Statistical Variable

Statistical Variable Meaning | Different Types of Variables in Statistics with Examples PDF

In the previous articles, we have explained what is Statistics and Statistical Data and here we will be discussing the statistical variable and their types. These variables in statistics are classified into different types of variables.

Kids can refer to this article for a better understanding of the statistical variable concept like definition, types of statistical variables, the differences between classified statistical variables with examples, etc. Let’s dive into this guide completely and be familiar with the concept for solving complex questions.

What is a Variable?

In Statistics, you will find two defining for a variable. They are like:

  • A variable is a characteristic that defines a person, place, thing, or idea. Also, it can be called a data item.

Examples of Variables are Age, sex, business income and expenses, country of birth, capital expenditure, class grades, eye color, and vehicle type.

  • The value of the variable can “vary” from one entity to another, and also it may alter in value after a while.

For instance, ‘income’ is a variable that can change between data units in a population and also change after some time for each data unit (i.e. income can go up or down).

Definition of Statistical Variable in Math

A Statistical Variable is a variable that holds discrete values that vary through random situations and when arranged in the order form a statistical distribution or array.

Different Types of Variables in Statistics

In the world of mathematical statistics, the algebraic term which represents the unknown value (ie., not fixed value) that is in numerical format is called a variable. These types of variables are used for many types of research to perform easy computations. Hence, you will see several different variable types that are applied in distinct domains available. A few of the types of statistical variables are as follows:

1. Independent variables
2. Dependent variables
3. Categorical variables
4. Continuous variables

Classification of Statistical Variables

On the basis of the nature of variables, statistical variables are mainly classified into two categories. They are

(i) Numerical (Quantitative )
(ii) Categorical (Qualitative)

The chart of statistical variables classification is given below:

statistical variable types classification

Quantitative Variable in Statistics (aka Numerical Statistical Variables)

A quantitative variable is a variable that includes quantitative data. Numerical variables are measurable or countable variables. Some of the quantitative variables examples are plant height, fruit weight, crop yield, number of petals, seeds, leaves in a plant, etc. Further, it is categorized into (a) Discrete variables and (b) Continuous variables. Look at the below modules and gain more knowledge about types of quantitative statistical variables.

Discrete vs Continuous Variables

The following table will make you understand the difference between two types of quantitative variables in statistics ie., Discrete & Continuous.

Type of variable What does the data represent? Examples
Discrete variables Counts of individual items or values.
  • Number of students in a class
  • Number of different tree species in a forest
Continuous variables Measurements of continuous or non-finite values.
  • Distance
  • Volume
  • Age

Qualitative Statistical Variables (aka Categorical Variables in Statistics)

A variable that includes qualitative data is called Qualitative Variable. Categorical data describe groupings. These variables are un-measurable variables. Examples of categorical statistical variables are the color of flowers, the shape of leaves, the shape of seeds, etc. Moreover, categorical type of variables is classified into three types. They are binary, nominal, and ordinal variables.

Difference Between Nominal and Ordinal Variables

Nominal Ordinal
Variable with two or more categories Variable with two or more categories
The order of the categories doesn’t matter The order of the categories does matter
Examples:

  • Name
  • Hair color
  • Nationality
Examples:

  • Satisfaction level
  • Product rating

FAQs on Variables in Statistics

1. Define Statistical Variable? 

The quantity whose different values during observation constitute the collection of data is called a statistical variable.

2. What are independent and dependent variables in statistics? 

Variable that denote the result of the experiment is called dependent variables whereas variable that manipulates to affect the outcome of an experiment is called the independent variable.

3. What is a continuous variable in statistics?

The variables that assess some number or quantity and don’t hold any limits are named as continuous variables and they can be divided into ratio or interval, or discrete variables.

Properties Questions on Arithmetic Mean

Properties Questions on Arithmetic Mean with Answers | Word Problems on Properties of Arithmetic Mean (Average) PDF

Students who are getting confused by understanding the AM Properties with proof. If that’s the concern then have a look at this Properties Questions on Arithmetic Mean article. In this article, you will see how to solve the different example questions on average/mean/AM by using properties of arithmetic mean.

Simply follow the steps covered in the solution of Problems Based on Average & understand the explanations for solving various complex questions on the arithmetic mean properties. Also, Improve your math proficiency by answering the various example word problems on Mean/Average/Arithmetic Mean in Statistics.

Do check:

Example Questions on Arithmetic Mean Properties with Solutions PDF

Example 1: 
The mean of nine numbers is 40. If four is subtracted from each number, what will be the new average.
Solution: 
Let the given numbers be x1, x2, x3,…., x9
Then, the mean of these numbers = (x1+x2+x3 + …. +x9)/9
Hence, (x1+x2+x3 + …. +x9)/9 = 40
⇒ (x1+x2+x3 + …. +x9) = 40 x 9
⇒ (x1+x2+x3 + …. +x9) = 360 ……(i)
The new numbers are (x1 – 4), (x2 – 4),…., (x9 – 4)
Mean of the new numbers = {(x1 – 4)+ (x2 – 4)+….+ (x9 – 4)} / 9
= [(x1+x2+x3 + …. +x9) – 36] / 9
= [360-36]/9, [using (i)]
= 324/9
= 36
Hence, the new mean is 36.

Example 2:
Find the average of the first five prime numbers.
Solution:
Given quantities are First five prime numbers
The first five prime numbers are 2, 3, 5, 7 and 11
The formula of Arithmetic mean or the Average = Sum of the quantities/Number of quantities
= 2+3+5+7+11 / 5
= 28/5
= 5.6

Example 3: 
Find the second term of the given A.M terms \(\frac{1}{log_{3}3} \), X , \(\frac{1}{log_{18}3}\)
Solution:
Given terms are \(\frac{1}{log_{3}3} \), X , \(\frac{1}{log_{18}3} and this can be rewritten as log3 3, X, log3 18
⇒ 2X = log3 3 + log3 21 = log3 3 + log3 (6×3)
⇒ 2X = 2 ( log3 3) +   2 log3 3
⇒ 2X = 2 ( log3 3) +   (2 x 1 )
⇒ 2X =  2(1) + 2
⇒ X = 4/2
⇒ X = 2

Example 4:
The average of 10 numbers is 5. If 2 is added to every number, what will be the new average?
Solution:
Let the given numbers be x1, x2, x3,…., x10
Then, the average of given numbers = x1, x2, x3,…., x10 / 10
Hence, (x1+x2+x3+…+x10)/10 = 5
⇒ (x1+x2+x3+…+x10) = 50 …….(i)
The new numbers are (x1 – 4), (x2 – 4),…., (x10 + 2)
Mean of the new numbers = (x1 + 2), (x2 + 2),…., (x10 + 2) / 10
= (x1+x2+x3+…+x10) + 20 / 10
= (50 + 20)/10 [Using (i)]
= 70/10
= 7.
Hence, the new mean is 9.

