Thursday, January 31

Difficult Logic Problems


Problem: - Which term will replace the question mark in the series:

ABD, DGK, HMS, MTB, SBL,?

Solution: Clearly, the first letters of the first, second, third, fourth, and fifth terms are moved three, four, five, six and seven steps forward respectively to obtain the first letter of the successive terms. The second letters of the first, second, third, fourth and fifth terms are moved five, six, seven, eight and nine steps forward respectively to obtain the second letter of the successive terms. The third letters of the first, second, third, fourth and fifth terms are moved seven, eight, nine, ten and eleven steps forward respectively to obtain the third letter of the successive terms.

Thus the missing term is ZKW             (Answer)
Difficult Logic Problems next Set

Problem: - A child is looking for his father. He went 90 meters in the east before turning to his right. He went 20 meters before turning to his right again to look for his father at his uncle’s place 30 meter from this point. His father was not there. From there, he went 100 meters to his north before meeting his father in a street. How far did the son meet his father from the starting point?

Solution: - Clearly the child moves from A 90m eastwards up to B, then turns right and moves 20m up to C, then turns right and moves 30m up to D. Finally, he turns right and moves 100m up to E.

difficult logic problem

Clearly, AB = 90m, BF = CD = 30m.

So, AF = AB- BF = 60m

Also, DE = 100m, DF = BC = 20m

So, EF = DE- DF = 80m.

Therefore, his distance from starting point A = AE = `sqrt[ (AF)^2 +(EF)^2]`  = `sqrt[(60)^2 + (80)^2]`

= `sqrt(3600 + 6400)` = `sqrt10000` = 100m    (Answer)

Problem: - Each odd digit in the number 5263187 is substituted by the next higher digit and each even digit is substituted by the previous lower digit and the digits so obtained are rearranged in the ascending order, which of the following will be the third digit from the left end after the rearrangement?

Solution: - After performing operation on the digit we get 6154278

Arranging the above number in ascending order we get 1245678

Here third digit from the left end is 4.                (Answer)
More Difficult Logic Problems

Problem: - In a certain code TEMPORAL is written as OLDSMBSP. How is CONSIDER written in that code?            (Answer: RMNBSFEJ)

Problem: - In a certain code language ‘how many goals scored’ is written as ‘5 3 9 7’; ‘many more matches’ is written as ‘9 8 2’ and ‘he scored five’ is written as ‘1 6 3’. How is ‘goals’ written in that code language?  (Answer: either 5 or 7)

Problem: - Reaching the place of meeting on Tuesday 15 minutes before 08.30 hours, Jack found himself half an hour earlier than the man who was 40 minutes late. What was the scheduled time of the meeting? (Answer: 8.05 hrs)

Wednesday, January 30

Number of Diagonals in a Pentagon


Polygon is any shape, which is enclosed by its sides. The line segments that connect any two vertices are called as diagonals. The names of the polygon are classified on the basis of their number of sides. Any polygons which possess five sides are called as pentagon. In this article, we shall discuss about the number of diagonals of pentagon. Also we shall solve problems regarding number of diagonals of pentagon.

Formula - Number of Diagonals of Pentagon:

The number of diagonals of any polygon can be obtained by using the formula,

 [n(n-3)]/2

where n is the number of sides of the polygon.

diagonals of pentagon

Now we shall determine the number of diagonals of pentagon.

The number of sides of a pentagon is five.

Therefore n = 5.

By substituting n = 5 in the above formula, we can determine the number of diagonals of pentagon.

Number of diagonals of pentagon = [5(5-3)]/2

= [5(2)]/2

= 10/2

= 5

Therefore the number of diagonals of pentagon is 5.


Example Problem - Number of Diagonals in a Pentagon:

Determine the number of sides and name of the polygon whose number of diagonals is 5.

Solution:

Given:

The number of Diagonals = 5

The formula to determine the number of diagonals is

[n(n-3)]/2 ,

where n is the number of sides of the polygon.

We have to find n:

Equate the both.

[n(n-3)]/2 = 5

Multiply by 2 on both sides:

2[(n(n-3))/2] = 2(5)

n(n - 3) = 10

Multiply by n within the bracket:

n2 - 3n = 10

Subtract 10 on both sides.

n2 - 3n - 10 = 0

n2 - 5n + 2n - 10 = 0

n(n - 5) + 2 (n - 5) = 0

(n - 5) (n + 2) = 0

n - 5 = 0 and n + 2 = 0

n = 5 and n = -2

The sides of polygon must not be negative. So ignore -2.

