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.