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.
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.
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

Exponential Growth Graph

The exponential growth graph is nothing but the exponential function occurs if the rate of growth is proportional to the functional value and the current value in the functional part. The exponential graph contains some of the equal intervals and can be called as the exponential growth or exponential decay. Now we are going to see about the exponential growth graph.
Problems on Exponential Growth Graph

Determine the exponential growth for the function $3000 and to double at 21/2 % continuously

Solution:

The exponential growth function can be calculated as,

A = Pert

The rate value which can be taken as 0.105

6000 = 3000 e 0.105t

We have to take natural log on both sides we get,

2 = e0.105t

ln 2 = ln e 0.105t

ln 2 = 0.105t (ln e)

ln 2 = 0.105t

By using the calculator the value can be found as,

0.693147 = 0.105t

t = 6.666

Thus it takes 6.66 years to double the money.

Graph:Graph
More Problems on Exponential Growth Graph:

Example 1:

Determine the value of ‘r’ where A = 50 at t = 6 years and P = 10

Solution:

The exponential growth can be calculated as,

A = Pert

50 = 10 e6r

5 = e6r

The logarithm for the above equation given as,

ln 5 = 6r

ln (5)/6 = r

r = 1.6094/6

r = 0.2682

The growth of exponential is 0.2682.

Example 2:

Determine the exponential growth for the function $4000 and to double at 24/2 % continuously

Solution:

The exponential growth function can be calculated as,

A = Pert

The rate value which can be taken as 0.12

8000 = 4000 e 0.12t

We have to take natural log on both sides we get,

2 = e0.12t

ln 2 = ln e 0.12t

ln 2 = 0.12t (ln e)

ln 2 = 0.12t

By using the calculator the value can be found as,

0.693147 = 0.12t

t = 5.77

Thus it takes 5.77 years to double the money.

Graph:

Graph

Relational Algebra Notation

The algebra is a mathematical system that consists of operands and operators. The operand is a variable or value from which new value can be constructed. The operator is a symbol denoted the procedures of that construct new value from given values. The relational algebra is the operand variable that represented relations and operator is designed to relation in the database. In this below details about the relational algebra notation.

Symbolic Notation in Relational Algebra Notation:

The relational algebra is whose operand is variable that represent in the relations and the operator is designed to do the most common things that we need to do with relations in a database. The algebra result can be used to query language for relations. There are following as relational algebra notation.

Symbolic notations are,

SELECT,

PROJECT,

PRODUCT,

JOIN,

UNION,

INTERSECTION,

DIFFERENCE,

RENAME,

Usage of notations in relational algebra notation:

1.      SELECT:

The SELECT notation represent in symbol as σ (sigma).The SELECT operation used to select the particular data details get in the database.

2.      PROJECT:

The PROJECT notation represent in symbol as π (pi). The PROJECT operation used to can provide itself to concision in the particular database.

3.      PRODUCT:

The PRODUCT is representing in symbol as x (times).The PRODUCT is performed combination of tuples in the data base.

4.      JOIN and UNION:

The JOIN is representing in symbol as |x|(bow-tie) and UNION is representing in symbol as U (cub). The JION   used to get particular data in that it has two tuples in restriction of their cartesian product based on the conditions.  The UNION is used to set the rotational attributes are same means at that data provided.

5.      INTERSECTION and DIFFERENCE,

The INTERSECTION and DIFFERENCE have symbols as ∩ and – (minus). The intersection is used to intersection of two relations and provides the common tuples in the relations. The difference used to when applied to relations which first relation has in the tuple but in the second relation.
Example of Relational Algebra Notation:

To find all employees in department cs.

SELECT depname = cs (employee), becomes σdepname = cs (employee).

To find all employees in department cs called jack.

SELECT depname = cs ^ jack = jack(employee), becomes σdepname = cs  ^ jack= jack(employee).

