Problem Solving Training

Introduction to problem solving training

In this article we are giving problem solving training and we can understand how to solving problems. It is very helping you to improve your problem solving skills. In mathematical terms we can see different types of problem solving. Here you can learn math terms, Exam practices test, problem solving and online test and our tutor will helps you and you can get free online tutor. Let us see Problem solving training. Please express your views of this topic An Equation with no Solution by commenting on blog.

Problem Solving:

Let us see few problems and their solving methods.

Problem solving training 1:

Solve:  `12 / 6 + 5 / 3`

Solution:

Above problem is showing fraction addition. So we should solve this problem in fraction addition operation.

Step 1: `12 /6 + 5 / 3`

Here numerator values are same, but denominators are different. So we should take LCM then only we can add both values. (We can take LCM if denominators are different).

Step 2:  `12 / 6 + 5 / 3`

Take LCM 6, 3 (therefore LCM is 6)

we have to change denominators values like 6.

Step 3: `(12*1)/(6*1) = 12/6 , (5*2)/ (3*2) = 10 / 6`

`12/6 + 10 / 6`

Now denominators are same so add both values

Step 4: `12 / 6 + 10 / 6`

`22 / 6`

Therefore `12/ 6 + 5 / 3 = 22 / 6.`

Problem solving training 2:

Solve:    20 ___ 5   = 25

Solution:

Step 1: Given 20 ___ 5   = 25.

Step2: here we find the symbols which are need for this operation.

Step 3: if we put the - (minus) symbols like 20 - 5 = 15 we can get 15.

So minus operation is not accept

Step4: + (plus) is correct operation for this problem.

20 + 5 = 25

Step 5: Therefore + (plus) symbol is making the number sentences true.

Problem solving training 3:

Divide the two fractions `40 -: 1/ 5`

Solution:

Above problem is showing fraction division. So we should solve this problem in fraction division operation.

Step 1: given` 40 -: 1/ 5`

Step 2: It denoted by `40 -: 1/5`

The right hand side denominator will be change like as 5/1 so

=   `40 * 5/1`

= `200`

Step 3: Therefore answer is 200. Is this topic formula chart for math hard for you? Watch out for my coming posts.

Practices Problems:

1) Solve this fraction `40/ 20`      answer: 2

2) Add `2 / 3 + 3/ 2 `                      answer: `13/6`

3) Find missing number 4 , 8  12 , ___ , 20 , 24 , ____ , 32      answer: 16 , 28

Problem Based Learning Math

Introduction :

Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions.

Example Problems for Problem Based Learning Math:

Problem based learning math – Example: 1

Find a function that has choral on the slip between the lines `y=-x+3, y=-x-3` that takes the values -50 and 10 on the lower and upper lines.

Solution:

Guess `\phi(x,y) = Ax + By + C`

Find the values of `A, B, C.`

`\phi(3,0) = 3A+C=10`

`\phi(-3,0) = -3A+C=-50`

`10 = -50 + 6A`

`A = 10`

`\phi(0,3) = 3B+C=10`

`\phi(0,-3) = -3B+C=-50`

`B = 10`

`\phi(3,0) = 30 + C = 10, C = -20`

The solution is

`\phi(x,y) = 10x + 10y - 20`

Problem based learning math – Example: 2

Find the partial fraction decomposition of `\frac{4z+4}{z(z-1)(z-2)^2}.`

Solution:

`\frac{4z+4}{z(z-1)(z-2)^2} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2} + \frac{D}{(z-2)^2}`

`4z+4=A(z-1)(z-2)^2 + Bz(z-2)^2 + Cz(z-1)(z-2) + D z(z-1)`

Plug in z=0 to get A=-1

Plug in z=1 to get B=8

Plug in z=2 to get D=6

Differentiate both sides of the equation once with respect to z and plug in z=2 to get C=-7.

Finally

`\frac{4z+4}{z(z-1)(z-2)^2} = \frac{-1}{z} + \frac{8}{z-1} - \frac{7}{z-2} + \frac{6}{(z-2)^2}`

Problem based learning math – Example: 3

If `u(x,y) = e^x\sin y`   find `f(x,y) = u(x,y) + i v(x,y)`   and check if it satisfies the Cauchy-Riemann equations.

