Monthly Archives: October 2022

The number which can be expressed as the sum of two cubes in different ways is said to be a Hardy – Ramanujan number.

1729 = 1728 + 1 = 12³ + 1³

1729 = 1000 + 729 = 10³ + 9³

As 1729 is the smallest such type of number so it is called the smallest Hardy-Ramanujan number. There is infinite such type of numbers. Like- 4104 (2, 16; 9, 15), 13832 (18, 20; 2, 24), etc.

Cubes

Cube is a 3-dimensional figure with all equal sides. If one cube has all the equal sides of 1 cm then how many such cubes are needed to make a new cube of side 2 cm?

8 such cubes are needed, and what if we need to make a cube of side 3 cm with the cubes of side 1 cm? The numbers 1, 8, 27 …etc can be shown below in the cube.

These are known as perfect cubes or cube numbers. This shows that we got the cube numbers by multiplying the number three times by itself.

Cubes of Some Natural Numbers

Number	Cubes	Numbers	Cubes
1	13 = 1	11	113 = 1331
2	23 = 8	12	123 = 1728
3	33 = 27	13	133 = 2197
4	43 = 64	14	143 = 2744
5	53 = 125	15	153 = 3375
6	63 = 216	16	163 = 4096
7	73 = 343	17	173 = 4913
8	83 = 512	18	183 = 5832
9	93 = 729	19	193 = 6859
10	103 = 1000	20	203 = 8000

This table shows that

• There are only 10 perfect cubes between 1-1000.
• The cube of an even number is also even.
• The cube of an odd number is also an odd number.

One’s digit of the Cubes

 One’s digit of the Cubes of a number having a particular number at the end will always remain same. Let’s see in the following table:

Unit’s digit of number	Last digit of its cube number	Example
1	1	113 = 1331, 213 = 9261, etc.
2	8	23 = 8, 123 = 1728, 323 = 32768, etc.
3	7	133 = 2197, 533 = 148877, etc.
4	4	243 = 13824, 743 = 405224, etc.
5	5	153 = 3375, 253 = 15625, etc.
6	6	63 = 216, 263 = 17576,etc.
7	3	173 = 4913, 373 = 50653,etc.
8	2	83 = 512, 183 = 5832, etc.
9	9	193 = 6859, 393 = 59319, etc.
10	20	103 = 1000, 203 = 8000, etc.

Some Interesting Patterns

1. Adding Consecutive Odd Numbers

This shows that if we add the consecutive odd numbers then we get the cube of the next number.

2. Cubes and their Prime Factors

Prime factorization of a number is done by finding the prime factors of the number and then pairing it in the group of three. If all the prime factors are in the pair of three then the number is a perfect cube.

Example

Calculate the cube root of 13824 by using prime factorization method.

Solution

First of all write the prime factors of the given number then pair them in the group of three.

Since all the factors are in the pair of three the number 13824 is a perfect cube.

Smallest Multiple that is a Perfect Cube

As we have seen that the group of three prime factors makes a number perfect cube, so to make a number perfect cube we need to multiply it with the smallest multiple of that number.

Example

Check whether 1188 is a perfect cube or not. If not then which smallest natural number should be multiplied to 1188 to make it a perfect cube?

Solution

1188 = 2 × 2 × 3 × 3 × 3 × 11

This shows that the prime numbers 2 and 11 are not in the groups of three. So, 1188 is not a perfect cube

To make it a perfect cube we need to multiply it with 2 × 11 × 11 = 242, so, it will make the pair of 2, 3 and 11.

 Hence the smallest natural number by which 1188 should be multiplied to make it a perfect cube is 242.

And the resulting perfect cube is 1188 × 242 = 287496 ( = 66³).

Cube Roots

Finding cube root is the inverse operation of finding the cube.

If 3³ =27 then cube root of 27 is 3.

We write it as ∛27 = 3

Symbol of the Cube Root

Some of the cube roots are:

Statement	Inference	Statement	Inference
13 = 1	∛1 = 1	63 = 216	∛216 = ∛63 = 6
23 = 8	∛8 = ∛23 = 2	73 = 343	∛343 = ∛73 = 7
33 = 27	∛27 = ∛33 = 3	83 = 512	∛512 = ∛83 = 8
43 = 64	∛64 = ∛43 = 4	93 = 729	∛729 = ∛93 = 9
53 = 125	∛125 = ∛53 = 5	103 = 1000	∛1000 = ∛103 = 10

Method of finding a Cube Root

There are two methods of finding a cube root

1. Prime Factorization Method

Step 1: Write the prime factors of the given number.

