Precalculus Practice

In American mathematics education, pre calculus, (or Algebra 3 in some areas) an advanced form of secondary school algebra, is a foundational mathematical discipline. It is also called Introduction to Analysis. In many schools, precalculus is actually two separate courses: Algebra and Trigonometry. Precalculus does not prepare students for calculus as pre-algebra prepares students for Algebra I. (Source: Wikipedia)

Example problemsfor precalculus practice

Precalculus practice problem 1:
 Find the roots for the given quadratic equation 3x² + 11x + 10.
Solution:
Given 3x² + 11x + 10
Factorize the given equation, we get
3x² + 11x + 10 = 3x² + 6x + 5x + 10
                       = 3x (x + 2) + 5 (x + 2)
                       = (3x + 5) (x + 2)
Equating the equation to zero, we get
   (3x + 5) (x + 2) = 0.
3x + 5 = 0, x + 2 = 0.
x = (- 5 / 3), x = - 2.
After solving, we get
The roots are (- 5 / 3), - 2.
Answer:
The roots are x = (- 5 / 3) and x = - 2.

Precalculus practice problem 2:
Solve the given factorial (`(6!) / (4!)`).
Solution:
 Given (`(6!) / (4!)`)
Formula: 
n! = n (n - 1) (n - 2).... .1.
(`(6!) / (4!)`) = `(6!) / (4!)`
6! = 6 * 5 * 4 * 3* 2 *1
    = 720.
4! = 4 * 3 * 2 * 1
    = 24
Therefore, we get
`(6!) / (4!)` = `(720) / (24)`
After solving, we get
       = 30
Answer:
The final answer `(6!) / (4!)` = 30

Precalculus practice problem 3:
Find the slope and equation of the straight line, it passes through the points (0,4) and (4, 6).
Solution:
Given points are (0,6) and (4, 10)
Here, x₁ = 0, x₂ = 4, y₁ = 6, and y₂ = 10

Formula for finding slope:
           Slope (m) = `(y_2 - y_1) / (x_2 - x_1)`
Substitute the given values in the above formula, we get
                          = `(10 - 6) / (4 - 0)`
                     m  = 1
Formula for line equation:
                   (y - y₁) = m (x - x₁)
Substitute the given values, we get
                   y - 6 = 1 ( x - 4)
                   y - 6 = x - 4
Add 6 on both the sides, we get
                         y = x + 2
Answer:
  Slope of the equation is (m) = 1
  Line equation is y = x + 2


My forthcoming post is on find the prime factorization of 36 and neet syllabus will give you more understanding about Algebra.

Practice problems for precalculus

Precalculus practice problem 1:

 Solving the given factorial and find its value (`(12!) / (10!)`)

Answer:

 The final answer is 132

Precalculus practice problem 2:

 Find the slope of the given straight line passes through the points (7, 8) and (9, 16)

Answer:

 The final answer is Slope m = 4

Precalculus practice problem 3:

Multiply the given expressions (2x + 17) (x² - 15)

Answer:
 The final answer is 2x³ - 17x² - 30x - 255

Roster Method in Math

Roster Method in Math:

The roster method in math is  called as tabular method. In roster method a set{} which denotes the set of data’s and by listing all the elements of the set, that elements will be separated by commas  like {1,2,5} A set containing 8,4,6,3,11,8 is Written as {8,4,6,3,11,8} in Roster method.

The two methods of the notation sets which can be given as follows,

Roster method
Builder method
Let us see about the roster method of math as given below.


Roster Method in Math:


The roster method of math is the list or diagram of the ordered elements, when the rule is the equivalent form which will be using the selection of elements from a domain matching a rule.


The Roster method is one of two ways of representing that the elements of a set using brackets, {}.For example, all even numbers under 14 would be represented as : {2,4,6,8,10,12}.


The roster method is often associated with 'roster and rule' this is a way of finding a rule that the elements of a set follow. Sets can be generally comprise any list of items or numbered lists.

Examples of Roster method in Math:


Example  Sets 1:

Write each of the following sets in the roster form

i) P =  Set of all factors  of 34

ii) Q= set of all odd numbers less than 15

iii)R  = Set of all even numbers  less than 20

iv)S= set of all letters in the Word,'TUTORVISTA'

Solution:

i)All the factors of 34 are 1,2,17

Therefore,P={1 ,2,17}

ii) Q = All odd numbers less than 15 are 1,3,5,7,9,11,13

Therefore,Q={1,3,5,7,9,11,13}

iii)R=All even numbers less than 20 are 2,4,6,8,10,12,14,16,18.

Therefore ,R= {2,4,6,8,10,12,14,16,18}

iv) The set of all letters in the word,'TUTORVISTA'

Therefore ,S ={T,U,O,R,V,I,S,A}

Example Set 2:

1)Set of all prime numbers between 10  to 20

Solution:

Prime numbers between 10 to 20 ⇒ 11,13,17,19

Using roster method ,In the form of set notation ,

Prime numbers between 10 to 20 ⇒   {11,13,17,19}

2)The set of all letters in the word,' ROSTER METHOD IN MATH'

Solution:

Note: The set contains the alphabets only once.

Therefore, The set of all letters in the word,' ROSTER METHOD IN MATH'

Set ⇒ {R,O,S,T,E,M,H,D,I,N,M,A}

Example in the form of Perfect roster Method in Math:

1)X =`{(1/2),(5/4),(4/6)}` is a set of fraction numbers.

2) Y   ={English,Math,Science,Biology} is  a set of subjects.

3) Z  = {Banana,Mango} is a set of  fruits .

Grade 3 Math Fractions

Introduction for Grade 3 math fractions:

This article we are discussing about grade 3 math fractions. A part of a whole is called fraction. Fractions can be consisting of minimum two digits. The numerator is specified as top digits. The denominator is specified as bottom digits.

Numerator
denominator

An arithmetical expression relating digits or two quantities, one divided by the other is called fraction. The digits may be whole numbers of this is a rational number.

For example, 3/6 is a fraction. The digit 3 is numerator and digit 6 is a denominator. Grade 3 math fractions are used to solve the simple addition fraction, multiplication fraction and subtraction fraction.


Examples problem for grade 3 math fractions:


Example 1: Add the fractions for given two fraction, `10/5` + `7/5` .

Solution:

The given two fractions are `10/5` + `7/5`

The same denominators of the two fractions, so

= `10/5` + `7/5`

Add the numerators the 10 and 7 = 10+7 = 17.

The same denominator is 5.

= `17/5`

The addition fraction solution is `17/5` .

Example 2: Subtract the fractions for given two fractions, `10/5` - `7/5` .

Solution:

The given two fractions are `10/5` - `7/5`

The same denominators of the two fractions, so

= `10/5` - `7/5`

Subtract the numerators the 10 and 7 = 10-7 = 3.

The same denominator is 5.

The addition fraction solution is `3/5` .

Example 3: Multiply the fractions for given two fraction, `10/6` * `7/6` .

Solution:

The given two fractions are `10/6` * `7/6`

The same denominators of the two fractions, so

= `10/6` * `7/6`

Multiply the numerators the 10 and 7 = 10*7 = 70.

Multiply the denominators the 6 and 6 = 6 * 6 = 36

= `70/36`

The addition fraction solution is `70/36` .

Example 4: Add the fractions for given two fraction `3/4 ` + `2/5`

Solution:

The denominator is different so we take a least common denominator (LCD)

LCD = 4 * 5 = 20

So multiply and divide by 5 in first term we get

`(3 * 5) / (4 * 5)`

=`15/20`

Multiply and divide by 4 in second terms

= `(2*5) / (5*4)`

= `10/20`

The denominators are equals

So adding the numerator directly

= `(15+10)/20`

Simplify the above equation we get

= `25/20`

Therefore the final answer is `5/4` .

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

Practices problems for grade 3 math fractions:


Problem 1: Add the two fraction `5/6` + `7/6`

Solution: `12/6`

Problem 2: Subtract two fractions `5/6` – `3/6`

Solution: `2/6`

Problem 3: multiply two fractions `5/6` * `5/6`

Solution: `5/6`

Teacher Help Math

Introduction about teacher help math

Teacher help math is that the topics and problems on math. Math is the basics of all the subject and it very useful in real life. Teachers help math for the students in all chapters by performing step by step process. Teacher help math in every classes. Teacher help math to their students by working out problems and giving homework on the same model so that students can remember the steps of certain types of problems. The students learn many things about the math by their teacher help math. From kinder grade to college grade in all stages we study math and teacher help math on every grade.


