How to solve a percent problem

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to solve percent problem.                                                                                                                                - Source from Wikipedia

Formula – How to solve percent problem:

To solve the percent problem convert the fraction else a decimal to a percent and then multiply it by 100:

To get the percent value multiply the fraction into hundred. For example

` 2 / 4` (100) = `2 / 4 ` * 100 = 0.5 * 100 = 50%

To convert the percent value to fraction divide it by hundred.

25% = `25 / 100 ` = `1 / 4`

Example: How to solve percent problem:

Problem 1:

What is the decimal value of 72%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `72 / 100` = 0.72.

Therefore the decimal value of 72% is 0.72.

Problem 2:

What is the decimal value of 86.3%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `(86.3) / 100` = 0.863.

Therefore the decimal value of 86.3% is 0.863.

Problem 3:

What is the decimal value 18% of 60?

Solution:

To find 18 percentages divide 18 by 100 and then multiply it by 60.

= `18 / 100` * 60 = 10.8

Hence 18% of 60 = 10.8

Problem 4:

What is the decimal value for 15.5% of 150?

Solution:

To find 15.5 percent divide 15.5 by 100 and then multiply it by 150.

=` (15.5) / 100` * 150 = 23.25

Hence 15.5% of 150 = 23.25

Problem 5:

The book cost is 5.00 dollars. Now it was 15% increased. What will be the new cost?

Solution:

Multiply the cost by 15%. That is `15 / 100` * 5.00.

= `15 / 100` * 5.00 = 0.75

Add this decimal value with original cost (5.00+0.75= 5.75)

Therefore the original cost of the book is 5.75 dollars.

Problem 6:

In a box there are 85 chocolates. Suppose if a boy took out 60% of chocolates, how many chocolates did he left in the box?

Solution:

The boy took out the chocolates from the box is 60% of 85 or else 60 / 100 × 85

`60 / 100` × 85 = 51%

So the box has 85 chocolates and the boy took out 51 chocolates. Hence the number of chocolates the boy left in the box is 85 – 51 = 34.
Therefore the boy left 34 chocolates in the box.

Problem 7:

A theatre contains 530 chairs. 470 of them are occupied. What percentages of the chairs not occupied?

Solution:

Number of chairs                        = 530

Number of chairs occupied       = 470

Number of chairs not occupied = 530 – 470 = 60

Number of chairs not occupied percentage = `60 / 530` × 100 = 11.32%
Practice problems – How to solve percent problem:

Problem 1:

What is the decimal value of 27%?

Solution:

27% = 0.27

Problem 2:

What is the decimal value 8% of 75?

Solution:

8% of 75 = 6

What is the Associative Property

The addition or multiplication of a collection of integer is the same separately from of how the numbers are grouped. A binary operation * is assumed to be associative if any three elements a, b, c of a set a *(b * c) = (a * b) * c. Multiplication of real integers are associative but the division of real integers are not assosiative. And addition is associative but subtraction is not.

What is the Associative Properties

What is the associative property?..The associative property is a property which always involve 3 or more numbers in calculations. The parenthesis indicates the terms that are consider one unit. The gathering (Associative Property) are within the parenthesis. Hence, the figures are 'associated' together. In multiplication, the result is always the same regardless of their combination.

There were two properties are involved in what is the associative property,

They were, Associative property in addition

Associative property in multiplication


Examples for what is the Associative Property (Addition and Multiplication)

Example for what is the Associative property in addition:

Let a, b, c be any real numbers, then

A + (B + C) = (A +B) + C

Let A = 5, B =9, C = 8

L.H.S = 5 + (9 + 8) = 5 + (17) = 22

R.H.S = (5 + 9) + 8 = (14) + 8 = 22

L.H.S = R.H.S

More Examples for addition:

When we change the groupings of addends, the sum does not change:

(8 + 5) + 9 = 22               (or)

8 + (5 + 9)  = 22

(9 + 1) +8 = 18                 (or)

9 + (1 + 8) = 18
Just consider that when the grouping of addends vary, the sum leftovers the same.