Step 2: Make the pair of three if possible.

Step 3: Then replace them with a single digit.

Step 4: Multiply these single digits to find the cube root.

Example

Find the cube root of 15625 by the prime factorization method.

2. Estimation Method

This method is based on the estimation. Let’s take the above example.

Step 1: If 15625 is the number then make the group of three digits starting from the right.

15 625

Step2: Here 625 is the first group which tells us the unit’s digit of the cube root. As the number is ending with 5 and we know that 5 comes at the unit’s place of a number only when its cube root ends in 5.

So the unit place is 5.

Step 3: Now take the other group, i.e., 15. Cube of 2 is 8 and a cube of 3 is 27. 15 lie between 8 and 27. The number which is smaller among 2 and 3 is 2. The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of 15625. Thus, 

Square Number

Any natural number ‘p’ which can be represented as y², where y is a natural number, then ‘p’ is called a Square Number.

Example

4 = 2²

9 = 3²

16 = 4²

Where 2, 3, 4 are the natural numbers and 4, 9, 16 are the respective square numbers.

Such types of numbers are also known as Perfect Squares.

Some of the Square Numbers

Properties of Square Numbers

• We can see that the square numbers are ending with 0, 1, 4, 5, 6 or 9 only. None of the square number is ending with 2, 3, 7 or 8.
• Any number having 1 or 9 in its one’s place will always have a square ending with 1.

Number	Square Number
1	1
9	81
11	121
19	361
21	441

• Any number which has 4 or 6 in its unit’s place, its square will always end with 6.

Number	Square Number
4	16
16	256
24	576
36	1296
44	1936

• Any number which has 0 in its unit’s place, its square will always have an even number of zeros at the end.

Number	Square number
10	100
50	2500
100	10000
150	22500
400	160000

Some More Interesting Patterns

1. Adding Triangular Numbers

If we could arrange the dotted pattern of the numbers in a triangular form then these numbers are called Triangular Numbers. If we add two consecutive triangular numbers then we can get the square number.

2. Numbers between Square Numbers

If we take two consecutive numbers n and n + 1, then there will be (2n) non-perfect square numbers between their squares numbers.

Example

Let’s take n = 5 and 5² = 25

n + 1 = 5 + 1 = 6 and 6² = 36

2n = 2(5) = 10

There must be 10 numbers between 25 and 36.

The numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.

3. Adding Odd Numbers

Sum of first n natural odd numbers is n².

Any square number must be the sum of consecutive odd numbers starting from 1.

And if any natural number which is not a sum of successive odd natural numbers starting with 1, then it will not be a perfect square.

4. A Sum of Consecutive Natural Numbers

Every square number is the summation of two consecutive positive natural numbers.

If we are finding the square of n the to find the two consecutive natural numbers we can use the formula

Example

5² = 25

12 + 13 = 25

Likewise, you can check for other numbers like

11² = 121 = 60 + 61

5. The Product of Two Consecutive Even or Odd Natural Numbers

If we have two consecutive odd or even numbers (a + 1) and (a -1) then their product will be (a²– 1)

Example

Let take two consecutive odd numbers 21 and 23.

21 × 23 = (20 – 1) × (20 + 1) = 20² – 1

6. Some More Interesting Patterns about Square Numbers

Finding the Square of a Number

To find the square of any number we needed to divide the number into two parts then we can solve it easily.

If number is ‘x’ then x = (p + q) and x² = (p + q)²

You can also use the formula (p + q)² = p² + 2pq + q²

Example

Find the square of 53.

Solution:

Divide the number in two parts.

53 = 50 + 3

53² = (50 + 3)²

= (50 + 3) (50 + 3)

= 50(50 + 3) +3(50 + 3)

= 2500 + 150 + 150 + 9

= 2809

1. Other pattern for the number ending with 5

For numbers ending with 5 we can use the pattern

(a5)² = a × (a + 1)100 + 25

Example

25² = 625 = (2 × 3) 100 + 25

45² = 2025 = (4 × 5) 100 + 25

95² = 9025 = (9 × 10) 100 + 25

125² = 15625 = (12 × 13) 100 + 25

2. Pythagorean Triplets

If the sum of two square numbers is also a square number, then these three numbers form a Pythagorean triplet.

For any natural number p >1, we have (2p) ² + (p² -1)² = (p² + 1)². So, 2p, p²-1 and p²+1 forms a Pythagorean triplet.