Teacher help math problem and solution


Jack and Joe agree to meet in Chicago for the weekend. Jack travels 114 miles and Joe travels 96 miles. If Jack rate of speed is 6mph faster than Joe’s then at what rate of speed does Jack travel?
Solution

Take Joe’s speed = x,

Distance travel by Joe = 96 miles

Jack’s speed = x+6,

Distance travel by Jack =114 miles

Distance/speed = time, here time is constant.

Now `114/(x+6) ` = `96/x`

Cross multiply so it becomes 114x = 96x+ (96*6)

Subtract 96 x on both side as common

114x -96x=96x-96x+576

18x = 576

Divide by 18 on both sides

`(18x)/18` = `576/18`

x= 32 mph = Joe’s speed

So Jack speed = 32 + 6 = 38mph.

2. The sum of the ages of Nancy and her daughter is 55 years. If the age of the daughter is 16 years what will be the age of Nancy?

Solution

Let the age of Nancy be x years.

The age of her daughter =16 years.

Sum of two ages =(x+16) years.

But this sum is given as 55years.

So  x +16 = 55.

x =55-16

x =39.

Thus the age of the Nancy is 39 years

Understanding Daily Compound Interest Formula is always challenging for me but thanks to all math help websites to help me out.

Teacher help math problem


1. Tim and Tom leave the school at the same time traveling to their home in opposite directions to each other. Both of them travel for three hours and are then 360 miles apart. If Tim travels 10 miles per hour faster than Tom, find the average rate of speed for Tim and Tom.

Answer Tom speed is 55 mph and Tom speed is 65 mph

2. The sum of the ages of Peter and his son is 59 years. If the age of the son is 14 years what will be the age of peter?

Answer Peter age is 45

Logic Problems Math

Introduction to logic problems in math:

Logic problems in math deals with out the symbols used in algebra.  We can define any mathematical equations as logic terms. Therefore, we can solve the logic problems in math more easily. This logic problems defines the solution of the problems are given as theoretical forms . Now let us discuss  the logic word problems.

Please express your views of this topic Sum of Arithmetic Series Formula by commenting on blog.

Problems based of math logic


Determine the logic problem if a set of 12 numbers is 32 .If we taken a number from that list means then the remaining number is  24 .Find the number which is removed ?
Solution:

Step 1: The removed number can be viewed from the given list of 12 numbers and the amount of remaining numbers 24.

The list of 12 numbers - Amount of remaining numbers 24

Step 2: Using the formula

Sum of terms = Average* Number of terms

The list of 12 numbers = 32 x12 = 384

The amount of remaining numbers = 24 x 12 = 288

Step 3: Using the formula from step 1

Number removed = The list of 12 numbers – The amount of remaining numbers 24

384 – 288 = 96

Answer: The number removed is 96.


Some more problems based on logic math


In a sports day, there are 800 peoples consist of men, women, and children.  There are six times as many men as children, and thrice as many women as children. How many of women are there?
Sol:

Let 'x' be the number of children.

Then the number of men = 6 x and the number of women = 3x

So, the equation is  x + 6x + 3x = 800

10x = 800

x = 800 / 10

x = 80

Therefore, the number of women = 3x = 3(80) = 240

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

There were 54 computers in each urn. 24 such urn of computers were distributed to a college. 14 computers were damaged and had to be removed. The un delivered computers were packed into the urns of 8 computers each. How many urns of computers were there?
Sol:

First determine the total number of computers distributed to a college

54 × 24 = 1296

The total number of computers distributed to the college is 1296.

Get the number of undelivered computers.

1296 – 14 = 1282

Get the number of urn that contains computers.

1282 ÷ 8 = 160.25

There were 160.25 urns of computers.

What is Median Math

Introduction to median in math:
In math,Median is nothing but the middle number in a sequence of numbers.To find the Median, set the numbers we are given in value order and find the middle number.For Example: Find the Median of {16, 23 , 10 , 11 , 26 , 13 and 15}. Put them in ascending order first: {10, 11, 13 , 15 ,16 ,23 ,26}, the middle number is 15, therefore the median is 15.


What is median math?

The median is the middle value of the distribution that is the median of a distribution is the value of the variable which divides it into two equal parts. It is nothing but the value of the given variable such that the number of annotations above it is equal to the number of annotations below it.

Median of a group or continuous frequency distribution = l+ ((n/2)-c*f)/f)*h

Where,

l indicates lower limit of the median class

n indicates number of observations.

f indicates frequency of median class.

h indicates size of the median class.

cf indicates cumulative frequency of the class preceding the median class.

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

Example problems - What is median math?


Example 1:

Find the median of the following: 89, 39, 71, 28, 46, 68, 41, 71, 30, and 54

Solution:

Arrange the given sequence in the ascending order.

28, 30, 39, 41, 46, 54, 68, 71, 71, 89
we know that Median = Middle score-most score
Median = (46+54/2)

Median =100/2
Median = 50

Example 2:

Find the median of prime numbers between 46 and 84.

Solution:

The prime numbers between 46 and 84 are 47, 53, 59, 61, 67, 71, 73, 79, 83
Median = Middle-most score
Median = 67

Example 3:

Find the median of the first 8 even numbers.

Solution:

The first 8 even numbers are {2, 4, 6, 8, 10, 12, 14, and 16}

Arrange the given sequence in the ascending order.

{2, 4, 6, 8, 10, 12, 14, and 16}

We know that Median = Middle score-most score

Median = (8+10)/2.

= 18/2.

= 9.

Median of grouped data- What is median math?
Example 4:

The distribution shown below gives the marks 50 of students in a class. Find the median Mark of the students.

Mark       Number of Student

0 -10           2

10 -20        12

20 - 30       22

30 – 40       8

40 - 50        6

Solution:

Marks (C. I.)     f       c.f

0 -10                  2       2

10 -20               12      14

20 - 30              22     36?

30 - 40              8        44

40 - 50              6        50

Here n / 2 = 50 / 2 = 25

So, 20 -30 will be the median class.

Now, l =20

h = 10

c.f = 14

f = 22

Median = 20 + [(25 - 14) / 22] x 10

= 20 + (11 / 22) x 10

= 20 + 5

= 25

This indicates 50% of the students got less than 25 marks and the other 50% got more than 25 marks.

Simple Form in Math

Introduction to simple form in math:

In mathematics simple form means simplest form or simplifying. Fractions are simplifying means dividing by the value of GCF for both numerator and denominator. Here GCF stands for greatest common divisor. Simple form in decimal means converting whole number or rounding the value. Therefore, simple form is a reducing of values for the given.


Example problems of simple form in math:


Simple form in math problem 1:

Writing the simple form: `18/12`

Solution:

Given `18/12`

Method 1:

First we are factoring the values of numerator and then factoring a denominator values.

Finally reduce the fraction by cancelling the common value.

Method 2:

Find the Greatest common divisor for the given fraction values and then simplifying them.

Therefore we can simply the given problem using the second method.

Greatest common divisor of 18, 12= 6

So, `(18-:6)/(12-:6)=3/2`

Answer: `3/2`

Simple form in math problem 2:

Writing the simple form: `28/16`

Solution:

Given `28/16`

Method 1:

First we are factoring the values of numerator and then factoring a denominator values.

Finally reduce the fraction by cancelling the common value.

Method 2:

Find the Greatest common divisor for the given fraction values and then simplifying them.

Therefore we can simply the given problem using the second method.

Greatest common divisor of 28, 16= 4

So, `(28-:4)/(16-:4)= 7/4`

Answer: `7/4`

Simple form in math problem 3:

Simplify  `(x(x+7)-4(x+7))/( x+7)`

Solution:

We can simplify the given fraction values by using the following steps:

First we are factoring the values of numerator and then factoring a denominator values.

Finally reduce the fraction by cancelling the common value.

Given `(x(x+7)-4(x+7))/( x+7)`

Here we can doing only reducing fraction. Because of already have factors for both numerator and denominator.

So reduce the fraction individually and then simplifying that,

`(x(x+7)-4(x+7))/( x+7)=(x(x+7))/(x+7)-(4(x+7))/(x+7)`

Here cancel the factor `x+7 ` on both numerator and denominator.

After simplifying, we get the final answer.

`(x(x+7)-4(x+7))/( x+7)= x-4`

Answer: ` x-4`


Practice problems of simple form in math:


Write the simple form: `55/22`
Simplify `(x(x+4)+3(x+4))/( x+4)`
Answer:

1)      `5/2`

2)    ` (x+3)`

Math Practice 6th Grade

Introduction to Math Practice 6th grade:

The topics involved in Math Practice 6th grade are  place values, addition, subtraction, rounding& estimation, multiplication, measurement, division, fractions, area, comparing and ordering numbers, surface area, volume, Geometric properties and expressions. In this article we shall discuss about the math practice 6th grade. The following are the examples involved in  math practice 6th grade.