Example 5: 
In the exams, the mean of marks scored by 30 students was calculated as 60. Next, it was identified that the marks of one student were wrongly copied as 57 instead of 75. Find the correct mean?
Solution:
Given that Mean of marks = [latex]\frac { Incorrect sum of marks of 30 students }{ 30 } \)
\(\frac { Incorrect sum of marks of 30 students }{ 30 } \) = 60
An incorrect sum of marks of 40 students = 60 x 30 = 1800
Considering that the marks of one student were wrongly copied as 57 instead of 75,
Then, the correct sum of marks of 30 students = 1800 – 57 + 75 = 1818
Finally, correct mean = 1818 / 30 = 60.6

Example 6: 
The mean marks of two batches of students having 60 and 40 students respectively are 35 and 65. Find the average marks of all the 100 students, taken together.
Solution:
Let x be the average marks of all 100 students taken together.
Batch – I (\(\overline{x}\))
Marks (x1) = 35
No. of students n1 = 60
Batch – II (\(\overline{x}\))
Marks (x2)= 65
No. of students n2 = 40
\(\overline{x}\) = \(\frac {n1[latex]\overline{x1}\) + n2\(\overline{x2}\)}{ n1+n2 } [/latex]
= \(\frac { 60×35 + 40×65 }{ 60+40 } \)
= \(\frac { 4700 }{ 100 } \)
= 47.
Therefore, \(\overline{x}\) = 47 marks.

Decimal Representation of Rational Numbers

Decimal Representation of Rational Numbers | How do you find the Decimal of a Rational Number?

Before getting into the actual concept of the Decimal Representation of Rational Numbers firstly, let us understand what are rational numbers. Rational Numbers are the numbers that can be expressed in the form of p/q where p q are integers and q is a non-zero number. Whenever we simplify the rational numbers the resultant is the decimals. In this article, we will cover everything on how to expand rational numbers in decimal form all explained with enough examples.

Decimal Representation of Rational Numbers

Decimal Representation of Rational Numbers is simply converting a rational number to a decimal form having the same mathematical value. We can express a rational number in its decimal form using the long division method. The quotient which we obtain in the long division process is known as the decimal representation of the rational number.  The two types of decimal representation of Rational Numbers are

  • Terminating
  • Non-Terminating or Repeating

In a long division method if you get the remainder as zero then decimal expansion of such number is called Terminating. And while dividing if the decimal expansion continues and the remainder doesn’t become zero then it is called Non-Terminating.
Decimal Representation of 1/5
Decimal Representation of 1 by 5

 

 

 

Here 1/5 = 0.2 is a terminating decimal since the remainder is 0.

Decimal Representation of 2/3
Decimal Representation of 2 by 3
2/3 =0.66666 the process repeats and the remainder isn’t zero thus 2/3 is a repeating or non-terminating decimal.

Also, Check:  Rational Numbers in Terminating and Non-Terminating Decimals

Decimal Representation of Terminating Rational Number

Terminating Decimal Expansion refers to decimal representation or expansion that terminates after a certain number of digits. In the Case of a Terminating Decimal Expansion, you will find the Prime Factorization of the Denominator has no other factors other than 2 and 5.

Decimal Representation of Terminating Rational Number

In the Case of Non-Terminating but Repeating Decimal Expansion, the decimal representation has an infinite number of digits. However, there is a pattern of numbers that repeats. Rational Numbers whose denominators have factors other than 2 or 5 don’t have a terminating decimal number as result.

Solved Examples on Decimal Representation of Rational Numbers

Example 1.
Express 1/12 in Decimal Form?

Solution:

Here Divisor = 12, Dividend = 1
Finding the Decimal Form of the given rational number using the long division method we have
Decimal Representation of 1 by 12

Thus 1/12 denoted in decimal form is 0.08333…. Therefore, it is a non-terminating or repeating decimal.

Example 2.
Express 11/24 in decimal form?
Solution:

Here Dividend = 24, Divisor = 11
Finding the Decimal Form of the given rational number using the long division method we have as beow
Decimal Representation of 11 by 24
thus 11/24 denoted in its decimal form is 2.1818….. Thus, it is a non-terminating or repeating decimal.

Example 3.
Express 5/10 in decimal form?
Solution:
Given rational number is 5/10
Divisor = 10, Dividend =5
Performing the Long Division Method we obtain the decimal form as such

Decimal Representation of 5 by 10

5/10 =0.50 It is a Terminating Decimal.

Rational Numbers

Rational Numbers – Definition, Facts, Properties, Types, Examples | How to Identify Rational Numbers?

Rational Numbers are a type of real numbers. Go through the article to learn in detail about what is a rational number, the types of rational numbers, properties of rational numbers explained. In addition to these, you will learn the standard form of a rational number, how to identify a rational number, the difference between a rational number and an irrational number. Check out solved examples on rational numbers available and master the concept.

Also, See: Rational Numbers in Terminating and Non-Terminating Decimals

List of Rational Numbers Topics

What is a Rational Number?

In Mathematics, a Rational Number is defined as any number that can be denoted in the form of a/b where b≠0. In other words, we can say that a fraction is said to be a rational number if both the numerator and denominator are integers and the denominator is not equal to zero. On dividing a rational number we get the result in decimal form and it can be either a terminating or a repeating decimal.

Examples of Rational Numbers are 1/3, 4/5, 6/7, etc.

How to Identify Rational Numbers?

You need to check the below-listed conditions to determine if a number is a rational number or not. They are outlined for your reference

  • Rational Number is represented in the form of a/b where b≠0
  • We can simplify and represent the ratio a/b in decimal form.

