Decimals and Fraction Solving

We are going to see about the topic of decimals and fraction solving with some related problems.  Decimal have point in between the numbers.  For example, 12.34 is a decimal number.  Fraction is formed by division method.  For example, `45/2` is a fraction.  Both the decimal and fraction are involved in the method of addition, subtraction, multiplication and division.

Decimal and fraction methods

Adding decimal:  Two decimal numbers are added.
For example,
                  1 2 . 3 4
                +   1 . 2 2
                -------------------
                  1 3 . 5 6
Subtracting decimal:  Two decimal numbers are subtracted.
For example,
                 1 2 . 3 4
                -   1 . 2 2
                ----------------
                  1 1 . 1 2

Multiplying decimal:  More than one decimal numbers are multiplied.
For example,
                1 . 2 × 1.2
               -------------------
                  1 . 4 4
Dividing decimal:  Divide any two decimal numbers.
For example,
                2)2.4   (1.2
                    1
                ----------
                       4
                       4
                ------------
                       0
In fraction,
Adding fraction:  More than one fraction numbers are added.  It has two method same denominator and different denominator.
For example,
               ` 5/12` + `3/12` = `8/12`
Subtracting fraction:  Subtract the fraction number from another fraction number.  It has two method same denominator and different denominator.
                `5/12` - `3/12 ` = `2/12`
Multiplying fraction:  More than one fraction numbers are multiplied.
                `5/12` * `3/12` = `5/48`
Dividing fraction:  Divide the fraction numbers.
                `5/12` % `5/4` = `1/3`

Example problems – Decimal and fraction solving

Example problem 1 – Decimal and fraction solving
Solving the fraction in addition and subtraction method:  `2/5` and `4/15` .
Solution:
The given fraction is `2/5` and `4/15`
First we are going to add the given fractions
`2/5` + `4/15`
In this given fraction, the denominators are not equal so we need to take the L.C.M
L.C.M of `2/5 ` and `4/15`
                = `2/5` + `4/15`
                = `(6 + 4)/15`
                = `10/15`
We can simplify the answer
                =` 10/15`
                = `2/3`
Now we are going to subtract the given fractions
`2/5 ` - `4/15`
In this given fraction, the denominators are not equal so we need to take the L.C.M
L.C.M of `2/5` and `4/15`
                = `2/5` - `4/15`
                =` (6 - 4)/15`
                = `2/15`
Answer for add and subtract is `2/3` and `2/15` .

Example problem 2 – Decimal and fraction solving
Solving the decimal in addition and subtraction 15.36 and 5.78
Solution:
First, we are going to add the given decimal numbers
Arrange the numbers one by one and then add the decimal number starts from right side,
                  1 5 . 3 6
                +   5 . 7 8
                  ---------------
                  2 1 . 1 4
Now, we are going to subtract the given decimal numbers
Arrange the numbers one by one and then subtract the decimal number starts from right side,
                  1 5 . 3 6
                -   5 . 7 8
                  ---------------
                     9 . 5 8
Answer for add and subtract is 21.14 and 9.58.

Practiced problems – Decimal and fraction solving

Practiced problem 1 – Decimal and fraction solving
Solving the fraction in multiplication method `32/5` and `55/64`
Answer:  `11/2`
Practiced problem 2 – Decimal and fraction solving
Solving the decimal in multiplication method 32.1 and 11.2
Answer:  359.52

Solving Expanded Binomials

In binomial is the distinct probability of the number of success in a series of n experiments, all of that defer success by probability p. Such experimentation is also identified a Bernoulli experiment. Actually, when n = 1, the binomial is a Bernoulli distribution. The binomial is the base for the accepted binomial analysis of statistical importance.

solving expanded binomials:

It is often used to form number of success in an example of size n since a population of size N. as the example are not self-sufficient the resultant distribution is hyper arithmetical distribution, not a binomial one. But, for N a large amount of n, the binomial distribution is a good estimate, and widely used.

Definition of binomial: (source: Wikipedia)

The binomial a2 − b2 can be factored as the product of two other binomials:

a2 − b2 = (a + b)(a − b).

an+1-bn+1 = (a-b)`sum_(k=0)^n a^kb^(n-k)`

This is a special case of the more general formula:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

A binomial increase to the nth power stand for as (a + b)n  know how to be expanded through way of the binomial theorem consistently. Taking a simple instance, the ideal square binomial (p + q)2 know how to be establish by squaring the first term, calculation twice the product of the first also second terms and finally calculation the square of the second term, to give p2 + 2pq + q2.

A simple however interesting application of the refer to binomial formula is the (m,n)-formula for make:

for m < n, let a = n2 − m2, b = 2mn, c = n2 + m2, then a2 + b2 = c2.