Therefore the number of sides of the polygon is 5.

Pentagon is the polygon which possess 5 sides.

Hence pentagon is the polygon which possess 5 sides and 5 diagonals in it.

Monday, January 28

Plane Distance Calculator


In a plane distance calculator we have the axes points such as (x,y) and this coordinates are placed on the quadrant, there are four quadrants for (x,y) axis. Origin is the center or starting point for the (x,y). In two dimension coordinate system we have the two x and y plane.We can measure the distance between plane by the formulas.In this article we have the formulas and the problems for finding the plane distance.

Plane Distance Calculator:

Distance between plane calculator can be measured by the following formualas and that can explained by the figur shown below. From the above figure we can clearly understand that the distance 'l' from the origin we can find the plane distance by the coordinates given.

D =  ` |d1-d2|/sqrt(A^2 + B^2 + C^2)`

distance between two plane

How to use plane distance calculator:

First seperate the d1,d2 and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Problems in Plane Distance Calculator:


Example 1:
Find the distance from the two plane 2x – 3y + 3z = 12 and –8x + 12y – 12z = 24.
Solution:

First seperate the d1,d2 and all othher parameters
Now, enter the parameters in the respective column
And get the result in the new column

2x – 3y + 3z = 12
–8x + 12y – 12z = 24.  by dividing equation by 4   we get 2x - 3y + 3z = -6.
First sepearte the
Formula for find the distance between the plane

D =  ` |d1-d2|/sqrt(a^2 + b^2 + c^2)`

Here, a =2, b= -3 and c=3  d1= 12 d2 =-6

= | 12 - (-6) | / √(4 + 9 + 9)

= 18/√22

Example 2:

Find the distance between the parallel planes z = x + 2y + 1 and 3x + 6y - 3z = 4.

Solution:

First seperate the a,b,c,d and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Formula for find the distance between the plane

D =  ` |ax_1+by_1+cz_1+d|/sqrt(a^2 + b^2 + c^2)`

Here, a =3, b= 6 and c=-3 d =-4

=  ` |3 xx 0+6xx0+(-3)xx1-4|/sqrt(3^2 + 6^2 + -3^2)`

= `7/sqrt54 `

Thursday, January 24

Calculating Areas of Figures


In day-to-day life,we often came across the word area.In math, Area is nothing but a region bounded by a closed curve. In differential geometry of surfaces, area is considered as an important invariant.

Calculating areas of different figures is an important and an interesting one. In this article of  calculating areas of figures, the areas of different figures are calculated using the formulas.

Triangle:            Area of Triangle    =    ½ b h

b ---> base

h ---> vertical height

Rectangle:        Area of  Rectangle  =  l w

l ----> length

w ----> width

Square:               Area of a square   =  a2

a ----> side length

Parallelogram: Area of Parallelogram = b × h


b ----> breadth

h ----> height

Circle:                       Area of a Circle = pi r2

r ----> radius
Worked Examples for Calculating Areas of Figures:

Example 1:

Find the area of a triangle with base of  13 m and a height of 6 m.

Solution:

Area of a triangle =  ½ b h

=  ½ (13) (6)

=  39 m2

Example 2:

Find the area of rectangle given the length is 10 cm and width is 5 cm.

Solution:

Area of a Rectangle  =  l * w

= 10 * 5

= 50 cm2

Example 3:

Find the area of a square of side length 21 cm

Solution:

Area of a square  = a2

= 212

= 441 cm2

Example 4:

Find area of a parallelogram through base of 23 cm and a height of 17 cm.

Solution:

Area of a Parallelogram = b h

= (23) · (17)

= 391 cm2

Example 5:

The radius of a circle is 27 inches. Find its area.

Solution:

Area of  Circle = pi r2

= 3.14 (27)2

= 3.14 (729)

= 2289.06 in2
Practice Problems for Calculating Areas of Figures:

1) Calculate the area of rectangle given the length is 10 m and width is 7 m.

Answer: 70 m2

2) Find area of a square of side length 22 cm.

Answer: 484 cm2

3) Find the area of triangle given base is 14 cm and height is 7 cm.

Answer: 49 cm2

4) Find area of a parallelogram through base of 35 cm and a height of 15 cm.

Answer: 525 cm2

Wednesday, January 23

Radical Math Problems and Solutions


Radical problems and solutions are defined as one of the important topic in mathematics. Basically, there are three values are present in the radical number. Those values are named as called the index number, radical number, and the another one is known as the radicand number. For example, root(4)(12) is denoted as radical numbers. In this example of radical number, 4 is called as the index number, 12 is called as the radicand number. Mainly square root and the cubic roots are present in the radical statement.