Understanding Formula for Area of Triangle is always challenging for me but thanks to all math help websites to help me out.

Math Practice 6th Grade Problems:


Example 1: Solve 2x + 5 = 15

Solution:Given: 2x + 5 = 15

Subtracting 5 on both sides,we get

2x + 5 – 5 = 15 – 5

2x = 10

Dividing 4 on both sides, we get

2x = 10
2      2

x = 5

Example 2: Find the area of a parallelogram of length 9 inches and height of 9 inches.

Solution: Formula : Area of parallelogram = length * height

Given: length = 9 inches, height = 9 inches.

So, area of parallelogram = 9 * 9 = 81 square inches.

Example 3: Find the volume of a cube with length of side equal to 13.

Solution: Formula to find the volume of a  cube = (length of the given side)3

Given: Length of the side = 13

So, Volume of a cube  = (13)3

= 13 * 13 * 13

=2197.

Example 4:  Find the area of a rectangle with base of length is 3 inches and height of 2 inches.

Solution:Area of a rectangle = length * height

So, Area of a rectangle =3 * 2

= 6 square inches.

Example 5: 95 is what percent of 100?

Solution: Perform the division operation from first number by the second number.

`95 divided by100`=>(`95/100` ) = 0.95

Multiply the answer(0.95) by 100

0.95 * 100 = 95

So 30 is 95% of 100.

Having problem with Chi Square Test Calculator keep reading my upcoming posts, i will try to help you.

Math Practice Problems 6th Grade:


Example 1: Solve x + 5 = 10

Answer: x = 5

Example 2: Find the area of a parallelogram of length 6 inches and height of 2 inches.

Answer:  12 square inches.

Example 3: Find the volume of a cube with length of side equal to 6.

Answer: 216

Example 4: Find the area of a rectangle with base of length is 8 inches and height of 3 inches.

Answer: 24 square inches.

Example 5: 73 is what percent of 100?

Answer:  73%

Word Problems Math Work Done

Introduction:
The numbers and variables in a problem which are given in the form of words are known as word problems. It is hard to translate the words into mathematical expression or equations. When it is done, then it is easy to solve those problems. But translation of those words to mathematical symbols or equation is hard. Understanding the word problem is more important than solving it. Without understanding the math word problems the solving work can’t be done. Understanding What is Arithmetic Mean? is always challenging for me but thanks to all math help websites to help me out.


Steps in solving Word problems:


The following are the steps to be considered when solving math word problems,

Step 1:  The first step in solving the word problems is, translating the given word data’s into numbers.

Step 2: After translation to numbers, read the problem completely and understand what is asked to find?

Step 3: Next step is to name the unknown values using variables like x, y z etc. This will help to solve the problem easily.

Step 4: After naming the unknown variables, work out the problem in an organized manner using basic arithmetic operations like addition, subtraction, multiplication and division.

For example, if a word problem consists of two equations, first find the value of one unknown variable and then substitute that value in any one of the equation to find the another  value.


Math word problem work done examples:


Example 1: At present, the ratio between the ages of Anand and Ram is 3: 4. After 5 years, Anand's age will be 29 years. What is the age of Ram at present?

Solution:

Let the age of Anand = 3x

Let the age of Ram = 4x

3x + 5 = 29

Subtract 5 on both sides,

3x + 5 – 5 = 29 – 5

3x = 24

Divide 3 on both sides,

3x/3 = 24/3

x = 8

Ram age = 4(8)

= 32 yrs

Ram age is 32 years.

Example 2: Raj has some hens and sheep’s. If the number of heads is 50 and the number of feet equals 144, what will be the number of hens?
Solution:

Let the number of hens = x

Let the number of sheep’ = y

x + y = 50        equation 1

2x + 4y = 144    equation 2

Multiply equation 1 with 2,

2x + 2y = 100

2x + 4y = 144   (subtracting)

-2y = -44

-2y = -44

Divide 2 on both sides,

-2y/2 = -44/2

y = 22

Plug in y value in equation 1

x + y = 50

x + 22 = 50

Subtract 22 on both sides,

x + 22 – 22 = 50 – 22

x = 28

The number of hens is 28.

World Math Day

Introduction to world math day:

World math day is not a day to celebrate for mathematics. It is an international competition in math. World math day was started in 2007 on March 14. The date can be written as 3.14.2007. The first four numbers is same as the value of pi. So this world math day can also be say as pi day.


Playing method:


The trademark names are world maths day and world math day. Usually the questions are based on the basic math multiplication, division, addition, subtraction and equations. The students are required to answer each game in one minute. When one student is playing a game the website will find the same level player and make that player to play against the player. If there is no such competitor was found then the player will start to play against the computer. After completing the game the result of both the competitors will be individually send.

The organizer of this event is an Australian. From 235 countries over 1.13 million students participated in 2010 math world day which was conducted on March 3rd. The entire student totally solved 479,732,613 math questions correctly.


Age category and levels:


There are three categories to play a game. The category is based on ages. The three categories are 5-8 ages, 9-13 ages and 14-18 ages. The student who answered correctly for large number questions is a world champion. The champion will awarded by minted gold medal. There is a separate medal for each age group topper.

The top 10 students in each group who answered more questions correctly will receive a gold medal. All the students who answered more than 4000 questions with correct solutions will receive a world maths day gold award. The top 50 students in each group who answered more questions correctly will receive a printed certificate gold award. If the whole class was participated then the top 10 classes will receive a world math day trophy in each group. The top 50 classes in each group who answered more questions correctly will receive a printed certificate gold award.

Fast Ways To Do Math

Introduction to fast ways to do math:

Mathematics provides the solutions to solve the problem by defining mathematical definitions. However during solving the complex problems it consumes more time and the process becomes tedious. Many mathematical experts provide the fast, easy ways to do math problems. Here we deal with the fast way to do the math problems in the following sessions. Please express your views of this topic Introduction to Mathematical Logic by commenting on blog.


Fast ways to do math:


Many mathematical experts provides us the easy fast ways to learn maths, in the following parts we will see some of the fast ways to do math problems,

Fast ways to do math multiplication of the two digit numbers using 11:

For a two digit number that multiplies with 11, follow the steps for fast multiplication,

For example: Take the number 45

45 * 11

Consider the number 45.
Add 4 +5 = 9
Now insert the 9 inbetween the 4 and 5, gives 495.
We got the final answer 495 fastly using this ways.

Another example:

Take the number 25

25 * 11

Consider the number 25.
Add 2 +5 = 7
Now insert the 7 inbetween the 2 and 5, gives 275.
We got the final answer 275 fastly using this ways.


Fast ways to do math multiplication of the two digit numbers using 16:


For a two digit number that multiplies with 16, follow the steps for fast multiplication,

For example: Take the number 44

44 * 16

Consider the number 44.
Multiply the given number with 10 gives 44*10 = 440.
Then take the half value of the given number, and multiply with 10.
Gives (44 * 1 /2  ) * 10

= 22 * 10

= 220

Now add all the computed terms in above steps with the given number,
= 440 + 220 + 44

= 704

Thus we got the answer as fast as possible using this ways.

We got the final answer 495 fastly.

Another example:

Take the number 64

64 * 16

Consider the number 64.
Multiply the given number with 10 gives 64*10 = 640.
Then take the half value of the given number, and multiply with 10.
Gives (64 * 1 /2  ) * 10

= 32 * 10

= 320

Now add all the computed terms in above steps with the given number,
= 640 + 320 + 64

= 1024

Thus we got the answer 1024 as fast as possible using this ways.

Story Problems in Math

Introduction about story problems in math:

Story problems are also known as word problems which are mainly used for kids to apply their knowledge of mathematics to the real world problem.  Story problems in math develop thinking power, problem-solving skills, and reasoning ability among students. Story problems in math are a sequence of expressions that fits into an equation. We will see some math story problems in this article.

Example Story problems in math:


Example 1: There are 25 pencils in the box. 13 pencils were used by the kids. How many pencils are there remaining?

Solution:

Total Number of Pencils in box = 25

Number of pencils used = 13

Remaining pencils = 25 -13

= 12

Example 2: Rahul, Joey, Nizam were playing Carom. Rahul scored 55 points, Joey scored 42 points and Nizam scored 30 points. How many points did they score together?

Solution:

Points scored by Ravi = 55

Points scored by John = 42

Points scored by Nizad = 30

Total points scored = 55+42+30

= 127 points.

Example 3: Joseph is 3 ½   cm Shorter than Sara.  Sara is 162 `3/10` cm tall.  How tall is Joseph?