Standard Form of Rational Numbers

The Standard Form of a Rational Number is defined as the number if it has no common factors between the dividend and divisor. For Example, 12/24 reduced in its simplest form is 1/2. Common factors between the divisor and dividend are only one thus it is said to be in its simplest form.

Positive and Negative Rational Numbers

A Rational number is in the form of p/q, in which p and q are integers and q should be non-zero integers. Rational Numbers can be either positive or negative. If a rational number is positive, then both p and q are positive integers. If a rational number takes the form of -(p/q), then either p or q takes the negative value. It means that

-(p/q) = (-p)/q = p/(-q).

Positive Rational Numbers Negative Rational Numbers
In the Case of Positive Rational Numbers, both the numerator and denominator will be of the same signs. In Negative Rational Numbers, numerator and denominator will be of opposite signs.
All are greater than 0 All are less than 0
Examples of positive rational numbers: 13/14, 9/10 and 3/4 Examples of negative rational numbers: -1/12, 7/-13 and -3/6.

Arithmetic Operations on Rational Numbers

Arithmetic Operations are the basic operations we perform on Integers. Below we will discuss how to perform these arithmetic operations on rational numbers.

Addition: While adding two rational numbers say a/b, c/d we need to make the denominators same. Thus we get (ad+bc)/bd. For Example 1/3+4/5 added gives (1*5+4*3)/15 = 17/15

Subtraction: In the case of subtracting rational numbers a/b, c/d we need to make the denominators the same and then subtract. For Example, 1/3-1/2 subtracted gives (6- 2)/6 i.e. 4/6 or 2/3

Multiplication: In the Case of Rational Numbers Multiplication we need to multiply the numerators and denominators respectively. If a/b * c/d is multiplied we have ac/bd. For Example 3/4*5/3 = (3*5)/(4*3) = 15/12

Division: If a/b÷c/d then we can denote the rational numbers division as (a/b)÷(c/d) = ad/bc

Multiplicative Inverse of Rational Numbers

Multiplicative Inverse of a Rational Numbers is nothing but the reciprocal of the given fraction. For Example, if 3/4 is a rational number then the multiplicative inverse of the given rational number is 1/(3/4) = 4/3

Rational Numbers Properties

As we all know Rational Numbers are a set of real numbers it obeys all the properties of real numbers. We have outlined a few of the important properties of rational numbers for your reference. They are as such

  • Whenever you add, subtract, multiply or divide a rational number with another rational number a result is always a rational number.
  • Rational Numbers will not change if we multiply or divide both the numerator and denominator with the same factor.
  • If we add zero to a rational number we will get the same number itself.
  • Rational Numbers are closed under addition, subtraction and multiplication.

Rational Numbers Vs Irrational Numbers

Fractions having non-zero denominators is called rational number. Irrational Numbers can’t be written in simple fractions instead we can represent them using decimals. Irrational Numbers have endless non-repeating digits next to the decimal point.

How to Find the Rational Numbers between Two Rational Numbers?

There are n number of rational numbers between 2 rational numbers. We can find rational numbers between two rational numbers using 2 different methods. The two methods are listed as such

Method 1:
Firstly, find the equivalent fraction for the given rational numbers and find out the rational numbers between them. The numbers obtained are the required rational numbers.

Method 2:
Obtain the mean value of the given two rational numbers. Mean value is the required rational number. To obtain even more rational numbers repeat the process with old and newly obtained rational numbers.

Solved Examples on Rational Numbers

Example 1.
Determine whether 1.25 is a rational number or not?
Solution:
Since the given number 1.25 is in the decimal format we need to change it to fraction form.  if the denominator of the fraction is non zero then the number is a rational number.
1.25=125/100
=5/4
Since denominator 4 is non zero given decimal number 1.25 is a rational number.

Example 2.
Identify whether the mixed fraction 1 4/2 is a rational number or not?
Solution:
Simplest Form of a Rational Number = 1 4/2
=6/2
Since denominator 2 is non-zero 6/2 is a rational number.

Properties of Arithmetic Mean

Properties of Arithmetic Mean | Merits and Demerits of Arithmetic Mean | AM Properties with Proof

Arithmetic Mean(AM) is the most important concept in statistics. It is one of the measures of central tendency that can be directly described as the sum of all quantities to be divided by the number of quantities. Every time we can’t apply the formula of AM to solve the problems on average or mean or arithmetic mean. So, we have explained the properties of arithmetic mean with proofs which aid students to calculate different types of questions on average with ease.

Do Check Some Other Related Articles:

What are the Properties of Arithmetic Mean?

Properties of AM are used to solve complex problems based on mean/arithmetic mean/average. Some of the important arithmetic mean properties that are used in solving the problems based on average are mentioned here briefly. Just take a look at them and be aware of all properties to use.

  • If \(\overline{x}\) is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is increased by y, the mean of the new observations is (\(\overline{x}\)+y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is decreased by y, the mean of the new observations is (\(\overline{x}\)–y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is divided by a non-zero number y, the mean of the new observations is (\(\overline{x}\)*y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is multiplied by a non-zero number y, the mean of the new observations is (y*\(\overline{x}\)).
  • If all the observations in the given data set have a value say ‘y′, then their arithmetic mean is also ‘y′.

Illustration of Arithmetic Mean Properties

Property 1: 

If \(\overline{x}\) is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.

Proof:

We know that

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Hence, (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))

= (x1 + x2 + x3 + . . . + xn) – n\(\overline{x}\)

= (n\(\overline{x}\) – n\(\overline{x}\)), [using (i)].

= 0.

Therefore, (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.

Property 2:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is increased by A, the mean of the new observations is (\(\overline{x}\) + A).

Proof: 

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1 + A), (x2 + A), …, (xn + A)

= {(x1 + A) + (x2 + A) + … + (xn+ A)}/n

= {(x1 + x2 + . . . + xn) + nA}/n

= (n\(\overline{x}\) + nA)/n, [using (i)].

= {n(\(\overline{x}\) + A)}/n

= (\(\overline{x}\) + A).

So, the mean of the new observations is (\(\overline{x}\) + A).