Example for what is the Associative property in multiplication:

Let A, B, C be any real numbers, then

A * (B * C) = (A *B) * C

Let, A = 5, B =9, C = 8

L.H.S = 5 * (9 * 8) = 5 * (72) = 360

R.H.S = (5 * 9) * 8 = (45) * 8 = 360

L.H.S = R.H.S

More Examples for Multiplication:

When we change the groupings of factors, the product does not change:

(1 x 9) x 5 = 45          or

1 x (9 x 5) = 45

(2 * 5) * 10 = 100      or

2 * (5 * 10) = 100

Just remember that when the grouping of factors changes, the product remains the same.

Simple Math Multiplication

In mathematics multiplication is used to multiple the numeric values or numbers. Multiplication can also be visualize as counting objects set in a rectangle or as result the area of a four-sided figure whose sides have given lengths. The area of a four-sided figure is not depend on which side is calculated first which illustrates that the order in which values are multiply as one does not matter

Multiplication examples:

Example 1:12 X 8

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 8 = 8.

Step 2: Multiply this 8 by 10 giving 80.

Step 3: Now multiply the 8 by the 2 of twelve giving 16. Add this to 80 giving 84.

Therefore 8 X 12 = 96

Example 2: 18 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 18. So 1 X 8 =1 8.

Step 2: Multiply this 18 by 10 giving 180.

Step 3: Now multiply the 18 by the 2 of twelve giving 36. Add this to 180 giving 216.

Therefore 18 X 12 = 216

Example 3 :12 X 9

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 9 = 9.

Step 2: Multiply this 9 by 10 giving 90.

Step 3: Now multiply the 9 by the 2 of twelve giving 18. Add this to 70 giving 108.

Therefore 9 X 12 = 108

Sample examples:

Example 4: 19 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 9 =1 9.

Step 2: Multiply this 19 by 10 giving 190.

Step 3: Now multiply the 19 by the 2 of twelve giving 34. Add this to 170 giving 228.

Therefore 19 X 12 = 228

Example 5 :23 X 7

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So, 1 X 7 = 7.

Step 2: Multiply this 7 by 10 giving 70.

Step 3: Now multiply the 7 by the 2 of twelve giving 14. Add this to 70 giving 161.

Therefore 7 X 23 = 161

Example 6: 20 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 20. So 2 X 10 = 20.

Step 2: Multiply this 20 by 10 giving 170.

Step 3: Now multiply the 17 by the 2 of twelve giving 34. Add this to 170 giving 240.

Therefore 17 X 12 = 240

Example 7 :12 X 10

Step 1: Multiply the 1 of the 12 by the

number we are multiplying by, in this case 7. So 1 X 10 = 10.

Step 2: Multiply this 12 by 10 giving 120.

Step 3: Now multiply the 10 by the 2 of twelve giving 20. Add this to 70 giving 84.

Therefore 12 X 10 = 120.

Example 8: 30 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 30 =30.

Step 2: Multiply this 30 by 12 giving 360.

Step 3: Now multiply the 30 by the 2 of twelve giving 60. Add this to 170 giving 360.

Therefore 17 X 12 = 360.

Complex Analysis Problems

Complex analysis is the functions of the complex numbers. Complex numbers have the real and also the imaginary parts. The complex numbers can be represented as x +i y, where x denoted as the real part and y denoted as the imaginary part. The complex numbers comprises the addition of two complex numbers, subtraction of two complex numbers, multiplication of two complex numbers and also the division of two complex numbers. The conception of the complex numbers is the reflection of the fact. This article has the functions of the complex numbers.

Examples for learn complex analysis problems:

Example 1 to learn complex analysis problems:

Compute the value for the complex number f (x) = (75+42i) + (90+72i).

Solution:

The given complex number is f (x) = (75+42i) + (90+72i).