Solution:

Given Sara height = 162 `3/10`

Joseph height = Sara height - 3½

= 162 `3/10 ` – 3 ½

= 162`3/10` – 7/2

= (1623 – 35)/ 10

= `1588/10`

= 158 `8/10`

Joseph height is 158 `8/10` cms.

Understanding Properties of Multiplication is always challenging for me but thanks to all math help websites to help me out.

More Example Story problems in math:

Example 4:

There are 5 tigers and 6 lions. The tigers each weigh 15 pounds. The lions each weigh 20 pounds. How much do they weigh altogether?

Solution:

Number of Tiger= 5

Number of Lion= 6

Weight of 1 Tiger = 15 pounds

Weight of 5 Tigers = 5 * 15

= 75 pounds

Weight of 1 Lion = 20 pounds

Weight of 6 Lions = 20*6

= 120 pounds

Total weight = 120+75

= 195 pounds

Example 5: Sanjana bought 25 Balloons for `$` 5. What is the cost of 1 balloon?

Solution:

Number of balloons= 25

Cost of 30 balloons = `$` 5

Cost of 1 balloon    = `5/25`

= `$` 0.2

Example 6: In a Conference, among the total number of people 12 people came to the meeting late which is the 24% of the total count. How many people are there in that Conference hall totally?

Solution:

Total number of people = x

Number of people arrived late =12 (4% of x)

Hence,

`(24/100) ` * x = 12

24x= 12 *100

24x = 1200

X= `1200/24`

= 50

Hence Total numbers of people are 50.

Assessment Of Math

Definition assessment:

Educational assessment is the process of documenting, usually in measurable terms, knowledge, skills, attitudes and beliefs. Assessment can focus on the individual learner, the learning community the institution, or the educational system as a whole. "Studies of a theoretical or empirical nature addressing the assessment of learner aptitude and preparation, motivation and learning styles, learning outcomes in achievement and satisfaction in different educational contexts are all welcome, as are studies addressing issues of measurable standards and benchmarks"

Source: Wikipedia


Teachers need to ask themselves with math assessment:

The following are two very important questions that teachers have to ask when teaching.

Is what I am doing helping children to develop a desire to learn mathematics?

Is what I am doing teaching children to become numerate?


Types of assessment in math:


There are two main ways in which to assess children

Formative assessment in math.

Summative assessment in math.

Formative assessment is assessing a learner while the learner is forming the new knowledge.

Example for formative assessment in math:

An example of formative assessment would be sitting with a learner while he or she is doing a task (say using a number line to count in groups), watching how the child goes about the task and asking the child to explain how and what he or she is doing. In this way, you find out what strategies the child is using and developing and what strategies you should be helping the child with; you are getting direct and instant feedback on hoe the child is coping and you are able to respond to the situation immediately through re-teaching and explaining again, asking anther learner to help, or planning another lesson on that need for the next day.

Summative assessment is assessing a learner at the end of the lesson, section, topic, quarter or year as a summing up of what the learner knows. Therefore, tests and exams are summative versions of assessment.

In math, when both formative and summative assessments are used, that is continuous assessment. In an outcomes-based education system, continuous assessment is used. The teacher studies the learning outcomes required of the learners and then plans lessons to teach to achieve these outcomes. During the lessons, the teacher observers what children are doing and saying and how children are doing a task. The teacher asks for explanations from the children as to what and how they are doing a take. The teacher helps those learners who are confused and continually monitors which learners are gaining control of the skills and concepts. Once a child can do something independently within the number range for that learner, a teacher can say that that child has learnt what was intended by the lesson and so can record that performance as a desired learning outcome for that child

What is Median in Math

Introduction to what is median in math:

Let us see what is median in math. In general, median in math is the middle value in a set of data. Statistically median is defined as a measure of central tendency which gives the value of the middle-most observation in the data. The median in math is obtained by arranging the given list of data in ascending order. Median separates the higher value and lower vale. This article deals with what is median in math and how to find median with examples.


How to find median in math:


To find median initially we have to arrange the data values of the observations in ascending order. Then,

If n = odd {n= number of data} ,

Median = `(n+1)/2` th observation

If n= even,

Median = `(((n/2)th data ) + ((n/2) +1)th data) / 2 ` {which means average of ((n/2) + (n/2)+1)th data }


Example Problems to find median in math:

Example 1:

What is the median of the given observations 50, 125, 10, 210, 50, 24, 175

Solution:

Ascending order of given data,

10,24,50,50,125,175,210

Here the number of observations= n= 7

n= odd,

Median = `(n+1)/2 ` th observation

= `(7+1)/2`

=` 8/2`

= 4th observation

Hence median is 50.

Example 2:

Find the median of the given observations 20, 7, 9, 12, 7, 9, 5,3,15, 15

Solution:

Ascending order of given data,

3, 5, 7, 7, 9,9,12,15,15,20

Here n= 10 (even) ,

Median = `(((n/2)^(th) data) + ((n/2) +1^(th) data)) / 2`



=   ((10/2)th data) + ((10/2)+1) thdata)) / 2

= `(5^(th) data + 6^(th) data) / 2`

= `(9+9 )/ 2`

= `18/2`

= 9

Hence the median is 9.

Example 3:

What is the median of this observation 20, 29, 28, 33, 42, 38, 43, 25?

Solution:

Ascending order of given data,

20, 25, 28, 29, 33, 38, 43, 42.

Number of observation = n= 8

N= even ,

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Median = `(((n/2)th data ) + ((n/2) +1)th data) / 2 `

=((8/2)th data +(8/2)+1 th data )/2

= `(4^(th) data + 5^(th) data) / 2`

= `(29+33)/2`

= 31.

Hence the median is 31.

Example 4:

Find the median of the observation 6 ,28, 24, 15, 2, 4, 1 ,20,5

Solution:

Ascending order of given data,

1,2,4,5, 6, 15,20,24,28

Here the number of observations = 9

n= 9 (odd)

Median =`(n+1)/2` th observation

= `(9+1)/2`

= `10/2`

= 5

Median is the 5th observation.

Hence median = 6.

New Math Division

Introduction for new math division:

In algebra basic arithmetic operation (addition, subtraction, multiplication, and division) widely used in day to day life. In these articles we are going to see about new math division. Division (÷) can be considered as repeated subtraction or equal distribution.

The following division methods are all based on the form Q = N / D where

• Q = Quotient

• N = Numerator (dividend)

• D = Denominator (divisor).

Specifically, if c time’s b equals a, written: c x b = a, Where b is not zero, then a divided by b equals c, written: a / b = c. In the above expression, a is called the dividend, b the divisor and c the quotient.

(Source: Wikipedia)


New math division and long division definition and steps:


Definition for division:

Division is defined as an arithmetic function, which is the opposed process of multiplication. From the process of division, the proportion or ratio of two numbers be capable of be calculated.

Otherwise, the process of decision how many periods of one number is included in a further one. Symbol of division is ‘/’ or ‘÷’. If we divide a number by another number, then

Dividend = (Divisor * Quotient) + Remainder

Long division:

Division plays a key role in our day to day activities. Long division is one of the usual methods of solving the division problems. Long division is usually studied in elementary classes. In this article, we are going to see about long division with remainders.

Steps for division math facts:

Step1. Division of two integers by the related signs resolve be positive sign

a) Positive ÷ positive = positive

b) Negative ÷ negative = positive

Step2. Division of two integers by the unlike signs will be negative

a) Positive ÷ negative = negative

b) Negative ÷ positive = negative.



Example problem for new math division:


Problem 1: Jack built a tower of blocks forty-nine inches high. Each block in the tower is seven inches tall. How many blocks were used to build the tower?

Solution:

Jack built a tower of blocks forty-nine inches high.

Each block in the tower is seven inches tall.

So, total blocks =   `49 / 7`

Therefore total blocks used to build the tower = 7 blocks

Problem 2: The school's Internet connection transferred thirty eight megabytes of data in two seconds. How many megabytes can it transfer in just one second?

Solution:

The school's Internet connection transferred thirty eight megabytes Data in two seconds.

So, therefore total megabytes can it transfer in just one second

=   `38 / 2`

= 19 megabytes.

Problem 3: There are four soft drink machines in the university. They hold sixty - four cases of soda altogether. How many cases does each machine hold?

Solution:

There are four soft drink machines in the university.

They hold sixty - four cases of soda altogether.

Total cases of soda machine hold =` 64 / 4`

= 16 Cases of soda.

Arithmetic Problem Examples

Introduction to Arithmetic:

Arithmetic is the most elementary branch of mathematics, used by almost everyone, for tasks ranging from simple day-to-day counting to advanced science and business calculations. It involves the study of quantity, especially as the result of combining numbers. In common usage, it refers to the simpler properties when using the traditional operations of addition, subtraction, multiplication and division with smaller values of numbers.