Volume Paraboloid

Paraboloid is defined as one of the most important mathematical shape. Paraboloid has three dimension shape. Paraboloid is defined as the combination of ellipse and parabola.  All the sections present in the paraboloid parallel to one coordinate plane is parabola and all the sections present in paraboloid parallel to one coordinate is ellipse. In this section, we are going to see about the volume of paraboloid in detail.
Explanation to Volume Paraboloid

The explanation to volume of paraboloid is given below the following section,

Formula:

Volume of paraboloid = `1/2` `Pi` a2 h

where,

h = height of the paraboloid

Example Problem to Volume Paraboloid

Problem 1: Find the volume of paraboloid, where, h = 10, a = 5.

Solution:

Step 1: The given values for finding the volume of paraboloid is as follows,

h = 10,

a = 5.

Step 2: To find:

Volume of Paraboloid

Step 3:The formula given for finding the volume of paraboloid is as follows,

Volume of Paraboloid = `1/2` `Pi` a2 h

Step 4: By substituting the values in the volume formula,

Volume of Paraboloid = `1/2` `Pi` a2 h

= `1/2` (3.14) (52 ) (10)

= `1/2` (3.14) (25) (10)

= `1/2` (3.14) (250)

=  `1/2` (3.14) (250)

= `1/2` (785)

= 392.5

Result: Volume of paraboloid = 392.5

Thus, this is the require answer for solving the volume of paraboloid.

Problem 2: Find the volume of paraboloid, where, h = 15, a = 6.

Solution:

Step 1: The given values for finding the volume of paraboloid is as follows,

h = 15,

a = 6.

Step 2: To find:

Volume of Paraboloid

Step 3:The formula given for finding the volume of paraboloid is as follows,

Volume of Paraboloid = `1/2` `Pi` a2 h

Step 4: By substituting the values in the volume formula,

Volume of Paraboloid = `1/2` `Pi` a2 h

= `1/2` (3.14) (62 ) (15)

= `1/2` (3.14) (36) (15)

= `1/2` (3.14) (540)

= `1/2` (1695.6)

= 847.8

Result: Volume of paraboloid = 847.8

Thus, this is the require answer for solving the volume of paraboloid.

Multiple Coefficients of Determination

In this topic we will discuss about multiple coefficients determination. The multiple coefficients are the coefficients that have the many term in an expression. In a variable, a coefficient is a number in front of a variable. For example, in the expression `2x^3-7x+19` , the coefficient of the `2x^2` is 1 and the coefficient of the x is -7. The third term, 19, is known as a constant. Here we see about multiple coefficients and problems.

Additional Information about Multiple Coefficients of Determination:

In an expression the coefficients may be positive or negative or zero. When addition or subtract the polynomials, just we want to add or subtract the coefficients in the same terms. The Coefficient in an expression is the number which is multiplied by one or more variables or powers of variables in the term. The following some of the related terms in multiple coefficients determination.

1) Variable

2) Expression

3) Polynomial

4) Term
Example Problem for Multiple Coefficients of Determination:

Multiple coefficient of determination – Example 1

Identify the number of terms in the expression, 2x2 + 4x + 7y + 6xy + 5xy2 + 3x2y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

2x2 the variable has the coefficient of 2.

4x the variable have the coefficient of 4.

7y this variable has the coefficient of 7.

6xy this variable’s having the coefficient of 6.

5xy2 this variable’s having the coefficient of 5.

3x2y this variable’s having the coefficient of 3.

Therefore, there are multiple terms in the expression.

Multiple coefficient of determination – Example 2

2) Identify the number of terms in the expression, x3 + 4x2 + 3x + 6y3 + 2y2 + 7y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

x3 the variable having the coefficient of 1.

4x2 the variable having the coefficient of 4.

3x this variable have the coefficient of 3.

6y3 this variable’s having the coefficient of 6.

2y2 this variable’s having the coefficient of 2.

7y this variable’s having the coefficient of 7.

Therefore, there are multiple terms in the expression.

Symmetric Equation of a Line

There are several ways to denote a distinct line in R3. Symmetric equation is one such way of representing distinct lines. Initially to deal with symmetric equation familiarity with parametric equation is necessary. Obviously this notation is considered compact too. Getting into the concept of symmetric equation, it sets each component of the line equal to a common parameter. Following this all the components are set equal to one another. Knowledge in vector equation of line and parametric equation would help us in solving tutorials.