Solution:

The Cauchy-Riemann equations are `u_x=v_y, v_x=-u_y.`

`u_x = e^x\sin y, u_y = e^x\cos y`

`v_y = e^x\sin y`

`v =-e^x\cos y+g(x)`

`v_x =-e^x\cos y+g'(x)`

For the CR equations to hold, we must have `g'(x)=0` so that `g(x)=c\isin{R}.`

`f(x,y) = e^x\sin y + i(-e^x\cos y + c)`

`=e^x(\sin y-i\cos y) + ic`

`=-i e^x(\cos y + \frac{1}{-i}\sin y) + ic`

` =-i e^x e^{iy} + ic = -i e^z+ic`

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Practice Problems for Problem Based Learning Math:

1. Show that if `\phi(x,y)`   is harmonic then `\phi_x - i \phi_y`   is analytic.

2. Find a function that is choral on the vertical slip from x = 1 to 2 and equals 20 and 30 at x = 1 and 2.

`Answer: \phi(x,y) = 10x+10 `

Surface Area of Prisms and Cylinders

Introduction about prism and cylinder:

Prism:

In geometry, an n-sided prism is a polyhedron made of an n-sided polygonal base, a translated copy, and n faces joining corresponding sides. Thus these joining faces are parallelograms

Cylinder:

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder.

(Source – Wikipedia)


Having problem with Find the Area of a Square keep reading my upcoming posts, i will try to help you.

Formula Used to Find the Surface Area of the Prism and Cylinder:

Rectangular prism:

Surface area of the rectangular prism (A) = 2(wh + lw + lh) square units

w – Width

h – Height

l – Length

Triangular prism:

Total Surface Area of the triangular prism (T.S.A)

T.S.A = L.S.A + 2 x Base Area

Total surface area T.S.A = (P x h + 2 A) sq. units

P - Perimeter

A - Area of the base

h - Height of the prism

Cylinder:

Surface area of cylinder (SA) = 2 p r^2 + 2 p r h square units

r – Radius

h – Height

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Surface Area of Prisms and Cylinders - Example Problems:

1. The rectangular prism has the length 25 cm, width 13cm and height 10 cm. find the surface area of the rectangular prism.

Solution:

Given:

Length (l) = 25 cm

Width (w) = 13 cm

Height (h) = 10 cm

Formula to find the surface area of the rectangular prism:

Surface area (A) = 2(wh + lw + lh) square units

= 2 (13 x 10 + 25 x 13 + 25 x 10)

= 2 (130 + 325 + 250)

= 2 (705)

= 1410

Surface area of the rectangular prism (A) = 1410 cm 2

2. The triangular prism has the base side length 3 cm, 4cm and 5 cm. its height is 10 cm. find total surface area of prism.

Solution:

Given:

Three side length of triangular prism = 3 cm, 4 cm, 5 cm

Height of the prism = 10 cm

Base of the triangle = 3 cm

Height of the triangle = 4 cm

The lateral surface area of the prism = p x h square units

= (3 + 4 + 5) x 10

=12x10
L.S.A =120 cm2

Now the area of the bases, A = 1/2 bh square units

h is the height of the triangle

= 1/2 x 3 x 4

A= 6 cm2

The total surface area of the prism = p h + 2 A square units

= 120 + 2 x 6

Total surface area of the prism  =  1440 cm2

3. The cylinder has radius r = 3 cm, h= 12 cm. Find the surface area of cylinder.

Solution:

Given:

r= 3 cm

h=12 cm

Geometric Formula:

The surface area of the cylinder  = 2 p r^2 + 2 p r h square units

= 2 x 3.14 x 32 + 2 x 3.14 x 3 x 12

= 56.52 + 226.08

The surface area of the cylinder = 282.6 cm2

Reasons Why Math is Important

Introduction :
Math is an integral part of life.  It has a definite role to play in every body's life.

It is in demand both by lay men and  educated people.  Lay men or service men use math to add,  subtract, multiply or divide.

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A grocer adds up all the price of the purchases.  When he has to give back some money, he subtracts.  When we buy more than a kilogram of tomatoes, say 2 kgs of tomatoes, he uses multiplication table.  When we want  half a kilogram of tomatoes, he divides the price of 1 kilogram by 2.