Property 3: 

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is decreased by p, the mean of the new observations is (\(\overline{x}\) – a).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1 – p), (x2 – p), …., (xn – p)

= {(x1 – p) + (x2 – p) + … + (xn – p)}/n

= {(x1 + x2 + . . . + xn) – np}/n

= (n\(\overline{x}\) – np)/n, [using (i)].

= {n(\(\overline{x}\) – p)}/n

= (\(\overline{x}\) – p).

Hence, the mean of the new observations is (\(\overline{x}\) + p).

Property 4:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is multiplied by a nonzero number p, the mean of the new observations is p\(\overline{x}\).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of px1, px2, . . ., pxn,

= (px1 + px2 + … + pxn)/n

= {p(x1 + x2 + . . . + xn)}/n

= {p(n\(\overline{x}\))}/n, [using (i)].

= p\(\overline{x}\).

Hence, the mean of the new observations is p\(\overline{x}\).

Property 5:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is divided by a nonzero number p, the mean of the new observations is (\(\overline{x}\)/p).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1/p), (x2/p), . . ., (xn/p)

= (1/n) ∙ (x1/p + x2/p + …. + xn/p)

= (x1 + x2 + . . . + xn)/np

= (n\(\overline{x}\))/(np), [using (i)].

= (\(\overline{x}\)/p).

Merits of Arithmetic Mean

  1. It is determined strictly.
  2. It is calculated on the basis of all observations
  3. Simple to solve and easy to understand
  4. It is responsive to mathematical treatment or properties.
  5. It is inspired by the value of every item in the observation series.

Demerits of Arithmetic Mean

  1. If any single observation is missing or lost then it’s unable to find the arithmetic mean of the data.
  2. By inspection or graphically, it is quite difficult to find the arithmetic mean.
  3. By the extreme values in the set of the data, the AM gets affected.
  4. In some situations, the arithmetic mean does not exemplify the original item.

Solved Problems on Mathematical Properties of Arithmetic Mean Proof

Example 1:
If the two variables x and y are related by 3x + 4y + 6 = 0 and x̄ = 10, then Arithmetic mean of “y” = (-6 – 3x̄) / 6, Find AM of y?

Solution:
Given x̄ = 10
Arithmetic mean of “y” = (-6 – 3x̄) / 6
Arithmetic mean of “y” = (-6 – 3×10) / 6
Arithmetic mean of “y” = (-6 – 30) / 6
Arithmetic mean of “y” = (-36) / 6
Arithmetic mean of “y” = -6

Example 2:
If a variable “x” assumes 7 observations, say 11, 22, 33, 44, 55, 66, 77 then x̄ = 44. Calculate the instance using property 1.
Solution:
Given 7 observations are 11, 22, 33, 44, 55, 66, 77 and it’s mean is 44
Now, by using property 1 ie., ∑(x – x̄) = 0.
The deviations of the observations from the arithmetic mean (x – x̄) are -33, -22, -11, 0, 11, 22, 33
Now, ∑(x – x̄) = -(33) + (-22) + (-11) + 0 +11 + 22 + 33 = 0.

Worksheet on Solving a Word Problem by using Linear Equation in One Unknown

Worksheet on Solving a Word Problem by using Linear Equation in One Unknown | Linear Equation on One Unknown Word Problems Worksheet

One can learn how to solve a word problem by using a linear equation in one unknown from here. Practice as many questions as possible from the Worksheet on Solving a Word Problem by using Linear Equation in One Unknown for a better understanding of the concept. Linear Equation on One Unknown Word Problems Worksheet includes step-by-step solutions for all the problems so that you will learn the concept on a deeper level. Enhance your math proficiency by answering the different models of questions framed on the topic in the worksheet.

See Similar:

Solving Linear Equation in One Unknown Worksheet PDF

Example 1.
The sum of the two numbers is 80. If one exceeds the other by 20, find the numbers?

Solution:

Let the number be x
From the given data one number exceeds the other by 20 the other number is x+20
Sum of two numbers = 85
x+x+20 =80
2x+20=80
2x=80-20
x=30
The other number is x+20
=30+20
=50
Therefore, the numbers are 30 and 50.


Example 2.
The three angles in a triangle are in the ratio of 4:3:5. Find the measure of each angle?

Solution:

We know Sum of the angles in a triangle = 1800
As per the given data, we can frame the equation as 4x+3x+5x=1800
12x=1800
x=1800/12
x=150
The three angles are in the measure of 4:3:5
Thus each angle is 4x = 4*150 =600
3x=3*150=450
5x=5*150=750
Thus, the three angles in a triangle are 600, 450, 750


Example 3.
The length of a rectangular plot exceeds its breadth by 6 meters. If the perimeter of the plot is 150. Find the dimensions of the plot?

Solution:

Let the breadth of a rectangular plot be x meters
Since length exceeds its breadth by 6 meters we have l = x+6
The Perimeter of the Rectangular Plot = 150 Meters
We know the formula of the perimeter of a rectangle = 2(l+b)
Substituting the known values in the formula of the perimeter we get
150=2(x+6+x)
150=2(2x+6)
150=4x+24
150-24=4x
136=4x
x=136/4
x=34
Thus, length = x+6
=34+6
=40 meters
Therefore, the length and breadth of the rectangular plot are 40 meters and 34 meters respectively.


Example 4.
If you add 1/2 to a number and multiply the result by 1/4, you get 1/5. What is the number?

Solution:

Let the number be x
(x+1/2)1/4 = 1/5
(2x+1)/2*1/4 = 1/5
(2x+1)/8 = 1/5
(2x+1)5=8
10x+5=8
10x=8-5
10x=3
x=3/10
Therefore, the number is 3/10


Example 5.
The base of an isosceles triangle is 3/4 cm. The perimeter of the triangle is 7 1/10 cm. What is the length of either of the remaining equal sides?

Solution:

Let x be each of the lengths of the remaining two sides of the isosceles triangle.
Thus, the sides of the triangle are x, x, 3/4 cm
Given Perimeter of the Triangle = 7 1/10 cm
x+x+3/4 = 7 1/10
2x+3/4 = 71/10
2x=71/10-3/4
2x= (142-15)/20
2x=127/20
x=(127*2)/20
=254/20
=127/10


Example 6.
A number subtracted from 7 is equal to 3 times the number. Find the number?