- Source: Wikipedia


Example Arithmetic Problems:


Addition:

Arithmetic Problem example 1: In a vegetable basket 18 tomatoes and 17 potatoes. How many total vegetables are there in the vegetable basket?

Solution:

No. of tomatoes in the basket = 18

Number of potatoes in the basket = 17

Total vegetables in the basket = 18 + 17

Total vegetables in the basket = 35 fruits.

Subtraction:

Arithmetic Problem example 2:  There are 67 seats in a compartment. 61 seats were occupied by the passenger. How many seats were not occupied?

Solution:

Total no. of seats in the class = 67

No. of seats occupied by the passengers = 61

No. of seats not occupied = 67 – 61

No. of seats not occupied = 6 seats.

Multiplication:

Arithmetic Problem example 3: A notebook cost 2 dollars. What is the price of 12 notebooks?

Solution:

Price of 1 notebook = 2 dollars

Price of 12 notebooks = 12 x 2

Price of 12 notebooks = 24 dollars.

Division:

Arithmetic Problem example 4:  5 trousers price is 25 dollars. What is the price of 1 trouser?

Solution:

Price of 5 trousers = 25 dollars

Price of 1 trouser = 25 / 5

Price of 1 trouser = 5 dollars.


Practice Problems in Arithmetic Problems:


Addition:

Problem: There are 15 pens in an Office. Davina is going to put 16 of them in the Office. Totally how many pens will be there?

Answer: 31.

Subtraction:

Problem: There are 25 students in a computer science class. 22 students got pass in computer science. How many students failed?

Answer: 3.

Multiplication:

Problem: George delivers 40 letters in a day. How many letters does she deliver in 5 days?

Answer: 200.

Division:

Problem: Thomas has 30 chocolates. If he presents his friends 5 chocolates each, how many friends can he share his chocolates with?

Answer: 6.

Math Definition for Kids

Addition:

Addition means sum of two quantities. The sign used to do addition is ‘+ ‘.

If Joe has 2 black pencil and 3 blue pencil means then the total number of pencil is

2+3 that is 5 pencils.

Subtraction:

Subtraction means difference of two quantities. The sign used to do subtraction is ‘– ‘.

If Kayla has 5 rupees and she spent 2 rupees in a store then the amount she has left in a hand is 5-2 that is 3 rupees.

Division:

Division means sharing a number into equal parts. The sign used to do division is ‘/’.

If a mom has 6 chocolates and then if she wants to give those chocolates to 2 children means, then it is 6/2 that is 3 .So she will give 3 chocolates to each child.

Multiplication:

Multiplication means a number is added to itself a number of times. The sign used to do multiplication is ‘x’ or sometimes ‘*’.

2+2+2=6 which is same as 3 times 2 which is equal to 6


Types of Angles Definition


Acute angle: The angle which is less than 90 degree is an acute angle. For instance, we can say 45 degree is an acute angle.

Right angle: The angle which is equal to 90 degree is right angle.

Obtuse angle: The angle which is greater than 90 degree is an obtuse angle. For instance, we can say 75 degree is an obtuse angle. I have recently faced lot of problem while learning cbse 10th sample papers sa2, But thank to online resources of math which helped me to learn myself easily on net.

Ascending order: Arranging a list of given elements from smallest number to greatest number.

Consider, the given list of numbers are 9,4,8,2 then the ascending order of a given list is

2,4,8,9.

Descending order:

Arranging a list of given elements from greatest number to smallest number.

Consider, the given set of numbers is 9,4,8,2 then the descending order of a given list is 9,8,4,2.

Twice:

Twice means two times.

The twice of 4 is 2 times of 4 that is 8.

Arithmetic Homework

Introduction to arithmetic homework:

Arithmetic  is the fundamental theory in mathematics. It was mostly used to work out mathematical function of addition, subtraction, multiplication and division. It is specially for the mathematical operations. An arithmetic operation is not just deals with integers,it also deals with arithmetical calculation, actual numbers and complex numbers.

Several times Number theory is known as higher arithmetic.
Congruence theory is known as Modular arithmetic.
Arithmetic performed on actual numbers is known as floating point arithmetic.
Let us do some arithmetic homework problems.


Example for Arithmetic homework help:


Addition:

Sum of two or more numbers is called Arithmetic. The addition symbol is + (plus).  We know how to do the addition function for the all kind of values. Addition will increase the values.

Homework Example:

8 + 3 = 11

Subtraction:

Subtraction is also one of the fundamental arithmetic operations. Minimizing two or more numbers is known as Subtraction. The subtraction symbol is – (minus). Subtraction problems reduce the values.

Subtraction can do in the following way,

C = A - B

Where,

C is difference.

A is the minuend value.

B is the subtrahend value.



Homework Example:

5 – 3 = 2

Multiplication:

Multiplication is the range of 2 numbers. X (product) is the symbol for multiplication. Multiplication also performs by recurrence of addition. The term frequently used for multiplication is “times”.

Homework Example:

8 times 2 means 8 multiply with 2.

8 * 2 = 16

The other way of multiplication frequent addition,

7 times 2 = 14

2 + 2 + 2 + 2 +2 + 2 +2 = 14

Rational multiplication:

16 / 4 * 8 / 8 =?

4 * 1 =?

= 4

Division:

Division is same like as multiplication function. But it is the reverse of multiplication. The sign of division is ‘/ ‘(slash). The most important function of division is decreasing the value then it is also repeat subtraction.

Homework Example:

12 / 2 = 6


Arithmetic homework practice problems:


Add:

22 + 61 = 83
58 + 68 = 126
Subtract:

75 – 20 = 55
58 – 10 = 48
Multiply:

4 * 2 = 8
7 * 5 = 35
Divide

25 / 5 = 5
60 / 5 = 12

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Arithmetic homework problems with solution:


Add:

16 + 22 =
29 + 46 =
Subtract:

52 – 48 =
19 – 53  =
Multiply:

8 * 9 =
12 * 5 =
Divide:

68 / 2 =
12 / 4 =
Answers:

Add: 38, 75

Subtract: 4, -34

Multiply: 72, 60

Divide: 34, 3.

Arithmetic Reasoning

Introduction for arithmetic reasoning:

Arithmetic reasoning is a branch of mathematics that deals with the study of number especially with regard to basic operations such as addition, subtraction, multiplication and division etc and applications to solution of problems. Generally, we use arithmetic in our day-to-day life. The literacy rate is calculated by the 3 R’s namely ‘Read’, ‘wRite’ and ‘aRithmetic’. so arithmetic is an important role in our daily life. Arithmetic reasoning example and practice problems are given below. 


Arithmetic reasoning example problems:


1. A Washing machine costs Rs.5, 00,000/–. If the value depreciates 15% the first year, 13½% the next year 12% the third year and so on. What will be its value at the end of 10 years, the entire percentages applying to the original cost?

Sol:  The percentage of depreciation in value in consecutive years form an A.P.
Total depreciation (in %) = 15 + 13½ + 12 + … to 10 terms, Here a = 15, d = – 1.5
Sn = n/2 [ 2a + (n-1)d]

Sn = 10/2 [ 30 - 13.5 ] in % = 82.5%
Value of the machine after 10 years = 100 – 82.5 = 17.5%
Original cost of the machine = Rs.5,00,000
Value of the machine after 10 years = Rs.5,00,000 ×17.5/100 = Rs.87,500

2. Convert the following into percentages
(a) 2/5 (b) 3/4
Sol:
(a) 2/5 =2/5 ×100/100 =40/100 = 40%
(b) 3/4 =3/4 ×100/100 =75/100 = 75%

3. The monthly salary of Rani is Rs. 4000. She spends 80% of her salary every month. How much does she save every month?
Sol:
Rani's monthly salary = Rs. 4000
Expenditure = 80% of 4000
= 80/100 × (4000) = Rs. 3200
Savings = 4000 – 3200 = Rs. 800


Having problem with Property of Multiplication keep reading my upcoming posts, i will try to help you.


Arithmetic reasoning problems to practice:


1. Gilly deposits Rs. 2000 in a bank. The bank pays interest at the rate of 4% per year. Find the interest received by him at the end of 3 years. Also, find the amount to be paid at the end of 3 years.

Ans: Rs. 2240.

2. In a room there are 25 boys and 15 girls. What is the ratio between boys and girls?

Ans: 5: 3

how to do math arithmatic

Introduction to math arithmetic:

The basic math arithmetic properties are addition, subtraction, multiplication and division. Let discuss about how to do work with these arithmetic properties. In addition the sum of two or more numbers is said to be addition. Subtraction – it is the inverse of the addition. Multiplication is one of the major progresses in the math arithmetic. I like to share this Arithmetic Mean Formula with you all through my article.


math arthmatic addition and subtraction


math arithmatic addition:

The sum of one or more numbers is said to be addition.

