Prime Factorization

Prime factorization is one of the most important concepts in mathematics. Before learning about prime factorization, let’s have a look at prime numbers and related concepts in this post. Prime number is a number that is exactly divisible by only 1 and itself. For example: Madhumita bought 5 action figures toys /b> for her children. Here, the number 5 in action figures toys is a prime number because it is divisible by only 1 and 5. 2, 3, 5, 7 etc. are examples of prime numbers. Now let’s move on to the concept of prime factorization hereafter.

Prime factorization is the process of finding the prime number multiplying which we can get a certain number. For example Sheena bought 8 brightest flashlight toys for her baby. She bought these brightest flashlight toys from online kids’ shop. Find the prime factors of 8: 8 = 2 * 2* 2. Here, 2 is the prime factor of 8. A number that doesn’t have any prime factors except itself is a prime number. An exact divisor of a number is called a factor. For example: Meenu bought 30 crib mobile toys for orphan kids. She bought these crib mobile toys in wholesale. Now, 30 is exactly divisible by 10 and therefore, 10 is a factor of 30. And 5 * 3 * 2 = 30, therefore, 5, 3 and 2 are the prime factors of 30.

Examples on Prime Factorization:
1. Maria bought 60 apples from the big market of the city. Find the prime factors of 60:
60 = 2 * 2* 3* 5
Here, 2 * 2 * 3 are the prime factors of 60
2. Rupa bought 18 chocolates from the corner shop to distribute among the kids. Find the prime factors of 18:
18 = 2 * 3 * 3
Here, 2 * 3* 3 are the prime factors of 18.
These are the basics about prime factors and prime factorization.

How to Use Algebra

Algebra:

Algebra word problems are very useful to answer real-life problems. Remember the famous words of Albert Einstein

Definition:

That branch of math which treats of the relations and properties of quantity by means of letters and other symbols.

A branch of mathematics in which signs, usually letters of the alphabet, represent numbers or members of a specified set and are used to represent quantities and to express general relationships that hold for all members of the set.


How to use Algebra:


An algebraic expression is every combination of variables and constants with mathematical operations (addition, subtraction, multiplication, division, roots and powers). Writing algebraic expressions is a method of determining and combining constants, variables, and math convention using the order of operations. The next shows how to put in writing algebraic expressions, with an example in parentheses.

Step 1:

Write down what the algebraic expression is to show. (A toy line has toys of 3 different sizes and both with a different price. Small costs $5, medium costs $10, and large costs $15. Write as an algebraic expression.)

Step 2:

Determine the constant(s). (5, 10, 15 for price of each size toy.)

Step 3:

Determine the variable(s). (x, y, z for the quantity of each size toy.)

Step 4:

Determine the operations that need to be done. (Multiply both prices by the quantity and find the sum total of all 3 sizes.)

Step 5:

Mingle the constants, variables, and operations using the conventions of the order of operations. This is the algebraic expression. (5x + 10y + 15z)

I have recently faced lot of problem while learning Division of Rational Numbers, But thank to online resources of math which helped me to learn myself easily on net.

Use Algebra expressions with detailed solutions:


Problem 1:

Answer the expression 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

-15x - 10 - x + 3 = -16x - 20 +13

-16x - 7 = -16x – 7

0 = 0

Problem 2:

Reduce the algebraic expression 2(a -3) + 4b - 2(a -b -3) + 5

Solution:

= 2(a -3) + 4b - 2(a -b -3) + 5

= 2a - 6 + 4b -2a + 2b + 6 + 5

= 6b + 5

Problem 3:

Reduce |x - 2| - 4|-6|

Solution:

= |x - 2| - 4|-6|

Alternate |x - 2| by -(x - 2) and |-6| by 6.

