Elementary Math Word Problem

Introduction to elementary math word problem:

In mathematics education, the term solving word problem is often used to refer to any mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems often involve a narrative of some sort, they are occasionally also referred to as story problems and may vary in the amount of language used. In this article we shall discuss elementary math word problems. (Source: Wikipedia)

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Elementary math word problem examples


Here we are going to solve elementary math word problem with detailed solutions.

Example:

In an office there are 90 employees work. If the number of gents employees is 37, frame the equations and find how many number of ladies work in office.

Solution:

Let the number of ladies work in office be x.

The number of gents works 37.

Total number of worker 90.

Therefore simple equation is x+37 = 90

x =90 - 37

x = 53 ladies in office.

Example:

The 12 orange costs is `$` 125, find the total cost of 16 orange.

Solution:

Step 1:

Consider the given problem as If the 12 orange costs is 125. Find the cost of 16 orange?

Step 2:

12 orange cost =125

16 orange cost = `16/12 ` x125

=167

Answer:

Therefore the cost of 16 orange is `$ ` 167

Problem:

In a class room there are 60 students taking a math exam. 20 students get an average score of 70. The left over students get an average score of 75. Find the average score of the entire class?

Solution:

To find the average score of the entire class, we need to follow the following steps these steps very helpful to find the average.

Step 1:

First we need to find the sum of weighted terms and multiply each average value by the number of students in the entire class had that average and sums.

70 × 20 + 75 × 40 = 1400 + 3000 = 4400

Step 2:

Number of total terms = Number of students in the entire class = 60

Step 3:

Here we need using the formula for weighted average formula to find the average

Average = sum of average terms
Total number of term

=` 4400/60`

= 73.3

Answer:

The average mark of the entire class is 73.3


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Elementary practice math word problem


Problem:

In an office there are 70 employees work. If the number of gents employees is 27, frame the equations and find how many number of ladies work in office.

Answer:

43 ladies in an office

Problem:

The 13 orange costs is `$ `130, find the total cost of 17 orange.

Answer:

The cost of 17 orange is` $` 170

Solving Elementary Math Test

Introduction to solving elementary math test:

Mathematics is used for throughout the entire world used in many areas like, engineering, science, and medicine. In elementary school the students receive first level of education is called as primary education. Elementary mathematics includes topics from algebra, arithmetic, number system and geometry etc. In this article we shall discuss solving elementary math test example problem.

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Solving elementary math test example problem


Here we are going to solving the elementary math test example problem with detailed solutions. It is very helpful for elementary math students.

Example:

Solving the equation: x-15 = 35

Solution:

Given x-15= 35

x = 35 + 15 (-15 in the LHS becomes +15 when it comes to the RHS)

Therefore y = 50

Example:

The cost of 20 pens is `$` 420. Find the cost of each pen.

Solution:

Consider the cost of one pen be x

Therefore the cost of 20 pen is 20 × x = 20x

But the cost of 20 pen            = `$` 420

The simple equation is 20x         = $420

To get the cost of one pen, solving the equation

20x = 420

x = 420 x 1/20

= 21

The cost of each pen is `$` 21

Example:

Solving the equation 6x – 4 = 26

Solution:

Given 6x – 4 =26

6x = 26 +4

6x = 30

x = 30 x `1/6`

Therefore x = 5

Example:

Solve the expression 12x + 5y - 4 + 6x + 6y + 8

Solution:

Find the like terms in the expression and then group and combine like terms:

• +12x and +6x are like terms, and combined to get +18x,

• +5y and +6y combine all like terms to get +11y, and

• -4 and +8 combine all like term and get +4.

So after simplify, the expression:

18x + 11y + 4

Example:

The sum of two numbers is 25. If one number is 14, find the other number.

Solution:

Consider the unknown number be x.

The sum of two numbers = 25

One number is 14.

Therefore the simple equation is            x = 25 - 14

x = 25 -14

x = 11

The other number is 11.

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Elementary math test practice problem


Problem:

The cost of 25 pens is `$` 500. Find the cost of each pen.

Answer:

The cost of each pen is `$` 20

Problem:

Solving the expression 8x + 6y - 5 + 6x + 6y + 10

Answer:

14x + 12y + 5

Problem:

The sum of two numbers is 22. If one number is 10, find the other number.

Answer:

The other number is 12.

Elementary Math Area

Introduction For elementary Math Area:

Elementary math area consist of area for square, rectangle ,triangle it is a menstruation part it is interesting in finding area of shapes and measuring length for given shapes. In menstruation area is an essential part for measuring the shape length. Measurement math consists also volume, surface area such as in higher grade for elementary math we have only fundamental shapes.

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Elementary Math Area For Square:


Area of a Square:
We know  square is a like rectangle in which its Length = Breadth.
The area of the square
= Length `xx ` Length
= (Length)2 = (Side)2
= a × a
= a^2

Example 1:
Find area of a square whose side is 12 cm.

Solution :
Side of the square, a = 12 cm
Area of the square, A = a^2
= a × a
= 12 × 12
= 144 sq.cm (or 144 cm2)
Example 2:
Find area of a square whose side is 14 cm.

Solution :
Side of the square, a = 14 cm
Area of the square, A = a^2
= a × a
= 14 × 14
= 196 sq.cm (or 196 cm2)

Area of Rectangle:
Length `xx` Breadth

Example 1 :
Length and breadth of a rectangle are 15 cms and 13 cms respectively. find its area.