For example:

Find the sum of the given values

22, 34

First arrange the numbers in the vertical format, and add the numbers at the right hand side.

22

34

_____

56

______

Another example for math arithmatic addition:

Find the sum of the given values

56, 78, 49, 34, 62.

Write the numbers in the vertical format.

Add the numbers from the right hand side. The remainder is placed in the head of the left side numbers.

(2) Balance

56

78

49

34

62

______

279

______

Subtraction:

Subtraction is the inverse form of the addition.

For example:

Find the difference between 78 and 32

Write the numbers in the vertical format

78

32

____

56

____

Another example for the subtraction:

Find the difference between the decimal values. 56.27 and 34.14.

Write the numbers in the vertical format.

56.27

34.14

_______

22.14

_______

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math arithmetic division and multiplication

Multiplication:

It is used to product of two numbers. It is another fundamental arithmatic operation. It is denoted as `xx` .

Example:

Find the product of the given values.

34 is multiply by 2

34 * 2

_______

68

_______.

Another example for the multiplication.

Find the product of the given values.

If 574 is multiplied by 6.

574 * 6

________

3444

_______



Division:

It is the opposite for multiplication. It is represented as /, `-:`

Example 1:

Find the values using division methods.

42 divided by 2.

21) 42 (2

4

____

02

2

______

0

______

Another example for the division.

The number 345 is divided by 5.

69) 345 (5

30

_____

45

45

______

0

_______

Calculating Half Life

Introduction to Original half life :-

The formula for Exponential decay is N = N0eKt

N0 is the initial population and the decay rate  is k

N is the final population after the time t with the decay rate k.

Half- Life :-

In half life `N = N_0/2 `

Plug in the the value of  N as `N_0/2` in the initial equation we get .

`N_0/2``=N_0 e^(kt)`

Now canceling N_0 in both side and solving we get .

`1/2 = e^(kt)`

`2^-1 = e^(kt)`

Take ln on both sides of above equation we get

` ln (2^-1) = ln (e^(kt))`

Solving it we get

-1 ln(2) = kt.

Solving for t we get

` t = -ln2/k`

Half life time ` (t) = -ln(2)/k`

Now lets  see some solved problems in the topic origianl half life.


Original Half Life Solved Example Problem

The original half-life of an plutonium particle is 18,000 years. If 30 gram are present now, find how long it will take until only 20% of the original atomic particle remains.

Solution To Half life Day One Solved Example Probblem 1:-

The half period time is 18,000 years.

The initial amount is 30 gram..

We need to find the time it will take to make the 20% of original plutonium particle.

First we need to find the decay constant k.

The formula used for decay constant is

Plug in the values in the formula we get.

`18, 000 = - ln (2)/ k`

Now arraigning the above expression to find the value k we get

`k = - ln (2)/ (18,000)`

The value of ln(2) is 0.6931.

`= -0.6931 /( 18,000)`

`(-0.6931) / (18 000) = -0.0000385`

Therefore the value of decay constant is -0.0000385.

Now we have to find the time when there will be only 20% of atomic particle remains.

`6 = 30 xx e^(-0.0000385 t)`

Now divide by 30 on both side of the above expression.

`6/ 30 = (30 xx e^(-0.0000385 t))/ 30`

`0.2 = e^(0.0000385 t)`

Taking ln on both side of the above expression we get

`ln (0.2) = ln (e^(-0.0000385 t))`

`ln (e^(-0.0000385 t))` can be written as `-0.0000385 t.`

`ln(0.2) = -0.0000385 t..`

Now divide by `- 0.0000385` on both side of the above expression we get .

`ln(0.1)/ -0.0000385 = (-0.0000385 t)/ -0.0000385.`

By solving the above expression we get

`ln(0.2)/ -0.0000385 = t`

By solving the above fraction we get

41,803 = t

20 percent of  30 gram plutonium particle remains after 41,803 years.

Original Half Life Practice Problem

The half-life of an Barium particle is 16,000 years. If 40 gram are present now, find how long it will take until only 10% of the original atomic particle remains.

Answer:-

10 percent of  40 gram plutonium particle remains  after 40,859 years.

Definition of Scale in Math

Introduction to definition of scale in math:

In math a drawing that displays a real thing with accurate sizes without they have all been strong or distended by a exacting amount called as scale.

In math the definition of scale is showing as the length of the drawing, next that a colon (":"), and then the identical length on the real thing. Let, we learn about the how to do scale drawing ratio. This article shows the clear definition of the scale and  the how to perform the scale drawing  ratio with its examples.

Definition of Scale Drawing Ratio in Math:
In real situation, the length of the van is assumed as 300 inches. But, the length of a copy or print paper that you could use to outline this van is a little bit lower than 12 inches.

As 300/10 = 25, you will require about 25 sheets of copy sheet to formulate the length of the original size of the van.

In order to expand just one paper, you could then use 1 inch on your draft to specify 25 inches on the real-life things.

Therefore, we can write this as in ratio form of 1:25 or 1/25 or 1 to 25. Is this topic Division Operations hard for you? Watch out for my coming posts.

Examples of Definition of Scale in Math:

Example 1: The distance amid two towns measures 9cm on a map. What is the true distance if the scale is 1 : 40 000

Solution:

The True Distance = 9cm x 40 000

= 3600 00cm

We can shorten this by dividing the 360000 by 100 which gives us 3600m

and by again dividing the 3600m by 1000 we can simplify this to 3.6km

So the True Distance amid the two towns is 3.6km.

Example 2: The drawing of an Aircraft uses a scale of 1 : 5000 and if the aircrafts wingspan is 100m what length on the drawing actually refers this?

Solution:

Drawing Length = 100m ÷ 5000

= 0.02 m ( to change 0.02 to mm X 1000)

= 20mm

Therefore, the length on the drawing that refers the wingspan is 20 mm.

Definition of Length Kids Math

Introduction to definition of length kids math:

Length is one kind of measurement in math geometry and it describes the object’s proportions. Occasionally the height of an entity derives the length. Length is one dimension measurement in math. The distance also identical as length and it contains metric units. Depends on measurement, the units are used. The definition of length is very much used for kids for doing any type of length problems. Let us see the more information about the definition of length kids math.

Explanation for Length Definition in Kids Math:

Definition of length in kids math:

Length is also known as distance and measure the entity displacement or space among two end peaks.

Units for length in math:

Based on the length definition, there are two systems used for units of length. They are,

Metric system
Customary units
Metric system for length measurement in math:

The unit metric system is described as,

10 mm = 1 cm
10 cm = 1 dm
10 dm = 1 m
1,000 m = 1 km
Customary units:

The unit customary is described as,

12 inches = 1 feet
3 feet = 1 yard
5,280 feet = 1 mile
1,760 yards = 1 mile
Conversion of length from large unit to small unit:

The following points are describing the conversion of length from large unit to small unit.

1 cm = 0.3937 inches
12 inches = 1 feet
1 inch = 2.54 cm
1 feet = 0.3048 m
3 feet = 1 yard
1 m = 3.28083 feet
1 km = 0.6214 miles
In supplementary section of math, the distance is calculated using formula in math geometry. The distance among two end peaks is calculated as, `d=sqrt((x2-x1)^2+(y2-y1)^2)`

Examples of Length Kids Math:

From the definition of length we can solve the kids length problems.

Problem 1: Length of road is 7,000 m and change it into km.

Answer:

We know the conversion rule that is 1,000 m = 1 km.

Hence, the road length is divided by 1000 as 7,000 / 1,000 = 7 km.

Problem 2: How many feets are used for change the 48 inches length of car.

Answer:

We know the conversion rule that is 12 inches = 1 feet

Hence, the 48 inches are changed as 4 feets.

Practice problem for length kids math:

1. Find the distance among the two points (6, 2) and (8, 4).

Solution: d = 2.82 cm.

2. Convert the 80 mm length of box into cm.

Solution: Length = 8 cm.

Present Value Calculation Formula

Learning present value calculation formula:
An annuity is a series of payments of a fixed amount of money at regular intervals of time.usually the interval is a year. But the interval may be half year, quarter year or monthly, etc., unless otherwise stated about this interval, it will be taken as a year. The payments may be for a fixed number of years or to continue forever. If it is for a fixed number of years then the annuity is called Annuity certain. If it is continued forever it is called perpetual Annuity, or perpetuity.

If each payment of an annuity is made at the end of each period the annuity is called Immediate annuity. If each payment is made at the beginning of each period, the annuity is called annuity due.

