Practice Percentages

Introduction for practice percentage:

The word “percent” is consequent from Latin. It was at first “per centum”, which means “by the hundred”. Thus the statement is frequently complete that “percent means hundredths”.

Percentage deals with the collection of decimal fraction whose denominators are 100 – that is, fractions of two decimal spaces Since hundredths be used so regularly, the decimal position was drop and the symbol % be located after the number and understand “percent”. Thus, 0.25 and 25% represent the same value, 25/100.  The first is read “25 hundredths”, and the second is read “25 percent”. Both mean 25 parts out of 100.

Originally, percent is used in discussing relative values. For example, 25 percent may convey an idea of relative value or relationship. To say “35 percent of the crew is ashore” gives an idea of what part of the crew is gone, but it does not tell how many of discussing the percentage.

I like to share this formula percentage change with you all through my article.

Practice percentage for steps and example problems:


Steps for practice percentage:

STEP 1: Start the percentage x/100 = is/of. X is the percentage (over 100 of course), "is" refers to fraction, and "of" refers to entire.

STEP 2: In the question "80 is to 40 percent of what number? x=40, is=40 ("80 is"), and of = the unknown ("of what number"). Therefore write 40/100=80/x.

STEP 3: Cross multiply. You will have a constant value on one side and multiply a variable on the other side. Here it is 40x=8,000.

STEP 4: Solve for x. Here, x = 8,000/40 = 400, So x = 400.

Example problems for percentage:

Example problem 1:

The $169.99 poodle I purchased was on sale for 10% off, what did I pay for my poodle?

Solution:

$ 169.99 x 10 / 100

= $ 169.99 / 10

= $16.999

So Pay for poodle = $169.99 - $16.99

Answer = $153

Example problem 2:

I got 40% off when I purchased a rare comic book regularly priced at $74.50. How much did I pay?

Solution:

$74.50 x (40 / 100)

= $74.50 x (4 / 10)

= ($74.50 x 4) / 10

= $298 / 10 = $29.8

Pay for books = $74.50 - $29.8

= $44.7

Example problem 3:

Our take out dinner was $74.95 but we got 20% off because we picked it up. What did our meal end up costing us? $59.96

Solution:

$74.95 * (20 / 100)

= $74.95 * (2 / 10)

= ($74.95 * 2) / 10

=$149.9 /10 = $14.99

End of meal cost = $74.95 - $14.99

Answer = $ 59.96


Practice problem for percentage:


Practice problem 1:

My flight was $199.99 but I got 30% off because the plane wasn’t full. What did I pay?

Answer: $139.99

Practice problem 2:

$40.50 video games were on sale for 30% off, how much are they now?

Answer: $28.25

Practice problem 3:

My new cell phone cost me $190.00 but when I signed a 2 year plan, I got 20% off so it only cost me?

Answer: $152.00

Practice Problem 4:

It’s usually $16.50 to go to the show, but if you go on Tuesday, there’s a 40% discount, what does it cost to go to the show on Tuesday?

Answer: $9.90

Place Value Practice

Introduction for place value practice:

Place value is a second important concept in mathematics. It refers to the value of the digit, relative to its location in the number. The place value can be identified for any given digit in a written number, in the number 7403, the digit 4 is in the hundreds place and stands for 400. The 3 in 34 is not the same as the 3 in 23, even though both digits are 3. Is this topic Statistics Problems hard for you? Watch out for my coming posts.

This is because the 3 in 34 is in the tens place and therefore worth 30, while the 3 in 23 is in the ones place and worth only 3. Child is developing an understanding of place value up to at least the thousands. (This also means that she needs to practice skip counting starting with any number up to one thousand: 326,336,346)

Example: The number 3725 is read as three thousand seven hundred twenty- five. In it, there are 3 thousands, 7 hundreds, 2 tens, and 5 ones. A number with only ones has only one digit, one with tens two digits, and one with thousands has four digits.

