In maths, we do come across topics on estimation.
This topics relate to our daily living very closely and is truly practical.
Example is when we go shopping and starts to count the expenses to be paid.
But mathematical estimation has one key concern.
To what extent or how accurate does one wish to?
There is no right or wrong to an answer when dealing with estimation. After all it is an ESTIMATED answer.
The basic requirement is thus to get as close to the true answer as possible.
Let's start with one simple example to demonstrate the concept.
Example:
y = 0.501 + square root(3.89)
What is y without using calculator ?
Answer 1:
y = 1 + square root (4)
y = 1 + 2 = 3
Answer 2:
y = 0.5 + square root (4)
y = 0.5 + 2 = 2.5
You can now see that both answers is close to the actual answer of 2.4733.
However, it is the gap or extent of the difference you wish for.
If possible, LOOK carefully at the numbers and give a best estimation closest to ability to compute the answer.
Here, from the above, you will notice the first term of 0.501 decides the outcome.
Estimate this 0.501 to what numeric value?
Think further and you will find that 0.501 to 1 will give you a bigger difference compared to 0.501 to 0.5.
If you can mentally handle 0.5 as the estimated value for computation, go for it since this should be closer to the final outcome.
Thus in conclusion, do look a bit closer to the numbers presented in your problem and simply do a brief calculation of the differences between estimated and raw data. Then you are one step closer to getting a good estimation.
Estimation, finally, boils down to how far you are to the actual answer. Nothing difficult.
Interesting? Any more suggestions?
.
Showing posts with label concept. Show all posts
Showing posts with label concept. Show all posts
Friday, 22 March 2013
Friday, 17 February 2012
Fear Over Maths
Maths is one of the crucial and necessary subject we have to learn in school. Everyone has to go through it.
Some like it, while some fear it. Have you wonder why?
Actually this fear is not only towards maths. The underlying reasons applies to anything we do.
It could be literature, Chinese language, dancing, or even driving and riding a bicycle.
What we have to do to reduce maths anxiety is to know why the fear occurs.
One of the issue links to the confidence level. This affects the comfort level. With things we are not comfortable, we tend to avoid. Avoiding make us do less of the subject.
We thus lack practices.
Learning maths involves 2 parts, namely:-
1) Procedural skills, and
2) Conceptual understanding.
Different stages / levels in the learning journey entails different focus of these 2 stages.
To have a better understanding, you may go to this maths site to read more.
It is rather enlightening.
Learning anything needs a plan. (Procedure ==> Conceptual ==> Application)
If fear is the barrier, and causes obtrusion to learning, seek out why.
We grow as a result of this process.
Maths is one target in life that we can use to challenge yourself, besides the tools and techniques picked up to solve mathematical problems.
It involves character development as well, if you border to analyse the learning process.
Finally, to motive any maths learners,
just know that "Maths Is Interesting!".
You will then like maths, and anything you set forth to master.
Cheers! :-D.
Some like it, while some fear it. Have you wonder why?
Actually this fear is not only towards maths. The underlying reasons applies to anything we do.
It could be literature, Chinese language, dancing, or even driving and riding a bicycle.
What we have to do to reduce maths anxiety is to know why the fear occurs.
One of the issue links to the confidence level. This affects the comfort level. With things we are not comfortable, we tend to avoid. Avoiding make us do less of the subject.
We thus lack practices.
Learning maths involves 2 parts, namely:-
1) Procedural skills, and
2) Conceptual understanding.
Different stages / levels in the learning journey entails different focus of these 2 stages.
To have a better understanding, you may go to this maths site to read more.
It is rather enlightening.
Learning anything needs a plan. (Procedure ==> Conceptual ==> Application)
If fear is the barrier, and causes obtrusion to learning, seek out why.
We grow as a result of this process.
Maths is one target in life that we can use to challenge yourself, besides the tools and techniques picked up to solve mathematical problems.
It involves character development as well, if you border to analyse the learning process.
Finally, to motive any maths learners,
just know that "Maths Is Interesting!".
You will then like maths, and anything you set forth to master.
Cheers! :-D.
Friday, 18 November 2011
Tips On Using Substitution
Maths entails the usage of our brain juice in solving problems. It is a good platform for stretching our imagination and creativity by using simple concepts learned to handle seemingly complex maths questions.
Let's look at a "complex" simultaneous equations maths problem, and its way of solving (suggested).
