places and then ten’s digit places. This is known as no regrouping.
Suppose the unit digit place value of the largest number is less than another unit digit place value then we can’t able to directly subtract.  For this we can borrow the value ten from the neighbor unit and then we can subtract the unit digit and ten’s digit place values. This method is called as regrouping.
This procedure can be followed when the sum of three digit and four digit numbers.

Regrouping Problems in Math

Problem 1:

Simplify 45 + 49 -65

Solution:

First we can consider 45 + 49

Here the sum of unit digit place value is 5 + 9 = 14

Now keep 4 as it is and then takes the remainder value is 1

Now we can add the ten’s digit place value is 4 + 4 =8

Now we can 8 to the remainder value 1.

It gives the ten’s digit place value is 9.

Therefore the addition solution is 94

Now consider 94-65

Here unit digit 4 is less than 5. So we can borrow 10 to neighbor digit 9.

Now the unit digit value is 14 -5 =9

Then ten’s digit be changed as 8-6 =2

Therefore the solution is 29.

Problem 2:

Simplify 98 +82-18

Solution:

These are the important regrouping problems in math.

Random Number Sample

Introduction to random number sample :

To understand random number sample by substituting the following definitions which has to be understood  in a large collection of individuals or numerical data is called a random number sample.  A population may possibly finite or infinite. For example: number of patients required admission in a hospital in the year 2009 is finite and the population of all possible outcomes heads or tails in successive or unlimited tosses of a coin is infinite. A finite subset of the population is called a random number sample. The selection of an item from the population in such a way that each has the same chance of being preferred is called random number sample.

Random Number Sample:

Examples:     

The process of selecting a sample from the population is called random number sample. In random number sampling with replacement, the things are drawn one by one and are locate back to the population previous to the next draw.  A finite subset of the population is called a random number sample. The process of selecting a sample from the population is called random number sampling The selection of an item from the population in such a way that each has the same chance of being selected is called as random number sampling,  In random number sampling with replacement , the substance are drawn one by one and are put back to the population before the next draw.

Example 1:

Find the probability of choosing an individual for factory from a team of 200 labors where men and women are equal in number if 50 labor where taken?

Solution:

Total number of student is 200

Here a group of 50 students were taken to sports

So the probability of receiving a chance = `50/200`

= `1/4`

Probability of the person = `1/4`

For every men and women receiving a chance is `1/4` .

Random Number Sample:

Examples for random number sample:

Consider a finite population and assign numbers to each member of the population, write these numbers on small pieces of cakes, place them in an box and draw numbers from the box, we comprise a choice of replacing or not replacing the number into the box before the second draw. If we put back the number before the second draw, the number can come up again and again

Another definition for random number sample, a large collection of individuals or numerical data is called random number sample  A population may be finite or infinite. For example: number of workers required  in a company in the year 1999 is finite and the population of all possible outcomes heads or tails in following or unlimited tosses of a coin is infinite. A finite subset of the population is called a sample.

Working with Negative Exponents

Introduction to working with negative exponents:

Negative exponents are one of the basis of mathematics in the exponents function. The power terms are also called as the negative exponents. If any of the function is having the negative power means, then we have to write that function in the fraction format. For example, x-1 is said to be negative exponents. Then this function can also be written as `1/x` .

Rules for Working with Negative Exponents
There are number of rules are followed for working with negative exponents. They are given as,

In the multiplication function of two numbers are having the exponents terms means, then we have to add the power given in the multiplicative functions. The rule is given below  the following,
Example: (x2) (x4), then we have to write this as (x2+4).

If the function having one power over the other power terms means, then we can also write this by using the multiplication function. The rule is given below the following,
Example: (x2 )3, then we have to write this as ( x2*3).

In the division format if the exponent function is given means, then we can write this for both the numerator function and the denominator function. The rule is given below the following,
Example:  `(x/y)^2` , this can also be written as .`(x^2)/(y^2)`


Example Problem for Working with Negative Exponents

Problem 1: Work the given problem of negative exponents, -35.

Solution:

- 35 = (- 3) `xx` (- 3) `xx`  (- 3) `xx`  (- 3)  `xx`  (- 3) 

= - 243

This is the required solution of working with negative exponents function.

Problem 2: Work the given problem of negative exponents, -46.

Solution:

- 46 = (- 4) `xx` (- 4) `xx`  (- 4) `xx`  (- 4)  `xx`  (- 4) `xx` (- 4) 

= - 4096

This is the required solution of working with negative exponents function.

Practice Problem for Working with Negative Exponents

Problem 1: Work the given problem of negative exponents, -24.