Radical Expressions Calculator

The explanation for radical math problems and solutions are given below the following,

We can do many of the operation by using the radical. They are called as,

Addition problems and solutions by using the radical.
Subtraction problems and solutions by using the radical.
Multiplication  problems and solutions by using the radical.
Division problems and solutions by using the radical.

Example Problems and Solutions for Radical Math

Addition problems and solutions by using the radical.

Example 1: Add the following radical numbers, 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) + 10sqrt(2) + 10sqrt(5)

= 12sqrt(5)+ 10sqrt(5)  + 12sqrt(2)  + 10sqrt(2)

= 22sqrt(5) + 22sqrt(2)

This is the answer for radical numbers addition.

Subtraction problems and solutions by using the radical.

Example 2: Subtract the following radical numbers, 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) - 10sqrt(2) - 10sqrt(5)

= 12sqrt(5) - 10sqrt(5)  + 12sqrt(2)  - 10sqrt(2)

= 2sqrt(5)  + 2sqrt(2)

This is the answer for radical numbers subtraction.

Problem 3: Multiply the following radical numbers, 12( sqrt(5) + sqrt(2) ) and  10( sqrt(2) + sqrt(5) )

Solution:

12( sqrt(5) + sqrt(2) )   xx   10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) xx   10sqrt(2) + 10sqrt(5)

= 12 sqrt(10) xx  10 sqrt(10)

= 120 sqrt(100)

= 120  xx 10

= 1200

This is the answer for radical numbers multiplication.

Example 4: Divide the following radical numbers 1/(sqrt(7) - sqrt(8)) .

Solution:

1/(sqrt(7) - sqrt(8))

1/(sqrt(7) - sqrt(8))  xx  (sqrt(7) + sqrt(8))/(sqrt(7) + sqrt(8))

(sqrt(7) + sqrt(8))/(sqrt(7)^2 - sqrt(8)^2)

(sqrt(7) + sqrt(8))/(7 - 8)

(sqrt(7) + sqrt(8))/ - 1

= - (  sqrt(7) + sqrt(8) )

= - sqrt(7)  - sqrt(8)

This is the answer for radical numbers Division.
Practice Problems and Solutions for Radical Math

Example 1: Add the following radical numbers, 24( sqrt(3) + sqrt(12) ) + 12( sqrt(12) + sqrt(3) )

Answer: 36 sqrt(3)  +  36  sqrt(3)

Example 2: Subtract the following radical numbers, 24( sqrt(3) + sqrt(12) ) - 12( sqrt(12) + sqrt(3) )

Answer: 12 sqrt(3)  +   12 sqrt(3)

Monday, January 21

Percent Return Formula


In math, how much of parts done in every hundred is called as percents. The percents are represented by the symbol ‘%’. In other words, how much of value is noted out of hundred in experiments. The formula is returned with 100. Now we are going to see about percent return formula.


I like to share this Formula for Permutation with you all through my article.

Explanations for Percents Return Formula in Math

Percents return formula:

The percents are represented as fraction with percentage symbol that is 32/100%. We can denote the percents in whole number also like 32%.T he formula for returns the percents are P = ( observed value / total value) x 100.

How to return the percents using formula:

The formula for percents is divide the observed value and total value. Then multiply the 100 with that resultant value. Now, we can say this value is percents with symbol ‘%’. Sometimes, the formula returns the decimal value.

How to returns the fraction into decimal value:

We can represent the percent value in fraction and if there is any possible, we can simplify the fraction. Then divide the numerator value with denominator value.

More about Percents Returns Formula

Example problems for percents return formula in math:

Problem 1: Return the percent value using formula for given expression.

The student got the marks 140 out of 200. What is the percent value of student?

Answer:

The percent return formula is P = (observed value / total value) x 100.

The observed value is 140.

Return the percent as (140/200) x 100 = 0.7 x 100 = 70%.

Therefore, the formula returns the percent value as 70%.

Problem 2: Return the percent value using formula for given expression.

The fruit seller has 1650 apples out of 300 fruits. What is the percent value of apple?

Answer:

The percent return formula is P = (observed value / total value) x 100.

The observed value is 165.

Return the percent as (165/300) x 100 = 0.55 x 100 = 55%.

Therefore, the formula returns the percent value as 55%.

Exercise problems for percents return formula:

1. Return the percent value using formula for 65/130.

Answer: The percent value is 50.