Example

Write a Pythagorean triplet having 22 as one its member.

Solution:

Let 2p = 6

P = 3

p² + 1 = 10

p² – 1 = 8.

Thus, the Pythagorean triplet is 6, 8 and 10.

6² + 8² = 10²

36 + 64 = 100

Square Roots

The square root is the inverse operation of squaring. To find the number with the given square is called the Square Root.

 2² = 4, so the square root of 4 is 2

10² = 100, therefore square root of 100 is 10

There are two square roots of any number. One is positive and other is negative.

The square root of 100 could be 10 or -10.

Symbol of Positive Square Root

Finding Square Root

1. Through Repeated Subtraction

As we know that every square number is the sum of consecutive odd natural numbers starting from 1, so we can find the square root by doing opposite because root is the inverse of the square.

We need to subtract the odd natural numbers starting from 1 from the given square number until the remainder is zero to get its square root.

The number of steps will be the square root of that square number.

Example

Calculate the square root of 64 by repeated addition.

Solution:

64 – 1 = 63

63 – 3 = 60

60 – 5 = 55

55 – 7 = 48

28 – 13 = 15

48 – 9 = 39

15 – 15 = 0

39 – 11 = 28

2. Prime Factorization

In this method, we need to list the prime factors of the given number and then make the pair of two same numbers.

Then write one number for each pair and multiply to find the square root.

Example

Calculate the square root of 784 using prime factorization method.

Solution:

List the prime factors of 784.

784 = 2 × 2 × 2 × 2 × 7 × 7

√784 = 2 × 2 × 7 = 28

3. Division Method

Steps to find the square root by division method

Step 1: First we have to start making the pair of digits starting from the right and if there are odd number of digits then the single digit left over at the left will also have bar .

Step 2: Take the largest possible number whose square is less than or equal to the number which is on the first bar from the left. Write the same number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide to get the remainder.

Step 3: Like a normal division process bring the digits in next bar down and write next to the remainder.

Step 4: In next part the quotient will get double and we will right in next line with a blank on its right.

Step 5: Now we have to take a number to fill the blank so that the if we take it as quotient then the product of the new divisor and the new digit in quotient is less than or equal to the dividend.

Step 6: If there are large number of digits then you can repeat the steps 3, 4, 5 until the remainder does not become 0.

Example

 Calculate the square root of √729 using division method.

Solution:

Thus, √729 = 27.

Square Roots of Decimals

To find the square root of a decimal number we have to put bars on the primary part of the number in the same manner as we did above. And for the digits on the right of the decimal we have to put bars starting from the first decimal place.
Rest of the method is same as above. We just need to put the decimal in between when the decimal will come in the division.

Example

 Find √7.29 using division method.

Solution:

Thus, √7.29 = 2.7

Remark: To put the bar on a number like 174.241, we will put a bar on 74 and a bar on 1 as it is a single digit left. And in the numbers after decimal, we will put a bar on 24 and put zero after 1 to make it double-digit.

174. 24 10

Estimating Square Root

Sometimes we have to estimate the square root of a number if it’s not possible to calculate the exact square root.

Example

Estimate the square root of 300.

Solution:

 We know that, 300 comes between 100 and 400 i.e. 100 < 300 < 400.
Now, √100 = 10 and √400 = 20.

 So, we can say that

10 < √300 < 20.

We can further estimate the numbers as we know that 17² = 289 and 18² = 324.
Thus, we can say that the square root of √300 = 17 as 289 is much closer to 300 than 324.

Introduction to Data Handling

Data handling means to collect and present the data so that it could be used in further studies and to find some results.

Data

Any information collected in the form of numbers, words, measurements, symbols, or in any other form is called data.

Graphical Representation of Data

The grouped data can be represented graphically for its clear picture and it is the easiest way to understand the data.

Types of Graph

1. Pictograph

When we represent the data through pictures or symbols then it is called Pictograph.

Here one tree represents 10 trees. And we can easily read the pictograph.

The graph shows that there are 30 trees of apple and so on.

2. Bar Graphs

In the bar graph, the information represented by the bars of the same width with equal gaps but the height of the bars represent the respective values.

Here, the names of pets are represented on the horizontal line and the values of the respective pets are shown by the height of the bars. There is an equal gap between each bar.

3. Double Bar Graph

To compare some data we can use the double bar graph as it shows the information of two quantities simultaneously.

Here, in the above graph, it represents the marks of the students in two different tests altogether. So we can compare the marks easily.