Answer: - 16

Problem 2: Work the given problem of negative exponents, -53.

Answer: - 125

Vector Line Integrals

vector line integrals:

In integral calculus, we have defined indefinite integral, definite integral, line integrals, surface integral and volume integral for a real valued function. If `vecf` (t) and `vecF` (t) be two vector valued functions, such that `d/dt` `vecF` (t) = `vecf` (t) then `vecF` (t) is called the integral of `vecf` (t) with respect to t and this is denoted in symbol by `int vecf` (t)dt = `vecF` (t). In general if `d/dt[vecF(t) + vecc] = vecf(t) ` where `vec c` is  an orbitrary vector independent of t, then `int vecf(t)dt = vecF(t) + vec c`

Here, `vecF` (t) is called indefinite integral of `vecf` (t) and `vecc` is an orbitrary constant of integration.  The definite integral for vector valued function is `int_a^bvecf(t)dt`  = `[ vecF(t)+vecc]^b_a`  = `vec F` (b) -` vecF` (a)

Explanation of Vector Line Integral:

Any integral which is to be evaluated along a curve is called a line integral . Let `vecF (t) = F_1veci + F_2 vecj + F_3 veck ` be a vector point function defined along a curve C.  Let `vecr = xvec i + y vecj + z veck` be the position vector of any point on this curve. Let the arc length along this curve be measured from a fixed point A. If s denotes the arc length from A to any point P(x, y, z) we know that `(dvecr)/(ds)`  = vect is a unit vector. along the tangent to the curve at P. The component of `vecF` along the tangent given by `vecF` `(dvecr)/(ds)`. The integral of this component along C measured from the point A to the point B is given by  `int_A^B vecF`  `(dvecr)/(ds)` ds. This integral is called the line integral of `vecF` along C. This integral is also called the tangential line integral of `vecF` along C.

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Scalar function:

The scalar function of line integral  is `int_c( vecF. (dvecr)/(ds))ds = int_c vecF. dvecr`

Note 1: if `vecF (t) = F_1veci + F_2 vecj + F_3 veck `

`vecr = xvec i + y vecj + z veck`

`dvecr = dx veci+dy vecj+dzvec k`

`vecF.dvecr = F_1dx+F_2dy+F_3dz`

So `int_c vecF.dvecr` = `int_c ` (F1dx + F2dy + F3 dz)

Note 2: if the equation of the curve is given in parametric form say x = x(t), y = y(t) and z = z(t) and the parametric values at A and B are t = t1 and t = t2 then

` int_c vecF. dvecr = int_(t_1)^(t_2)(F_1(dx)/(dt) + F_2 (dy)/(dt) + F_3 (dz)/(dt)) dt`

Application of Vector Line Integral;

F is a force acting upon a particle which moves along a curve C in space and r be the position vector of the particle at a point on C. Then work done by the particle at C is F.dr and the total work done by F in the displacement along a curve C is given by the line integral `int_c F.dr`

Solving Calculas Problems

Introduction to solving calculus problems:

Solving calculus problem is one of the great achievement in mathematics which concentrate on the functions, limits, integrals, derivatives, and infinite series. Solving calculus problem has two branches,the first one is differentiation and the second one is integration. The differential solving calculus problem helps us to locate out the rate of change of an amount wherever as the Integral solving calculus problem helps to find out the quantity where the rate of change is known.

Solving Calculus Problems-sample Problems:

Problem 1:

Solving the following calculus problem for the derivative of f(x) = 3x 3

Solution:

let c = 3

and g(x) = x 3,

then f '(x) = c g '(x)

= 3 (3x 2)

= 9 x 2 

Solving Calculus Problems-problem 2:

If F(x) is an anti derivative of f(x), then
(1/a) F(ax) is an anti-derivative of f(ax). True or false.

Solution:

Let u = a x and Differentiate (1/a) F(ax) w.r. to x

d/dx( (1 / a) F(a x) )

= (1 / a) d(u) / dx dF/dU

= (1/a) a f(u)

= f(a x)

Hence the given statement is True.

Problem 3:

The sum of two non negative numbers is 9 and so that the maximum number is the product of one number and the square of the other. Find those two non negative number.

Solution:

The sum of the two variables x and y is given to be 9 = x + y ,

so that, y = 9 - x

Maximize the product,

P = x y2 .

Substitute for y, we get

P = x y2

= x ( 9-x )2 .