Example for solving expand binomials:

Example 1:

Expanded (4+5y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (4+5y)2

Step 2: `sum_(k=0)^2 ((2),(k))4^(2-k)(5y)^k`

Step 3: `((2),(0)) 4^2(5y)^0 + ((2),(1)) 4^1(5y)^1 +((2),(2)) 4^0(5y)^2`

Step 4:  16 + 20y + 25y2.

Example 2:

Expanded (6+3y)3 to solving the binomial equation.

Solution:

Step 1: given the binomials are (6+3y)3

Step 2: `sum_(k=0)^3 ((3),(k))6^(3-k)(3y)^k`

Step 3: `((3),(0)) 6^3(3y)^0 + ((3),(1)) 6^2(3y)^1 +((3),(2)) 6^1(3y)^2 + ((3),(3)) (3y)^3`

Step 4:  216 + 108y + 54y2+27y3.

My forthcoming post is on taylor series cos x and cbse sample papers for class 12th will give you more understanding about Algebra.

Example 3:

Expanded (2+3y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (2+3y)2

Step 2: `sum_(k=0)^2 ((2),(k))2^(2-k)(3y)^k`

Step 3: `((2),(0)) 2^2(3y)^0 + ((2),(1)) 2^1(3y)^1 +((2),(2)) 2^0(3y)^2`

Step 4:  4 + 6y + 9y2.

Subtracting Fractions with Common Denominators

In this article we are going to discuss about Subtracting fractions with common denominator or  like denominators.

A fraction involves two numbers. The top number is said to be numerator and the  bottom number is said to be denominator.

Fraction    =  (Numerator of a fraction) / (Denominator of a fraction)

To subtract the fractions with the common denominator, first subtract the numerators and then put that difference over that common denominator.

How to Subtract fractions with like denominators

Below are the examples on Subtracting fractions with common denominators:

Example problem 1:

Simplify the given fraction, 5/3 – 7/3

Solution:

= 5/3  – 7/3

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (5-7)/3

=(-2)/3

So the result is -2/3

Example problem 2:

Simplify the given fraction, 10/7 – 9/7

Solution:

= 10/7 – 9/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (10-9)/7

=(1)/7

So the result is 1/7

Example problem 3:

Simplify the given fraction, 13/9  – 10/9

Solution:

= 13/9 – 10/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (13-10)/9

=(3)/9

Equivalent fraction is 1/3

So the result is 1/3 .

Example problem 4:

Simplify the given fraction, 7/9 – 3/9

Solution:

= 7/9 – 3/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (7- 3)/9

= (4)/9

Equivalent fraction is 4/9

So the result is 4/9 .

Example problem 5:

Simplify the given fraction, 8/7 – 3/7

Solution:

= 8/7 – 3/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-3)/7

= (5)/7

Equivalent fraction is 5/7

So the result is 5/7 .

I am planning to write more post on unseen passages for class 7 and unseen passages for class 8. Keep checking my blog.

Example problem 6:

Simplify the given fraction, 8/13 – 5/13

Solution:

= 8/13 – 5/13

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-5)/13

= (3)/13

Equivalent fraction is 3/13

So the result is 3/13 .


Practice problems

Here are some practice problems on Subtracting fractions with like denominator

Problem1:  9/13 –5/13

Answer: 4 /13

Problem2:  7/10  – 5/10

Answer: 1/5

Problem3:  11/12 – 5/12

Answer: 1/2

Problem4:  6/7 – 5/7

Answer: 1/7

Problem5:  2/5 – 1/5

Answer:1/5

Figure Measurements

Measurements are used to measure the length of a cloth for stitching, the area of a wall for white washing, the perimeter of a land for fencing and the volume of a container for filling. The measurements consist of lengths, angles, areas, perimeters and volumes of plane and solid figures. If a student wants to know about the measurements of figure, they can be referring the following examples.

Figure Measurements – Examples 1:

These are the examples for measurements of a figure.

figure measurements

Find the area of the triangle whose height is 10cm and base is 6cm.

Solution:

Given base = 6cm

height = 10cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (6 xx 10)`

= `1 / 2 (60)`

= 30

Therefore, the area of the triangle is 30cm2

Find the area of the triangle whose height is 12cm and base is 8cm.

Solution:

Given base = 8cm

height = 12cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (8 xx 12)`

= `1 / 2 (96)`

= 48

Therefore, the area of the triangle is 48cm2

Figure Measurements – Examples 2:

These are the examples for measurements of a figure.

Measurements of Figure 1:

Three angles of a triangle are x + 34˚, x + 40˚ and x + 46˚. We have to find x for the triangle.

Solution:

x + 34 + x +40 + x + 46 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 120 = 180˚

3x = 180˚ - 120˚

= 90˚

x = `60/3`

= 20˚

Measurements of Figure 2:

The triangle has a three angles  x + 20˚, x + 10˚ and x + 30˚. We have to find x for the triangle.

