Prime Numbers in Grade V

Hello friends, today in math help session, we will learn about the few interesting topics of Grade V. algebra 1 (also you can play number sets algebra 1 worksheets) is taught to students of grade V and in this we learn several interesting topics. As you all know the building block of anything must be strong. If we have strong base then you can set up the strong building. The same case impels with mathematics if your basic concepts are good then no one can defeat you in the this subject. Math is all game of concepts and once you come up with all the concepts you become the master of the subject. In this article we will talk about the few interesting topics of mathematics that you study in Grade V. Let’s start with the very first topic that is, relationship between data. In this students learn how to determine the relation between the different information given in any problem.

What do you understand by Prime Numbers and Composite numbers? Start with prime numbers; it is defined as a number that can be evenly divided only by one and number itself. The other condition for this is that the number must be whole number and greater than one. In other words we can also define the prime number as a whole number that has only two factors in which one is number itself and other is one. On the other side a composite number is that which has other factors also in addition to one and itself. The two numbers zero and one are neither considered as prime number nor as composite number. Zero and one are out of the family of prime and composite numbers. All Even numbers are divisible by two and so set of all the even numbers greater than two. Composite numbers are that numbers which end with five and zero are divisible by five. The prime numbers between 2 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.
A positive integer is a composite number if it has a positive divisor other than one or itself. It can also be explained as any positive integer than one that is not on group of prime numbers. In mathematical language n>0 is an integer and there are integers 1< a, b<n in such a way that n = a X b, then ‘n’ is composite number. So, according to definition of prime numbers and composite number, every integer greater than one is either a prime number or a composite number.


For instance the number 14 is a composite number. How? Are you thinking this then relax I will explain how 14 is be a composite number, 14 is having one and itself as its factor. Apart from this, it has other factors also, 14 can also be written as: 7 X 2. The number 14 is now having four factord including 1, 2, 7 and 14 so we can easily say that 14 is a composite number. Now who will tell me what is 2? Ya friends, 2 is a prime number because it has only two factors one and itself. So, 2 is a prime number.

Now, take few more examples of composite and prime numbers so that you can easily understand the concept of the problem. Now, what is 5 a prime number or a composite number? 5 is a prime number because it has two factors only that includes: one and number itself. And in composite more than these factors should be present. Now one more example of this, is 25 a prime number or composite number? As 25 can be factored as 25 = 5 X 5. The 25 is having more than two factors so it is considered as composite number not a prime number. The number 25 is having 1, 25 and 5 as its factor, which is satisfying the condition of composite number.

Now, move to next topic that is “variable”, it is a “symbol” or “name” that stands for a value. For example: a + b, here ‘a’ and ‘b’ both are variables. In other terms we can define the same variable as a value that may change within the scope of a given problem or set of operations. In opposite of this a constant is used, which defines a constant value. A constant is a value that remains unchanged, in unknown and undetermined terms. Both these concepts are fundamental to many areas of mathematics and its different applications. Students let’s clear the example with help of previous studies that you had done. In the primary classes when teacher asks you to solve different questions of finding the unknown, what you use to do? You try to find the unknown value by applying different operations that are defined in the problem. Like, _+ 4 = 8,
In this problem you use to calculate the value of blank space by changing the position of constants like,
_ = 8 – 4,
_ = 4,
Thus in blank place we have to put 4, in order to get the answer,
4 + 4 = 8.
In the same way we use to solve the problems involving variables. In grade V the blank space is replaced with help of variables. The variables can easily be used in place of numeric value. In the same problem variable can be put as: x + 4 = 8,
In this blank is taken by variable x, and we can solve this same as of above problem.
Take constants on one side and variables on the other side. Generally, we use to place variables on left side and constants on right side. Whenever we change the position of constants and variables we used to change the sign of the constant and variables. Like if on side side any number is positive than on left side it is negative. In above example:
X + 4 = 8, take 4 on the left side and when we take 4 on other side we have to change the symbol of the 4 here it is positive on the other side it is negative.
X = 8 -4,
X = 4. Thus answer is 4. Or value of variable is 4. Now, let’s see few more examples of the same problem, like
2x + 4 = 14 – 4x,
In this problem we have one variable x, but it is defined on both the sides, so take the x values one side and constants on the other side of the equal sign. On doing so we get,
2x + 4x = 14- 4,
6x = 10,
X = 10/6,
X = 5/3
In this the value of variable x is equal to 5/3. One more example of the same type problem.
X + 2x + 3x – 4 + 8 – 5 = 0 in this, problem all the terms are given on the same side, no need to take tension follow the same strategy that I have explained in this article, i.e. of taking constants on one side and variables on the other side of the equal sign.
X + 2x + 3x = 5 +4 – 8,
6x = 9 -8,
6x = 1
And, x = 1/6. In this way you can simply calculate the value of the variable or variables by simply moving the numbers and constants from one side to other and performing operations on them.
Now, move to the other important topic of grade V i.e graphs. Graphs are the simple way of representation of data with the help of graph you and easily place the data on the graph and the other person can easily understand the problem. In the graph we have two axes one is defined as x- axis and other is defined as y- axis. The horizontal line is the x- axis and the vertical line is defined as y- axis. Whenever we have to plot the values we place the values on both the axes according to the requirement. Usually we divide the graph in 4 quadrants, the two are of x and two y. first is + x quadrant, next is + y quadrant, third is –x and the last is – y. With the help of graph you can easily understand the problem.
In grade V students learn the graphs of data management and probability, geometry and spatial geometry, number sense and numeration, pattering and algebra.
Kids it is quite difficult to draw the graph here, but I have given you an overview which will help you while solving this type of problems. In short I can only say that graph representation is the best way to solve any problem as in this we can easily determine the data that what is given to us and what we have to calculate.
This is all about all the three topics from my side, if you face any problem and you can ask your problems and I am here to answer all your queries and to help you in learning mathematics.

