Do Basic Division

Introduction to do basic division:

Division is one of the arithmetic operations. Division is the inverse form of multiplication. The term division is sometimes said to be sharing. Division is denoted by the symbol `-:`. For example: When M is divided by N.

M/N = O

M – Dividend.
N – Divisor.
O – Quotient.
For example:

The value 120 divided by12.

120/12 = 10

In this,

120 – Dividend.

12 – Divisor.

10 – Quotient.


Do basic division


In this division there are divided, divisor, remainder and quotient will appear in the basic way. It is very easy to solve. This division is very easily understandable by the kids. Let see some examples of the basic division and do some practice problems for this division method.

Example problems for basic division:

Calculate the value 245 is divided by 15.

Solution:

Steps to solve the given problem

First step:

To find the number of times do 15 go into the 240?

Second step:

24 is divided by 15

24/15 = 1

The remainder is 9

Third step:

Then the remainder is put in front of the value 0.

90 is divided by 15

90/15 = 6.

The remainder is said to be zero (0).

Solution:

15) 240 (16

15

_______

90

90

_______

0

________

Divisor = 15

Dividend = 240

Quotient = 16

Remainder = 0

Another example for basic division:

The value 418 is divided by 16.

Dividend = 418.

Divisor = 16.

Solution:

16) 418 (24

32

_________

99

90

__________

9

_________

Answer:

Quotient = 24

Remainder = 9.

Another example for basic division:

The value 16724 is divided by 12.

The dividend = 16724

The divisor = 12.

Solution:

12) 16732 (1311

12

_______

47

36

_______

13

12

_________

12

12

_________

0

__________

Answer:

Quotient = 1311

Remainder = 0.


Some practice problems for the basic division

Problem 1:

Find out value 450 is divided by 5.

Answer:

The correct answer is 90.

Problem 2:

Find the value of 43978 is divided by 14.

Answer:

The correct answer is

Quotient = 3141

Remainder = 4

Problem 3:

Find the value of 12450 is divided by 10

Answer:

The correct answer is

Quotient = 1245

Remainder = 0.

Problem 4:

Find the value of 29726 is divided by 2.

Answer:

The correct answer is

Quotient = 14863

Remainder = 0

Area Math 4th Grade

Introduction to 4th grade math area:

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron. Area is an important invariant in the differential geometry of surfaces. In this article we shall discuss about 4th grade math area problem. (Source: Wikipedia)

4th grade math area example problem

Here we are going to discuss some 4th grade math area problems with detailed solutions.

Example 1:

Find the area of the rectangle, the rectangle base value 23m and height of rectangle is 10 m.

Solution:

Let base b = 23 and rectangle height h = 10.

Then the area of the rectangle formula = b × h

The area of the rectangle = 23 × 10

= 230 sq. meters.

Therefore the area of rectangle = 230 sq. meters.

Example 2:

Find the area of the circle, the circle radius value 12.

Solution:

Area of the circle formula = pi r2

Pi = 22/7 or 3.14

The area of the circle = 22/7× 12x12

= 452

Answer:

Therefore the area of circle = 452

Example 3:

Find the area of the square; the square one side of length value is 22units.

Solution:

Formula for finding area of square = a2 of (Side) 2

The side value of the area = 22

Therefore (22) 2

That is 22 * 22 = 484

Area of the square = 484units.

Example 4:

Find the area of triangle, the triangle base value=20cm and the height of the triangle = 12cm

Solution:

Area of the triangle formula = `1/2` b x h

Height of the triangle = 12

Base of the triangle = 20

Therefore area of triangle = `1/2` x 20 x 12

The area of triangle = 120cm

Example 5:

Find the area of the rectangle, the rectangle base value 22m and height of rectangle is 15 m.

Solution:

Let base b = 22 and rectangle height h = 15.

Then the area of the rectangle formula = b × h

The area of the rectangle = 22 × 15

= 330 sq. meters.

Therefore the area of rectangle = 330 sq. meters.

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4th grade math area practice problem

Problem 1:

Find the area of the rectangle, the rectangle base value 20m and height of rectangle is 10 m.

Answer:

The area of rectangle = 200 sq. meters.

Problem 2:

Find the area of the square; the square one side of length value is 20 units.

Answer:

Area of square = 400 units

Practice Test for Basic Math

Introduction to mathematics:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. There is debate over whether mathematical objects such as numbers and points exist naturally or are human creations. (Source: Wikipedia)

Example problems of practice test for basic math

Basic math test example problem 1:

Add the given two values 35 and 84.

Solution:

Given numbers are 35 and 84

(35 + 84) = 35
84  ( + )
____
119
____

Answer:

The final answer is 119

Basic math test example problem 2:

Find the area of the triangle with the base length is 20 m and its height is 12 m.

Solution:

Given base length (b) = 20 m and height (h) = 12 m

Area of the triangle = `(1 / 2)` * (base length) * (height)

= `(1 / 2)` * 20 m * 12 m

= 120 m^2

Answer:

The final answer is 120 m^2

Basic math test example problem 3:

The sum of the two numbers is 23. Smaller number is three less than that of larger number. Find out the two number values.