Math  is important before a child is born on earth.  Doctors calculate the month of pregnancy to give the right medicine to the expectant mothers.  The growth of the child in the womb is measured.  After birth, the  injections and polio drops are to be given at the right time in the right doses all of which involves math calculation.

One of the reason why math is important is in agriculture.  Every nation has to feed its growing population.  The population of the world grows in exponential terms but the agricultural land is the same.  Hence  the math calculation involves  such  terms as how to get more yield quality wise from the same land to feed the growing population.

Reason Why Math is Important in School
Math is important for a parent when he puts his ward in school.  He has to pay tuition fees and it involves certain calculations like how  much should he save monthly to pay a term fees.  It also involves transportation fee,  cost of books and uniform.

It involves the basic  four principles of addition, subtraction, multiplication and division.
School also teaches math to prepare the student for their future life.  It teaches them, beside the four fundamental operations, fractions, decimals, measurements, conversions, square of a number, cube of a number, square root of a number , cube root of a number, profit and loss, investments, geometry  and statistics. The process of educations starts from knowing the numbers to using the numbers  with intelligence so that the students will become prosperous citizens.

Reasons Why Math is Important in Science, Astronomy Etc

Math is important for  calculations in science like the multiplication of cells in botany, the calculations involved in physics, the chemical formulas, the astronomical calculations like the distance of the planets from the Sun, the rotation period and the revolution periods, the appearance of certain asteroids, the  placement of the zodiacal  signs  in a year, the navigation of the ships, satellites, the places where the artificial satellites are to be placed to cover the whole earth, the study of the deep sea and  the depth of the earth, the depth at which good drinking water is available inside the earth, the gold mines, the coal mines, the depth where the diamond mines can be seen, and all the other important elements of the earth are situated.

Math is important  today in the telecommunication world, where conferences can be had between many countries sitting in their office desks.  Here math calculation of distances, speed of communication, clarity of sound etc are involved.

It can be shortly said that without math, there can be no  purposeful life.

Trigonometry Calculating Range

Introduction :
In mathematics trigonometry is one of the divisions. Trigonometry is derivative from the word Greek. In trigonometry, mainly study about the right triangle. Angle and side of the triangle is deal is deals with trigonometry.  The spherical trigonometry is major division of the trigonometry.  Interval of the given value (higher value and lower values) is said to be range. In the following we see detailed about trigonometry calculating range.

Range of Trigonometry Function:

The different between input values is said to be range. Range between the smaller value and the higher value.

Sine function:

[-1, 1] is the range of the sine function

Cosine function:

[-1, 1] is range of the cosine function.

Tangent function:

All real number is the range of the tangent function.

Cotangent:

All the real number is range of the cotangent.

Secant function:

Range of secant function is the [-`oo` ,-1] and [1,+`oo` ] is ;

Cosecant function:

The range of cosecant function is [-`oo` , -1] and [1, + `oo` ]

Example1: Trigonometry Calculating Range

Find the range of the given function:

P=13sin`(x+(Pi)/(8))+5`

Solution:

In this problem given function is:

P=13sin`(x+(Pi)/(8))+5`

For

P=sin`theta`

Common range value of the given sin function is:

-1`<=sintheta<=+1`

`theta``(x+(pi)/(8)) `

Substitute the `theta` value.

-1`<=sin(x+(pi)/(8))<=+1`

Multiply 13 from the both sides:

-13`<=13sin(x+(pi)/(8))<=+13`

Adding +5 from both side:

-13+5`<=13sin(x+(pi)/(8)+5)<=+13+5`

-8`<=13sin(x+(pi)/(8)+5)<=+18`

Range of the given problem is [-8, 18]

Example2: Trigonometry Calculating Range

Find the range of the given function:

P = 9sin`(x+(Pi)/(7))+3`

Solution:

In this problem given function is:

P = 9sin`(x+(Pi)/(7))+3`

For P = sin`theta`

Common range value of the given sin function is:

-1`<=sintheta<=+1`

`theta``(x+(pi)/(7)) `

Substitute the `theta` value.