= |x - 2| - 4|-6| = -(x - 2) -4(6)

= -x -22

Problem 4:

Estimate f (2) - f (1), f(x) = 6x + 1

Solution:

f (2) - f (1) = (6*2 + 1) - (6*1 + 1)

= 6

Problem 5:

Answer the equation |-2x + 2| -3 = -3

Solution:

|-2x + 2| -3 = -3

|-2x + 2| = 0

x = 1

Finding Intercepts

Introduction for finding intercepts:

In coordinate geometry, the y-intercept is the y-value of the point where the graph of a function or relation intercepts the y-axis of the coordinate system. If the curve in question is given as y = f(x), the y-intercept is found by calculating f(0). Functions which are undefined at xy-intercept. Some 2-dimensional mathematical relationships such as circles, ellipses, and hyperbolas can have more than one y-intercept. Because functions associate x values to no more than one y value as part of their definition, they can have at most one y-intercept.                                                  Source – Wikipedia.



The general rules for finding intercepts:


Rule for finding X intercepts:

To find an x-intercept, put y = 0 in the equation.

Your x-intercept will be written as a point (x, 0).

Rule for finding Y intercepts:

To find a y-intercept, x = 0 in the equation. Your y-intercept will be written as a point (0, y).

There are two special types of lines are to be considered these are horizontal lines and vertical lines.

1. A horizontal line is in the form y = k; that is, the y-value is always a constant value.

2. A vertical line is in the form x = h; that is, the x-value is always a constant value.


Examples for finding X and Y intercepts:

Examples for finding X and Y intercepts are given below:

1) find the x Intercept in an equation y = 5x-1.

Solution:

Let the given equation is y = 5x-1.

We have to put y = 0 in the above equation

Y = 5x-1

0 = 5x-1

Add 1 on the both sides

0+1 = 5x-1+1

1 = 5x

1/5 = x

Therefore the x-intercept is (1/5, 0)

Example 2

Find the x intercept for x=2y+3

To find the x - intercept:

Solution:

We have to put y=0 in the above equation

X = 2y+3

X = 2(0) + 3

X = 0 + 3

X = 3

Therefore the x-intercept is (3, 0)

3. To find the y - intercept for y= 5x-1

Solution:

We have to put x = 0 in the above equation

Y = 5x-1

Y = 5(0)-1

Y = 0-1

Y = -1

Therefore the y-intercept is (0,-1).

4)find the y – intercept for x = 2y + 3

Solution:

We have to put x = 0 in the above equation

X = 2y+3

0 = 2y+3

-3 = 2y+3-3

-3 = 2y

Y = -3/2

Therefore the y-intercept is (0,-3/2).

Free Algebra Practice

Introduction to free algebra practice:

Algebra comes from the Arabic word al-gebra. It represents as the “reunion of the broken words”. In mathematics, Letters like a, b, c, x are been used to represent the numbers, values, complex numbers, integers and these unite with the general operations to form the algebraic expressions. If two algebraic expressions are equated we get the algebraic equations. In free algebra practice we shall discuss about the examples and practice problems.

Solved examples - free algebra practice:

Example problem: 1

Solve the two consecutive numbers have a sum of 81. find the  numbers?

Solution:

Let x = The first Consecutive Number

Since the numbers are consecutive, meaning one number comes right after the other, the second number must be one more than the first. So, x + 1 equals the second number.

Let x + 1 is a 2nd consecutive

The problem says that the sum of the two numbers is 81. This can be shown in the equation like the following:

x + (x + 1) = 81

The equation which you just wrote can be solved as follows:

Initial Equation

x + (x + 1) = 81

After combine like terms

2x + 1 = 81

After subtracting 1 from each side

2x = 80

After dividing each side by 2

x = 40

Answer : x = 40


Solving linear equations - free algebra practice:

Solve the following linear equations: x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11.

Solution:

x + 2y + 3z = 14 (1)

3x + y + 2z = 11 (2)

2x + 3y + z = 11 (3)

Consider the equations (1) and (3)

(1) ? x + 2y + 3z = 14

(3) × 3 ? 6x + 9y + 3z = 33 (subtracting)

–5x – 7y = –19

5x + 7y = 19 (4)

Consider the equations (2) and (3)

(2) ? 3x + y + 2z = 11

(3) × 2 ? 4x + 6y + 2z = 22 (subtracting)

–x – 5y = –11

x + 5y = 11 (5)

Consider the equations (4) and (5)

(4) ? 5x + 7y = 19

(5) × 5 ? 5x + 25y = 55 (subtracting)

–18y = –36

? y = 2

Substitute y = 2 in (5) we get

x + 5(2) = 11

x + 10 = 11

x + 10 - 10 = 11 - 10

? x = 1

Substitute x = 1, y = 2 in (3) we get

2(1) + 3(2) + z = 11

2 + 6 + z = 11

8 - 8 + z = 11 - 8

? z = 3

The solution is x = 1, y = 2, z = 3.