2. Return the percent value using formula for 87/150.

Answer: The percent value is 58.

Friday, January 18

Probability of Rolling Doubles


Probability is the chance of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 3 when rolling a dice is ` 1/6` . In this article we will discuss about probability problems using dice.


Rules of Probability Doubles - Example Problems

Example 1: If rolling two dice, what is the probability of getting doubles?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, n(A) = 6

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Example 2: If rolling two dice, what is the probability of getting doubles or primes?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

n(A) = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} = 6

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Let B be the event of getting primes.

n(B) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5,2), (5, 6), (6, 1), (6, 5)} = 15

P(B) = `(n(B))/(n(S))` = `15/36` = `5/12`

P(A or B) = P(A) + P(B) = `1/6` + `5/12` = `7/12`

P(A or B) = `7/12`

Therefore probability of getting doubles or primes is `7/12`

Example 3: If rolling two number cubes, what is the probability of getting doubles or sum of 7?

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

n(A) = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} = 6

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Let B be the event of getting sum of 7.

n(B) = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} = 6

P(B) = `(n(B))/(n(S))` = `6/36` = `1/6`

P(A or B) = P(A) + P(B) = `1/6` + `1/6` = `1/3`

P(A or B) = `1/3`

Therefore probability of getting doubles or sum of 7 is `1/3`

Probability of Rolling Doubles - Practice Problems

Problem 1: If rolling two dice, what is the probability of getting a sum of 5 or 6?

Problem 2: If rolling two number cubes, what is the probability of getting 6 or 7?

Answer: 1) `1/4` 2) `11/36`

Thursday, January 17

Solving Double Number Identities


Trigonometry is arrived from the Greek word, trigonon = triangle and metron = measure. The father of trigonometry is Hipparchus. He designed the first trigonometric table. Identity is defined as an equation that is true for all probable values of its variables. In online, many websites provide online tutoring using tutors. Double number trigonometric identities problems are easy to solve. Solving double number trigonometric identities problems are easy.  Through practice, students can learn about solving trigonometric identities. Through online, students can practice more problems on trigonometric identities. In this topic, we are going to see about; solving double number identities.

Solving Double Number Identities: - Double Number Identities

The list of double number identities are given below,

sin `2theta` = 2sin `theta` cos `theta `

cos `2theta` = 2cos2 `theta` -1

cos `2theta` = 1- 2sin2 `theta`

cos `2theta` = cos2 `theta` – sin2 `theta`


Solving Double Number Identities: - Examples

Example 1:

Evaluate sin 60 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 60 = (2x30)

= 2 sin 30 cos 30

= 2 (0.5) (0.866)

= 2*0.433

= 0.866

The answer is 0.866



Example 2:

Evaluate sin 90 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 90 = (2x45)

= 2 sin 45 cos 45

= 2 (0.707) (0.707)

= 2*0.499

= 1

The answer is 1



Example 3:

Evaluate cos 50 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 50 = cos (2x25)

= cos2 25 - sin2 25

= (0.906)2 - (0.423)2

= 0.820 - 0.179

= 0.641

The answer is 0.641

Example 4:

Evaluate cos 90 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 90 = cos (2x45)

= cos2 45 - sin2 45

= (0.707)2 - (0.707)2

= 0.499-0.499

= 0

The answer is 0



Example 5:

Find cos 2y if sin y = -15/16 and in 3rd quadrant

Solution:

It is given that sin 2y is in 3rd quadrant,

Use the double angle identities

cos 2theta = 1- 2sin2 theta

cos 2y = 1 -2sin2 y

= 1 – 2`(-15/16)` 2

= 1 – 2 `(225/256)`

Taking LCM, we get

= `(256-450)/256`

= `-194/256`

= -97/128

The answer is `-97/128`

Wednesday, January 16

Sigma Algebra Examples


In mathematics, an σ-algebra is a technological concept for a group of sets satisfy certain properties. The main advantage of σ-algebras is in the meaning of measures; particularly, an σ-algebra is the group of sets over which a measure is distinct. This concept is important in mathematical analysis as the base for probability theory, where it is construed as the group of procedures which can be allocated probabilities. Now we will see the properties and examples.
Properties - Sigma Algebra Examples

Take  A  be some set, and 2Aits power set. Then a subset Σ ⊂ 2A is known as the σ-algebra if it satisfies the following three properties:

Σ is non-empty: There is as a minimum one X ⊂ A in Σ.
Σ is closed below complementation: If X is in Σ, then so is its complement, A \ X.
Σ is closed under countable unions: If X1, X2, X3, ... are in Σ, then so is X = X1 ∪ X2 ∪ X3 ∪ … .