More about Symmetric Matrix Equation of a Line:

The parametric equation of the line is stated as follows:

`(x-a_x)/(m_1)=(y-a_y)/(m_2)=(z-a_z)/(m_3)`

To arrive at this equation, each expression in the parametric equation of the line should be treated equal. For example

`x: a_x + (t)(m_1)`

It should be taken as

`x=a_x + (t)(m_1)`

The expression should be solved for each value of t and ultimately the resultant expression would be equal to the constant t and therefore would be equal to each other. Interestingly this expression of line violates one of the common doctrines of mathematics which limits us to use only one equality sign per expression. It is this factor that helps us to use this mode of representation in fewer or more dimensions. In case we are to use it in R2 form then the z expression would be ignored thus containing only one equality sign per expression. Similarly if we are expected to express a line that is parallel to a co ordinate plane then the term of the axis to which it is parallel to is also ignored.
Example Problems – Symmetric Equation of a Line:

Example 1 – Symmetric equation of a line:

Find the symmetric equation of the line through point (6,-7,20) & perpendicular to the plane 2x+3y-6z-8=0.

Solution:

Let us consider a normal vector n

2x + 3y - 6z - 8 = 0 is n = <2 -6="-6" 3="3">

The vector n is also the directional vector of the line thru point P(6, -7, 20) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <6 -7="-7" 20="20"> + t<2 -6="-6" 3="3">

L = <6 -7="-7" -="-" 20="20" 2="2" 3t="3t" 6t="6t" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 6 + 2t

y = -7 + 3t

z = 20 - 6t

Symmetric form of equation.

`t` =` (x - 6)/2` = `(y + 7)/3` = `(z - 20)/-6`


Example 2 – Symmetric equation of a line:

Find the symmetric equation of the line through point (2, 3, 4) & perpendicular to the plane x+2y-3z-4=0.

Solution:

Let us consider a normal vector n

x+2y-3z-4=0 is n = <1 -3="-3" 2="2">

The vector n is also the directional vector of the line thru point P(2, 3, 4) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <2 3="3" 4="4"> + t<1 -3="-3" 2="2">

L = <2 -="-" 2t="2t" 3="3" 3t="3t" 4="4" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 2 + t

y = 3 + 2t

z = 4 - 3t

Symmetric form of equation.

`t` = `(x - 2)/1` = `(y - 3)/2` = `(z - 4)/-3`

Solving for x With Polynomials

Polynomial defined as the function p(x) of the form p(x) =a0 +a1x+a2x2+……….anxn. Where a0, a1…an real numbers and n is the non negative integer is called polynomial in x over reals.For example 4x2-7x+3 is a polynomial over integers. If one of the powers of x in p(x) is either a negative integer of a fraction (either positive or negative), then p(x) is not a polynomial. For example x+2/x  is not a polynomial. The highest exponent of the variable in polynomial is called the degree of the polynomial.. Here we are going to study about how to solving for x with polynomial and its example problems.

Solving for X with Polynomials - Example Problems

Example: 1

Solve for x in the following polynomial expression 3x+5+6x +7 = 3x+4

Solving steps:    In the left hand side combine the like term first

3x+ 6x+ 7 + 5 = 3x + 4

9x + 12 = 3x + 4

Add both sides -4 we get

9x + 12 - 3 = 3x + 4 - 4

In right hand side 4 - 4 will be cancelling

9x+ 9 = 3x

Add both sides -3x

9x -3x + 9 = 3x -3x

6x + 9 =0

Add both sides -9 we get

6x + 9 -9 = -9

6x = -9

Divide both sides 6

x = - 9/6

The simplest form is - 3/2

Therefore the value of x = - 3/2

Solving for X with Polynomials - Example: 3

Solve for x in the following polynomial x2 + 9x +18 =0

Solving steps:

First we have to find the factor for a given polynomial

We can write

x2 + 9x +18  = (x+3)(x+6)

These are the two factors the equation

Now we solve the both equation.

Both terms equating to zero we get

First x+3=0

Add both sides -3 we get

x = -3

Next term is x+6 = 0

Add both sides -6

x = -6

Therefore the value of x is -1,-6