-1`<=sin(x+(pi)/(7))<=+1`

Multiply 9 from the both sides:

-9`<=9sin(x+(pi)/(7))<=+9`

Adding +3 from both side:

-9+3`<=9sin(x+(pi)/(7))+3<=+9+3`

-6`<=9sin(x+(pi)/(7)+3)<=+12`

Range of the given problem is [-6, 12]

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Example3: Trigonometry Calculating Range

Find the range of the given function:

P = 6sin`(x+(Pi)/(4))+2`

Solution:

In this problem given function is:

P = 6sin`(x+(Pi)/(4))+2`

For P = sin`theta`

Common range value of the given sin function is:

-1`<=sintheta<=+1`

`theta``(x+(pi)/(4)) `

Substitute the `theta` value.

-1`<=sin(x+(pi)/(4))<=+1`

Multiply 6 from the both sides:

-6`<=6sin(x+(pi)/(4))<=+6`

Adding +2 from both side:

-6+2`<=6sin(x+(pi)/(4))+2<=+6+2`

-4`<=6sin(x+(pi)/(4))+2<=+8`

Range of the given problem is [-4, 8]

Solve Histogram Revision

Introduction to solve histogram revision:

A Histogram is a diagrammatic representation of frequency distribution through the four-sided figure whose width represents class intervals and corresponding frequencies. The class interavels and corresponding frequencies are directely propotional to each other. These histograms are two dimentional images. The histogram is used for constant data graphing. The histogram will be plotted continuously. Here we are going to solve histogram revision. I like to share this Histogram Definition with you all through my article.

Steps to Solve Histogram Revision:

Building a histogram is very easy. We normally use discrete data's for drawing the histograms. Here we are going to help for using of histogram

Step 1 :- If the given frequency distribution which we have is in inclusive form then we have to convert it into an exclusive form.

Step 2 :- Taking suitable scales are more important,and then mark the class-intervals along x-axis and frequencies on y axis.  Note that the scales chosen for both the axes need not be the same.

Step 3 :- Construct rectangles with class intervals as bases and the corresponding frequencies as heights. This is how we build a histogram. These are the steps to solve histogram revision. Please express your views of this topic 4th grade math problems online by commenting on blog.

Examples to Solve for Histogram Revision:

Solve a histogram  from the following frequency table.

Class Interval	Frequency
0 – 10	12
10 – 20	19
20 – 30	22
30 – 40	20
40 – 50	15

A two dimensional frequency density diagram is called a histogram. A histogram is a diagram which represents the class interval and frequency in the form of a rectangle. There will be as many adjoining rectangles as there are class intervals.

(1) The class intervals is taken on the X-axis and frequencies are taken on the Y-axis

(2) The scales for both the axes are marked. In the y axis the scales are marked 0, 10, 15, 20, 25.

(3) while we are marking the class intervals we have to be more perfect, so that we may not make any errors.

(4) Draw rectangle bars with class intervals as bases and the corresponding frequencies as heights.

(5) This is how we build a histogram by solving the given question

Property of Opposites

Introduction of Property of Opposites:

The property of opposites is the number changing its sign, that is positive being change to negative and negative being change to positive. I like to share this Associative Property Addition with you all through my article.

For Example: Let the number be ‘’a’’ and the property of opposites of the number ‘’a’’ is ‘’-a’’.

Let we see this property of opposites are applicable for addition, subtraction, multiplication and division. In this article, we see about how the property of opposites used for solving Problems.

Property of Opposites:

Case 1: Property of Opposites in Addition

Statement: Add the number and its property of opposites is zero

Example: A + (-A) = 0

Case 2: Property of Opposites in Subtraction

Statement: difference between number and its opposites is equal to the twice of that number.

Example: A – (-A) = A + A = 2A

Case 3: Property of Opposites in Multiplication

Statement: Multiply the number and its opposites, we get negative square of that number.

Example: A × (-A) = - A2

Case 4: Property of Opposites in Division

Statement: Divide a number by it opposites or vice versa, we get -1

Example:

A ÷ (-A) = -1

(-A) ÷ (A) = -1

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Example Problems – Property of Opposites:

Example 1:

Write the value of the sum of the number 5 and its opposites.