Solution:

Let the number be x
From given data 7-x =3x
7=3x+x
7=4x
x=7/4
Therefore, the number is 7/4


Example 7.
A farmer cuts a 300-foot fence into two pieces of different sizes. The longer piece should be five times as long as the shorter piece. How long are the two pieces?

Solution:

Let the shorter piece be x
From given data Longer Piece is 5 times long as shorter piece = 5x
Given longer piece +shorter piece = 300 foot
5x+x=300
6x=300
x=300/6
=50 foot
Longer Piece = 5x
=5*50
=250 foot
Therefore longer piece and shorter piece are of 250 foot and 50-foot length.


Example 8.
The perimeter of an equilateral triangle is 45 meters. How long is each side?

Solution:

Let us assume the side of the equilateral triangle be x meters
The Perimeter of Equilateral Triangle = 3a
3a=45
a=45/3
a=15 meters
Thus, each side of the equilateral triangle is 14 meters long.


Example 9.
If 5 blocks weigh 30 ounces, how many blocks weigh 90 ounces?

Solution:

Let us denote the weight of the block be x
5x=30 Ounces
x=30/5
=6 Ounces
How many Blocks weigh 90 Ounces =?
= 90/6
=15 blocks
Therefore, 15 blocks weigh 90 ounces


Example 10.
Ina class of 44 students, the number of girls is one-third the number of boys. Find the number of boys and girls?

Solution:

Total number of students in the class = 44
Consider Number of boys = x
Number of girls = 1/3(x)
=x/3
Sum of boys and girls = 44
x+x/3 =44
4x/3 =44
4x=44*3
4x=132
x=132/4
x=33
Number of girls = x/3
=33/3
=11
Therefore, the number of boys and girls in the class is 33 and 11 respectively.


Arithmetic Mean

Arithmetic Mean – Definition, Formula, Symbol, Properties | How to Find Arithmetic Mean in Statistics?

Arithmetic Mean(AM) is the most common measure of central tendency. In the terms of a layman, the mean of the data set demonstrates an average of the given data set. A few real-life statistics examples in terms of AM are the average rainfall of a place, the average income of the employees in the firm, and more. In this article, we will be concentrating more on Arithmetic mean like definition, formula, properties, best methods to find Mean, and Questions on average (arithmetic mean).

Definition of Arithmetic Mean (Average)

Arithmetic Mean portrays a number that is gained by dividing the sum of the elements of a dataset by the total number of values in the dataset. You can use the term Average or use the word Arithmetic Mean for making it a bit fancier. The AM can be either

  • Simple Arithmetic Mean
  • Weighted Arithmetic Mean

Arithmetic Mean Formula

In short, the arithmetic mean is knowns as the average or the mean of the given numbers.

If any data set comprising of the values x1, x2, x3, …., xn, then the Arithmetic Mean A is defined as:

data set of x

Also, it is denoted as:

Arithmetic Mean Formula

If you are finding the mean when the frequency of the observations is given, in such a way x1, x2, x3,… xis the recorded observations, and f1, f2, f… fn is the corresponding frequencies of the observations then;

mean formula of a frequencies observations

Also, it is presented like:

formula of arithmetic mean

This formula is used when the Arithmetic Mean is calculated for the ungrouped data. In order to calculate the mean or average of grouped data, we calculate the class mark. Here, the calculation of midpoints of the class intervals can be done as:

class mark formula

Once, you are done with calculating the class mark, the mean is solved as explained above. This method of solving the arithmetic mean is known as the direct method.

Arithmetic Mean Formulas

Properties of Arithmetic Mean

A few basic properties of the arithmetic mean are as such:

  • The sum of deviations of the items from their arithmetic mean is always zero, i.e. ∑(x – X) = 0.
  • The sum of the squared deviations of the items from Arithmetic Mean (A.M) is minimum, which is less than the sum of the squared deviations of the items from any other values.
  • If all the observations in the given data set have a value say ‘n’, then their arithmetic mean is also ‘n’.
  • If every item in the arithmetic series is replaced by the mean, then the sum of these replacements will be equivalent to the sum of the specific items.
  • In case each value in the score increases or decreases by a fixed value, then the mean also increases/decreases by the same number. Let the mean of x1, x2, x3,… xn be X̄, then the mean of x1+k, x2+k, x3+k,… xn+k will be X̄+k.

How to Find the Arithmetic Mean?

There are three ways to find the arithmetic mean. They are as follows:
Direct method
Short-cut method
Step Deviation method

Direct Metod for finding the Average (Arithmetic Mean)

Let x1, x2, x3,… xn be the observations with the frequency f1, f2, f… fn. Then, the calculation of mean can be done by using the formula:

x̄ = (x1f1+x2f2+……+xnfn) / ∑fi

Here, f1+ f2 +….fn = ∑fi demonstrates the sum of all frequencies.

Short-cut Method for solving the Arithmetic Mean

An assumed mean method or change of origin method is also known as the short-cut method. The steps that should be followed to calculate the arithmetic mean are as follows:

  1. First, find the class marks (mid-point) of each class (xi)
  2. Let A represent the assumed mean of the dataset.
  3. Calculate deviation (di) = xi – A
  4. Now, apply the formula of the mean i.e., x̄ = A + (∑fidi/∑fi)

Step Deviation Method for Obtaining the Arithmetic Mean

The other names of this method are the change of origin or scale method. Follow the below steps and easily calculate the mean of the given data:

  1. Find the class marks of each class (xi).
  2. Let’s A be the assumed mean of the data.
  3. Calculate ui=(xi−A)/h, where h is the class size.
  4. At last, use the formula ie., x̄ = A + h × (∑fiui/∑fi)

Arithmetic Mean Examples with Solutions

Example 1:
Study the following example table data and find the arithmetic mean using the step-deviation method.

