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October 15, 2008

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This is a part of the interactive online tutorial for Geometry learning with the help of audio visual lessons. It has been created to help you learn the concept, perception with explanations and includes solution to practice questions from the topic of ‘Word problems’. The mini-lessons here include: Getting started, Percent mixture problems, Work word problems, Other common types.

- Mathematics > General
- Mathematics > Algebra
- Mathematics > Applied Mathematics
- Mathematics > Problem Solving

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This is a part of the interactive online tutorial for Geometry learning with the help of audio visual lessons. It has been created to help you learn the concept, perception with explanations and includes solution to practice questions from the topic of ‘Word problems’. The mini-lessons here include: Getting started, Percent mixture problems, Work word problems, Other common types.

This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Getting started.

When it comes to solving word problems, generally we've to break the problem down into smaller parts and solve one part at a time. It is necessary to understand that certain words indicate certain mathematical operations, e.g. "more than", "decreased by", "times" and "ratio of" mean addition, subtraction, multiplication and division respectively. To take a specific example, the ratio of 6 more than x to x would mean (x + 6)/x. A few other things to keep in mind:

= is used for is, are, and equal

< is used for is less/lower than

? is used for is less than or equal to

> is used for is greater/larger than

? is greater than or equal to

()^{2 } is used squared

()^{3 } is used for cubed.

When it comes to solving word problems, generally we've to break the problem down into smaller parts and solve one part at a time. It is necessary to understand that certain words indicate certain mathematical operations, e.g. "more than", "decreased by", "times" and "ratio of" mean addition, subtraction, multiplication and division respectively. To take a specific example, the ratio of 6 more than x to x would mean (x + 6)/x. A few other things to keep in mind:

= is used for is, are, and equal

< is used for is less/lower than

? is used for is less than or equal to

> is used for is greater/larger than

? is greater than or equal to

()

()

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Algebra I. Click on the video below to go through it. If you like it, you can buy our online course in Algebra I by clicking here.

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This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Getting Started - Right triangle and trigonometric functions.

This mini-lesson introduces and walks you through the basic concepts of the trigonometric functions, right triangle and their relationship. You'll learn it with the help of some examples, practice questions with solution, using video and explanation in own handwriting by the instructor that brings in an element of real-class room experience.

The three primary trigonometry functions are: sin*x*, cos *x*, tan *x*. The input value usually represents an angle. The length of three sides of a right triangle, are simply termed as the ‘opposite’. ‘adjacent’ and ‘hypotenuse’. The values for the trigonometric functions are defined as the value that you get when divided one side by the other side i.e. ratio of one side to the other. E.g. sin *x* = opposite / hypotenuse.

Further you’ll explore, how you can use the ratios of side-length of right triangles to determine the measures of sides and angles. We’ll apply the Pythagorean Theorem, concept of ratio and properties of sines, cosines, and tangents to solve the triangle, that is, to find unknown parts in terms of known parts. E.g. In a right triangle*ABC*, with sides *a*, *b* and *c*, you need to remember:
*a* = 10 and *b* = 24, then

*c*^{2} = *a*^{2} + *b*^{2} = 10^{2} + 24^{2} = 100 + 576 = 676.

The square root of 676 is 26, so*c* = 26.

This mini-lesson introduces and walks you through the basic concepts of the trigonometric functions, right triangle and their relationship. You'll learn it with the help of some examples, practice questions with solution, using video and explanation in own handwriting by the instructor that brings in an element of real-class room experience.

The three primary trigonometry functions are: sin

Further you’ll explore, how you can use the ratios of side-length of right triangles to determine the measures of sides and angles. We’ll apply the Pythagorean Theorem, concept of ratio and properties of sines, cosines, and tangents to solve the triangle, that is, to find unknown parts in terms of known parts. E.g. In a right triangle

- Pythagorean theorem:
*a*^{2}+*b*^{2}=*c*^{2} - Sines: sin
*A*=*a*/*c*, sin*B*=*b*/*c* - Cosines: cos
*A*=*b*/*c*, cos*B*=*a*/*c* - Tangents: tan
*A*=*a*/*b*, tan*B*=*b*/*a*

The square root of 676 is 26, so

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Geometry. Click on the video below to go through it. If you like it, you can buy our online course in Geometry by clicking here.