Organizing Data

Any data which is available in the unorganized form is called Raw Data.

This raw data is arranged or grouped in a systematic manner to make it meaningful which is called the Presentation of Data.

Terms Related to Data Organizing

1. Frequency

Frequency tells us the no. of times a particular quantity repeats itself.

2. Frequency Distribution Table

Frequency can be represented by the frequency distribution table.

The above table shows the no. of times a particular colour repeat in the bag of skittles.

Frequency can also be shown by the tally marks. A cut over four lines represents the number 5.

1. Grouping Data

If we have a large number of quantities then we need to group the observation and then make the table. Such a table is called a Grouped Frequency Distribution Table.

Some Important terms related to grouped Frequency Distribution Table

• Class Interval or Class: When all the observations are classified in several groups according to their size then these groups are called Class Interval
• Lower-class Limit: The lowest number in every class interval is known as its Lower-class Limit.
• Upper-class Limit: The highest number in every class interval is known as its Upper-class Limit. 
• Width or Size or Magnitude of the Class Interval: The difference between the upper-class limit and the lower class limit is called the Size of the Class Interval. 

Example

There is a list of marks of 40 students in a school. Arrange this in grouped frequency distribution table.

Solution

As we can see that the lowest number in the above data is 27 and the highest number is 78, so we can make intervals if 20 – 30, 30 – 40 so on.

Remark: As number 30 comes in two class interval but we cannot count it in both the intervals. So it is to remember that the common observation will always be counted in the higher class. Hence 30 will come in 30-40, not in 20-30.

Histogram

Basically, the bar graph of the grouped frequency distribution or continuous class interval is called Histogram.

The class intervals are shown on the horizontal line and the frequency of the class interval is shown as the height of the bars.

There is no gap between each bar.

Example

Draw a histogram for the wages of 30 workers in a company. The wages are as follows: 830, 840, 868, 890, 806, 840, 835, 890,840, 885, 835, 835, 836, 878, 810, 835, 836, 869, 845, 855, 845, 804, 808, 860, 832, 833, 812, 898, 890, 820.

Solution

Make the grouped frequency distribution of the given data.

Class Interval	Frequency
800 – 810	3
810 – 820	2
820 – 830	1
830 – 840	9
840 – 850	5
850 – 860	1
860 – 870	3
870 – 880	1
880 – 890	1
890 – 900	4

Draw the histogram by taking the class interval on the horizontal line and the frequency on the vertical line.

Remark: As the class interval does not start from zero, so we will put a jagged line which shows that there is no number between 0 – 800.

Circle Graph or Pie Chart

If we represent the data in a circle form then it is said to be a pie chart. This graph shows the relationship between the whole and its part. We have to divide the circle into sectors and each sector is proportional to its respective activity.

We use it when we have information on percentage or fraction.

Drawing of a Pie Chart

If we have the information in percentage then we need to calculate the respective angles to show them in the pie chart.

As we know that a complete circle is of 360°, so we need to calculate the fraction of 360° for every sector.

Example

Draw a pie chart of the following percentage of genres of movies liked by the public.

Genres of Movie	Percentage of the no. of people
Comedy	27%
Action	18%
Romance	14%
Drama	14%
Horror	11%
Foreign	8%
Science fiction	8%

Solution

To draw the pie chart first we need to calculate the angle by taking the fraction of 360°.

Genres of Movie	Percentage of the no. of people	In fractions	Fraction of 360°
Comedy	27%	27/100	27/100 × 360° = 97.2°
Action	18%	18/100	18/100 × 360° = 64.8°
Romance	14%	14/100	14/100 × 360° = 50.4°
Drama	14%	14/100	14/100 × 360° = 50.4°
Horror	11%	11/100	11/100 × 360° = 39.6°
Foreign	8%	8/100	8/100 × 360° = 28.8°
Science fiction	8%	8/100	8/100 × 360° = 28.8°

By using these angles draw a pie chart.

Chance and Probability

Probability tells the degree of uncertainty. It measures the likelihood that an event will occur.

Random Experiment

If the result of the experiment is not known then it is known as a random experiment.

Example

If we throw a dice then the result could be any number from 1 – 6.

Outcomes

When we do an experiment then there could be different results, these possible results of the random experiment are called outcomes.

Example

There are two possible outcomes when we toss a coin i.e. head and tail.

Equally Likely Outcomes

If every outcome has the same possibility of occurring these outcomes are called Equally Likely Outcomes.

