Which Fraction is Larger

 Yes, which fraction is larger or smaller? The question has great importance. As in many of the competitive exams, we are asked to arrange a set of given fractions. Before learning, how to compare the fractions, we'll learn the basics of fractions. Then we'll learn two methods, that will help us to tell "which fraction is larger".

Fraction: A number of the form `a/b` , where a and b are integers and b`!=` 0,  is known as fraction. Here, a is called numerator and b is called denominator.

Example: `3/5` is a fraction, where Numerator = 3, and Denominator = 5.

7 is also a fraction, because it can be written as `7/1`

1st Method to find which fraction is larger


Method to compare two fractions: This method is known as Cross multiplication method. Steps are as follows:

Let `a/b` and  `c/d` be the two given fractions, that we have to compare and decide which fraction is larger?

Find the product of numerator of the first fraction and the denominator of the 2nd fraction. So, we get ad and bc.

i) If ad > bc, then `a/b`  >  `c/d` .

ii) If ad = bc, then   `a/b` = `c/d` .

iii) If ad < bc, then `a/b`  <  `c/d`

Ex: Compare, which fraction is larger `3/5` and `5/8`?

Products are 3X8= 24 and 5X5=25.

Here, 24 < 25,  So `3/5`` < ``5/8`

Now, we can say very easily, which fraction is larger.

Hence, `5/8` is larger fraction among the two.


2nd Method to know which fraction is larger:


This method is used to compare more than two fractions:

Step I: Find the LCM or LCD of the given fractions. Let it be m.

Step II: Convert all the given fractions into like fractions, each having m as denominator.

Step III: Now we compare the numerators. The fraction having greater numerator is greater and vice- versa.

Ex: Arrange the fractions  `2/5` , `3/10` and `9/14` in ascending order.

LCM of 5, 10 and 14 is 70.

Now, let us change each of the given fractions into an equivalent fraction having 70 as denominator.




Practice problems of which fraction is larger


1. Which fraction is larger: `5/6` or `6/7` ?

2. Which fraction is larger: `21/25` or `3/5` ?

Study graph functions

Graphs of functions f is the collection of the all ordered pairs (x, f(x)). In particular, if x is a real number, a graph means the graphical representation of this collection, in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. Graphing on the Cartesian plane is sometimes are called curve sketching. If the function input x is an ordered pair (x₁, x₂) of real numbers, the graph is the collection of all ordered triples (x₁, x₂, f(x₁, x₂)), and its graphical representation is a surface .

Example problems for study graph functions :

Study linear functions graph:
If a function f   : R → R is defined in the form f(x) = dx + e then the function is called a linear function. Here d and e are constants.

Problem 1: 

Draw the graph of the linear function f : R → R defined by f(x) = 7x + 1.
Solution:
Draw the table of some pairs (x, f(x)) which satisfy f(x) = 7x + 1.

x	-2	-1	0	1	2
f(x)	15	8	1	-6	-13

Plot the points and draw the curve passing through these points. Note that, the curve is a straight line.



Study functions of graph:

Graph of a functions: The graph of a function f is a graph of the equation y = f(x)

Problem 2:
Draw the graph of the function f(x) =2x²

Solution:
Draw a table of some pairs (x, y) which satisfy y = 2x²

x	-3	-2	-1	0	-1	-2	-3
f(x)	18	8	2	0	2	8	18

Plot the points and draw the smooth curve passing through the plotted points.



Note:

Note that if we draw a vertical line to the above graph, it meets the curve at only one point
i.e. for every x there is a unique y

Example problem for study logarithmic graph functions :

Draw the graphs of the logarithmic functions (1) f(x) = log ₅2x (2) f(x) = log ₁₀ex (3) f(x) = log ₅3x
Sol:

The logarithmic function is the defined only for the positive real numbers. i.e. (0, ∞)
Domain: (0, ∞) Range: (− ∞, ∞)

Study Online Exponentiation

Exponentiation should be the operation, which is written as the form of an. Where a and n is said to be base and exponent as well as n is any positive integer. In general, exponentiation means that repetitive multiplication. Otherwise, exponentiation an is the product of n factors of a. Online study should be the topic-oriented or else technical help for clarifying doubts that are convey through the computer software. Let us study properties and example problems for exponentiation.


Properties - Study Online Exponentiation

We are having seven number of exponentiation properties that used for solving problems. In this properties, a, m and n are any integer values.

Product of like bases:

The product of powers with the same base means we can add the powers and keep the common base.

am an = am+n

Quotient of like bases:

To divide the powers with similar base, we can subtract the exponents and remain the common base.

`a^m/a^n` = am-n

Power to a power:

In case of raising the power to power, we need to keep same base and multiply the exponent values.

`(a^m)^(n)` = amn

Product to a power:

In case of raising the product to power, we need to raise each factor to the power value.

(ab)m = am bm

Quotient to a power:

In case of raising the quotient to power, we need to raise the numerator and denominator to the power value.