Question:
---- (A)
--- (B)
Solve for y and x.
How do you go about it?
Look scary, right?
But like what I said, looks can be deceiving. Use the brain to go around the issue!
Tips: The structure of the simultaneous equations looks similar to the conventional type.
(Conventional type:-
Ax + By = nn
Cx + Dy = kk )
So what we have to do can be to simply substitute
by m, and 
by h (or any variable name).
What we thus convert to is:
---- (A)
--- (B)
Will this simultaneous equations be more comfortable to solve?
Hence, a simple twist to the former mathematical questions can result in a totally familiar situations where we have solve many a times.
Thus, the technique and usefulness of substitution cannot be under-estimated.
It can be powerful at times to reveal a beautiful mathematical expression for user to resolve.
Maths Is Interesting!
Treasure our brain and our thinking.
:-)
Let's look at a "complex" simultaneous equations maths problem, and its way of solving (suggested).
Question:
---- (A) --- (B)Solve for y and x.
How do you go about it?
Look scary, right?
But like what I said, looks can be deceiving. Use the brain to go around the issue!
Tips: The structure of the simultaneous equations looks similar to the conventional type.
(Conventional type:-
Ax + By = nn
Cx + Dy = kk )
So what we have to do can be to simply substitute
by m, and by h (or any variable name).What we thus convert to is:
---- (A) --- (B)Will this simultaneous equations be more comfortable to solve?
Hence, a simple twist to the former mathematical questions can result in a totally familiar situations where we have solve many a times.
Thus, the technique and usefulness of substitution cannot be under-estimated.
It can be powerful at times to reveal a beautiful mathematical expression for user to resolve.
Maths Is Interesting!
Treasure our brain and our thinking.
:-)
Saturday, 5 March 2011
Explanation of the Elimination Method
Solving of Simultaneous equations may require one common technique called "Elimination" method.
From the name, we know that it has to eliminate or remove something from the equations.
The target is one selected variable or unknown in the mathematical equations.
However, when approaching this method, you noticed that it involved the subtraction (or addition) of equations.
The question is "Can equations be subtracted?".
And "What is the real meaning of subtracting equations?"
My answers:-
Yes, equations can of course be subtracted. Equations are like other items, e.g. apples, chairs.
The real meaning of subtracting equations is not that apparent.
The true and desired wish to subtract equations boils down to commonising a certain coefficient of a variable.
With this common coefficient, it will then be able to remove this mathematical unknown.
(It is not really the direct processing of equations, and the magical removal of variable as a result!)
We commonise the coefficient of the selected variable first before subtracting the equations in order that same items are eliminated.
Hope this clarify some doubts of new learners to simultaneous equations solvers.
Concepts have to be learned upfront without pending questions for complete understanding and smooth follow-up learning in the later stage. Seek to clarify any doubts as far as possible.
It will reduce maths anxiety and allow you to enjoy maths as a result. The reward of clearing any doubts cannot be spelled out in words but through actual working and practice with proper analysis.
I believe you support this notion.
Cheers to maths.
.
From the name, we know that it has to eliminate or remove something from the equations.
The target is one selected variable or unknown in the mathematical equations.
However, when approaching this method, you noticed that it involved the subtraction (or addition) of equations.
The question is "Can equations be subtracted?".
And "What is the real meaning of subtracting equations?"
My answers:-
Yes, equations can of course be subtracted. Equations are like other items, e.g. apples, chairs.
The real meaning of subtracting equations is not that apparent.
The true and desired wish to subtract equations boils down to commonising a certain coefficient of a variable.
With this common coefficient, it will then be able to remove this mathematical unknown.
(It is not really the direct processing of equations, and the magical removal of variable as a result!)
We commonise the coefficient of the selected variable first before subtracting the equations in order that same items are eliminated.
Hope this clarify some doubts of new learners to simultaneous equations solvers.
Concepts have to be learned upfront without pending questions for complete understanding and smooth follow-up learning in the later stage. Seek to clarify any doubts as far as possible.
It will reduce maths anxiety and allow you to enjoy maths as a result. The reward of clearing any doubts cannot be spelled out in words but through actual working and practice with proper analysis.
I believe you support this notion.
Cheers to maths.
.
Sunday, 16 January 2011
Simultaneous Equations - Decimal Numbered
In the learning of maths, questions grow challenging as one progress upwards.
One such example is the solving of simultaneous equations.
The easy type:
Solve for x and y using elimination method.