Eg:

Thus, if X = {w, x, y, z}, one possible sigma algebra on X is

Σ = { ∅, {w, x}, {y, z}, {w, x, y, z} }.
Examples - Sigma Algebra

Example 1

X={1,2,3,4}. What is the sigma algebra on X?

Solution:

Given set is X={1,2,3,4}

So Σ = { ∅, {1,2}, {3,4}, {1,2,3,4}}.

Example 2

What is the sigma algebra for the following set ? X={2,4,5,9,10,12}

Solution:

Given set is X={2,4,5,9,10,12}

So   Σ = { ∅, {2,4}, {5,9}, {10,12},{2,4,5,9,10,12}}.

Example 3

11x+2y+5x+12a. Simplify the given equation in basic algebra.

Solution:

The given equation is  11x+2y+5x+12a

There are two related groups are available. So join the groups.

The new equation is,

(11x+5 x)+2y+12a

Add the numbers inside the bracket. We get 16x+2y+12a.

Arrange the numbers and we get the correct format.

=12a+16x+2y

We can divide the equation by 2.

So the equation 6a+8x+y.

These are the examples for sigma algebra.

Friday, January 11

Generating Function of Exponential Distribution


In mathematics, generating function of exponential distribution is one of the most interesting distributions in probability theory and statistics. The random variable X has an exponential distribution along with the parameter λ, λ> 0. If its probability density function is given by

f(x) = lambdae^(-lambdax) ,         x >= 0

Otherwise, f(x) = 0, x < 0. The following are the moment generating functions and example problems in generating function of exponential distribution.

Generating Function of Exponential Distribution - Generating Function:

Generating function (or) Moment generating function:

Moment generating function of exponential distribution function can be referred as some random variable which contains the other definition for probability distribution. Moment generating function give the alternate way to analytical outcome compare and also it can be straightly working with the cumulative and probability density functions. Moment generating function can be denoted as Mx(t) (or) E(etx). Here we help to calculate the moment generating function for the exponential distribution function.

Mx(t) = E(etx) =int_-oo^ooe^(tx)f(x)dx

= int_0^ooe^(tx)(lambda e^(-lambdax))dx

= λ int_0^ooe^(tx)(e^(-lambdax))dx

= λ int_0^ooe^((t-lambda)x)dx

= λ [e^(((t- lambda)x)/(t- lambda))]^oo_0

= λ [e^(((t- lambda)oo)/(t- lambda)) - e^(((t- lambda)0)/(t- lambda)) ]

Here we use this eoo = 0, and also we use this e0 = 1

= λ [ 0 – 1/(t- lambda) ]

= λ [ -(1/-( lambda-t)) ]

= lambda [1/ (lambda-t)]

= lambda/( lambda-t)

Mx (t) = E (etx) = lambda/(lambda-t)

Generating Function of Exponential Distribution - Example Problem:

Example 1:

If X has probability density function f(x) = e-5x, x > 0. Determine the moment generating function E[g(x)], if g(x) = e11x/15

Solution:

Given

f(x) = e-5x

g(x) = e11x/15

E(g(x)] = E[e11x/15]

= int_0^ooe^((11x)/15)f(x)dx

=  int_0^ooe^((11x)/15) (e^(-5x))dx

= int_0^ooe^(-x(5-(11/15)))dx

= [e^((-x(5-11/15))/-(5-11/15))]^oo_0

=[(e^-oo- (e^0)/-(5-11/15)) ]

= [0+ 1/(5-11/15) ]

= [1/((75-11)/15) ]

= [1/(64/15) ]

= 15/64

E [g(x)] = 0.2344

Answer:

E [g(x)] = 0.2344

Thursday, January 10

Pre Algebra Geometry


Geometry is one of a division of mathematics. Geometry is concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially the body of the practical knowledge is concerning with the lengths, areas, and volumes. In this article we shall discuss about pre algebra geometry problem. (Source: wikipedia)
Sample Problem for Pre Algebra Geometry:

Problem 1:

Find the diameter of the given circle. A given radius value of the circle is 16 cm.

Solution:

Given:

Radius of the circle is r = 16 cm

Diameter of the circle is d = 2r

d = 2 * 16

d = 32

So, the diameter of the circle is 32 cm2.

Problem 2:

Find the diameter of the circle. A given radius value of the circle is 20 cm.