In upcoming posts we will discuss about Linear Functions in Grade V. Visit our website for information on 12th biology Maharashtra board syllabus

Solve fifth grade algebra using PEMDAS

Friends, today I am going to discuss about algebra solver, fifth standard mathematics topics like variables, equations etc. Students may understand how to perform basic math functions like subtraction, addition, division, decimals etc. But another possibility is that students might not know how to put all the functions together to solve a complex equation. If students start doing tasks in random manner or in improper manner then students can end up with completely incorrect results or answers. Whenever students opt to sole equations, they just remember the mnemonic device that is “ M-DAS,” which refers to Maths, Division, Addition and Subtraction. (you can also play solving equations with variables on both sides worksheet to improve your skills)

Fifth grade mathematics topics like equations, expressions etc. consists of the following parts : addition, subtraction, multiplication and division, and along with decimals and fractions. Starting from the basic what we need to do is to understand the problem and what steps required to solve a given problem. Few of the ways are:

First way is: Convert all of your decimals to fractions or convert all of your fractions to decimals, if the equation comes to solve contains a mix of both. If students want to convert fractions to decimals, what he/she needs to do is simply divide the numerator by the denominator. While converting decimals to fractions what requires is, ignore any zeros on the far right for example ignore zero in the given value 6.40. Now make a fraction with all of the numbers in the decimal over a number corresponding to the last decimal. Now the final step is to represent fraction in terms of tenths if the last value or number of the fraction is in the tenth place, this will vary according to place of the last number. For example
6.50 can be represented as 65/10 in the fraction form.

Second method: The second way is to evaluate or solve all the multiplication and division tasks. What we need to do is to mention product or quotient in place of the multiplication or division part of the equation.

Lets take an example to elaborate the above statement in the mathematical manner. In the equation
4 x 10 + 4.4 / 0.4 - 3.6 - 5,
First multiply 4 x 10 which give us 40. Then divide 4.4 by 0.4, getting 11.
After you replace "4 x 10" with "40" and "4.4 / 0.4" with "11,"
Your equation should read: 50 + 11 - 3.6 - 4.

Third way: Question is how to perform all the addition tasks. The most suitable manner is to replace the addition task with the acquired sum. Let’s take the same above example, the above formed equation can be further simplified as we all know that 50 + 11 = 61, so replace 50 + 11 with 61. Now the equation becomes 61 – 3.6 - 4

Fourth way: now the final task is to perform all subtractions. If the equation consists of more than one subtraction task, move from left to right. In the above example, first solve and perform subtraction task that is 51 – 3.6 which give us 47.4. Then subtract 4 from 47.4 , getting the final number that is 42.4.
Following the same example, first perform the subtraction task: 61 - 3.6, which equals 57.4. Then subtract 5 from 57.4, getting the final number, 52.4.