Solution:

Given: addition of the number 5 and its opposites

Expression: 5 + (-5)

We know that sum of any number and its opposites we get 0

5 + (-5) = 0

Answer: 0

Example 2:

Multiply the number 3 and its opposites

Solution:

Given: 3 × (-3)

Formula:

A × (-A) = - A2

3 × (-3) = - 32 = -9

Answer: 3 × (-3) = -9

Practice Problems – Property of Opposites

Problem 1:

What is the value of difference between the number 6 and its opposites?

Answer: 12

Problem 2:

What is the value when the number and its opposites comes under the division operation?

Answer: -1

Uniform Probability Distribution Learning

Introduction to uniform probability distribution learning:

In mathematics, uniform distribution is one interesting topics in probability theory and statistics. A continuous random variable X are said to be follow a uniform distribution, if it assumes only a values between a to b and its probability density function is represented by

f(x) =`1/(b-a)` , for a < x < b

Otherwise, f(x) = 0, then the values of x is not in the interval a and b.

In this article, we shall learning about the uniform probability distribution. The following are the example problems in uniform probability distribution learning.

Uniform Probability Distribution Learning - Formulas:

Some of the uniform probability distribution formulas for calculating mean, variance and standard deviation are

Mean (or) Expected value:

E(x) = `(b + a)/2`

Variance:

σ2 = `((b - a)^2)/12`

Standard deviation:

σ = `sqrt(((b-a)^2)/12)`

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Uniform Probability Distribution Learning - Example Problems:

Example 1:

If X is a continuous random variable for uniform probability distribution in the interval of 0 < x < 25. Estimate the expected value, variance and standard deviation.

Solution:

Given

0 < x < 25

f(x) = `1/(b -a)`

Here a = 0 and b = 25

f(x) = `1/(25 - 0)`

= `1/25`

f(x) = `1/25`

Expected value (or) Mean:

E(x) = `(b + a)/2`

= `(25 + 0)/2`

= `25/2`

= 12.5

E(x) = 12.5

Variance:

σ2 = `((b - a)^2)/12`

= `((25 - 0)^2)/12`

= `(25^2)/12`

= `625/12`

= 52.08

σ2 = 52.08

Standard deviation:

σ = `sqrt(((b -a)^2)/12)`

= `sqrt(((25 - 0)^2)/12)`

= `sqrt((25^2)/12)`

= `sqrt(625/12)`

= `sqrt(52.08)`

= 7.22

σ = 7.22

Answer:

f(x) = `1/25`

Mean (or) Expected value E(x) = 12.5

Variance σ2 = 52.08

Standard deviation σ = 7.22

Example 2:

If X is a continuous random variable for uniform probability distribution in the interval of 1 < x < 15. Estimate the expected value, variance and standard deviation.

Solution:

Given

1 < x < 15

f(x) = `1/(b -a)`

Here a = 1 and b = 15

f(x) = `1/(15 - 1)`

= `1/14`

f(x) = `1/14`

Expected value (or) Mean:

E(x) = `(b + a)/2`

= `(15 + 1)/2`

= `16/2`

= 8

E(x) = 8

Variance:

σ2 = `((b - a)^2)/12`

= `((15 - 1)^2)/12`

= `(14^2)/12`

= `196/12`

= 16.33

σ2 = 16.33

Standard deviation:

σ = `sqrt(((b -a)^2)/12)`

= `sqrt(((15 - 1)^2)/12)`

= `sqrt((14^2)/12)`

= `sqrt(196/12)`

= `sqrt(16.33)`

= 4.04

σ = 4.04

Answer:

f(x) = `1/14`

Mean (or) Expected value E(x) = 8

Variance σ2 = 16.33

Standard deviation σ = 4.04

Numbers in Expanded Form

Introduction to Numbers in Expanded Form:

The numbers are the main source of math. The number can be split up by showing each digit in the number represents. In other words, the expanded form is pulling a number out and it can be expressed it as a sum of the digits in a whole number. The simplest and easiest way for identifying the expanded form is to write the numbers in terms of words. Let us see about the numbers written in expanded form in this article.