`(a/b)^n` = `a^n/b^n`

Zero exponent:

Any number that is raised with zero power should be equivalent to ‘1’.

a0 = 1

Negative exponent:

a-n = `1/a^n` or `1/a^(-n)` = an

These are the properties that are used for exponentiation problems in study math online.


Example Problems - Study Online Exponentiation


Example 1:

Solve 32 34.

Solution:

Given, 32 34.

This is in the structure of am an, so we need to use am an = am+n property.

Here, m = 2 and n = 4 and a = 3.

Thus, 32 34 = 32+4

= 36

= 3 × 3 × 3 × 3 × 3 × 3

= 9 × 9 × 9

= 729

Hence, the answer is 32 34 = 729.

Example 2:

Shorten the following `6^7/6^4` .

Solution:

Given, `6^7/6^4` .

This is in the structure of `a^m/a^n` , so we need to use `a^m/a^n` = am-n property.

Here, m = 7 and n = 4 and a = 6.

Thus, `6^7/6^4` = 67-3

= 64

= 6 × 6 × 6 × 6

= 36 × 36

= 1296

Hence, the answer is `6^7/6^4` = 1296.

That’s all about the study online exponentiation.

Simple Algebra Problems

Algebra is a branch of mathematics dealing with variables, equations , expressions.Here we are going to solve some of the simple algebra problems including some equations, expressions and algebraic identities.

Solve Simple algebra problems on numbers:

Ex 1: Subtract  – 4 from – 10.

Sol :
=– 10 – (– 4)

= – 10 + (additive inverse of – 4)

= – 10 + 4

= – 6 (see addition of two integers)

Ex 2: Find the product of  (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)

Sol : 

(+ 6) × (– 5) = – 30 (positive × negative= negative)

(– 12) × (+ 12) = – 144 (negative × positive = negative) 

(– 15) × (– 4)= + 60 (negative × negative = positive) 

Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Sol:

Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)

= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

  The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2 


Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)

= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.

Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5 

Ex 5: Find the product of   given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.

Sol:

(x3 – 2x2 – 4) (2x2 + 3x – 1)

= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)

= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)

= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4

= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4

= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.


Algebra Identities to solve some simple problems :


 Algebraic identity for (x + a)(x + b)

 By using the distributive properties of numbers,

(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab

= x2 + ax + bx + ab= x2 + (a + b)x + ab.

 (x + a)(x + b) ≡ x2 + (a + b)x + ab

(x – a)(x + b) ≡ x2 + (b – a)x – ab

(x + a)(x – b) ≡ x2 + (a – b)x – ab

(x – a)(x – b) ≡ x2 – (a + b)x + ab

(a + b)2 ≡ a2 + 2ab + b2

(a – b)2 ≡ a2 – 2ab + b2

(a + b)(a – b) ≡ a2 – b2

(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3

(a − b)3 ≡a3−3a2b+ 3ab2−b3

a3 + b3 ≡ (a + b)3−3ab(a + b)

a3 − b3 ≡ (a −b)3+ 3ab(a−b)

Solve the following simple algebra problems using above identities:


Pro 1: Find the product:  (x + 3) (x + 5)

Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.


Pro 2:  Find the product:  (p + 9) (p – 2)

Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2

= p2 + 7p – 18.


Solve simple algebra problems on the cubic identity


Algebra problems on cubic identity:

(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3

= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3.

 (ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3

= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3

= 8x6– 36x4y + 54x2y2 – 27y3


Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.

Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)

= 64 – 12

= 52.


Pro 2: If a – b = 4 and ab = 2, find a3 – b3.

Solution:a3 – b3 = (a–b)3+3ab(a–b)

= (4)3+3(2)(4) =64+24

=88.

Median Test Learning

Median Definition :If the generally number of values within the case is even, next the median is the mean of the two center numbers. The median is a supportive number within cases wherever the distribution contain extremely large huge values which would or else twist the data. Represent the center value of a given set of values is called median.

To locate of values set within lower value toward higher value (rising order). If the collections of data contain an odd number of ways; the median is the middle value of the place behind arrangement the listing into rising order.


Median test learning


Method for median computation

`M=(n+1)/2`

Wherever,

M- Median,

n – Entirety number of set

But here is an odd number of values, then the median is the center value of the set.

If an even number to find median first

For example the given set{2,6,8,10}

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=6

Third value =8

Mean of second and third value is the Median of this set

Median`M=(14)/2`

`M=(7)/2`

M=3.5


Examples for median test learning


Example 1 for median test learning

Compute the median of this set {20, 10, 12, 40, 60,50,30}.

Solution:

First values can be arranged in order

Therefore the given set={10, 12, 20, 30, 40, 50, 60}

Here, total number of value (n) =7.

Median`M=(n+1)/2`

`M=(7+1)/2`

`M=(8)/2`

M=4

Therefore the fourth value is median that is 30.

Example 2 for median test learning

Compute the median of this set {10, 12, 13, 17, 18, and 20}

Solution:

Total number of value (n) =6

Median`M=(n+1)/2`

`M=(6+1)/2`

`M=(7)/2`

M=3.5

Consequently median must appear among Third and Fourth value.