Class Intervals 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total
Frequency 5 7 8 10 15 11 4 60

Solution:
In order to calculate the mean of the given data, we need to get the value for class marks and then decide A (assumed mean).
Let A = 35,
Here h (class width) = 10

C.I. xi fi ui= xi−A/h​​ fiui
0-10 5 5 -3 5 x (-3)=-15
10-20 15 7 -2 7 x (-2)=-14
20-30 25 8 -1 8 x (-1)=-8
30-40 35 10 0 10 x 0= 0
40-50 45 15 1 15 x 1=15
50-60 55 11 2 11 x 2=22
60-70 65 4 3 4 x 3=12
Total ∑fi=60 ∑fiui=12

By using the arithmetic mean formula:
x̄ = A + h × (∑fiui/∑fi)
= 35 + 10 x (12/60)
= 35 + 2
= 37.

Example 2:
The runs scored by Sachin in 3 test matches are 100, 167, 215 respectively. Find the mean.
Solution:
Given that Sachin scored runs in 3 test matches are 100, 167, 215
Total number of test matches = 3
Mean = Sum of total runs / Number of test matches
= 100 +167 + 215 / 3
=  482 / 3
= 160

FAQs on Average (Arithmetic Mean)

1. What is Arithmetic Mean in Statistics?

In statistics, Arithmetic Mean is the ratio of all observations to the total number of observations in a data set.

2. What is the formula to find the Arithmetic Mean?

The formula for solving the mean problems is the sum of all observations divided by a number of observations.

Arithmetic mean formula = Σ \(\frac { Xi }{ n } \), where ‘i’ ranges from 1 to n

3. How do you solve the word problems on Arithmetic mean or average?

Arithmetic mean can be simple arithmetic mean or weighted arithmetic mean. The simple arithmetic mean calculations vary for different data sets like for individual observations, for discrete series, and for continuous series. Go with this page and collect deep knowledge on how to calculate the arithmetic mean with examples.

Worksheet on Forming of Linear Equations in One Variable

Worksheet on Forming of Linear Equations in One Variable | Forming Linear Equations in One Variable Worksheet with Solutions

Worksheet on Forming of Linear Equations in One Variable come with numerous questions framed on the topic and gives you ample practice. Solve the Questions on Forming of Linear Equations in One Variable and improve your logical skills as well as speed while attempting the exams. Step by Step Solutions makes it clear for you to understand the topic. Download the free accessible Forming Linear Equations in One Variable Worksheet and resolve all your queries in no time.

Do Check:

Forming Linear Equations in One Variable Worksheet

Example 1. One of the numbers is two times the other. The sum of these two numbers is 45. Form the equation to find the numbers using a linear equation in one variable?

Solution:

Let the number be x
The Other Number is 2 times the given number i.e. 2x
From the given condition, the sum of these two numbers is 40
We can write the equation x+2x=45
3x=45
x=45/3
x=15
The other number is 2x i.e. 2(15)=30


Example 2.
The perimeter of a rectangular swimming pool is 140 meters. Its length is 4 m more than twice its breadth. What are the length and breadth of the pool?

Solution:

Let the breadth of the rectangular swimming pool be x
Since length is 4m more than twice its breadth we can have length = 2x+4
Perimeter of a Rectangular Swimming Pool = 2(l+b)
140 =2(2x+4+x)
140=2(3x+4)
140=6x+8
140-8=6x
132=6x
x=132/6
x=22
Length l =2x+4
=2(22)+4
=44+4
=48
Thus, the breadth and length of the swimming pool are 22m and 48m respectively.


Example 3.
The sum of three consecutive odd numbers is 45. Find the numbers?

Solution:

Let the three odd consecutive numbers be x, x+1, x+2
As per the given condition sum of three consecutive odd numbers is 45
x+x+1+x+2=45
3x+3=45
3x=45-3
3x=42
x=42/3
x=14
x+1=14+1=15
x+2=14+2=16
Therefore, the three consecutive numbers are 14, 15, 16


Example 4.
A sum of Rs. 4500 is to be given in the form of 90 prizes. If the prize is of either Rs. 100 or Rs. 25, find the number of prizes of each type?

Solution:

Let us assume the type of 100Rs prizes be x
Since the total number of prizes is 90 the number of 25 Rs. prize is 90-x
As per the given data in the question
100*x+(90-x)=4500
100x+90-x=4500
99x+90=4500
99x=4500-90
99x=4410
x=4410/99
~44
Therefore 25Rs. Prizes are 90-44 = 46


Example 5.
A dealer sold a television set for Rs. 12,000 and earned a profit of 15%. Find the cost price of the television set?

Solution:

Selling Price of the Television = Rs. 12,000
Let us assume the cost price = x
Profit earned = 15%
Cost Price of the television set CP = (100  / ( 100 + percentage profit))*SP
=(100/(100+15)*12000
=(100/115)*12000
=Rs. 10434
Therefore, the dealer bought the television set for a cost price of Rs. 10434


Example 6.
Twenty years from now Rahul’s age will be 5 times his current age. What is his current age?

Solution:

Let us consider the current age of Rahul as x
Twenty Years from now his age would be x+20
As per the given condition in the question we have x+20=5x
20=5x-x
20=4x
20/4 =x
x=5
Therefore current age of Rahul is 5 Years.


Example 7.
Solve 2y -10 = 4

Solution:

Given Expression is 2y-10=4
Transfering constants to one side and variables to the other side we have 2y =4+10
2y=14
y=7


Example 8.
Rajesh is a cashier in a State bank. he has notes of denominations of Rs. 100, 50, and 20 respectively. The ratio of the number of these notes is 4:3:2 respectively. The total cash with Rajesh is 4,72,000. How many notes of each denomination does he have?

Solution:

Let us assume the numbers of notes be 4x, 3x, and 2x respectively based on the ratio of notes
100*4x+50*3x+20*2x =4,72,000
400x+150x+40x=4,72,000
590x=4,72,000
x=4,72,000/590
=800
No. of 100 Rs Notes Rajesh has = 4x
=4*800
=3200
No. of 50 Rs Notes Rajesh has = 3x
=3*800
=2400
No. of 20 Rs Notes Rajesh has = 2x
=2*800
=1600


Example 9.
Solve for 8x + 40 = 4x +100?

Solution:

Given Expression is 8x + 40 = 4x +100
Transferring the constant terms to one side and variables terms to another side we have
8x-4x=100-40
6x=60
x=10


Example 10.
Amar thinks of a number and subtracts 3/2 from it. She multiplies the result by 7. The final result is 4 times her original number. Find the number?