Solution:

Given:

Radius of the circle is r = 20 cm

Diameter of the circle is d = 2r

d = 2 * 20

d = 40

So, the diameter of the circle is 40 cm2.

Problem 3:

A triangle has a perimeter of 34. If the two sides are equal and the third side is 4 more than the equivalent sides, what is the length of the third side?

Solution:

Let x = length of the equal side of a triangle

A formula for perimeter of triangle is

P = sum of the three sides of triangle

Plug in the values from the question

34 = x + x + x+ 4

Combine like terms

34 = 3x + 4

Isolate variable x

3x = 34 – 4

3x = 30

x = 10

The question requires length of the third side.

The length of third side is = 10 + 4 = 14

Answer: The length of third side triangle is 14.
Practice Problem for Pre Algebra Geometry:

Find the diameter of the circle. A given radius value of the circle is 14 cm.

Answer: d = 28 cm2

Find the radius of the circle. A given diameter value of the circle is 11 cm2.

Answer: 5.5 cm

Wednesday, January 9

Solving Word Percent Problems


In mathematics, a percentage is a way of expressing a number as a fraction of 100. It is often denoted using the percent sign, "%", or the abbreviation "pct". Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. Source: Wikipedia.
Solving Word Percent Problems

Solving percent problems - Example

Example 1: What percent of 30 is 60?

Solution:

30 × `x/100` = 60

30x = 6000

x = 200

Therefore 60 is 200 percent of 30.

Example 2: What is 45% of 250?

Solution:

`45/100`  × 250 = x

45 × 250 = 100x

x = `11250/100` = 112.5

Therefore 45 percent of 250 is 112.5.

Example 3:  25% of what is 15?

Solution:

`25/100` × (x) = 15

25x = 1500

x = 60

Therefore 25 percent of 60 is 15.

Example 4: What is the percentage increase from 14 to 24?

Solution:

Increase = 24 - 14 = 10

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `10/14` ×100% = 71.42%

Therefore 71.42% increase from 14 to 24.

Example 5: Last month, Anita earned $300. This month she earned $450. Calculate the percentage increase in her earnings.

Solution:

Increase = $450 - $300 = $150

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `150/300` × 100% = 50%

Therefore percentage increase in her earnings is 50%.

Example 6: What is the percentage decrease from 30 to 14?

Solution:

Decrease = 30 – 14 = 16

Percentage decrease = (Change in value / Original value) × 100%

Percentage decrease = `16/30` × 100% = 53.33%

Therefore 53.33% decrease from 30 to 14.

Example 7: Last month, Anita earned $400. This month she earned $340. Calculate the percentage decrease in her earnings.

Solution:

Decrease = $400 - $340 = $60

Percentage decrease = Change in value / Original value × 100%

Percentage decrease = `60/400` × 100% = 15%

The percentage decrease in her earnings is 15%.
Solving Word Percent Problems

Solving percent problems - Practice

Problem 1: What is 28% of 150?

Problem 2: Harry earned $270 in last month. He earned $370 this month. Calculate the percentage increase in his earnings.

Problem 3: Wilson earned $340 in last month. He earned $250 in this month. Calculate the percentage decrease in his earnings.

Answer: 1) 42 2) 37.03 3) 26.47

Monday, January 7

Number of Sides in a Pentagon


In geometry, a pentagon is a polygon with five sides. In a simple pentagon the sum of internal angles are about 540°. For example pentagram is a self-intersecting pentagon. Pentagon may be classified into regular and irregular. A pentagon that contains equal sides and equal internal angles are said to be regular pentagon otherwise the pentagon is irregular.pentagon


Number of Sides in a Pentagon:

The term penta indicates 5 .Hence the number of sides in a pentagon are 5 and the number of angles in a pentagon is 5.

A pentagon contains 712.694 million separate parallel lines.

Pentagon doesn’t have parallel lines.

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

Where perimeter of a polygon = 5 x side.

The following are the other polygonal shapes with their sides.

Tetragon - 4 sides

Hexagon- 6 sides

Heptagon- 7 sides

Octagon- 8 sides

Nonagon Enneagon- 9 sides

Decagon-10 sides

Undecagon- 11 sides

Dodecagon- 12 sides

Properties of pentagon:

Number of diagonals:

Number of diagonals in a pentagon is 5

The number of different diagonals possible from all vertices.

Number of triangles:

Number of triangles in a pentagon is 10.

The number of triangles formed by sketching the diagonals from a given vertex.

Sum of interior angles:

Sum of interior angles of a pentagon is 540° in general 180(n–2) degrees.