Variables are generally used to explain the general situations or real time situations and they can also be used to solve problems that in anyway would be very difficult or even impossible to solve. Whenever we are going to solve algebraic expressions or any kind of word problems, we will see the use of these applications.

Till now we have done expressions, playing with fractions, doing basic mathematical operations like addition, subtraction, multiplication and division. Now moving towards I am going to discuss about fifth grade Linear equations. In fifth grade linear functions or equations, variables are raised to power one and the equation does not contain any variable in denominator, no variables to any power other than one and no variable with under root sign.

In linear equation what we are doing is to solve the equation or getting a value of variable let’s take “x”. Solve for x means to isolate x on one side of the equation and move everything else on the other side.
X = number
Here number is the desired value which satisfies the original equation. That is when the number is substituted for the value x, the equation is true.
Let’s take an example to understand it better: in the given equation 2x + 5 = 1 we need to calculate the value for x then what we need to do is: we just need to place all the constants on one side and variables on the other side:
2x = 1 -5
2x = -4
x = -2.
Here x = -2 is the solution as we can check it in following way : 2(-2) + 5 = 1 is a true statement. For any other number, the equation or statement would be failed or results in false. For confirmation, if we mention the value of x = 3, the equation would be 2(4) + 5 = 1, which is false.
Another important thing to understand is that not every equation will have a solution. For example
x + 2 = x + 4 has no solution. It can be explained as that there is no number that can be added to two and be the same quantity as when it is added to 4. If student opt to solve for x, he/she would end up with the false statement 2 = 4.

Now I am going to discuss about how to solve an equation. In order to solve equations and to verify solutions whatever you get, first priority is that you must know the order of operations. The following order is required: Named as PEMDAS or we can say that BODMAS(B = Bracket, O = Off)

P—parentheses is the first to solve
E—exponents (and roots) are the second to perform
M—multiplication is the third
D—division third (multiplication and division should be done together, working from left to right)
A—addition is the fourth operations to be done.
S—subtraction is the fourth (addition and subtraction should be done together, working from left to right)

Another important thing is to understand the basic operations required to solve an equation. When performing with fractions you should always think of numerators and denominators as being in parentheses.
Eighth grade equations can be solved in many ways, but the general method to solve a fifth grade equation will be :
First: Both the sides of equation need to be simplified
Second: Collect all terms with variables in them on one side of the equation and all non-variable terms on the other (this is done by adding/subtracting terms).
Third: Factor out the variable.
Fourth: Finally what is required is division. Divide both the sides of the equation by the variable's coefficient.

Let’s take an example to understand it better: 2(a – 3) + 7 = 5a – 8
2a – 6 + 7 = 5a – 8
2a + 1 = 5a – 8
1 = 3a – 8
3a = 9
a = 9/3
a = 3.

Another method to solve an equation for the unknown is that we need to use inverse operations which help us to isolate the variable. These inverse operations undo all the operations that have been done to the variable. This shows that inverse operations are basically used to move quantities across the equal signs.

Let’s take an example to understand it better: In the given equation: 4x = 8
Here x is multiplied by the 4, so to move 4 across the equal sign, what we need to do is to un-multiply the 4. Therefore what is required is to divide both the sides of the equation by 4 or we can say that we can multiply each side of the equation by 1/4. In the above equation 4 + x = 8, if we want to move 4 across the equal sign or opposite side then what we need to do is to un-add 4. It can be simply stated as subtract 4 from both the sides of the equation.
In simple mathematical manner we can say that whatever we added  must be subtracted and whatever we subtract must be added. Similarly whatever we multiplied be divided and what is divided must be multiplied.

Now moving forward we all are going to learn, How can we simplify Fractions and using the Associative Property to solve Linear Equations problems. What we need to perform, find the LCD (Lowest common division) of all fractions and multiply both the sides of the equation by this number. Then, distribute this quantity on each side of the equation. In next chapter we are going to elaborate it.

In upcoming posts we will discuss about Syllabus of Grade Vth. Visit our website for information on West Bengal council of higher secondary education syllabus