Solution:

Let the number be x
From the given statement we can infer 7(x-3/2)=4x
7x-21/2=4x
(14x-21)=2*4x
14x-8x=21
6x=21
x=21/6
Therefore, the number is 21/6


Problems Based on Average

Problems Based on Average | Free & Printable Average Problems with Solutions PDF Worksheet

Problems Based on Average Worksheet PDF aid students to solve all types of easy to complex average problems with ease. Practicing from average word problems & solutions pdf can excel in the concepts of Statistics like mean, arithmetic mean, median, mode, etc.

Also, you can simply rely on the answers solved here and check out the mistakes that you have made while practicing. Access and download the Average practice questions worksheet with solutions PDF easily without any charge and prepare any time anywhere.

Do Refer:

How To Solve Average Word Problems?

The calculation of the Average or Arithmetic Mean or Mean of a number of quantities can be done by using the formula ie., the Average of a number of quantities of the same units is equal to their sum divided by the number of those quantities.

Arithmetic average is utilized for all averages such as Average income, average profit, average age, average marks, etc. However, to solve the sum of observations, they must be in the same unit. Check out the below sections to know some quick tips & tricks to solve and practice with the questions based on average.

Easy Average Problems Tricks | Common Average Aptitude Formulas

The list of easy tricks to solve average problems pdf is given here for your reference. Simply memorize them daily and practice well from average questions for competitive exams with solutions pdf.

  1. Average = \(\frac { Sum of quantities }{ Number of quantities } \)
  2. Sum of quantities = Average * Number of quantities
  3. The average of first n natural numbers is \(\frac { (n +1) }{ 2 } \)
  4. The average of the squares of first n natural numbers is \(\frac { (n +1)(2n+1 ) }{ 6 } \)
  5. The average of cubes of first n natural numbers is \(\frac { n(n+1)2 }{ 4 } \)
  6. The average of first n odd numbers is given by \(\frac { (last odd number +1) }{ 2 } \)
  7. The average of first n even numbers is given by \(\frac { (last even number + 2) }{ 2 } \)
  8. The average of squares of first n consecutive even numbers is \(\frac { 2(n+1)(2n+1) }{ 3 } \)
  9. The average of squares of consecutive even numbers till n is \(\frac { (n+1)(n+2) }{ 3 } \)
  10. The average of squares of consecutive odd numbers till n is \(\frac { n(n+2) }{ 3 } \)
  11. If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2(m-1)
  12. If the number of quantities in two groups be n1 and n2 and their average is x and y respectively, the combined average is \(\frac { (n1x+n2y) }{ (n1+ n2) } \)
  13. The average of n quantities is equal to x. When a quantity is removed, the average becomes y. The value of the removed quantity is n(x-y) + y
  14. The average of n quantities is equal to x. When a quantity is added, the average becomes y. The value of the new quantity is n(y-x) + y

Word Problems on Average | Average Practice Questions and Answers

Example 1:
The average mark of 50 students in a class is 70. Out of these 50 students, if the average mark of 25 students is 65, what is the average mark of the remaining 25 students?
Solution:
Average of 50 students = 70
Therefore, total marks of all 50 students = 50 x 70 =  3500
Average of 25 students = 65
Therefore, total marks of 25 students = 25 x 65 = 1625
So, the total marks of remaining 25 students = 3500 – 1625 = 1875
Hence, the average of remaining 25 students out of 50 students = 1875 / 30 = 62.

Example 2:
The mean of 15 observations is 45. If the mean of the first observations is 10 and that of the last 15 observations is 45, find the 15th observation.
Solution:
Given that,
Mean of the first 15 observations = 10
Then, Sum of the first 15 observations = 10 x 15 = 150
Mean of the last 15 observations = 45
Then, Sum of the last 15 observations = 45 x 15 = 675
Mean of the 15 observations = 45
Then, Sum of all 15 observations = 45 x 15 = 675
Now, find the 15th observation
= 150 + 675 – 675
= 150
Therefore, the 15th observation is 150.

Example 3: 
The mean of 7 numbers is 35. If one of the numbers is excluded, the mean gets reduced by 5. Find the excluded number.
Solution:
Given Mean of 7 numbers = 35
Sum of the 7 numbers = 35 x 7 = 245
Mean of remaining 6 numbers = 35-5 = 30
Sum of remaining 6 numbers = 30 x 6 = 180
Now, find the excluded number to do that;
Excluded number = (sum of the given 7 numbers) – (sum of the remaining 6 numbers)
= 245 – 180
= 65
Hence, the excluded number is 65.

Example 4: 
Find the average of 2, 4, 6, 8, 10, 12?
Solution:
Here, you will know that the given numbers are even in numbers ie., a total of 6 numbers. These kinds of questions are solved by looking at the difference in numbers (the constant difference is 2).

If the number were given odd in numbers means 2, 4, 6, 8, 10, then the answers will always be the middle term ie., 6. However, if the given numbers are in even count ie, six numbers like 2, 4, 6, 8, 10, 12 then the answer is always Sum of middle terms/Sum of opposite terms divided by 2 or Sum of quantities divided by total quantities.
ie., (2+12)/2 or (4+10)/2 or (6+8)/2. (the average is 7)
The answer is the same always.
As per the average formula = Sum of quantities divided by total quantities
= 2+4+6+8+10+12 / 6
= 42 / 6
= 7
Hence, the average of 2, 4, 6, 8, 10, 12 is 7.

Some More Example Questions & Problems Based on Average

Example 5:
The average height of a group of four boys is 5.6 inches. The individual height (in inches) of three of them are 5.2, 5.4, 5.5. What is the height of the fourth boy?
Solution:
Given Average height of 4 boys = 5.6 inches
Total height of 4 boys = (5.6 x 4) in = 22.4 in
Total height of 3 boys = (5.2 + 5.4 + 5.5) in = 16.1 in
Height of the 4th boy = (total height of 4 boys) – (total height of 3 boys)
= (22.4 – 16.1) in
= 6.3 in
Hence, the height of the fourth boy is 6.3 in

Example 6: 
The average of 5 numbers is 30. If each number is increased by 2, what will the new average be?
Solution:
Given that the average of 5 numbers = 30
Sum of the 5 numbers = 30 x 5 = 150
If each number is increased by 2, the total increase = 2 x 5 = 10
Then, the new sum = 150+10 = 160 and new Average = 160/10 = 16
Hence, the new average is 16.