Solution:

x + 20 + x +10 + x + 30 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 60 = 180˚

3x = 180˚ - 60˚

= 120˚

x = `120/3`

= 40˚

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Mixed Number Fractions and Strategies for Addition. I am sure they will be helpful.

Measurements of Figure 3:

The measurements of the angles whose triangle are in the ratio 2:1:3. Calculate the angles of the given triangle values.

Solution:

Ratio of the angles of a triangle = 2:1:3

Total ratio = 2 + 1 + 3

= 6

Sum of the three angles of a triangle is 180˚. Therefore,

First angle = `2/6 xx 180`

= 60˚

Second angle = `1/6 xx 180`

= 30˚

Third angle = `3/6 xx 180`

= 90˚

Math Symbols Terms

There are many terms and symbols used in math. The symbols can be used to perform some operations such as addition, multiplication, subtraction, division; relationship symbols such as less than, greater than, equal, not equal, etc. The math terms can be used to describe the topics such as algebra, arithmetic, array, and axis, etc. Let us see about math symbols and terms in this article.

I like to share this Symbol for Correlation with you all through my article. 

Some of the Math Symbols

Addition:
It can be represented as +. It is used for addition operation and also for logical OR purpose. It can be read as plus or logical or.
Subtraction:
It can be represented as –. It is used for subtraction operation. It can also spelled out as minus.
Multiplication:
Multiplication can be represented as ×. It is used for multiply the given terms.  It can also spelled out as into.
Division:
It can be represented as ÷. It is used for dividing the given terms.  It can also spelled out as divide by.
Percentage:
It can be represented as %. It is used for calculating the ratio that compares to the number 100.  It can also spelled out as percent.
Summation:
Summation can be represented as `sum`   . It is used for calculating the sum of many or infinite values.  It can also spelled out as sum.
Dot Product:
It can be represented as (.). It is used for calculating the scalar (dot) product of two vectors.  It can also spelled out as dot.
Cross Product:
It can be represented as (×). It is used for calculating the vector (cross) product of two vectors. It can also spelled out as cross.



Algebra is widely used in day to day activities watch out for my forthcoming posts on Fibonacci Number Sequence and number to words converter. I am sure they will be helpful.

Some of the Math Terms

Algebra:
An algebraic equation represents the scale.
Algorithm:
A step-by-step problem solving technique is used in math computations.
Area:
It can be used for representing the two-dimensional shapes such as polygon, octagon, hexagon, triangle and circle.
Array:
A set of numbers can be occurred in a specific size of pattern. Matrix or array consists of columns and rows.
Axis:
Axis consists of horizontal and vertical axis having co-ordinates in a plane.
Attribute:
Describing the object with data consists of shape, color, or size.
Arc:
A circumference of the circle or the portion of a segment draws with a compass.

Maxima and Minima

Maxima and Minima are the largest value (maximum) or smallest value (minimum), that a function can take at a point either within a given boundary (local) or on the whole domain of the function in its entirety (global). In general, maxima and minima of a given set are the greatest and least values in that set. Together, Maxima and Minima are called extrema (singular: extremum). We need to learn minima to determine the nature of the curve or the function and various other applications like projectiles, astrophysics to microphysics, geometry etc.

Learning analytical definition of minima:

A function f(x) is said to have a local minima point at the point x*, if there exists some ε > 0 such that f(x*) ≤ f(x) when |x − x*| < ε, in a given domain of x. The value of the function at this point is called minima of the function.

A function f(x) has a global (or absolute) minima point at x* if f(x*) ≤ f(x) for all x throughout the function domain.

We need to learn minima points of a curve by observing the involved function.

learning prologue regarding minima:

To learn Minima & Maxima, one needs to have a basic knowledge of calculus. The following points are some bare necessary (maybe not sufficient) definitions.

A function, y = f(x) is a mathematical relation such that each element of a given set ‘x’ (the domain of the function) is associated with an element of another set ‘y’ (the range of the function).

Closed interval of a domain is defined as an interval that includes its endpoints, as opposed to open interval which is an interval that does not include its endpoints.

A function, f(x) is said to be continuous at a given interval if it can assume all values within the interval i.e. the function is not broken anywhere inside the interval. Mathematically we determine this by ensuring the function has a finite value at the given point and taking the limit on both sides of the point and checking if they both exist and are equal (L.H.L. = R.H.L.).

Differentiability of a function is out of the scope of this article, but simply put, a function is said to be differentiable at a point if the curve at that point is smooth i.e. there is no drastic change of slope. Mathematically this is achieved by checking if both the left hand derivative and the right hand derivative of the function at the given point finitely exist and are equal (incidentally this common value is the value of the derivative of the function at the given point).