Solution:

Let us consider,

x is the larger number and y is the smaller number

Given, sum of the two numbers is 23

x + y = 23 ------- (1)

Smaller number is three less than larger number, we get

y = x - 3 --------- (2)

Substitute equation 2 in equation 1, we get

2x - 3 = 23

After simplifying, we get

x = 13

Substitute the value of x in equation 2, we get

y = 10

Finally, that two numbers are x = 13, y = 10

Answer:

The final answer is x = 13, y = `10`

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Practice problems of practice test for basic math

Basic math test practice problem 1:

Subtract 45 and 34

Answer:

The final answer is 11

Basic math test practice problem 2:

Multiply the given values 31 and 10

Answer:

The final answer is 310

Basic math test practice problem 3:

Solve: x = 2x - 7

Answer:

The final answer is x = 7

Basic math test practice problem 4:

Solve:

3y - 10 = 2

Answer:

The final answer is y = 4

Basic Algebra Calculations

Introduction to basic algebra calculations:

Basic algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. The part of basic algebra calculations called elementary algebra calculations is often part of the curriculum in secondary education and introduces the concept of variables representing numbers. The variables are manipulated using the rules of operations that apply to numbers, such as addition. (Source: Wikipedia).

Examples for basic algebra calculations:

Example 1 for basic algebra calculations:

Find the value for x in (x+2) +(x-22) +(x+45) =0.

Solution:

The given expression is(x+2) +(x-22) +(x+45) =0.

In this above expression first add the constants separately and then add the variable x.

(x+2) +(x-22) +(x+45) = (x+ x+ x) + (2-22+45)

(x+ x+ x) + (2-22+45) =3x+25

3x+ 25=0

3x =-25

x=-25/3

The value for x in (x+2) +(x-22) +(x+45) =0 is -25/3.

Example 2 for basic algebra calculations:

Find the value for the quadratic equation x ^2+6x +8=0.

Solution:

The given quadratic equation is x ^2+6x +8=0.

We have to find the roots for the above quadratic equation.

This can be solved by the factoring by middle term.

A=co-efficient 0f x ^2=1

B= co- efficient of x=6

C= constant=8

We split the 6 as (2+4), and then only we get 2 x4 =8.

x ^2+6x +8= x ^2+2x+4x +8

Take x as common in first two terms and 4 as common in next two terms.

x ^2+6x +8= x(x+2)+4(x+2)

x ^2+6x +8= (x+4) (x+2)

(x+4) (x+2) =0

x+4 =0 and x+2 =0

In the equation x+4 =0, subtract 4 on both sides.

x+4 =0

x+4-4 =0-4

x=-4

In the equation x+2 =0, subtract 2 on both sides.

x+2 =0

x+2-2 =0-2

x=-2

The roots for the equation x ^2+6x +8=0 is x=-2,-4.

Is this topic Answers to Word Problems hard for you? Watch out for my coming posts.

Practice problem for basic algebra calculations:

Find the value of x for the equation (x+12)+( x-22)=0
Answer: x=5.

Find the roots for the quadratic equation x ^2+ 11x+30=0.
Answer: x=-5,-6.

Basic Proportionality Theorem

Introduction of basic proportionality theorem:
The Basic proportionality theorem is the theorem associated to triangle's property. This theorem says that if a line intersects two sides of one triangle and is parallel to the third side, then it divides the first two sides proportionally. That’s the Basic Proportionality Theorem shows the relation between the triangle and a straight line which bisects the two sides and parallel to another side.

Explanatory of basic proportionality theorem:

If a line is drawn parallel to the triangle, thus the line intersect two sides of the triangle parallel to the other (third) side in distinct points, then it cuts(divide) the two sides proportionally in the same ratio. This is known as proportionality theorem. It is otherwise called as Thales theorem.

Now we are going to prove the theorem and illustrative the statements

Given: A triangle LMN, S and T are any other two points on PQ and PR respectively such that ST parallel to QR.

To prove:

LS / SM = PT / TN

Construction:

Join MT and NS. From TU perpendicular to LM.

Proof:

In triangle LMN

ST is parallel to MN

And

If LS/MS = k (Where k is constant)

Thus LT / NT is also equal to k

Or we can say

LS / MS = LT / NT

We are given LMN,

ST is parallel to MN.

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Basic proportionality theorem:

Proportionality Theorem says that, A line parallel to one side of a triangle divides the other two sides into parts of equal proportion.

DATA:

In triangle ABC, a line PQ drawn parallel to BC cuts AB and AC at P and Q respectively.

To prove:

AP/ PB = AQ /QC

Construction:

Let the point P divide AB in the ratio of l: m where l and m are natural numbers. Divide AP into 'l' and PB into 'm' equal parts. Through all these points on AB, draw lines parallel to BC to cut AC.

Proof:

Statement                                                                           Reason

1. AP is cut in l equal parts and lines are drawn through                    Construction

these points parallel to BC.

2. Therefore, AQ is cut l equal parts.                                                        Intercept  theorem

3. Similarly,                                                                                                    Statement 1

QC is cut in m equal parts.