When an annuity is payable after a lapse of a given period, it is called as Deferred Annuity.

Present Value Definition and Calculation Formula:

The present value of an annuity is the sum of all present values of various instalments of the annuity.

Let us learn the present value calculation formula for different annuities.

Present value of an immediate annuity = `(a)/(i)[ 1 - (1+i)^(-n)]` where $a is the present value to be paid at the end of first year at the rate of i per dollar.

Note 1:

If the annuity is the annuity due, then

P = `(a(1+i))/(i)[1- 1/(1+i)^n]`

Note 2:

In case of immediate annuity if the instalments are paid k times a year, then

P = `(a)/(i)[ 1 - 1/(1+i/k)^nk]`

Present value of deferred annuity is `(a)/(1+i)^d.1/i[1-(1+i)^-n]`

Present value of perpetuity deferred for d years = `(1)/(1+i)^d xx 1/i`

Understanding Difference Quotient Formula is always challenging for me but thanks to all math help websites to help me out.

Problems Using Present Value Calculation Formula

Find the present value of an annuity of $5000 per annum for 12 years, the interest being 4% per annum compounded annually.
a = $5000, n = 12 years, i = 0.04

P = `(a)/(i)[ 1 - 1/(1+i)^n]`

Substituting the values we get, P = $46925

A man retires at the age of 60 and earns a pension of $8700 a year. He wants to commute one third of his pension. Find the amount he will receive, if the expectations of life at this age be 10 years, and the interest is compounded at 4% per annum.
Solution:

Annual pension = $8700

Commuted amount = 1/3 of the pension

So, a = 2900

n = 10 years

i = 0.04

Substituting the values in the formula we get P = $23519

Main Sequence

Introduction to main sequence:

A set of numbers whose domain is a real number is called a SEQUENCE and sum of the sequence is called a SERIES.

Those main sequences whose terms follow certain patterns are called progressions.

For example

·         1, 4, 7, 10, 13 …

·         7, 4, 1, – 2, – 5 …

·         1, 2, 4, 8, 16 …

·         8, 4, 2, 1, ½ …

Types of Main Sequence:

The main sequences can be classified as

There are three different progressions

·         Arithmetic Progression (A.P)

·         Geometric Progression (G.P)

Arithmetic Progression (A.P.)
It is a series in which any two consecutive terms have common difference and next term can be derived by adding that common difference in the previous term.

Geometric Progression
A series in which each preceding term is formed by multiplying it by a constant factor is called a Geometric Progression . The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it.

Properties of Main Sequence:

Properties of the Main Sequence:
If each term of a main sequence is increased, decreased, multiplied or divided by the same non-zero number, then the resulting series will also be a sequence. This can be explained as  3, 5, 7, 9, 11… is a  main sequence and let each term of the sequence is increased by 5    then the new sequence will be 8, 10, 12, 14, 16 ...  will also be a new main sequence.
The difference between the two consecutive terms of the main sequence that is in AP will always remain constant.
If each term of a main sequence, which is in Geometrical progression, is multiplied or divided by the same non-zero  quantity, then the resulting sequence is also a Geometric    progression and can be classified as main sequence.
For example: For G.P. is 2, 4, 8, 16, 32… If each of the term is multiplied by 2 we get the sequence 4, 8, 16, 32,                                 64 … which is also in Geometrical progression. Please express your views of this topic math formula sheet by commenting on blog.

Questions for Practice on Main Sequence:

What will be the next two term of the main sequence 6, 17, 28, 39, __ ,  __  ?
If we add to an arithmetic progression then the new sequence will also be in AP?   (True  / False  )
The difference between the consecutive terms of main sequence which is in AP is always constant?   (True  / False)

Relative Percent Difference Formula

Introduction to relative percent difference formula:

Percent difference or relative percent difference (RPD) among two numbers is the difference among them as a percent of one of them. It is frequently used as a quantitative indicator of quality pledge and quality control for frequent measurement where the result is expected to be the similar. The common obligation for choosing two values to be contrast is that the user of this method expects the two values to be numerically equal. This article defines the relative percent difference formula and its example problems.

Relative Percent Difference Formula:

Let, we see about the formula used to find the relative percent difference.

The difference percent of two experimental values x1 and x2, can be determined by dividing the absolute difference of the two data by the average value of the similar two values as described in the below formula:

% Difference:

` |(x1-x2)/(((x1+x2))/2)|xx100`

Here, x1 and x2 should have equal units order to be compared properly with one another. And as state before, a zero percent difference is best and the higher the percent data, the low precision of the two datas.

Examples Using Relative Percent Difference Formula:

Example 1:

If the value of x1 = 20 and the value of x2 = 15 then find the relative percent difference using the formula.

Solution:

Given x1= 20 and x2= 15

Formula:

` |(x1-x2)/(((x1+x2))/2)|xx100`

Solve:

% diff:

`= |(20-15)/(((20+15))/2)|xx100`

`= |(5)/(((35))/2)|xx100`

`= |(5)/17.5|xx100`

`= |0.2857|xx100`

= 28.57

Which is the required relative percent difference.

Example 2:

If the value of x1 = 30 and the value of x2 = 10 then find the relative percent difference using the formula.

Solution:

Given x1= 30 and x2= 10

Formula:

` |(x1-x2)/(((x1+x2))/2)|xx100`

Solve:

% diff:

`= |(30-10)/(((30+10))/2)|xx100`

`= |(20)/(((40))/2)|xx100`

`= |(20)/20|xx100`

`= |1|xx100`

=100

Therefore, 100 is the required relative percent difference. Is this topic how to measure diameter hard for you? Watch out for my coming posts.

Practice Problems Using Relative Percent Difference Formula:

Problem 1:

If the value of x1 = 10 and the value of x2 = 8 then find the relative percent difference using the formula.

Solution:

22.22

Problem 2:

If the value of x1 = 100 and the value of x2 = 50 then find the relative percent difference using the formula.

Solution:

66.66

Mean Diameter Formula

Introduction to mean diameter formula:

In mathematics, mean diameter is mainly used for the defining the shape of the circle. The other name given for the mean diameter is known as neutral axis. There are two terms present in the mean diameter. Two terms are called as the inside diameter and the thickness. In this article, we are going to study about mean diameter in detail with some examples problems.

Explanation to Mean Diameter Formula

The explanation to mean diameter formula is given below the following section,

Formula:

1)Mean Diameter:

Mean Diameter = Inside Diameter + Thickness

2)Another Mean Diameter:

Mean Diameter = Outside Diameter - Thickness

3)Mean circumference:

Mean Circumference  = Mean Diameter `xx` `Pi`

Please express your views of this topic integer number line by commenting on blog

Example Problems to Mean Diameter Formula

Problem 1: Find mean diameter for the cylinder, where thickness is 25mm, inside diameter is 650mm.

Solution:

Step 1: Given:

Thickness is 25mm

Inside diameter is 650mm

Step 2:To Find:

Mean Diameter

Step 3: Formula:

Mean Diameter = Inside Diameter + Thickness

Step 4: Solve:

Mean Diameter = Inside Diameter + Thickness

MD = 650 + 25

= 675

Result: Mean Diameter = 675mm.

Problem 2: Find mean diameter for the cylinder, where thickness is 25mm, outside diameter is 650mm.

Solution:

Step 1: Given:

Thickness is 25mm

Outside diameter is 650mm

Step 2:To Find:

Mean Diameter

Step 3: Formula:

Mean Diameter = Outside Diameter - Thickness

Step 4: Solve:

Mean Diameter = Outside Diameter - Thickness

MD = 650 - 25

= 625

Result: Mean Diameter = 625mm.

Problem 3: Find mean circumference for the cylinder, where thickness is 25mm, outside diameter is 650mm.

Solution:

Step 1: Given:

Thickness is 25mm

Outside diameter is 650mm

Step 2:To Find:

Mean Circumference

Step 3: Formula:

Mean Circumference  = Mean Diameter  `xx`

Step 4: Solve:

Mean Circumference  = Mean Diameter  `xx`

Mean Diameter = Outside Diameter - Thickness

MD = 650 - 25

= 625

Mean Circumference  = Mean Diameter  `xx`

MC = 625  `xx`  3.14

= 1962.5

Result: Mean Circumference  = 1962.5mm

Simply Math and Reading

Introduction for Simply Math and Reading:

Mathematics are also called as math. Math is study of calculation, quantity. It has different types of symbols, formulas and numbers. Using math symbols, formulas and numbers, we can easily solve the math problems. Some of the symbols are given below;

Addition (+),

Subtraction (-),

Multiplication (×) and

Division (÷)

In this article we shall discuss about simply math and reading. The following are the examples involved in simply math and reading.