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Problems for practice - free algebra practice:

Problem 1:

The quadratic polynomial is divided by (x–1), (x+1) and (x–2) leaves the remainders 2, 4, 4 respectively, find the quadratic polynomial.

solution:The required quadratic polynomial is q(x) = x2 – x + 2.

Problem: 2

Solve the value of a and b if ax3 + bx2 + 7x + 9 and x3 + ax2 – 2x + b – 4 when divided by x + 2 leave remainders –13 and –16 respectively.

Solution: a –1, b = –4

Learning Practice of Statistics

Introduction to learning practice of statistics :

Statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data. Information is collected, presented and organized in the form of tables, graphs etc analyzed and then inferences are drawn from them. Statistical method is a technique used to obtain, analyze, summarise, compare and present the numerical data.Learning Statistics in the singular is a science which investigate the statistical methods and deals with their application. I like to share this Degrees of Freedom Statistics with you all through my article.


Definition for some statistical terms for learning practice of statistics

Data:

The collection of a particular type of information in the form of numerical figures is called, a set of data. This set of data obtained in the original form is called a set of raw data. Each numbered figure in the set of data is called an observation.

Ex: 35, 28, 26, 30, 32, 35, 26, 31, 36, 28

Array:

It is very difficult to draw any inference from this raw set of data. So we arrange it in ascending or descending order of size. The above set of data arranged in ascending order is:

Ex: 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 30 ,31.

Mean:

The arithmetic mean in statistics is the same as 'average' in arithmetic.

Mean of Ungrouped or Raw data:

The mean of a set of data is found out by dividing the sum of all the observations by the total number of observations in the date. We denote the mean by x' (read “x bar”).

Mean = sum of observations / number of observations

Median:

The median of a set of numbers is the middle number when all the numbers are arranged in order of size. ie. In descending or ascending order.

Mode:

The mode of a set of numbers is the number which occurs most frequently in the set. If no number occur more than once, the set of data Is said to have no mode. If different numbers occur the same number of times, the set of data has more than one mode. Having problem with how to long divide decimals keep reading my upcoming posts, i will try to help you.

Let us see some practice problems on these.


learning practice of statistics - Example problems


1) find the mean and median of the following number 39, 37, 38, 28, 30, 35, 36.

Solution:

Mean = 39+37+38+28+30+35+36 / 7

= 34.7

Median = 39, 38, 37, 36, 35, 30, 28

= 36

2) find the mode of the following number 52, 58, 58, 58, 65, 73, 73, 73.

Solution:

two modes: 58 and 73.

Maths Tricky Questions

Introduction to maths tricky questions

Math is a study of quantity. Here is an article that provides some tricky questions in math that will help you in cracking various placement tests as well as other class tests.Lets see some of the tricky questions on math and their solutions. I like to share this Solve Math Problems Free with you all through my article.

Solutions to maths tricky questions

1. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 years while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?

Solution: Let the ratio be k:1

Then, k * 16.4 + 1 * 15.4 = (k+1) * (15.8)

16.4 k + 15.4         = 15.8k + 15.8

16.4k - 15.8k        = 15.8 - 15.4

0.6k        = 0.4

k   = 0.4 / 0.6

or   k = `(2)/(3)`

So, required ratio = `(2)/(3)`  : 1 = 2 : 3  Answer

2. A sum of Rs. 1000 is lent to be returned in 11 monthly instalments of Rs. 100 each, interest being simple. What is the rate of interest?

Solution:    Rs. 100 + SI on Rs. 1000 for 11 months

= Rs. 1000 + SI on Rs. 100 for (1 + 2 + 3 + 4 + ..... + 10) months

= Rs. 1000 SI on Rs. 100 for `(110)/(2)`  months

= Rs 1000 + SI on Rs. 100 for 55 months

SI on Rs. 100 for 55 months = Rs. 100

So, Rate =`(100 * 100 * 12)/(100 * 55)`  %

= 21.82 % Answer

3. Ramlal is four times as old as his son. Four years later the sum of their ages will be 43 years. What is the present age of the son?