Example problem for place value practice:

5 ones =             5

2 tens =            20

7 hundreds =     700

3 thousands =  3000

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Example problem for place value practice:


Read the numbers, and then tell the place and value of the underlined digit; Up to 10 millions.

1. 5,760,000

What place is the underlined digit in?

Ten thousands place

What is the value of the underlined digit?

60,000

2.   86,003,040

What place is the underlined digit in?

Millions place

What is the value of the underlined digit?

6,000,000

3.   4,954,560

What place is the underlined digit in?

Tens place

What is the value of the underlined digit?

60

4.    5,520,254

What place is the underlined digit in?

Millions place

What is the value of the underlined digit?

5,000,000

Practice Area Problems

Introduction

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron. (Source: From Wikipedia).

Here we are going to solve some practice problems, to find the area of basic regular shapes.

Practice problems to find the area

Square

The area of a square is given by the formula, A = a ^2 square units

Practice problem 1

Find the area of a square, with side lengths equal to 4 ft.

Solution

Area of the square = a ^2 square units

= 4^2

= 16 square feet

So teh area of the square = 16 square feet.

Rectangle

The area of a rectangle can be calculateed by the following formula,

Area of  a rectangle, A = lb square units

Where, l = length and b = breadth of the rectangle.

Practice problem 2

What is the area of a 4 in by 6 in rectangle.

Solution

Area of  a rectangle, A = lb square units

= 6 * 4 square inch

= 24 square inch

Answer: The area of the rectangle = 24 square inch

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Few more practice problems to find the area


Circle

The area of a circle = `pi` r^2 square units

Here, r is the radius of the circle.

Practice problem 3

Calculate the area of a circle with radius 10 centimeter.

Solution

Area of a circle = `pi` r^2 square units

= 3.14 * 102 square centimeter

= 314 square centimeter

So the area of the circle is 314 square centimeter.

Triangle

Area of a triangle = `1/2` b h square units

Where, b = base of the triangle

h = perpendicular height of the triangle

Practice problem 4

Find the area of a triangle whose base length is 5 meter and height is 10 meter.

Solution

Area of a triangle =  `1/2` b h square units

= square meter

= 25 square meter

The area of the given triangle = 25 square meters

Solving Probability Practice

Introduction to Probability:
Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.For solving probability practice problems we must know the following formula,

Probability of an event   =    number of times an event occurs / total number of outcomes

Consider an observation with 'n' possible ways and out of them in 'm' ways if the event 'A' occurs,then the probability of occurrence of the event 'A' is given by P(A) = m/n.

Let us workout some practice problems on solving probability in the following sections.

I like to share this Probability Problems and Solutions with you all through my article.

Probability on a coin problem:


Here Tossing a coin is a random experiment. When you toss a coin, you may get head or tail.Let us see how to  solving the coin problem.

1: What is the probability of getting Head when you toss a coin?

We can have 2 out-comes for tossing a coin :  Head or tail.

Here the event is getting Head. So we have 1 head.

So probability of getting head is one out of two outcomes.

Probability of getting Head    = number of head event occurs / total number of outcomes

= 1/2

= 0.5

So the probability of this practice problem is 0.5


Probability on Card problem:


In 52 cards, there are 13 spades, 13 clovers, 13 diamonds and 13 hearts , 26 cards are red in color (diamonds and heart) and 26 are black (spade and clover).Let us see how to  solving the card problems.

(1)  What is the probability that you get a red card?

The event is getting a red card. There are 26 red cards.

Total sample spaces are 52(52 cards).

Probability of getting a red cards = number of red cards / sample space

= 26/52

=1/2

= 0.5

(2) What is the probability of getting a spade when you draw from a well shuffled pack of 52 cards?

Number of Spade cards = 13

Total Sample space       = 52

Probability of getting a spade card = number of spade / total sample space

= 13/52

= 1/4

= 0.25

(3) What is the probability of getting a black queen in a pack of 52 cards?

In a pack of 52 cards, there are 2 Black Queens (spade 1, clover 1)

Total Sample space = 52

Probability of getting a black queen = number of black queen / total sample space

= 2/52

= 1/26

= 0.04

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Practice Problem:


The following practice problems on solving probability will helps for good understanding.