On differentiating the above function, we get

P' = x (2) ( 9-x)(-1) + (1) ( 9-x)2

= ( 9-x) [ -2x + ( 9-x) ]

= ( 9-x) [ 9-3x ]

= ( 9-x) (3)[ 3-x ]

= 0 [for x=9 or x=3]

Note that since both x and y are non-negative numbers and their sum is 9, it follows the order that 0<= x <= 9.

If x=3 and y=6 ,

then P= 108

which is the largest possible product.

Ways of Analyzing Data

Introduction to ways of analyzing data:

The analysis of data is the process of  inspecting and modelling the data which is used to highlight the useful information.and used to suggest some conclusion and it used to help in decision making.the analysis of data includes multiple methods to collect data and it goes through several phases of  analysing.

The pre-stage of analysis data integration..The data analysis includes Exploratory and confirmatory data analysis.

Ways of Analyzing Data-process of Data Analysis

We can distinguish the data analysis in to ,

Data cleaning
Initial data analysis
Main data analysis
Final data analysis
Data cleaning:

It is the process of cleaning the data,that is the process of including the preferable data and avoiding errors in the data.,and if possible correcting the data.This process can be done in the stage of data entry itself.

Initial data analysis:

In this stage we are refraining the data depends up on whether the data is having ht equality enough,depends upon the quality of measurements.

And it is in this stage ,checks for the missing data.and checked for any data whether it disturbs  the collected data.

For the analysis of data uni variate and bi variate ,and graphical methods are used ,to see whether the data collected are relevant.

And checks for whether the intension of the data collections are met,with the data.

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Ways of Analyzing Data-final Stage

For the final analysis of data computer simulations and data envelopment strategies are used.

For the final analysis of data some decision support systems are also used .For which some separate decision theory is also there.

As per the part of the final analysis some influence diagrams are made to view a clear picture for the datas going to be documented.

As a last step of analysing the data,the pareto analysis and the stakeholder analysis is also done.

In this step ,the stakeholders are examines whether the data collected are relevant and whether it will meet the objective of the datacollection.

During the final stage the datas are documented.

Pre Algebra B Calculator

Introduction to pre algebra b calculator:

Pre algebra is a branch of mathematics. Pre algebra plays an vital role in our day to day life. Pre algebra b calculator will do the four basic operations such as addition, subtraction, multiplication and division. The most important terms are variables, constant, coefficients, exponents, terms and expressions are handled by pre algebra b calculator. By pre algebra b calculator, we will know the use the symbols and alphabets in the place of unknown values, to form a statement. Hence, pre algebra b calculator gives the leads of Arithmetic. Therefore, students are getting pre algebra b calculator for their studies.

Examples by Pre Algebra Calculator:

Example 1:

Solve the equation x + 22 = 235 for x.

Solution:

x + 22 = 235 ( Now we have to add -22 on both sides. So we get,)

x + 22 –22 = 235 – 22

x = 213

Example 2:

Solve the equation x – 29 = 240 for x.

Solution:

x – 29 = 240 (Now we have to add 29 on both sides. So we get,)

x – 29 + 29 = 240 + 29

x = 269

Example 3:

Solve the equation 24x = 232 for x.

Solution:

24x = 234 (Now we have to divide both sides by 24. So we get,)

`(24x)/24` = `234/24`

x = 9.75

Example 4:

Solve the equation x ÷ 25 = 49 for x.

Solution:

x ÷ 25 = 49 ( This statement can be written as below.)

x/25= 49 ( Now we have to multiply both sides by 25, so we get)

`x/25` × 25 = 49 × 25

x = 1225

Example 5:

Solve the equation (x `xx` 23) – 22 = 25 for x.

Solution:

(x `xx` 23) – 22 = 25

( First we have to evaluate the expressions within the parenthesis. So, x `xx` 23 becomes 23x. Therefore, the given equation becomes like below.)

23x - 22 = 25 ( add 23 on both sides. So we get,)

23x - 22 + 22 = 25 + 22

23x = 25 ( divide both sides by 23)

`(23x)/23` = `25/23`

x = 1.09

Algebra is widely used in day to day activities watch out for my forthcoming posts on algebraic expressions and algebra 1. I am sure they will be helpful.

Practice Problems to Pre Algebra B Calculator:

Problem 1:

Solve the equation x + 23 = 28 for x.

Solution is, x = 5

Problem 2:

Solve the equation x - 23 = 29 for x.

Solution is, x = 52

Problem 3:

Solve the equation 3x = 242 for x.

Solution is, x = 80.67

Problem 4:

Solve the equation x ÷ 2 = 252 for x.