4. AP / PB = 1 /m                                                                                          Construction

5. AQ / QC = 1/ m                                                                                         Statement 2 and 3

6. Therefore AP / PB = AQ / QC                                                                 Statement 4 and 5.

Simplify Monomials

An algebraic expression consisting of only one term is called a monomial. It can be a product/quotient of variable and numerals or only variables or even constant terms.

Some of the examples are 3x^2, -p/2, 4xy, 63, -7z etc. Operations addition and subtraction cannot be used to represent a mono form as they are used when there are more than one term in an algebraic expression also monomials cannot have negative exponents. For instance 5-y, 7x+3, ab-3 are not mono forms.

The sum of the exponents of all the variables would be the degree of the same.
The degree of such a constant term is zero. To simplify monomials we would be using law of exponents, multiplication of mono forms and division of mono forms accordingly wherever required.

The order of operations is also followed while solving
The combination of these expressions which are given inside the parenthesis have to be simplified first.
If we have a power of an exponential algebraic expression then law of exponents is used. For example, given an expression (3a2b3)2 can be simplified by using the law of multiplying the exponents [(am)n=amn] giving 3a4b6

Multiplication of Monomials:

To multiply the given expression the exponents of the like terms are to be added using the law of exponents, am.an=am+n
Example: Given an expression 3x2y4. 4xy3 then we simplify applying the law of exponents am.an=am+n, we get 3.4 x2+1y4+3= 12x3y7
Example: -3(m2n).(-4mn3)
Coefficients are multiplied first and then the variable part
(-3.-4)(m2.m1)(n1.n3)=(12)(m2+1)(n1+3)= 12m3n4

Dividing monomials:

While dividing exponential expressions, the exponents of like terms are subtracted
Example: 9x3y4/3x2y this expression can be simplified applying the law of exponents am/an=am-n
We get, 3x3-2y4-1= 3xy3

Let us now learn the steps involved in solving monomials with the help of few examples as given below,
Example: -18a^3b^2c^5/3ab^2c
Here first the numeral part is simplified, -18/3= -6
Each of the like terms are simplified using the law of exponents,
a^3/a= a^3-1= a^2
b^2/b^2 = b^2-2=b0=1
c^5/c= c^5-1=c^4
Writing all the terms together we get, -6a^2c^4 which is the simplified form

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Example: Simplify (3p^2q3/3p) . (9p^2q^2/q^5)
First step would be to multiply the numerators and add the exponents of like terms
Numerator= (3.9)(p^2+2.q3+2) = 27p4q5
Then multiply the denominators and add the exponents of like terms if any
Denominator= (3.1)(p1q^5)  [when the exponent is not given it is considered as 1]
Now divide the coefficients of the expression if possible
27/3 = 9
Next step would be to subtract the exponents of like terms
P4-1q5-5=p3.q0=p3.1=p3
The final answer is (3p^2q^3/3p).(9p^2q^2/q^5)= 9p^3

Basic Algebra For

Introduction of Basic Algebra For:
Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics. The part of algebra called elementary algebra.

Source – Wikipedia.

Basic Algebra for Addition and Subtraction:

The followings are some of the examples for basic algebra of addition and subtraction.

Addition of fractions:  `4/9 ` +` 4/9` .

Solution:

From the fraction denominators are same.

By adding the numerator and same as in the denominator:

= `4/9` +` 4/9`

=` (4 +4)/9`

=` 8/9`

By Simplify and adding the fraction:

= `8/9`

= `4/3`

Subtraction of fractions: ` 2/6` - `4/6` .

Solution:

From the fraction denominators are same.

By subtract the numerator and same as in the denominator:

= `2/6 ` - `4/6`

= `(2- 4)/ 6`

=` -2/6`

By Simplify and subtract the fraction:

=` -2/6`

= `-1/3`

Algebraic expression of addition: 10x + 7y + 3x + 5a.

Solution:

The given expressions are 10x + 7y + 3x + 5a

The like terms in expression are 10x and -3x. We cannot do anything the 7y or 5a.

Then group the terms and add.

= (10x + 3x) + 5a + 7y

= 5a + 13x + 7y

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Basic Algebra for Multiplication and Division:

The followings are some of the examples for basic algebra of multiplication and division.

Multiply the integers: 5 × 9.

Solution:

Given integers are (5) × (9)

= |5| × |9|

= 5 × 9

= 45 is the solution for the given integers.

Multiply the integers: 4 × (-5).

Solution:

Given integers are 4 × (-5)

= 4 × -5

= 4 × -5

= -20 is the solution for the given integers.

Divide: 8a^2 – 40 by 8

Solution:

The given polynomials are 8a^2 – 40 ÷ 8

= `(8a^2-40)/8`

By dividing them we get

= `(8a^2)/ 8` * `-40/8`

= a^2 - 5

Divide: 12a^2 – 24 by 4

Solution:

The given polynomials are 12a^2 – 24 ÷ 4

= `(12a^2-24)/4`

By dividing them we get

= `(12a^2)/4` *` -24/4`

= 3a^2 - 6