Practice problem 1: What is the probability of getting Tail when you toss a coin?

Practice problem 2: What is the probability of getting a diamond when you draw from a well shuffled pack of

52 cards ?

Answer key:

Practice problem 1:  0.5

Practice problem 2: 0.25

Study Practice Algebra Problems

Introduction of Study Practice Algebra Problems:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, algebra is one of the main branches of pure mathematics. (Source: Wikipedia)

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Study Example Algebra Problems:


Study practice algebra problem 1:

Simplify the equation 5(a -3) + 4b - 7(a -b -3) + 5.

Solution:

5(a -3) + 4b - 7(a -b -3) + 5

Multiply factors.

5a - 15 + 4b -7a + 7b + 21 + 5

Group like terms.

-2a + 11b + 11

Answer:   - 2a + 11b + 11

Study practice algebra problem 2:

To find the 4x - 8y = 9 equation on x-intercept.

Given:

4x - 8y = 9

Solution:

4x - 8y = 9

To find the x intercept of y = 0 and solve for x.

4x - 0 = 9

Solve the value of x.

x = 9 / 4

The x intercept is at the point (9/4, 0).

Answer: x intercept is at the point (9/4, 0).

Study practice algebra problem 3:

16x + 2y = 32.

y + 8 = 16x

Find the x and y value.

Solution:

16x + 2y = 32 (equation 1)

y + 8 = 16x   (equation 2)

Step 1: To choose the equation where the coefficient of the variable is 1.Choose equation 2 to isolate the variable y

y = 16x – 8 (equation 3)

Step 2: From equation 3, we know that y-value is the same as the 16x – 8. We can to substitute the variable y-values in the equation 1 with 16x – 8.

16x + 2 (16x – 8) = 32

Step 3: Remove the brackets by using the distributive property

16x + 32x – 16 = 32

Step 4: To combine the terms

48x – 16 = 32

Step 5: To Isolate the variable of x

48x = 48

x = 48 / 48

x = 1

Step 6: Substitute x = 1 in the equation 3 to get the y-value.

y = 16 (1) – 8

=16 – 8

y = 16 - 8

y = 8

Step 7: To check the answer with the equation (1)

16(1) + 2 (8) = 16+ 16 = 32

Answer: x = 1 and y = 8

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Study Practice Algebra Problems:


Practice Problem 1:

Find the x and y value 12x + 2y = 20, y + 8 = 12x

[Answer: x = 1 and y = 4]

Practice problem 2:

Solve for x and y for the following equations 5x + y = 8, 6x + y = 7

[Answer: x = -1 and y = 13]

Practice Problem 3:

Simplify the equation 7(x -3) + 5y - 3(x -y -3) + 5.

[Answer:   4x + 8y – 7]

Practice Algebra Class

Introduction to practice algebra class:

Algebra class deals with the Pre-algebra, Functions and Graphs, Integers, Rational Numbers, Equations in One Variable, Equations in Two Variables, Simultaneous Equations, Problem Solving etc. in arithmetic, we use digit like 1,2,3, etc to represent numbers. In algebra we use numbers as well as letters of the alphabet such as a, b, c, etc for any numerical values we choose. The basic thought of the algebra is balancing the algebraic equations on both sides of the give equations in the problems. Algebra will perform the four fundamental operations such as addition, subtraction, multiplication and division. In this article we shall discuss with practice algebra class


Sample problem on practice algebra class


Problems of algebra class on linear equations:

Example 1:

Solve the linear equation for algebra class

11 x - 15 = 19 x - 23

Solutions:

11 x - 15 = 19 x - 23

Subtract 19 x from both sides of the above equation, we get the below term

-8x – 15 = -23
Add 15 to both sides of the above equation, we get the below term

-8x – 15 + 15 = - 23 + 15

-8x = - 8
Divide both sides by -8 of the equation, we get

X = 1

The answer is x = 1

Test the solution for the given equation. Replace with in the above given equation for x. If the left side of the equation sum value is equal to the right side of the equation sum value after the replacement of x value, you have got the exact answer.