Step 1: (75+42i) + (90+72i) = (75+90) + (42i + 72i)

Step 2: (75+42i) + (90+72i) = 165 + (42i + 72i)

Step 3: (75+42i) + (90+72i) = 165 +114i

The value for the complex number f (x) = (75+42i) + (90+72i) is f (x) = 165 +114i.

Example 2 to learn complex analysis problems:

Resolve the value for the complex numbers f (x) = (13+i) x (11-i).

Solution:

The given complex numbers are f (x) = (13+i) x (11-i).

Step 1: (13+i) x (11-i) = 13(11 - i) + i (11 -i)

Step 2: (13+i) x (11-i) = (143- 13 i) + (11i - i2)

Step 3: (13+i) x (11-i) = 143 - 13i + 11i - (-1) (where i2= -1)

Step 4: (13+i) x (11-i) = (143+1) + (-13i +11i)

Step 5: (13+i) x (11-i) = 144 - 2i

The value for the complex number f (x) = (13+i) x (11-i) is f (x) = 144 - 2i.

Example 3 to learn complex analysis problems:

Calculate the value for the complex number f (x) = (180+53i) - (82+ 7i).

Solution:

The given complex number is f (x) = (180+53i) - (82+ 7i).

Step 1: (180+53i) - (82+ 7i) = (180-82) + (53i - 7i)

Step 2: (180+53i) - (82+ 7i) = 98 +46i

The value for the complex number f (x) = (180+53i) - (82+ 7i) is f (x) = 98 +46i.

Practice problem for learn complex analysis problems:

Compute the value for the complex numbers f (x) = (45+29i)-(2+7i).

Answer: f (x) = 43- 22i

Predict the value for the complex numbers f (x) = (145+88i) + (33+39i).

Answer: f (x) = 178+127i

Compute the value for the complex numbers f (x) = (5+3i) x (2+i).

Answer: f (x) = 7+11i

Calculate Ratio Math

In mathematics, a ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient.

Example:

For every Spoon of sugar, you need 2 spoons of flour (1:2)

Source: Wikipedia

Definition- calculate ratio math:

The ration of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the terms of the ratio.

Concept - calculate ratio math:

The numeric ratio of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the conditions of the numeric ratio.

Types of ratio- calculate ratio math:

Compounded ratio in math.

Duplicate ratio in math.

Triplicate ratio in math.

Define compounded ratio in math:

Regular format for compounded numeric ratio is

p/q *r/s=(pr)/(qs)

Example for compounded ratio in math:

How to calculate ratios: 7/5*3/3

Solution:

=7/5*3/3

=(7*3)/(5*3)

=21/15

=7/5 or 7:5

Define duplicate ratio in math:

Regular format for duplicate ratio is

(r/s)*(r/s)=(r^2)/(s^2)

Example for duplicate ratio in math:

How to calculate ratios: 6/5*6/5

Solution:

=6/5*6/5

=(6^2)/(5^2)

=36/25 or 36:25

Define triplicate ratio in math:

Regular format for triplicates ratio is

r/sr/sr/s=(r^3)/(s^3)

Example for triplicate ratios in math:

How to calculate ratios: 5/7 * 5/7 * 5/7

Solution:

=(5*5*5)/(7*7*7)

=(5^3)/(7^3) or 5^3:7^3

=125/343   or 125:343

More about ratios- calculate ratio math:

Define Inverse ratio in math:

It is often want to estimate the numbers of a ratio in the inverse order. To perform this, we simply switch over the numerator and the denominator. Therefore, the inverse of 21:7 is 7:21. When the terms of a ratio are switch over are called  the INVERSE NUMERIC RATIO results.

Example problem for inverse ratios in math:

In math, how to calculate inverse numeric ratios:  7/21*7/2

Solution:

=7/21*7/2    (7 Divided by both the numerator and denominator)

=(7*7)/(21*2)

=49/42

=7/6 or 7:6

Answer is 7/6 or 7:6

Function Examples Math

Let f be a function from X to Y.  Then it have to satisfy the following two conditions.

(i) No two different ordered pairs in f have the same first element.

(ii) All the elements of X will occur as the first elements in f.