Example

If we throw a dice then there is an equal chance of every no. to come while doing the random experiment. i.e. a dice has the same possibility of getting 1, 2, 3, 4, 5 and 6.

Linking Chances to the Probability

Example

What is the chance of getting 3 when we throw a dice?

Solution

There is only one chance to get 3 in one throw and the total possible outcomes are 6.

Hence the probability of getting =1/6.

Outcomes as Events

Each outcome or collection of outcomes of an experiment is known as an event.

Example

If we throw a dice then getting each outcome 1, 2, 3, 4, 5 and 6 are events.

Example

What is the event of getting odd numbers when we throw a dice?

Solution

The probability of getting an odd number is 3(odd numbers are 1, 3, 5)

The total number of outcomes is 6.

The probability of getting an odd number = 3/6.

Example

What is the probability of spinning yellow?

Here number of chance to come yellow while spinning is 3.

The total number of outcomes is 8.

A quadrilateral has some measurements like – 4 sides, 4 angles and 2 diagonals.

We can construct a unique quadrilateral if we know the five measurements.

1. If the four sides and a diagonal of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which AB = 5 cm, BC = 7 cm, CD = 6 cm, DA = 6.5 cm and AC = 8 cm.

Solution

Step 1: ∆ABC can be constructed using SSS criterion of the construction of triangle.

Step 2: Here we can see that AC is diagonal, so D will be somewhere opposite to B with reference to AC.

AD = 6.5 cm so draw an arc from A as the centre with radius 6.5 cm.

Step 3: Now draw an arc with C as the centre and by taking radius 6 cm so that it intersects the above arc.

Step 4: The point of intersection of the two arcs is point D. Now join AD and DC to complete the quadrilateral.

Hence, ABCD is the required quadrilateral.

2. If  two diagonals and three sides of the quadrilateral are given

Example

Construct a quadrilateral ABCD if the two diagonals are AC = 6.5 cm and BD = 8 cm. The other sides are BC = 5.5 cm, AD = 6.5 cm and CD = 6 cm.

Solution

First of all, draw a rough sketch of the quadrilateral by using the given measurements. Then start constructing the real one.

Step 1: We can see that AD, AC and DC are given so we can construct a triangle ΔACD by using SSS criterion.

Step 2: Now, we know that BD is given so we can draw the point B keeping D as the centre and draw an arc of radius 8 cm just opposite to the point D with reference to AC.

Step 3: BC is given so we can draw an arc keeping C as centre and radius 5.5 cm so that it intersects the other arc.

Step 4: That point of intersection of the arcs is point B. Join AB and BC to complete the quadrilateral.

ABCD is the required quadrilateral.

3. If three angles and two adjacent sides of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which the two adjacent sides are AB = 4.5 cm and BC = 7.5 cm. The given three angles are ∠A = 75ᵒ, ∠B = 105ᵒ and ∠C = 120ᵒ.

Solution

Draw a rough sketch so that we can construct easily.

Step 1: Draw AB = 4.5 cm. Then measure ∠B = 105° using protractor and draw BC = 7.5 cm.

Step 2: Draw ∠C = 120°.

Step 3: Measure ∠A = 75° and make a line until it touches the line coming from point C.

ABCD is the required quadrilateral.

4. If the three sides with two included angles of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which the three sides are AB = 5 cm, BC = 6 cm and CD = 7.5 cm. The two included angles are ∠B = 105° and ∠C = 80°.

Solution

Draw a rough sketch.

Step 1: Draw the line BC = 6 cm. Then draw ∠B = 105° and mark the length of AB = 5 cm.

Step 2: Draw ∠C = 80° using protractor towards point B.

Step 3: Mark the length of CD i.e.7.5 cm from C to make CD = 7.5 cm.

Step 4: Join AD which will complete the quadrilateral ABCD.

Hence ABCD is the required quadrilateral.

Some Special Cases

There are some special cases in which we can construct the quadrilateral with less number of measurements also.

Example

Construct a square READ with RE = 5.1 cm.

Solution

Given Re = 5.1 cm.

As it is a special quadrilateral called square, we can get more details out of it.

a. All sides of square are equal, so RE = EA = AD = RD = 5.1 cm.

b. All the angles of a square are 90°, so ∠R = ∠E = ∠A = ∠D = 90°

Step 1: Draw a rough sketch of the square.

Step 2: To construct a square, draw a line segment RE = 5.1 cm. Then draw the angle of 90° at both ends R and E of the line segment RE.