4x + 3y = 10 ---- (1)
3x + 4y = 11 ---- (2)
For the above problem can be solved easily by selecting a coefficient to be commonised.
The post of elimination method is reference here for review.
But moving on (higher) with more challenging maths question ... we may get the below.
0.4x + 0.3y = 1 ---- (A)
3x + 4y = 11 ---(B)
What should we do next?
Equation (A) may seems unusual. It is in decimal form!
But as the blog title claims "Maths Is Interesting!", we should not be worried.
This type of question is actually not new in concept or tricky as it seems.
It is there to test you understanding by being "different".
We have to remove the "catch", which is to change the decimated number to integer.
How we do it here is simply multiplying the coefficients by 10.
This makes equation (A) to be 4x + 3y = 10 (back to the original first set of simultaneous equations at the start of this post.
The above example serves to illustrate the simplicity of changing numbers to suit the condition for easy solving. (Other questions may be multiply by another decimal number, or integer).
Just have a clear mind and a confidence attitude will be enough to allow you to solve most of the maths questions.
Try it and you will believe what I say (or write).
Maths is interesting!
.
One such example is the solving of simultaneous equations.
The easy type:
Solve for x and y using elimination method.
4x + 3y = 10 ---- (1)
3x + 4y = 11 ---- (2)
For the above problem can be solved easily by selecting a coefficient to be commonised.
The post of elimination method is reference here for review.
But moving on (higher) with more challenging maths question ... we may get the below.
0.4x + 0.3y = 1 ---- (A)
3x + 4y = 11 ---(B)
What should we do next?
Equation (A) may seems unusual. It is in decimal form!
But as the blog title claims "Maths Is Interesting!", we should not be worried.
This type of question is actually not new in concept or tricky as it seems.
It is there to test you understanding by being "different".
We have to remove the "catch", which is to change the decimated number to integer.
How we do it here is simply multiplying the coefficients by 10.
This makes equation (A) to be 4x + 3y = 10 (back to the original first set of simultaneous equations at the start of this post.
The above example serves to illustrate the simplicity of changing numbers to suit the condition for easy solving. (Other questions may be multiply by another decimal number, or integer).
Just have a clear mind and a confidence attitude will be enough to allow you to solve most of the maths questions.
Try it and you will believe what I say (or write).
Maths is interesting!
.
Wednesday, 8 December 2010
Decimal Number Simplification
.
Algebraic expressions and equations normally come in integer or fraction form.
Examples:
1) 4x - 3 = x
2) (3/4)x - 3x = 1/(3x)
Simplification of the above examples will not pose much of a problem except maybe in the challenge of bringing the numbers and unknowns over the "equal" sign.
But algebraic equations can come in decimal form too.
Example:
0.3(0.2x - 1) = 0.1x
How do we go about solving the above "decimated" algebraic equation easily?
A simple trick that I can think of (or maybe too simply a technique to call it 'trick").
What I would do is to multiply the expression on both sides by 10.
The idea is to bring the decimal number into the integer range.
BUT do note that the expression on the left side has two decimal numbers.
As such I would have to "x 10" twice.
This means that there is a "x 100" on the left and right side.
The new equation will thus be:
3 (2x - 10) = 10 x
==> 6x - 30 = 10x
==> -30 = 10x - 6x = 4x
==> x = -30 / 4 = -7.5
Conclusion:
Decimal can be seen to be intimidating when in the decimal form. However, it can be elevated to the familiar integer form through simple multiplication.
However, do take note of how many decimal number has been multiplied.
Left and right sides of the equation has to have the same number of multiplication (or division) to stay equal and valid.
Maths is not that frightening.
It can be interesting, if the method to "attack" it is properly done.
:-)
Algebraic expressions and equations normally come in integer or fraction form.
Examples:
1) 4x - 3 = x
2) (3/4)x - 3x = 1/(3x)
Simplification of the above examples will not pose much of a problem except maybe in the challenge of bringing the numbers and unknowns over the "equal" sign.
But algebraic equations can come in decimal form too.
Example:
0.3(0.2x - 1) = 0.1x
How do we go about solving the above "decimated" algebraic equation easily?
A simple trick that I can think of (or maybe too simply a technique to call it 'trick").
What I would do is to multiply the expression on both sides by 10.
The idea is to bring the decimal number into the integer range.
BUT do note that the expression on the left side has two decimal numbers.