Simply Math and Reading Example: 1

Solve the sum and find the value of ‘x’

`x/9` =-90

Multiply both sides by 9:

`(x*9)/9` = -90 * 9

Simplify both sides:

x   = - 810

Simply Math and Reading Example: 2

Solve the sum and find the value of ‘x’

9x + 90 = -900

Subtract 90 from both sides:

9x + 90 – 90 = -900 - 90

Simplify both sides:

9x = -990

Divide both sides by 9:

`(9x)/9`  = `-990/9`

Simplify both sides:

x   =   -110

Simply Math and Reading Example: 3

Solve the sum and find the value of ‘x’

`(90x)/40` = 90

Multiply both sides by 40:

`((90x)*40)/40` = 90*40

Simplify both sides:

90x   = 3600

Divide both sides by 90:

`(90x)/90`    =   `3600/90`

Simplify both sides:

x   =   40

Simply Math and Reading Example: 4

Solve the sum and find the value of ‘x’

`(90(x-90))/40` = 90

Multiply both sides by 40:

`((90(x-90))*40)/40` = 90*40

Simplify both sides:

90(x - 90)   =   3600

Divide both sides by 90:

`(90(x - 90))/90`    =   `3600/90`

Simplify both sides:

x - 90 =   40

Add 90 to both sides:

x - 90 + 90 =   40 + 90

Simplify both sides:

x   =   130

Simply Math and Reading Example: 5

Solve the sum and find the value of ‘x’

99x + 90 = 90x + 9

Subtract 99x from both sides:

99x + 90 - 90x =   90x + 9 - 90x

Simplify both sides:

9x + 90   =   9

Subtract 90 from both sides:

9x + 90 - 90 =   9 - 90

Simplify both sides:

9x   =   -81

Divide both sides by 9:

`(9x)/9`    =   `-81/9`

Simplify both sides:

x   = - 9

Open Ended Math Word Problems

open ended math word problems:

Open ended math word problems defines the math problem that are given in the word format.For example instead of using the math representation of math symbols they express the problem  in the word format.The open ended math problem exist when the problem happens in the real time.The open ended math problem is analyzed and the solution is extracted.Some of the problems are solved in the following.

Open Ended Math Word Problems

open ended math word problems Example 1:

Two farmers bought together 75 mangoes one bought 40 mangoes.How many mangoes did the other farmer buy?

Step 1:

It is important that when the word problem is converted to mathematical problem it is essential that its meaning should not be changed.

step 2:

The given problem exists between the two farmers.The two farmers bought together 75.

step 3:

Let the other farmer  who bought unknown mangoes be x.

In the open ended math word problems  the problem analyzed is between the two farmers.

step 4:

Hence the equation formed is

` 40 + X = 75`

step 4:

To find the other farmer who bought the unknown mangoes is solved

` x=75-40`

` x = 35`

step 5:

The other farmer  who bought mangoes will be 35.

step 6:

It satisfies the given equation and hence one farmer got 40 mangoes and the other farmer got 35 mangoes total of 75 mangoes

Open Ended Math Word Problems

open ended math word problems Example 2:

Two teachers bought together 62 gifts one teacher bought 22 gifts.How many gifts did the other teacher buy?

Step 1:

It is important that when the word problem is converted to mathematical problem it is essential that its meaning should not be changed.

Step 2:

The given problem exists between the two teachers.The two teachers bought together 62.

Step 2:

Let the other teacher  who bought unknown gifts be x.

In the open ended math word problems  the problem analyzed is between the two teachers.

Step 3:

Hence the equation formed is

`62+X=22`

Step 4:

To find the other teacher who bought the unknown gifts is solved

`X=62-22`

`X=40`

Step 5:

The other teacher who bought gifts will be 40.

Step 6:

It satisfies the given equation and hence one teacher got 22 gifts and the other teacher got 40 gifts total of 62 gifts.

Divisor Math

Introduction to Divisor Math:

An integer which is divided by another number, the dividend, is called as divided. The divisor is otherwise called as a factor while dividing an integer; it evenly divides the dividend without leaving a remainder. For example: 12 ÷ 3 = 4, here 12 is the dividend, 3 is the divisor and 4 is a quotient. Let us see about divisor math in this article.

Worked Examples to Divisor Math

Example 1 to Divisor Math:

Find the quotient for the problem using divisor.

84 ÷ 4

Solution:

Step 1:

Write the divisor 4 out of the division brackets and dividend 84 inside the division bracket.

______
4) 84

Step 2:

Find the divisor 4 goes into first digit of the dividend 8. The multiplicity of 4 and 2 gives 8.

____
4 ) 84  ( 2
8
-----------
0

Step 3:

Bring down the next digit 4 of the dividend. Continue dividing 4 by 4 we get 1.

21
-----------
4) 84
8
-------------
04
04
------------
0
-------------

Hence, the problem with divisor 4 divides 84 gives the quotient of 21.

Example 2 to Divisor Math:

Find the quotient for the problem using divisor.

144 ÷ 3

Solution:

Step 1:

Write the divisor 3 out of the division brackets and dividend 144 inside the division bracket.

-----------
3) 144

Step 2:

Find the divisor 3 goes into first digit of the dividend 1. It does not go into 1.

____
3) 144  ( 0
0
-----------
14

Step 3:

Bring down the next digit of the dividend. Take two digits as a dividend as 14 and the divisor 3 go into 14 for 4 times gives the result as 12.

____
4) 144 ( 4
0
-----------
14
12
------------

Step 4:

Subtract the 12 from 14 gives 2. Bring down the next digit 4 of the dividend.

____
4) 144  (
0
----------
14
12
------------
24
------------

Step 5:

The multiplicity of 8 and 3 gives the result 24. And, then subtract the result leave a zero remainder.

____
4) 144 ( 48
0
-----------
14

12

------------

24
24
------------
0
-------------

Hence, the problem with divisor 4 divides 84 gives the quotient of 21.

Practice Problems to Divisor Math

Problem 1:

Find the quotient for the problem using divisor.

72 ÷ 2

Problem 2:

Find the quotient for the problem using divisor.

174 ÷ 3

Solutions:

1. Quotient is 36

2. Quotient is 58

Mental Arithmetic Answers

Introduction to mental arithmetic answers

In this topic "mental arithmetic answers", we will see some questions on arithmetic, where the answer has to be obtained through mental calculations. This topic is a kind of mental exercise on arithmetic. You can calculate and check your answers with the answers given below every mental arithmetic questions. I like to share this What Does Arithmetic Mean with you all through my article.

Problems on Mental Arithmetic Answers

1. Write five prime numbers between 50 and 75.

[ Answer: 53, 59, 61, 67, 71]

2. Every prime number is odd except

[ Answer: 2]

3. What should be added to 40.09 to make it 51 dollars?

[ Answer: 10.91 dollars]

4. What is `9/50` of a dollar?

[ Answer: 0.18 dollars]

5. How many cm in 7 `1/4` m?

[ Answer: 725cm]

6. Give a number which is neither prime nor composite.

[ Answer: 1]

7. Provisions last for 14men for 2 weeks; how long will it last for one man?

[ Answer: 196 days]

More Problems on Mental Arithmetic Answers

8. Take away the sum of 99 and 99 from 1000.

[ Answer : 802]

9. 3/5 of a dollar + 1/8 of 2 dollar

[ Answer : 0.85 dollar or 85 cents]

10. 1 `1/2` hour after 4.15 pm

[ Answer : 5.45pm]

11. Which is greater `7/8 ` or `11/12` ?

[ Answer : `11/12` ]

12. How many grams in 6 `1/8` kg?

[ Answer : 6125 grams]

13. Simplify: 3 + `3/1000` + `2/10` + `4/100` .

[ Answer : 3.243]

14. 3/7 of 42 seconds.

[ Answer : 18 seconds]

Between, if you have problem on these topics Percentage Calculator Online, please browse expert math related websites for more help on prime numbers less than 1000.

More Problems on Mental Arithmetic Answers

15. A motor car travels 11 km in 1 liter of petrol. How many liters of petrol will be needed for a journey of 297 km?

[ Answer : 27 liters]

16. Two boys share 15 dollars. One of them takes 2/3 of it. How much the other get?

[ Answer : 5 dollars]

17. `2/5` of an hour is how many minutes?

[ Answer : 24 minutes]

18. What is the difference, in grams, between 2kg and 2900 grams?

[ Answer : 900 grams]

19. 2 men digging a hole in a road take 8 days to complete the work. How long would 4 men take?

[ Answer : 4 days]

20. A bus traveled for 4 hours at an average speed of 30 km/hr. How far has it travelled?

[ Answer : 120 km]

I believe that there  would have been some mental exercise would have happen after going through this material.