Example Problem- Number of Sides in a Pentagon:

Example 1:

Find the perimeter of a regular pentagon whose side is 5ft.

Solution:’

Given that, side = 5ft.

For a regular pentagon all the sides are equal.

Therefore the perimeter of a regular pentagon = 5 x side.

= 5 x 5 =25ft.

Example 2:

Find area, from the apothem and the perimeter of a polygon is 4ft and 20ft.

Solution:

Given that, apothem = 4ft.

Perimeter = 20ft

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

= (20 x 4) ÷2.

= 10 x 4.

= 40ft2

Friday, January 4

Boolean Algebra


Boolean Algebra is a branch of mathematic logics  whose use of symbols and  theory, set to represent the logical operations in the form of mathematics. This is the first logic which uses algebra and different methods for combining symbols used in proofs as well as deduction.

A Boolean Algebra is defined as:

It is  a set, having two special elements i.e, 0 and 1.
Algebra having three types of operations , which are
sum of two elements ("+"),
product sum of two elements  ("*") and
complement sum of two elements (" ' " or "prime")

these operations need to satisfy the Commutative axiom, Distributive axiom , Identity axiom (not including the boundedness identities) and Complement axiom.

These above axioms are almost equal to commutative property ,distributive property, identity property and complement property. Here we call them axioms because they are assumptions.

Boolean Algebra contains:

A  set of all propositions
The special characteristic elements - True  (1) i.e T  and False (0) i.e, F.
Three operations are
AND (product),
OR (sum) and
NOT (complement).

Laws of Boolean Algebra Axioms

To do any kind of operations using real numbers, they  depends on commutative axiom, associative axiom, and distributive axiom. In algebraic form these axioms  are expressed with letters or symbols, which are used to indicate an unknown number.

Commutative axiom

The commutative axioms explains that, numbers can be used for addition or multiplication in any manner.

Commutative axiom of Addition:

a + b = b + a        ( using addition law )

Commutative axiom of Multiplication:

a(b) = b(a)           ( using multiplication law )

Associative axiom

The associative axioms explain that, numbers which are used in addition or multiplication, also it can be grouped or regrouped in anyorder.

Associative Law of Addition:

a+(b+c) = (a+b)+c      ( using addition law )

Associative Law of Multiplication:

a(bc) = (ab)c                ( using multiplication law )

Distributive axiom

The distributive axioms are used for  both addition as well as  multiplication and state the following.

Distributive axiom for addition :

a(b + c) = ab + ac        ( using addition law )

Distributive axiom for multiplication :

(a + b)c = ac + bc             ( using multiplication law )

Identity axiom

Identity axiom for multiplication :

x · x = x                 ( using multiplication law )

Identity axiom for addition :

x + x = x                ( using addition law )



Zero Property in Boolean algebra axioms

0 · x = 0                  ( using multiplication law )

0 + x = x                   ( using addition law )

One Property  in Boolean algebra axioms

1 + x = 1                  ( using addition law )

1. x = x                     ( using multiplication law )
Examples on Boolean Algebra:

1)  `5xx(8+9)` = `5xx8 + 5xx9` ( USING DISTRIBUTIVE AXIOM  FOR MULTIPLICATION )

= `40 + 45`

=` 85`

= 5 x 8+5 x 9

2)  `3+7 = 7+3` ( BY USING COMMUTATIVE AXIOM FOR ADDITION)

3) `9xx5 = 5xx9 (` BY USING THE COMMUTATIVE AXIOM FOR MULTIPLICATION)

4)` 6xx1 = 6 ` (BY USING PROPERTY FOR ONE)
Practice Problems on Boolean Algebra:

1) Prove that  C+(A×B)=(C+A)×(C+B) by using Boolean algebra axioms

2) Prove that B+(C×A)=(B+C)×(B+A) by using Boolean algebra axioms

Thursday, January 3

Practical uses for Algebra


Algebra been used in all parts of our life activities either directly or indirectly.  If we want to find some interest which may recur with amount, we can apply progressions.  To see some different arrangements or combinations between available sources, we apply permutation and combination. To prove a generalized statement, we can apply mathematical Induction so on.

Let us see few problems of this kind.
Example Problem on Practical Uses for Algebra:

Ex 1: Peter buys a truck for dollar 120,000.  He pays half of the amount by cash. That is, dollar 60,000.  The remainingdollar 60,000, he pays in 12 annual installments of dollar 5000 each.  If the rate of interest is 12% and he pays with the installment the interest due on the unpaid amount, find the total cost of the truck.