Step 3: As all the sides of the square READ are equal, draw the arc of 5.1 cm from the vertex R and E to cut the lines RD and EA respectively.

Step 4: Join A and D to make a line segment AD.

READ is the required square.

Algebraic Expressions

Any expression involving constant, variable and some operations like addition, multiplication etc is called Algebraic Expression.

• A variable is an unknown number and generally, it is represented by a letter like x, y, n etc.
• Any number without any variable is called Constant.
• A number followed by a variable is called Coefficient of that variable.
• A term is any number or variable separated by operators.

Equation

A statement which says that the two expressions are equal is called Equation.

Linear Expression

A linear expression is an expression whose highest power of the variable is one only.

Example

2x + 5, 3y etc.

The expressions like x² + 1, z² + 2z + 3 are not the linear expressions as their highest power of the variable is greater than 1.

Linear Equations

The equation of a straight line is the linear equation. It could be in one variable or two variables.

Linear Equation in One Variable

If there is only one variable in the equation then it is called a linear equation in one variable.

The general form is

ax + b = c, where a, b and c are real numbers and a ≠ 0.

Example

x + 5 = 10

y – 3 = 19

These are called linear equations in one variable because the highest degree of the variable is one and there is only one variable.

Some Important points related to Linear Equations

• There is an equality sign in the linear equation. The expression on the left of the equal sign is called the LHS (left-hand side) and the expression on the right of the equal sign is called the RHS (right-hand side).
• In the linear equation, the LHS is equal to RHS but this happens for some values only and these values are the solution of these linear equations.

Graph of the Linear Equation in One Variable

We can mark the point of the linear equation in one variable on the number line.

x = 2 can be marked on the number line as follows-

Solving Equations which have Linear Expressions on one Side and Numbers on the other Side

There are two methods to solve such type of problems-

1. Balancing Method

In this method, we have to add or subtract with the same number on both the sides without disturbing the balance to find the solution.

Example

Find the solution for 3x – 10 = 14

Solution

Step 1: We need to add 10 to both the sides so that the numbers and variables come on the different sides without disturbing the balance.

3x – 10 +10 =10+14

3x = 24

Step 2: Now to balance the equation, we need to divide by 3 into both the sides.

3x/3 = 24/3

x = 8

Hence x = 8 is the solution of the equation.

We can recheck our answer by substituting the value of x in the equation.

3x – 10 = 14

3(8) – 10 = 14

24-10 = 14

14 = 14

Here, LHS = RHS, so our solution is correct.

2. Transposing Method

In this method, we need to transpose or transfer the constants or variables from one side to another until we get the solution. When we transpose the terms the sign will get changed.

Example

Find the solution for 2z +10 = 4.

Solution

Step 1: We transpose 10 from LHS to RHS so that all the constants come in the same side.

2z = 4 -10 (sign will get changed)

2z = -6

Step 2: Now divide both the sides by 2.

2z/2 = – 6/2

z = – 3

Here z = -3 is the solution of the equation.

Some Applications of Linear Equation

We can use the concept of linear equations in our daily routine also. There are some situations where we need to use the variable to find the solution. Like,

• What number should be added to 23 to get 75?
• If the sum of two numbers is 100 and one of the no. is 63 then what will be the other number?

Example

What is the height of the rectangle whose perimeter is 96 cm² and the length is 12 cm?

Solution

Let the height of the rectangle be ‘s’.

Area of rectangle = Length × Breadth

96 = S × 12

Now, this is a linear equation with variable s.

We need to divide both sides by 12 to find the solution.

96/12 = 12s/12

s = 8

Hence the height of the rectangle is 8 cm.

Solving Equations having the Variable on both Sides

As the equation can have the variable on both the sides also so we should know how to solve such problems.

In this type of problems, we need to bring all the constants on one side and all the terms having variables on the other side. Then they can be solved easily.

Example

Find the solution of 2x−3 = 6 − x.

Solution

Step 1: Bring all the terms including variable x on LHS and the constants on the RHS.

2x + x = 6 + 3 (sign will change while changing the position of the terms)

Step 2: Solve the equation

3x = 9

Step 3: Divide both the sides by 3 to get the solution.

3x/3 = 9/3

x = 3

Hence the solution of the equation is x = 3.

Some More Applications

Example

Renu’s age is four times that of her younger brother. Five years back her age was 9 times her brother’s age. Find their present ages.