Now let us discuss few examples that satisfies these conditions.


Example problems on functions:

Ex 1: Check whether the following ordered pair represents a function:

Here X = {1, 2, 3, 4}

Y = {a, b, c}

(i) {(1, a), (2, b), (3, c), (4, c)}.

(ii) {(1, a), (2, c), (3, b)}.

Sol: Given: (i) {(1, a), (2, b), (3, c), (4, c)}

Here all the first values are from x.  All the first values are different.  Therefore it is a function.

(ii) {(1, a), (2, c), (3, b)}.

Here 4 of X is not there in the ordered pair.  Therefore, this cannot be a function.

Ex 2: Let X = {7, 8} and Y = {1, 2}. Which one of the following is not a function of

f from X to Y.

(i) {(7, 1), (8, 2), (7, 2)}

(ii) {(7, 1), (8, 2)}

(iii) {(7, 1), (8, 1)}

Sol: (i) {(7, 1), (8, 2), (7, 2)}.  This is not a function from X to Y.  Because 7 is repeated in the first element.

(ii) {(7, 1), (8, 2)}.  This satisfies the definition of the function. Hence it is a function.

(iii) {(7, 1), (8, 1)}. This satisfies the definition of the function.  Hence it is a function.


More example problems on functions:

Ex 3: Let X = { 1,2,3,4,5,6,7,8,9} and  Y = {3,6,9}

R be a relation “is three times of”, find R on X.

Can this be a function {(3, 1), (6, 2), (9, 3)}?

Sol: R = {(3, 1), (6, 2), (9, 3)}.

Since here {(3, 1), (6, 2), (9, 3)} the first values are different, this can be a function from Y to X.

Continuous Probability Distribution

In statististics if we have the finite number of set and the probability of the set is called Discrete probability distribution. For an example for a cars on a road, deaths by cancer, tosses until a die shows a first 6. If we measure anything it is called as continuous probability distribution function. For an example we can to say the electric voltage, rainfall, and hardness of steel. In both cases we can determined the distribution by the following distribution function

F(x) = P(X ≤ x)

This is the probability that X will assume any value not exceeding x.

Continuous random variables and probability distribution:

Discrete random variables appear in experiments which is having finite set (defectives in a production, days of sun shines in Chicago, customers standing in a line etc.). Continuous random variables is appear in the experiments which we can't count but we can measure (length of screws, voltage in a power etc.). If we are find the the value probability for  the continuous  variable then it is called continuous probability distribution. By the definition of a random variable of X and its for the distribution are of continuous type will be defined as the following integral.

F(x) = `int_-oo^xf(v)dv`

it is called density of the distribution, is non negative, and is continuous perhaps except for finitely many x-values.

The continuous random variables are simpler than the discrete ones with respect to intervals. Indeed  in the continuous case the continuous probability distribution of the four probabilities corresponding to a < X ≤ b, a < X < b, a ≤ X < b, a ≤ X ≤ b with any fixed a and b (> a) are all same.

Sample problem for probability distribution:

Continuous probability distribution problem 1:

Let X have the density function f(x) = .75(1 - x2) if  -1 ≤ x ≤ 1 and zero otherwise. Find the value of distribution function. Find the continuous probability distribution of P(-1/2 ≤ X ≤ 1/2). Find x such that P(X ≤ x) = 0.95.

Solution:

F(x) = `int_-1^x`(1 - v2) dv= 0.5 + 0.75 x2 -.25x3              if  -1 ≤ x ≤ 1,

And F(x) = 1 if x > 1

P(-1/2 ≤ x ≤ 1/2) = F(1/2)  - F(-1/2) =  `int_(-1/2)^(1/2)`(0.5 + 0.75 x2 -.25x3) dx =  68.75%

Because for continuous probability distribution    P(-1/2 ≤ x ≤ 1/2) =    P(-1/2 < x ≤ 1/2)

Continuous probability distribution P(X = x) = F(x) = 0.5 + 0.75x -1.25x3  = 0.95

If we solve this we will get x = 0.73