Third value=13

Fourth value =17

Mean of third and fourth value is the Median of this set

Median`M=(13+10)/2`

`M=(30)/2`

M=15

Example 3 for median test learning
Compute the median of this set {26, 32, 36, and 55}

Solution:

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=32

Third value =36

Mean of second and third value is the Median of this set

Median`M=(32+36)/2`

`M=(68)/2`

M=34

Learning Weighted Median

In calculation of weighted median, the importance of all the items was considered to be equal. However, there may be situations in which all the items under considerations are not equal importance. For example, we want to find average number of marks per subject who appeared in different subjects like Mathematics, Statistics, Physics and Biology. These subjects do not have equal importance. Learning the formula to find weighted median by giving Median.


Learning Median Definition:


The arithmetic median computed by specific importance of every object is known weighted arithmetic median. To consider the every importance object, we can assume number known as weight to every object is directly proportional to its specific importance.

Weighted Arithmetic Median is computed by using following formula:

`sum` Yw = `(sum ty) / (sum t)`

Where:
Yw Stands for weighted arithmetic median.
y   Stands for values of the items and
t   Stands for weight of the item

Learning the important three methods of weighted median:

1.Arithmetic Median

2. Weighted Average

3. Average speed.

The formula for weighted average is:

Weighted Average = Sum of weighted objects / total number of objects


Learning to solve examples of Weighted median:


Ex 1:A student obtained 30, 40, 50, 60, and 35 marks in the subjects of Math, Statistics, Physics, Chemistry and Biology respectively. Assuming weights 1, 3, 5, 4, and 2 respectively for the above mentioned subjects. Find Weighted Arithmetic Mean per subject.

Solution:

Subjects	Marks Obtained       y	Weight    t	`sum` ty
Math	30	1	30
Statistics	40	3	120
Physics	50	5	250
Chemistry	60	4	240
Biology	35	2	70
Total		`sum` t = 15	`sum` y= 710

Now we will find weighted arithmetic mean as:
`sum` Yw = `( sum ty)/(sum t) = 710/ 15`   =  47.33marks/subject

Ex 2: A class of 30 students took a math test. 15 students had an average (arithmetic median) score of 90. The other students had an average score of 70. What is the average score of the whole class?

Solution:Step 1: To get the sum of weighted terms, multiply each average by the number of students that had that average and then sum them up.

90 × 15 + 70 × 15 = 1350 + 1050 = 2400

Step 2: Total number of objects = Total number of students = 30

Step 3: Using the formula

Weighted Average = Sum of Weighted objects / Total Numbers of objects.

=  `2400 / 30`

= 80.

Answer: The average score of the whole class is 80.

Learn Linear Functions in Real Life

In real life learning, linear function has single degree polynomial in its equation. The degree of the linear function always one. It cannot be increased. Linear function is  very useful in real life. In learning, linear function does not dependent any variable degrees. The linear function with the degree of one, always gives the straight line. In real life, linear function has the different variable equations. 'In learning of linear function in real life', we learn about linear equations and its degrees.

Forms of Linear function in real life


The function, which has straight line graph and that line function is known as linear function. In learning, linear functions are in three main forms.

1. Slope-Intercept Form is given by y = mx +b.

Example: y = 3x + 2

Here, slope m = 3 and y intercept b = 2

2. Point Slope Form is given by m = (y - y1) / ( x – x1)

Example: (y – 3) = 2(x – 1)

3. General Form is given by ax + by + c = 0

Example: 2x + 3y -1 = 0

Examples of set of real life linear function:

Example 1:

Solve for x and y. where, y = 3x + 4 and y = 7x.

Solution:

Plug, y = 3x + 4 in y = 7x

3x + 4 = 7x

x = 1

Now plug, x = 1 in y = 3x + 4

y = 3 + 4

y = 2

Answer:

The solutions are x = 1 and y = 2.

Example 2:

5x - y + 11 = 0 and x - y - 5 = 0, solve for x and y.

Solution:

5x - y + 11 = 0     (1)

x - y - 5 = 0         (2)

Subtract Equation (2) form Equation (1), we get,

4x + 16 = 0

4x = - 16

x = - 4

plug x = - 4 in equation (1)

- 4 – y + 13 = 0

- y + 9 = 0

y = - 9

The solutions are x = - 4 and y = - 9.


Learn linear function in real life - Practice problems:


Problem 1:

3x - y = 9 and 4x + y = 5. Solve for x, y.

A) (2, 3) B) (- 2, 3) C) (2, -3) D) (3, -2)

Answer: C

Problem 2:

2x + y = 12 and y = 3x - 3. Solve for x, y.

A) (3, 6) B) (3, - 6) C) (4, 5) D) (6, - 3)

Answer: A

Problem 3:

Solve the linear function 2x - y = 12 and y = 4x - 2

A) (- 5, 22) B) (5, - 22) C) (- 5, - 22) D) (5, 22)

Answer: C