Solution

Let the Renu’s brother age = x

Renu’s age = 4x (as her age is 4 times that of her younger brother)

Five years back her age was = 9(x – 5) which is equal to 4x – 5

9(x – 5) = 4x – 5

9x – 45 = 4x – 5

9x – 4x = – 5 + 45 (by transferring the variable and constants on different sides)

5x = 40

x = 40/5 = 8

Renu’s brother age = x = 8 years

Renu’s age = 4x = 4(8) = 32 years.

Reducing Equations to Simpler Form

When linear equations are in fractions then we can reduce them to a simpler form by-

• Taking the LCM of the denominator
• Multiply the LCM on both the sides, so that the number will reduce without the denominator and we can solve them by the above methods.

Example
Solve the linear equation

Solution
As the equation is in complex form, we have to reduce it into a simpler form.

Step 1: Take the L.C.M. of the denominators, 2, 3, 4, and 5, which is 60.
Step 2: Multiply both the sides by 60,

30x −12 = 20x + 15 + 60

Step 3. Bring all the variables on the LHS and all the constants on the RHS

30x − 20x = 15 + 12 + 60

10x = 87

Step 4: Dividing both the sides by 10

x = 8.7

Equations Reducible to the Linear Form

Sometimes there are some equations which are not linear equations but can be reduced to the linear form and then can be solved by the above methods.

Example

Solution

This is not a linear equation but can be reduced to linear form

Step 1: Multiply both the sides by (2x + 3).

Now, this is a linear equation.

Step 2: Multiply both the sides by 8.

8(x + 1) = 3(2x + 3)

8x + 8 = 6x + 9

8x – 6x = 9 – 8

2x = 1

x = 1/2

Hence the solution for the equation is x = 1/2.

Rational Numbers

 A number is called Rational if it can be expressed in the form p/q where p and q are integers (q > 0). It includes all natural, whole number and integers.

Example: 1/2, 4/3, 5/7,1 etc. 

Natural Numbers

All the positive integers from 1, 2, 3,……, ∞.

Whole Numbers

All the natural numbers including zero are called Whole Numbers.

Integers

All negative and positive numbers including zero are called Integers.

Properties of Rational Numbers

1. Closure Property

This shows that the operation of any two same types of numbers is also the same type or not.

a. Whole Numbers

If p and q are two whole numbers then

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + q will also be the whole number.	p – q will not always be a whole number.	pq will also be the whole number.	p ÷ q will not always be a whole number.
Example	6 + 0 = 6	8 – 10 = – 2	3 × 5 = 15	3 ÷ 5 = 3/5
Closed or Not	Closed	Not closed	Closed	Not closed

b. Integers

If p and q are two integers then

Operation	Addition	Subtraction	Multiplication	Division
Integers	p+q will also be an integer.	p-q will also be an integer.	pq will also be an integer.	p ÷ q will not always be an integer.
Example	– 3 + 2 = – 1	5 – 7 = – 2	– 5 × 8 = – 40	– 5 ÷ 7  = – 5/7
Closed or not	Closed	Closed	Closed	Not  closed

c. Rational Numbers

If p and q are two rational numbers then

Operation	Addition	Subtraction	Multiplication	Division
Rational Numbers	p + q will also be a rational number.	p – q will also be a rational number.	pq will also be a rational number.	p ÷ q will not always be a rational number
Example				p ÷ 0= not defined
Closed or Not	Closed	Closed	Closed	Not closed

2. Commutative Property

This shows that the position of numbers does not matter i.e. if you swap the positions of the numbers then also the result will be the same.

a. Whole Numbers

If p and q are two whole numbers then 

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + q = q + p	p – q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example	3 + 2 = 2 + 3	8 –10 ≠ 10 – 8 – 2 ≠ 2	3 × 5 = 5 × 3	3 ÷ 5 ≠ 5 ÷ 3
Commutative	yes	No	yes	No

b. Integers

If p and q are two integers then

Operation	Addition	Subtraction	Multiplication	Division
Integers	p + q = q + p	p – q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example	True	5 – 7 = – 7 – (5)	– 5 × 8 = 8 × (–5)	– 5 ÷ 7 ≠ 7 ÷ (-5)
Commutative	yes	No	yes	No

c. Rational Numbers

If p and q are two rational numbers then

Operation	Addition	Subtraction	Multiplication	Division
Rational numbers	p + q = q + p	p –q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example
Commutative	yes	No	yes	No

3. Associative Property

This shows that the grouping of numbers does not matter i.e. we can use operations on any two numbers first and the result will be the same.