First Derivative is defined as the differentiation of a function, y = f(x), once, with respect to ‘x’. It is denoted by dy/dx or f’(x) and simply put, it gives the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the function w.r.t. ‘x’ at any given value of ‘x’.

Second Derivative is defined as the differentiation of a function, y = f(x), twice, with respect to ‘x’. It is denoted by d2y/dx2 or f’’(x) and simply put, it gives the slope of the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the slope of the function w.r.t. ‘x’ at any given value of ‘x’.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

Tests for Minima:

Local Minima can be found by Fermat's theorem, which states that they must occur at critical points.

If f(x) has a minima on an open interval, then the minimum value occurs at a critical point of f(x).

If f(x) has a minima value on a closed interval, then the minimum value occurs either at a critical point or at an endpoint.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test or second derivative test.

learning minima -First Derivative Test

Suppose f(x) is continuous at a critical point x*.

If f’(x) <0 an="" and="" extending="" f="" from="" interval="" left="" on="" open="" x="">0 on an open interval extending right from x*, then f(x) has a relative minima at x*.

If f’(x) >0 on an open interval extending left from x* and f’(x) <0 a="" an="" at="" extending="" f="" from="" has="" interval="" maxima="" on="" open="" p="" relative="" right="" then="" x="">
If f’(x) has the same sign on both an open interval extending left from x* and an open interval extending right from x*, then f(x) does not have a relative extremum at x*.

An interesting point to NOTE:

Differentiability is not a criterion for the first derivative test. Suppose f(x) is continuous but not differentiable at x*, i.e. f’(x*) does not exist. Still the above holds true since the test is done in open intervals on the left and right sides of the point in consideration [see Figure below]. So the criteria is only that f(x) is continuous at x* and that f’(x) exists in the neighbourhood of x*.

In summary, relative minima occur where f’(x) changes sign.

learning minima -The Second Derivative Test:

Suppose that x* is a critical point at which f’(x*) = 0, that f’(x) exists in the neighbourhood of x*, and that f’’(x*) exists.

f(x) has a relative minima at x* if f’’(x*)>0.

f(x) has a relative maxima at x*if f’’(x*) <0 .="" p="">
f(x) does not have an extremum at x* if f’’(x) = 0.

NOTE:

Differentiability at the critical point is a criterion for the second derivative test as opposed to the first derivative test. Also, if f’’(x*) = 0, the test is not informative [see Figure below], it actually means there is no change of sign of f’(x) on going from the left to right of the given critical point (these points are called the points of inflection).

learning Absolute Minima and Maxima

For any function that is defined piecewise, one can learn minima (or maxima) by finding the minimum (or maximum) of each piece separately; and then seeing which one is smallest (or biggest).

Visualization to learn minima

max-min

Least Common Denominator Tutoring

Least common multiple of two rational facts a and b is the least positive rational number, that is an integer multiple of a as well as b. because it is a multiple, it can be divided by a and b without a remainder. Tutoring not only help students by giving answers, but also help students in their problem solving with step by step solutions.  In this article we shall discuss about least common denominator tutoring. The following are the examples and steps involved in least common denominator tutoring.

Least common denominator tutoring:

Least common denominator:

The least common denominator of two or more numbers is the least number which is a multiple of each of the given number.

Technique: Least common denominator tutoring

To find the l.c.d.

Step (1) Write the multiples of first number

(2) Write the multiples of second number

(3) Write the common multiples

(4) Write the least common denominator.

Example problems least common denominator tutoring:

Example 1: Find the L.C.D of `1/4` and `1/3`

Here least common denominator is 12

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

= `(1*3)/12` + `(1*4)/12` = `(4+3)/12`

So the final result is `7/12` .

Example 2: Find the L.C.D of `1/5` and `1/4`

Here least common denominator is 20

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

=  `(1*4)/20 ` + `(1*5)/20`

So the final result is 9/20.

Example 3: Find the L.C.D of `1/25` and `1/15`

Here least common denominator is 75

To find the l.c.d.

Step (1) Write the multiples of 25: 25, 50,75

Step (2) Write the multiples of 15: 15, 30, 45, 60, 75

Step (3) common multiple is 75

Step (4) least common denominator is 75.

= `(1*3)/75` + `(1*5)/75` =`(5+3)/75`

So the final result is` 8/75` .

Practice problem for least common denominator tutoring:

Example 1: Find the L.C.D of `1/6` and `1/3`

Here least common denominator is 6

So the final result is` 3/6` .

Algebra is widely used in day to day activities watch out for my forthcoming posts on Least Common Multiples. I am sure they will be helpful.

Example 2: Find the L.C.D of `1/6 ` and `1/4`

Here least common denominator is 12

So the final result is `5/12` .

Example 3: Find the L.C.D of `1/4` and `1/2`

Here least common denominator is 4

So the final result is `3/4` .