Solution: Let present age of son = x years

Present age of Ramlal = 4x years

Four years later:

Son's age = x + 4

Ramlal's age = 4x + 4

According to the given condition:

x + 4 + 4x + 4  = 43

5x + 8 = 43

5x= 43 - 8

5x= 35

x= `(35)/(5)`  = 7

So, present age of son = 7 years Answer

4. A man can row at 5 km/hr in still water and the velocity of the current is 1 km/hr. It takes him 1 hr to row to a place and back. How far is the place?

Solution: Distance = 1 x `(5^2 - 1^2)/(2 * 5)` 52 - 12 / 2 x 5

= `(24)/(10)`   = 2.4 km  Answer

Having problem with online math word problem solver keep reading my upcoming posts, i will try to help you.

Maths Tricky Questions (Continued)


Some more maths tricky questions with the solutions :-

5. Bucket P has thrice the capacity has bucket Q. It takes 60 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?

Solution: Let capacity of P be x litre

Then, capacity of Q = x/3 litre

capacity of drum = 60x litres

Required number of turns = `(60x)/(x + x/3)`

= `(3 * 60x)/(3x+x)`

= `(180x)/(4x)`  = 45 turns Answer

6. Rajan invests 65% of his intended investment in machinery and 20% on raw material in his business. If he has a balance of $6000 what was the intended investment?

Solution: Let intended investment be x

Investment spent on machinery = 65%

Investment spent on raw materials = 20%

So, total percentage of investment spent = 65% + 20% = 85%

Percentage of investment left = (100% = 85%) = 15%

According to the statement:

15% of x = 6000

`(15x)/(100)`   = 6000

15x = 6000 x 100

x = `(600000)/(5)`  = 40000

So, intended investment = $40000 Answer

7. By selling an article at $310 a merchant loses 7% on his outlay. Find the percentage of profit or loss when he sells the article at $350 ?

Solution: When article is sold at $310, percentage lost = 7%

So, remaining = 93%

Now, 93 : x   =  310 : 350

93 x 350 = 310 * x

x    = 105 %

i.e. Profit = (105 - 5) % = 5% Answer

8. Two trains each 300 m in length are running on the parallel linesin opposite directions with the speed of 70 km/hr and 50 km/hr respectively. In how much time will they cross each other completely?

Solution: t = `(300)/((70+50) * 5/18)`

= `(300)/((120) * 5/18)`

= 9 sec Answer

9. One year ago the ratio between father's and son's age was 4:1. The ratio of their ages after 4 years will be 3:1. Find the ratio of their ages after 15 years.

Solution: Let present age of father = x

present age of son = y

According to the statement:

x-1 : y -1  = 4:1

So, 4(y - 1) = x - 1

4y - 4  = x - 1

4y - x  = - 1 + 4

-x + 4y = 3 -----> (1)

Also, x + 4 : y + 4 = 3:1

So, 3 (y + 4) = x + 4

3y + 12  = x + 4

3y - x      = 4 - 12

-x + 3y     = -8  -------> (2)

Subtracting (2) from (1), we get,

4y - x - (-x + 3y) = 3 - (-8)

4y - x + x - 3y     = 11

y = 11

Substituting the value of y in equation (1)

-x + 4y = 3

-x + 44 = 3

-x = 3 - 44

- x = -41   or x = 41

So, present age of father = x = 41 years

present age of son = y    = 11 years

Note: If you require any more help in Maths Tricky Questions, our group of expert math tutors will help you in doing do.

rct math practice

Introduction to RCT math practice

Rct math practice includes aspects of Algebra, ratio, proportion, percent, rational numbers, probability, permutation, Statistics, geometry and Consumer Mathematics. In this article we shall discuss about some important rct math practice problem s. let us we discuss about algebra problems, permutation problem and combinations problems.


RCT math practice example problems

Example 1:

How many three-digit numbers can be formed by using the digits 1, 2, 3, 4, 5.