Solution is, x = 504

Steps to Solving Percentages

Introduction to solving percentages:

Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. Percentage was can represent the number in the form of percentage (%). For example 5% mean it was can written like as 5/100=0.05.

(Source –Wikipedia)

In this article steps to solving percentages, we see about some percentages example problems with detail steps.

I like to share this Adding Percentages with you all through my article.

Example Problem for Solving Percentages with Steps:

Example problem for solving percentages with steps:

Example problem 1:

solving 24 percentages of 120?

Solution: Steps for fining percentages

Step 1: Given 24% of 120

Step 2: 24% it can written like as =>24/100

Step 3: 24% of 120=> 24/100*120.here we have to multiply 24/100 and 120

Step 4: 0.24(120)

Step 5:  24 percentages of 120 is 28.8



Example problem 2:

What percents of 43 is 9?

Solution:

Step 1: Given x% of 43=9

Step 2: Here Find the value of x

Step 3: x% of 43=9.It can written as like  x/100(43)=9

Step 4: x/100*43=9

Step 5: X=9*(100/43)

Step 6: X=900/35

Step 7: X=22.71%

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Example problem 3: Solving decreased percentages problem

A mobile was last year sold at the rate of  35000.now current year glass was sold at 2100 rate .solving the reduced (amount)percentage of mobile?

Solution:

Step 1: Given data:

Price of mobile (old price) =35000

Current price of mobile is (new price) =21000

Step 2: Reduced amount=old price of mobile – new price of mobile

Step 3: Reduced price= 35000-21000

Reduced price=14000

Step 4: Reduced percentage= (14000/35000)*100=40%

Step 5: 40% of the amount was reduced in the New Year.

Example Problem for Solving Percentages with Steps:

Example problem 4:

A glass shop pays some amount $70 to client, and then sells glass at the price of $120.Find the percentage of markup rate?

Solution:

Step 1:Given data paid amount =$70

Sales price =$120

Step 2: First we have to find the absolute amount( difference)

Markup amount=120-70=50

Step 3: Let us consider as x is percentage (unknown value)

Step 4: Now we have to find the x% of 70 is 50

Step 5: That amount is original markup rate of glass

X% of 70=>50

Step 6: x/100*70=>50

Step 7: x=50* (100/70)

Step 8: x=71.42%

Percentage of mark rate= 71.42%

Two Equations Two Unknowns

Introduction to two equations and two unknowns:

Two linear equations in the same two variables (unknowns) are called a pair of linear equations in two variables. The most general form of a pair of linear equations is

a1x+b1y+c1=0

a2x+b2y+c2=0

An example of linear system involves two equations and two unknowns:

x+ y=3 and x - y=2

There are the methods used to solve two unknowns with two equations:

Substitution method

Elimination method

Graphing method

Here, we are going to see the problems on two equations and two unknowns by substitution and elimination method.

Two Equations Two Unknowns-solving

Example problem 1:

Solve for the two variables x and y from the following two equations:

3x+y=5
y+5x=2

Solution:

Here, we have to solve the pair of equations with two unknowns by substitution method.

Step 1: We pick any one of the equations and write one variable in terms of the other.

Let us consider the Equation (1):

3x+y=5

Subtract 3x on both sides of the equation

3x+y-3x=5-3x

y=5-3x------------------------Equation (3)

Step 2: Substitute the value of y in Equation (2). We get

y+5x=2

5-3x+5x=2

5+2x=2

Subtract 5 on both sides

2x=2-5

2x=-3

Divide by 2 on both sides of the equation

2x/2=-3/2

x=-1.5

Step 3: Plugging this value of x in Equation (3), we get

y=5-3x

y=5-3(-1.5)

y=5+4.5

y=9.5

So, the solution of two unknowns is (-1.5, 9.5).

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Two Equations Two Unknowns- by Elimination Method:

Example problem 2: Solve for the two variables x and y from the following two equations:

9x – 4y = 2000----------Equation (1)

7x – 3y = 2000----------Equation (2)

Solution:

We have to solve the pair of equations by Elimination method.

Step 1: Equation (1) is multiplied by 3 and Equation (2)is multiplied by 4 to make the coefficients of y equal. Then we get the equations:

27x – 12y = 6000------Equation (3)

28x – 12y = 8000------Equation (4)

Step 2: Equation (3) is subtracted from Equation (4) to eliminate y, because the coefficients

of y are the same. So, we get

(28x – 27x) – (12y – 12y) = 8000 – 6000

i.e., x = 2000

Step 3: Substituting this value of x in (1), we get

9(2000) – 4y = 2000

i.e., y = 4000

So, the solution of two unknowns is (2000, 4000).