Example 7: 
The average age of four boys is 25 years and their ages are in proportion 2:4:6:8. What is the age in years of the youngest boy?
Solution:
Given, Average age of four boys = 25 years
Sum of four boys = 25 x 4 = 100
Ages = 2x + 4x + 6x + 8x
2x + 4x + 6x + 8x = 100
20x = 100
x = 100/20
x = 5 years
Age of the youngest boy = 2x years = 2(5) years = 10 years.

Worksheet on Linear Equation in One Variable

Worksheet on Linear Equation in One Variable | Linear Equation in One Variable Worksheet PDF

Worksheet on Linear Equation in One Variable present here are designed in a way that they encourage students to think rather than simply identify a pattern to the solutions. You can refer to them anytime as our Linear Equation in One Variable Worksheet comes with solutions too. If you are unable to solve a particular question you can simply rely on the answers provided to know where you went wrong and improvise accordingly.

Get an opportunity to solve a wide range of problems on the concept and build a strong mathematical foundation. Step by Step Solutions provided for all the Linear Equation in One Variable Questions makes it easy for you to understand and solve them with accuracy in exams. Download the easily accessible Linear Equations in One Variable Worksheet with Solutions PDF for free and prepare any time.

Read Similar:

Linear Equation in One Variable Worksheets with Solutions

Example 1.
Solve the following
(i)3x – 9 = 0
(ii)6x – 2 = 8 + x
(iii)8 –3x = 5 – 4x

Solution:

(i)3x – 9 = 0
Given Equation is 3x – 9 = 0
3x=9
x=3
(ii)6x – 2 = 8 + x
Given equation 6x-2=8+x
Transferring constants to one side and variables to another side we have
6x-x=8+2
5x=10
x=10/5
x=2
(iii)8 –3x = 4x-5
Given Equation 8 –3x = 4x-5
Transferring constants to one side and variables to another side we have
8 –3x = 4x-5
8+5=4x+3x
14=7x
x=14/7
x=2


Example 2.
Solve the linear equation 10(y – 2) – 2(y – 3) – 5(y + 3) = 0

Solution:

Given Linear Equation is 10(y – 2) – 2(y – 3) – 5(y + 3) = 0
Simplifying it further we have
10y-20-2y+6-5y-15=0
3y-29=0
3y=29
y=29/3


Example 3.
Simplify 4x + 2(x+3) = 10 – (2x – 5)

Solution:

Given Linear Equation is 4x + 2(x+3) = 10 – (2x – 5)
Moving Constants to one side and Variables to another side we have
4x+2x+6=10-2x+5
6x+6=15-2x
6x+2x=15-6
8x=9
x=9/8


Example 4.
Solve for m in \(\frac { 3m+8 }{ 2 } \) =4m-6?

Solution:

Given \(\frac { 3m+6 }{ 2 } \) = 4m-6
3m+8=2(4m-6)
3m+8 =8m-12
8+12=8m-3m
20=5m
m=20/5
m=4


Example 5.
Solve \(\frac { 5x }{ 3 } \) = \(\frac { 3x }{ 2 } \)+\(\frac { 1 }{ 4 } \)?

Solution:

Given \(\frac { 5x }{ 3 } \) = \(\frac { 3x }{ 2 } \)+\(\frac { 1 }{ 4 } \)
\(\frac { 5x }{ 3 } \) – \(\frac { 3x }{ 2 } \) = \(\frac { 1 }{ 4 } \)
\(\frac { 10x-9x }{ 6 } \) = \(\frac { 1 }{ 4 } \)
4(10x-9x)=1*6
4x=6
x=\(\frac { 6 }{ 4 } \)
x=\(\frac { 3 }{ 2} \)


Example 6.
Solve the equation 0.18(3x – 4) = 0.2x + 0.8

Solution:

Given equation 0.18(3x – 4) = 0.2x + 0.8
0.54x-0.72=0.2x+0.8
0.54x-0.2x=0.8+0.72
0.34x=1.52
x=\(\frac { 1.52 }{ 0.34} \)


Example 7.
Simplify the Linear Equation and get the Value of the Variable \(\frac { x+3 }{ x-3 } \) = \(\frac { 5 }{ 4 } \)?

Solution:

Given Equation \(\frac { x+3 }{ x-3 } \) = \(\frac { 5 }{ 4 } \)
(x+3)4=5(x-3)
4x+12=5x-15
12+15=5x-4x
27=x


Example 8.
Five added to four times a whole number gives 37. Find the number?

Solution:

Let the whole number be x
As per the given condition 5+4 times whole number = 37
5+4(x)=37
5+4x=37
4x=37-5
4x=32
x=32/4
x=8


Example 9.
One-fourth of a number is 15. What will be 25% of that number?

Solution:

Let the number is x
From the given condition \(\frac { 1 }{ 4 } \)(x)=15
x=15*4
x=60
Since we are asked 25% of the number we have 25%(60)
=\(\frac { 25*60 }{100 } \)
=\(\frac { 1500 }{100 } \)
=15


Example 10.
There are 450 students in a school. If the number of girls is 104 more than the boys, how many boys are there in the school?

Solution:

Let the number of boys = x
Then, number of girls = x + 104
As per the given condition x + (x + 104) =450 
2x + 104 = 450
2x = 450 – 104= 346
x = 346/2=173
Hence, the number of boys = 173
And, the number of girls = (x + 104)
= 173 + 104
= 277


Example 11.
The Sum of 2 consecutive numbers is 64. Find the numbers?

Solution:

Let the 2 consecutive numbers be x and x+2
As per the given condition x+x+2=64
2x+2=64
2x=64-2
2x=62
x=62/2
x=31
the Other Consecutive Number is x+2 i.e 33


Example 12.
If a rectangle possesses a width of 4 inches and has a perimeter of 16 inches, then what is the length?

Solution:

We know the Perimeter of a Rectangle Formula is p = 2(l+w)
Substituting the given data in the formula we have 16 =2(l+4)
Simplifying further we have 16=2l+8
16-8=2l
8=2l
l=8/2
l=4
Therefore, length of the rectangle = 4 inches