We have to determine the total number of three digit numbers formed by using the digits 1, 2, 3, 4, 5.

Clearly, the repetition of digits is allowed.

A three digit number has three places viz. units, ten’s and hundred’s. Unit’s place can be filled by any of the digits 1, 2, 3, 4, 5. So unit’s place can be filled in 5 ways.

Similarly, each one of the ten’s and hundred’s place can be filled in 5 ways.

Total number of required numbers

= 5 × 5 × 5 = 125

Example 2:

Solve for x:

6 x - 4 = 2 x – 10

Subtract 2x from both sides of the equation

6x – 2x – 4 = 2x – 2x – 10

4x – 4 = - 10

Add 4 to both sides of the equation

2x – 4 + 4 = - 12 + 4

2x = - 8

Divided by 2 both side of the equation

x = - 4

The answer x is – 4

Example 3:

If nC3 = nC6, find 12Cn

Solution:

nC4 = nC6   n = 3 + 6 = 9

Now 12Cn = 12C9

= 12C (12 ? 9) = 12C3

=12 × 11
1 × 2× 3

= 22

Example 4:

Find the number of different 4-letter words with or without meanings that can be formed from the letters of the word ‘NUMBER’

Solution:

There are 6 letters in the word ‘NUMBER’.

So, the number of 4-letter words

= the number of arrangements of 6 letters taken 4 at a time

= 6P4

= 360

Please express your views of this topic math tutoring for free by commenting on blog.

RCT math practice problems

Question 1:

In how many ways can 5 gentlemen and 5 ladies sit together at a round table, so that no two ladies may be together?

Answer:2880

Question 2:

How many different signals can be made by hoisting 6 differently colored flags one above the other, when any number of them may be hoisted at one time?

Answer:1956

Solving Joint Variation

Introduction to solving joint variation:

Joint variation is like a direct variation in this it has two or more variables. The variables which are in right side of the equal sign vary equally. The value of the dependent variable increases when the values of the independent variable increases and the value of the dependent variable decreases when the value of the independent variable decreases. Having problem with Type of Functions keep reading my upcoming posts, i will try to help you.

General Form – Solving Joint Variation:

The general form of the joint variation is y = kxz , where k is a constant and x and z are variables.

Example Problems – Solving Joint Variation:

Example 1 – Solving joint variation:

Identify whether the given expression a=kbc is joint variation with the values b=2, c=3 and b=4 and c=5 with k = 1.

Solution:

The given expression is `a= kbc` .

Now substitute the given values in the given expression

When `b` =2 and `c` = 3 then `a` = `1*2*3 ` `=>` `a` = `6` .

When `b` = 4 and `c` = 5 then `a` = `1*4*5` `=>` `a` = `20`

When the value of `b` and `c` jointly directly varies then the value of `a` also varies directly. Please express your views of this topic Least Common Multiple Definition by commenting on blog.

Example 2 – Solving joint variation:

Identify whether the given expression y=kxz is joint variation with the values x =5, z=6 and x=6 and z = 7 with k = 2.

Solution:

The given expression is `y=kxz.`

Now substitute the given values in the given expression

When `x` =5 and `z` = 6 then `y` = `2*5*6` `=>` ` y` = `60` .

When `x` = 6 and `z` = 7 then `y` = `2*6*7` `=>`` y` = `84`

When the value of x and z jointly directly varies then the value of y also varies directly.

Example 3 – Solving joint variation:

Find the value of constant for the equation a=kbc where a = 3, b=4 and c = 1.

Solution:

The given expression is `a = kbc`

Now the above expression can be written as `k` = `a/(bc)`

`k` = `a/(bc)`

`k ` = `3/(4*1)`

`k` = `3/4`

`k` = 0.75

The value of `k` is 0.75.

Example 4 – Solving joint variation:

Find the value of z for the equation y = kxz where k = 3, y=4 and x = 1.