a. Whole Numbers

If p, q and r are three whole numbers then

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + (q + r) = (p + q) + r	p – (q – r) = (p – q) – r	p × (q × r) = (p × q) × r	p ÷ (q ÷ r)  ≠ (p ÷ q) ÷ r
Example	3 + (2 + 5) = (3 + 2) + 5	8 – (10 – 2) ≠ (8 -10) – 2	3 × (5 × 2) = (3 × 5) × 2	10 ÷ (5 ÷ 1) ≠ (10 ÷ 5) ÷ 1
Associative	yes	No	yes	No

b. Integers

If p, q and r are three integers then

Operation	Integers	Example	Associative
Addition	p + (q + r) = (p + q) + r	(– 6) + [(– 4)+(–5)] = [(– 6) +(– 4)] + (–5)	Yes
Subtraction	p – (q – r) = (p – q) – r	5 – (7 – 3) ≠ (5 – 7) – 3	No
Multiplication	p × (q × r) = (p × q) × r	(– 4) × [(– 8) ×(–5)] = [(– 4) × (– 8)] × (–5)	Yes
Division	p ÷ (q ÷ r) ≠ (p ÷ q) ÷ r	[(–10) ÷ 2] ÷ (–5) ≠ (–10) ÷ [2 ÷ (– 5)]	No

c. Rational Numbers

If p, q and r are three rational numbers then

Operation	Integers	Example	Associative
Addition	p + (q + r) = (p + q) + r		yes
Subtraction	p – (q – r) = (p – q) – r		No
Multiplication	p × (q × r) = (p × q) × r		yes
Division	p ÷ (q ÷ r)  ≠ (p ÷ q) ÷ r		No

The Role of Zero in Numbers (Additive Identity)

Zero is the additive identity for whole numbers, integers and rational numbers.

	Identity		Example
Whole number	a + 0 = 0 + a = a	Addition of zero to whole number	2 + 0 = 0 + 2 = 2
Integer	b + 0 = 0 + b = b	Addition of zero to an integer	False
Rational number	c + 0 = 0 + c = c	Addition of zero to a rational number	2/5 + 0 = 0 + 2/5 = 2/5

The Role of one in Numbers (Multiplicative Identity)

One is the multiplicative identity for whole numbers, integers and rational numbers.

	Identity		Example
Whole number	a ×1 = a	Multiplication of one to the whole number	5 × 1 = 5
Integer	b × 1= b	Multiplication of one to an integer	– 5 × 1 = – 5
Rational Number	c × 1= c	Multiplication of one to a rational number

Negative of a Number (Additive Inverse)

	Identity		Example
Whole number	a +(- a) = 0	Where a is a  whole number	5 + (-5) = 0
Integer	b +(- b) = 0	Where b is an integer	True
Rational number	c + (-c) = 0	Where c is a rational number

Reciprocal (Multiplicative Inverse)

The multiplicative inverse of any rational number

Example

The reciprocal of 4/5 is 5/4.

Distributivity of Multiplication over Addition and Subtraction for Rational Numbers

This shows that for all rational numbers p, q and r

1. p(q + r) = pq + pr

2. p(q – r) = pq – pr

Example

Check the distributive property of the three rational numbers 4/7,-( 2)/3 and 1/2.

Solution

Let’s find the value of

This shows that

Representation of Rational Numbers on the Number Line

On the number line, we can represent the Natural numbers, whole numbers and integers as follows

Rational Numbers can be represented as follows

Rational Numbers between Two Rational Numbers

There could be n number of rational numbers between two rational numbers. There are two methods to find rational numbers between two rational numbers.

Method 1

We have to find the equivalent fraction of the given rational numbers and write the rational numbers which come in between these numbers. These numbers are the required rational numbers.

Example

Find the rational number between 1/10 and 2/10.

Solution

As we can see that there are no visible rational numbers between these two numbers. So we need to write the equivalent fraction.

2/10 = 20/100((multiply the numerator and denominator by 10)

Hence, 2/100, 3/100, 4/100……19/100 are all the rational numbers between 1/10 and 2/10.

Method 2

We have to find the mean (average) of the two given rational numbers and the mean is the required rational number.

Example

Find the rational number between 1/10 and 2/10.

Solution

To find mean we have to divide the sum of two rational numbers by 2.

3/20 is the required rational numbers and we can find more by continuing the same process with the old and the new rational number.

Remark: 1. This shows that if p and q are two rational numbers then (p + q)/2 is a rational number between p and q so that

p < (p + q)/2 < q.

2. There are infinite rational numbers between any two rational numbers.