Sol:  First installment = 5000 + 12% of 60,000

= 5000 + `12/100 xx` 60000

= $ 12,200.

Second Installment = 5000 + 12% of 55,000   = 5000 + `12 /100 xx` 55,000

= $ 11600.

Third Installment   = 5000 + `12/100 xx` 50,000

= $ 11,000

Since the common difference of the amount is $600,

Total cost of the shop = 60,000 + sum of the 12 installments.

= 60,000 + `12/2` [2` xx` 12,200 + (12 – 1) (- 600)]

= 60,000 + 6 [24,400 – 6,600]

= $ 1, 66,800.
More Example Problem on Practical Uses for Algebra:

2. A father is three times a old as his son.  In 12 years time, he will be twice as old as his son.  Find the present ages of father and son.

Solution: Let x and y be the present ages of father and his son respectively.

Given:  x = 3y `implies` x – 3 y = 0 -------- (1)

Also, x + 12 = 2 (y + 12)` implies` x + 12 = 2y + 24

`implies` x – 2y = 12 -------------------------- (2)

Therefore (1) – (2) `implies` x – 3y = 0

- x + 2y = - 12

`implies` - y = - 12 `implies` y = 12.

Therefore (1)` implies` x – 3 (12) = 0

`implies` x = 36.

Therefore the age of the father is 36 and his son’s age is 12.

Wednesday, January 2

Steps to Solving Quadratic Equations


In mathematics, an equation with second degree and one variable is called quadratic equation. The general equation for quadratic equation is given by,

A x2 + B x + C =0

Here x represents the variable and A, B, and C are constants, with a ≠ 0. (When a = 0, the equation is named as linear equation.)

The constant A, B, and C are expressed as quadratic coefficient, linear coefficient and the constant expression or release expression.

Steps to Solving Quadratic Equations Example 1:

5x² - x – 6 = 0, solving the factor for the given quadratic equation.

Solution:

Now, we can find the factor for the given quadratic equation

5x2 + 5x - 6x - 6 = 0

Now get the value for 5x from the primary term and 6 from secondary term.

5x (x + 1) - 6(x + 1) = 0

Now we can combine the similar term (x +1)

(5x - 6) (x + 1) = 0

To get the value for x we can associate the factor to zero

x + 1 = 0   or   5x – 6 = 0

x = - 1        or   5x = 6

x = 6 / 5

Thus, the factors  x1 and x2 are -1, 6/5.
Steps to solving quadratic equations Example 2:

3x² - 6 = -3x solving the factor for the given quadratic equation and solving the sum and product of quadratic equation.

Solution:

Fetch the -3x over: 3x² + 3x - 6 = 0

Separate 3x as 6x and -3x,

3x² + 6x - 3x - 6 = 0

Take out 3x from first two terms and 3 as common from next two terms

3x(x + 2) – 3(x + 2)  = 0

Thus, the factors are: (3x - 3) (x + 2) = 0

Modulate both expressions to zero: 3x - 3 = 0 and x + 2 = 0

3x - 3 = 0            x + 2 = 0

3x = 3                   x = - 2

x =  3/3

x = 1

So, the factors for x are  1, -2

Sum of the roots:

To find sum of roots consider the factor as x1 and x2

The sum of the roots  = x1 + x2 = (1) + (-2)

Sum of the roots = -1

Product of the roots:

To find product of roots consider the factors as x1 and x2

The product of the roots is given by x1x2 = (1)(-2)

Product of roots = -2


Quadratic Equation

ax2 +bx + c = 0


Values of

‘a’, ‘b’ and ‘c’


One Root

(x1)


Other Root

(x2)


Sum of Roots

(x1 + x2)


Product of Roots

(x1x2)

3x² +3x - 6 = 0


a = 3, b = 3, c = -6


1


-2


-1


-2

Steps to Solving Quadratic Equation Example 3:

Solving the value for x, to the given quadratic equation x2 + 5x + 4 = 0.

Solution:

Steps 1: To find the factor for the given quadratic equation, find the multiplicative value for the 4 and the sum of root value for 5

x2 + 1x + 4x + 4 = 0

Steps 2: Now obtain x as similar from the primary term and 4 as similar from last two term.

x (x +1) + 4(x + 1) = 0

Steps 3: Now we can unite similar term (x +1)

(x + 1) (x + 4) = 0

Steps 4: To obtain the value of x we can associate the factor to zero

x + 1 = 0   or   x + 4 = 0

x = - 1        or   x = -4.

Steps 5: Thus, the factors are x are -1, - 4.