Solution:

The given expression is `y = kxz`

Now the above expression can be written as `z` = `y/(kx)`

`z` = `y/(kx)`

`z` = `4/(3*1)`

`z ` = `4/3`

`z` = `1.33`

Problems in Esl

Introduction to problems in esl:

ESL stands for English as second language spots in math problems. Research shows that students who build up proficiency in math in their primary language will have greater achievement in becoming proficient in math in the second language. Whenever possible, it is a superior strategy to teach math skills in ESL students’ main language for their great success and at the same time that they are learning these concepts in English. Math problems while presented in English can be discussed in the home language in order to enlarge knowledge acquirement in the primary language. Having problem with Projectile Motion Calculator keep reading my upcoming posts, i will try to help you.

Problems in Esl - Example Problems::

The most important thing is to simplify the language, not the math, of the story problems. Use short simple sentences, rather than complex sentences by adding conjunctions.

Example 1:

Ben had 15 books and he gave 10 of them to his friend Tom.

Solution:

Ben had 15 books. He gave 10 to Tom.

Example 2:

John has 8 cookies and Steve has 12 cookies. How many more cookies does Steve have than John?

Solution:

John has 8 cookies. Tom has 12 cookies. Who has extra’s? How many extra’s?

Example 3:

A mathematical problem in math notation:

Solve for G:

G = S - 20

G + 5 = (S + 5)/2

might be presented in a word problem as follows:

George is twenty years younger than Sarah, and in five years' time he will be half her age. What is George age now?

Solution:

The solution to the word problem is that George is 15 years old. While the answer to the math problem is G = 15 (and S =35). Understanding Common Logarithm is always challenging for me but thanks to all math help websites to help me out.

Problems in Esl - Practice Problems::

Word problems commonly include math questions, where facts and information about a certain system is given and a student is required to develop a model.

Problem 1:

Daisy has $5 and she use $3 to buy some that she needs. How much amount does she have now?

Problem 2:

A mathematical problem in a math notation:

Solve for M:

M=6+b

M+4=2(b+4)

Find the value of m?

Michael is six years older than his brother.  He will be twice as old as his brother in 4 years. What is the age of Michael?

Solution:

1. Two dollars

2. Michael age is 12 and his brother age will be 6 in next 4 years.

Practice and Apply

Introduction to practice and apply:

In this article we are going to discuss about how to practice and how to apply. We can practice all the math problems at home of with the help of online and we can apply the same concepts to other same problems from what we have learnt from the tutors. We can see lot of chapters in math to practice. Let we see some practice problems to practice and apply.

Practice Problems to Learn and Apply

The following problems are used to learn the concepts and we can apply the concept to other same kind of problems.

Problem1) Solve for x , 2x+5y+12 = 42, where y =2

Solution:

Here we need to find the value of x.

Apply the y value in the above equation, we get

2x+5y+12 = 42

2x+5(2) +12 =42

2x +10 +12 = 42

2x +22 = 42

Add -22 on both sides, we get

2x +22 -22 = 42-22

2x = 20

Divide by 2 on both sides, we get

2x/ 2 = 20/2

x = 10.

The value of x =10

Problem2) Find the area of a circle whose radius is 9 cm

Solution:

Here we need to find the area of a circle

Area of circle = `pi` r2

Here pi = 3.14 , r =9cm

Apply those values in the formula,we get

= 3.14 *92

= 3.14*9*9

= 3.14*81

=254.34 cm2

Understanding The Practice of Statistics Answers is always challenging for me but thanks to all math help websites to help me out.

More Practice Problems to Learn and Apply:

Problem) Solve by order of operation method, A= 5(15/3)+12+(2*3)3-25

Solution:

Here we need to find the A value by using the method of order of operation.

Before going to solve the problem let you know the solving method of order of operation concept

PEMDAS => Parentheses,Exponents,Multiplication,Division,Addition,Subtraction.

Step1) Check whether the expression has any parenthesis

A =  5(15/3)+12+(2*3)3-25

Step2) Check for exponents, If there is no exponents , then go for multiplication.

= 5(5) +12 +(6) 3-25

= 25+12+18-25

Step3) IF there is no division part then for addition

= 25+12+18-45

= 55-45

Step4) Subtract the values.

= 55-45

= 10

The value of A =10.

Practice problems:

Problem1)  Find the area of a circle whose radius is 11 cm

Ans: Area = 379.94

Problem2) Find the value of S , 5S+28 = S+40

Ans: S= 3