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Curriki Rating**'C'** - Curriki rating C**'C'** - Curriki rating

This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Percent mixture problems.

There are two important facts in the percent mixture problems: the percent in each mixture and the amount /quantity of each mixture. You'll typically be taking two mixtures and putting them into one pot to make a third mixture, and will be asked to calculate either the percent or the quantity of one of the mixtures. E.g. Find out how many gallons of a 15% acid solution must be mixed with 5 gal of a 20 % acid solution to make an 18% solution?

This problem can be written in the form of the equation:

.15 x + 1 =.18(5 + x)

When we simplify it, value of x = 33.3 i.e. we'll need 33.3 gallons of the 15% acid solution.

There are two important facts in the percent mixture problems: the percent in each mixture and the amount /quantity of each mixture. You'll typically be taking two mixtures and putting them into one pot to make a third mixture, and will be asked to calculate either the percent or the quantity of one of the mixtures. E.g. Find out how many gallons of a 15% acid solution must be mixed with 5 gal of a 20 % acid solution to make an 18% solution?

This problem can be written in the form of the equation:

.15 x + 1 =.18(5 + x)

When we simplify it, value of x = 33.3 i.e. we'll need 33.3 gallons of the 15% acid solution.

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Algebra I. Click on the video below to go through it. If you like it, you can buy our online course in Algebra I by clicking here.

Member Rating

Curriki Rating**'P'** - This is a trusted Partner resource P**'P'** - This is a trusted Partner resource

This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Work related problems.

This mini-lesson explains work related problems involving situations such as a group of people working together on a job. E.g. you might be told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together. The solution for work problems is not obvious and may be a little tricky to work on. You may have to think of the problem in terms of how much each person / machine / whatever does per unit of time.

For example: If a painter can paint the entire house in twelve hours, and the second painter takes eight hours. How long would it take the two painters together to paint the house?

If you think about the problem in terms of how much of the house does each of the painters paints in one hour, you'll realize that the problem can be stated as: t(1/12)+ t(1/8) = 1 where t is the amount of time it will take both of them to paint the house together.

When we simplify it, value of t = 24/5 hrs i. e. they will take 4^{4}/_{5}hrs.

Once you go through our instructor's explanation in the video, these concepts will become even clearer.

This mini-lesson explains work related problems involving situations such as a group of people working together on a job. E.g. you might be told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together. The solution for work problems is not obvious and may be a little tricky to work on. You may have to think of the problem in terms of how much each person / machine / whatever does per unit of time.

For example: If a painter can paint the entire house in twelve hours, and the second painter takes eight hours. How long would it take the two painters together to paint the house?

If you think about the problem in terms of how much of the house does each of the painters paints in one hour, you'll realize that the problem can be stated as: t(1/12)+ t(1/8) = 1 where t is the amount of time it will take both of them to paint the house together.

When we simplify it, value of t = 24/5 hrs i. e. they will take 4

Once you go through our instructor's explanation in the video, these concepts will become even clearer.

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Algebra I. Click on the video below to go through it. If you like it, you can buy our online course in Algebra I by clicking here.

Member Rating

Curriki Rating**'P'** - This is a trusted Partner resource P**'P'** - This is a trusted Partner resource

This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Other common types.

This mini-lesson deals with other word problem formats, such as time-distance problems, and explains them with the help of several solved examples. Typically, in time-distance problems, you need to use the idea of rate or speed. Speed is simply distance over time, i.e.

speed = distance / time.

Conversely, distance can be calculated multiplying speed and time. For example, if a passenger traveled for 4 hrs at 30 miles per hour, then the total distance covered will be 120 miles.

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Algebra I. Click on the video below to go through it. If you like it, you can buy our online course in Algebra I by clicking here.

This mini-lesson deals with other word problem formats, such as time-distance problems, and explains them with the help of several solved examples. Typically, in time-distance problems, you need to use the idea of rate or speed. Speed is simply distance over time, i.e.

speed = distance / time.

Conversely, distance can be calculated multiplying speed and time. For example, if a passenger traveled for 4 hrs at 30 miles per hour, then the total distance covered will be 120 miles.

Member Rating

Curriki Rating**'P'** - This is a trusted Partner resource P**'P'** - This is a trusted Partner resource

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