Solution:
= Length `xx` Breadth
= l `xx` b
= 15 cm. `xx` 13 cm.
= 195 Sq. cm.

Example 2:
Find the area  of a rectangle whose length is 2m and breadth is 70 cm.

Solution :
Length of the rectangle, l = 2 m or 200 cm
Breadth of the rectangle, b = 70 cm
Area of the rectangle, A = l × b
= 200 cm × 70 cm
Area = 14000 sq.cm.

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Elementary Math Area For Triangle:


Area of a  triangle:
`1/2` × base × height

Example 1 :
In a triangle the length of sides containing  angles are 20 cm and 21 cm. Find its area.

Solution :
Let b = 20 cm and h = 21 cm
Area of the right triangle, A =1/2 bh
= `1/2` × 20 cm × 21 cm
= 210 sq.cm

Example 2 :
In a right triangle the length of sides containing the right angles are 15 cm and 20 cm. Find its area.

Solution :
Let b = 15 cm and h = 20cm
Area of the right triangle, A = `1/2` bh
=`1/2` × 15 cm × 20 cm
= 150 sq.cm

Elementary Math Geometry

Introduction to elementary math geometry :

Geometry plays their important part in the mathematics in which it deals with the shapes like cone, circle, square, rectangle and their positions in the spaces and the measurement of the shapes using parameters like area, volume and perimeter.  Elementary geometry problems includes solving simple basic level math problems. Here we will solve some elementary math problems. In this article we will brief about elementary math geometry. I like to share this Equation Parabola with you all through my article.
 
Elementary Math Geometry:

Elementary math geometry involves in preparing geometry math problems. Here we will see some sample problems as below,
Example problem 1- Elementary math geometry

What is the area of the square whose sides measures 7cm?

Solution:

Area of the square = side 2

= 7 2

= 49

Area of the square = 49 cm 2

Example problem 2:

Find out the diagonal of the rectangle whose length is l =15 cm and breadth is b = 6 cm?

Solution:

Diagonal of the rectangle  = `sqrt(l^(2) + b^(2))`

= `sqrt(15^(2) + 6^(2))`

= `sqrt(225+ 36)`

= `sqrt(261)`

= 16.15

Diagonal of the rectangle = 16.15 cm

Example problem 3:

What is the area of the trapezoid whose side measures a = 9 cm and b = 7 cm and height measures h =6 cm?

Solution:

In the trapezoid, consists of  4 sides; out of the four sides 2 sides are parallel.

Area of the trapezoid = `1 / 2` (a + b) * h

= `1 / 2` (9 + 7) * 6

= `1 / 2(` (16) * 6

= 8 * 6

Area of the trapezoid = 48 cm2

Example problem 5:

The angle of a triangle is 120. Then what is the measure of its supplementary angle?

Solution:

Supplementary angle is 180.

Given angle of the triangle is 120.

Supplementary angle to measure for 120= 180 – 120

= 60


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Practice Problem: Elementary Math Geometry

Practice problem 1 - Elementary math geometry
What is the area of the square whose sides measures 8cm?

Answer: 64cm 2

Practice problem 2 - Elementary math geometry

The angle of a triangle is 160. Then what is the measure of its supplementary angle?

Answer: 20

Elementary Math Tutorial

Introduction to elementary math tutorial:

The elementary math work consists of mathematics topics taught in the primary and secondary schools. The most important in elementary mathematics are arithmetic, geometry, algebra, number work and special functions. In secondary school the topics are trigonometry, Calculus etc. In elementary math tutorial, the basic operations and problems are clearly taught by the tutors with the examples solved in steps. Now we are going to see about elementary math tutorial.

About Elementary Math Tutorial:

Now we see the elementary mathematics with the help of the tutorial problems.

1. Number system:

The number system consists of the natural, decimal, fractional, rational numbers etc. Let us see one example for the rational number in number system.

Tutorial Example:

Find the fraction for the given decimal number 0.66

Solution:

The given decimal can be simplified as follows,

0.66 = 6 tenths + 6 hundredths

= 6 × `(1/10)` + 6 × `(1/100)`

= `(6/10)` + `(6/100)`

= `66/100`

= `33/50` .

2. Arithmetic:

In arithmetic, we can see about the housing finance, speed, time, distance, ratio etc.

Tutorial example:

Find the ratio of 5 Kg to 850 g

Solution:

The ratio can be determined and explained below

5 Kg = 5 × 1000 = 5000 g

The ratio can be given as

Required ratio = 5000: 850

= 500: 85

= 100: 17

Thus, the ratio is given as 100: 17

Other Elementary Math Tutorials:

1. Algebra:

Algebra is the part of mathematics where it consists of many expressions such as variables, constants and coefficients. Now we see an example

Tutorial example:

Determine and expand the given expression: (2a + 2b + 2c)2

Solution:

Let us take the equation with ( x + y + z)2

x = 2a, y = 2b, z = 2c

(2a + 2b + 2c)2 = (2a)2 + (2b)2 + (2c)2 + 2(4ab + 4ac + 4bc)

= 4a2 + 4b2 + 4c2 + 8ab + 8ac + 8bc

2. Geometry:

Geometry is about the learning of various size, shape and properties of the geometrical solid figures.

Tutorial example:

Find the diameter of the circle where the circumference of circle is about 8 cm.

Solution:

The circumference of the circle determined as follows,

C = π × d

8 = 3.14 × d

d = `8/3.14`

d = 3.18cm.

The diameter of the circle is determined as 2.54 cm.