Left side of the equation:
11(1) - 15 = - 4

Right side of the equation:
19(1) - 23 = - 4

Practice word problems of algebra class

Example 2:

A bag contains ten, five and two rupees currencies. The total number of Currencies is 30 and the total value of money is Rs.135. If the second and third currencies are interchanged the value will decrease Rs.8. Find the number of currency in each sort.

Solution :

Let x, y, z be the number of Rs.10, Rs.5 and Rs.2 currencies respectively.

The total number of currencies is 30 => x + y + z = 30--------(1)

If the total value of money is Rs.135 => 10x + 5y + 2z = 136--------(2)

If the II and III of currencies are interchange the value will be decreased by Rs.8

=>10x + 2y + 5z = 135 – 8 = 127----------------(3)

10x + 2y + 5z = 127

10 x (1) – (2) =>     5y + 8z = 106------------(4)

(2) – (3)        =>     3y - 3z = 9

y – z = 3------------------(5)

Let us solve (4) and (5)

(4) =>      5y + 8z = 106

8 x (5)   ⇒ 8y – 8z = 24

13y = 130

y = 10

Substituting y = 10 in (5) we get

z = 7

Substituting z = 7, y = 10 in (1) we get

x + 10 + 7 = 30 => 30 – 10 – 7 or x = 13

Number of Rs.10, Rs.5 and Rs.2 currencies are respectively 13, 10 and 7.

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Practice problem for algebra class:

Problem 1:

2x + 11 = 20 - x

Answer: x = 3

Problem 2:

7x + 11 = 19 + 3x


Answer: x = 2

Statistics Practice Exam

Introduction to statistics practice exam:

Statistics is the science of making effective use of numerical data relating to groups of individuals or experiments. It deals with all aspects of this, including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments.

A statistician is someone who is particularly versed in the ways of thinking necessary for the successful application of statistical analysis. Often such people have gained this experience after starting work in any of a number of fields. There is also a discipline called mathematical statistics, which is concerned with the theoretical basis of the subject.

Let us see several statistics based solved problems and statistics practice exam. Statistics practice exam with solutions which would be helpful for students to evaluate their performance.

Concepts on statistics practice exam:


Before going for a statistics practice exam, let us recall how to find mean deviation. The steps to find the mean deviation are as follows.

Step 1: find the mean for given data
`barx` = (sum of all variables) /( given number of  variables).

Step 2: find the deviation of the respective observations from mean `barx` ,  in other words  xi -  `barx`
Step 3: find the modulation of deviation ,     | Xi - `barx` |
Step 4: mean deviation of the mean,
M.D.(`barx` ) = `sum` n i-1 | xi - x| / n

Example1:

Find the mean deviation about the mean for the following data:

12, 5, 16, 14, 11, 5, 10, 15

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(12+5+16+14+11+5+10+15)/8` = `88/8` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

12–11,5–11,16–11,14–11,11–11,5–11,10–11,15–11,

or 1,–6,5,3,0,–6,–1,4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

1,6,5,3,0,6,1,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(1+6+5+3+0+6+1+4)/8` = `26/8` = 3.25.

Example2:

Find the mean deviation about the mean for the following data:

15, 5, 12, 16, 12, 4, 9, 16,7.

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(15+5+12+16+12+7+9+16+7)/9` = `99/9` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

15–11,5–11,12–11,16–11,12–11,7–11,9–11,16–11,7-11.

or 4,–6,1,5,1,–4,–2,5,-4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

4,6,1,5,1,4,2,5,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(4+6+1+5+1+4+2+5+4)/8` = `32/8` = 4.

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Practice on statistics practice exam:

Statistics Practice exam:

1. Find the mean deviation for the mean of the following data:
6, 7, 10, 12, 13, 4, 8, 12

2. Find the mean deviation for the mean of the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5

Answers for the above questions of statistics practice exam:

1.   2 75

2.   6.2