Steps for Writing the Numbers in Expanded Form

Write the numbers both in numbers and words first. This represents the place value of each digit that holds in numerals.
Example: 2468 = 2 thousands, 4 hundreds, six tens and 8 ones.
Express the number in expanded form using given numbers such as writing the same number above, 2000 + 400 + 60 + 8.
If it is decimal, write the numbers as it is above for whole numbers but for decimals, first digit after the decimal point is tens and the second digit is hundreds and third digit is thousands.I have recently faced lot of problem while learning Subtracting Whole Numbers and Fractions, But thank to online resources of math which helped me to learn myself easily on net.

Example Problems to Numbers in Expanded Form

Example 1:

Write the number 827 in expanded form

Solution:

Step 1:

Write the number is words as well as in numbers.

827 = 8 hundreds, 2 tens and 7 ones

Step 2:

Write the numbers in expanded form representing the sum of the place values of each digit.

The value of 8 is 8 × 100 = 800

The value of 2 is 2 × 10 = 20

The value of 7 is 7 × 1 = 7

827 = 800 + 20 + 7

The solution for the number 827 in expanded form is 800 + 20 + 7.

Example 2:

Write the number 39845.76 in expanded form

Solution:

Step 1:

Write the number is words as well as in numbers.

39845.76 = 3 ten thousands, 9 thousands, 8 hundreds, 4 tens and 5 ones, 7 tenths and 6 ones

Step 2:

Write the numbers in expanded form representing the sum of the place values of each digit.

The value of 3 is 3 × 10000 = 30000

The value of 9 is 9 × 1000 = 9000

The value of 8 is 8 × 100 = 800

The value of 4 is 4 × 10 = 40

The value of 5 is 5 × 1 = 5

The value of 7 is 7/10 = 0.7

The value of 6 is 6/100 = 0.06

39845.76 = 30000 + 9000 + 800 + 40 + 5 + 0.7 + 0.06

The solution for the number 39845.76 in expanded form is 30000 + 9000 + 800 + 40 + 5 + 0.7 + 0.06

Problems to Practice for Numbers in Expanded Form

1. Write the number 7895 in expanded form.

Key: 7000 + 800 + 90 + 5

2. Write the number 45786.213 in expanded form.

Key: 40000 + 5000 + 700 + 80 + 6 + 0.2 + 0.01 + 0.003

Divisor

The most confusing words in math’s are in long division is dividend and divisor when we study long divisions specially is it not? So let us clear out the terms correct. So what is a divisor? The divisor is the number being divided by and on the other side when we are dividing something; and the dividend is the number that is being divided. Let us say 435 is our dividend and 25 be a divisor.  Now if we want to divide these two, we shall take our dividend first. So, 435 divided by 25. If we want to solve this using in fraction form we put the dividend on top as numerators and in the bottom the divisor as the denominator.

There is a quick trick to remember as these two words are very much alike. Divisor definition is that, they are divided in alphabetical order, so 432 divided by 25 or dividend divided by the divisor. So definition of divisor is nothing but the number we divide by.

Define divisor contains some rules, look how divisibility helps. Divisible by 2. All the even numbers can be divisible by 2. That is all numbers ending in 0, 2, 4, 6 or 8 are divisible by 2. Let us take a look, 452, 678, 420,366 all these numbers having the ending numbers in even number categories. So they are divisible by 2 and there remainders will come to zero. Divisible by 3. If we add up all the number and find out what is the total is or sum is. If the sum or the total is divisible by three, so is the number. For example, 12123 we add up all and get sum as 9, which is divisible by 3. 3 times 3 is equal to 9.  So we can conclude that 12123 is divisible by 3. Hence we can say here if divisor is 3 and the dividend is 12123 yes 3 can divide 12123 easily. When a divisor is say 4 what exactly we should keep in mind, the basic rule is we need to look at the last two digits in the number for example, 4720 here the 20 are the last two numbers which can get divided by 4 .Divisible by 5.  Say when the numbers ending in a 5 or a 0 are always divisible by 5. Numbers divided by 6 can be done when number divisible by 2 and 3 at the same time for example 36. Understanding the fundamental theorem of calculus is always challenging for me but thanks to all math help websites to help me out.

Thus these are few rules to keep in mind before we start using divisor in divisibility.