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

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

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

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.

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.

Solve Algebra Squared Fraction

Fractions:

In algebra, A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

In this article we are going to see how to solve algebra squared fraction and some example solved problems on algebra squared fraction.

Solve Algebra Squared Fraction :

Steps to solve algebra squared fraction:

Step 1: We know that ( a/b)^n = a^n / b^n . So we can take square for the numerator and denominator.

Step 2: If it is possible we can simplify it.

Let us see the example problems.
Example Problems on Solve Algebra Squared Fraction :

Problem 1:

Solve (4/5)^2

Solution:

Given , (4/5)^2

We need to squared the given fraction .

(4/5)^2

We know that ( a/b)^n  = a^n / b^n .

We can apply the above rule,

(4/5)^2 = 4^2 / 5^2

= 16 / 25

Answer: (4/5)^2 = 16 / 25

Problem 2:

Solve (3x / 4xy) ^ 2

Solution:

Given, (3x / 4xy) ^ 2

We need to find the squared the given fraction,

We know that (a/b)^n = a^n / b^n .

(3x / 4xy) ^ 2 = ((3x) ^ 2) / ( (4xy)^2)

= (9x^2) / ( 16x^2y^2)

now we can simplify it,

Divide by x^2 on both numerator and denominator,

(9x^2) / ( 16x^2y^2) = (9x^2)/x^2 / ( 16x^2y^2) / x^2

= 9 / 16y^2

Answer: (3x / 4xy) ^ 2   = 9 / 16y^2

Problem 3:

Solve (( x^2 - 4 ) / ( x + 2) )^2

Solution :

Given , (( x^2 - 4 ) / ( x + 2) )^2

We need to find the squared the given fraction,

We know that ( a/b)^n = a^n / b^n .

(( x^2 - 4 ) / ( x + 2) )^2 = ( x^2 - 4 )^ 2 / ( x + 2)^2

Before that we can simplify the given fraction,

(( x^2 - 4 ) / ( x + 2) )^2 = (((x+2)(x-2))/ ( x + 2))^ 2

= ((x - 2)^2)

= x^2 - 2x + 4

Answer: (( x^2 - 4 ) / ( x + 2) )^2 = x^2 - 2x + 4

Results and Discussion Section

Here we are going to see about the result and discussion, whether the solution to the given equation is correct or not. By simplifying the equation we find the value for the particular variable. when we substitute the value in the equation the equation balances on both sides.

For example: x – 4 = 0.

Here    x – 4 = 0

Add 4 on both sides. We get

x = 4

The solution is x = 4

According to the topic we have to check whether x = 4 is a solution to the given equation or not. By substituting the value the in equation we can see that both sides have same value.
Example Problem for Result and Discussion Section:

Given equation is discussion section  x + 7x + 5x + 3x – 5 + x = 29

Solution:

Given question is x + 7x + 5x + 3x – 5 + x = 29

Step 1:

Arrange the given question as per the x term and the numbers

x+ x + 3x + 5x + 7x – 5 = 29

Step 2:

Add the x term first

2x + 3x + 5x + 7x – 5 = 29

Step 3:

Add all the  ' x ' term in the equation

17x – 5 = 29

Step 4:

Add both sides by 5.Then we get

17x = 29 + 5

Step 5:

17 x = 34

Divide by 17 on both sides.

`(17x)/17` = `34/17`

x = 2

Discussion section about the result:

The result is 2. Here we are going to discuss about the result for given equation, whether the solution x =2 is correct or not.

Given question is x + 7x + 5x + 3x – 5 + x = 29

The result is x =2

Here we need to substitute the x value in the equation if that both side value is equal means the solution is correct. Otherwise the solution is not correct.

x + 7x + 5x + 3x – 5 + x = 29

Substitute the value x = 2 in the given equation

2 + 7( 2) + 5( 2) + 3(2) – 5 + 2 = 29

2 + 14 + 10 + 6 – 5 + 2 = 29

34 – 5 = 29

29 = 29

Both the sides are equal, so given solution is correct for the equation.
One more Example Problem for Result and Discussion Section:

Solve discussion section  16x - 12 = 20

Solution:

Add 12 on both sides. we get

16x - 12 + 12 = 20 + 12

16x  = 32

Divide by 16 on both sides.

`(16x )/ 16` = `32 / 16`

x = 2

whe we substitute x =2 in the equation 16x - 12 = 20. we get

16(2) - 12 = 20

32 - 12 = 20

20 = 20.

Thus the equation satisfies.