As such I would have to "x 10" twice.
This means that there is a "x 100" on the left and right side.
The new equation will thus be:
3 (2x - 10) = 10 x
==> 6x - 30 = 10x
==> -30 = 10x - 6x = 4x
==> x = -30 / 4 = -7.5
Conclusion:
Decimal can be seen to be intimidating when in the decimal form. However, it can be elevated to the familiar integer form through simple multiplication.
However, do take note of how many decimal number has been multiplied.
Left and right sides of the equation has to have the same number of multiplication (or division) to stay equal and valid.
Maths is not that frightening.
It can be interesting, if the method to "attack" it is properly done.
:-)
Saturday, 10 April 2010
Number of Answers | Common mistake
Maths can be tricky when you are not careful.
This is not to frighten you, though.
This post is just to remind you of the wonderful aspect of maths in covering all areas.
Below is an example of what I meant.
Let's take the quadratic eqaution solving as a starting point
x2 = 5x
x = 5x / x = 5 (Answer)
At first, this looks pretty fine. The answer, when substituted back, produces match of equation.
But this is actually not complete.
Those doing quadratic equation will know 2nd order (x2) equation evaluates to 2 answsers.
The answers may be the same though.
Now, if we approach it using another method, let's see the different.
x2 - 5x = 0
==> x (x - 5) = 0 , after factorising
==> x = 0 and (x - 5) = 0
==> x = 0 and x = 5
There are two answers now.
We had the x = 5 initially, but what about this new x = 0.
We have missed out on the x = 0 with the first mehtod. It looks OK then.
What happen?
It may be due to lack of experience handling this form of maths question.
The concept in solving quadratic equation is actually not limited to second order.
The hidden message is depending on the order, the number of answers will follow suit.
What I meant is :
2nd order gives 2 answers,
3rd order gives 3 answers,
4th order gives 4 answers, etc.
It is this verry message that maths learner should capture. Otherwise you will be tricked to give only one answer which leads you to "mistakes" of being incomplete.
I agree that this is tricky, but within reasonable argument.
If a student practice hard (and smart), he will not fall prey to this type of simple math problem.
Do not get con again.
Enjoy maths. It's fun and interesting.
:D
This is not to frighten you, though.
This post is just to remind you of the wonderful aspect of maths in covering all areas.
Below is an example of what I meant.
Let's take the quadratic eqaution solving as a starting point
x2 = 5x
x = 5x / x = 5 (Answer)
At first, this looks pretty fine. The answer, when substituted back, produces match of equation.
But this is actually not complete.
Those doing quadratic equation will know 2nd order (x
The answers may be the same though.
Now, if we approach it using another method, let's see the different.
x2 - 5x = 0
==> x (x - 5) = 0 , after factorising
==> x = 0 and (x - 5) = 0
==> x = 0 and x = 5
There are two answers now.
We had the x = 5 initially, but what about this new x = 0.
We have missed out on the x = 0 with the first mehtod. It looks OK then.
What happen?
It may be due to lack of experience handling this form of maths question.
The concept in solving quadratic equation is actually not limited to second order.
The hidden message is depending on the order, the number of answers will follow suit.
What I meant is :
2nd order gives 2 answers,
3rd order gives 3 answers,
4th order gives 4 answers, etc.
It is this verry message that maths learner should capture. Otherwise you will be tricked to give only one answer which leads you to "mistakes" of being incomplete.
I agree that this is tricky, but within reasonable argument.
If a student practice hard (and smart), he will not fall prey to this type of simple math problem.
Do not get con again.
Enjoy maths. It's fun and interesting.
:D
Monday, 5 April 2010
Simultaneous Equations | Re-write equations
*
Simple simultaneous equation problem comes as 2 straight forward mathematical expressions.
Example 1:
3x + y = 4
x + 2y = 3
But some may come in odd expressions (since life is always the case, which makes learning maths more exciting!)
Example 2:
(12 - x)(1 + y) = 15
(8 - x) (1 + y) = -15
Here you will notice that the unknowns are biased towards one side.
Approach 1:
Multiply the 2 factors to get something like example 1.
Using elimination method, remove one of the unknown.
Solve for the only one unknown left.
Using the result found, compute the other unknown.
Approach 2: (The focus of this post)
Re-write the expression to make it look simpler.
The example 2 can be re-written into below simpler form:
(12 - x) = 15 / (1 + y) ===> (A)
(8 - x) = -15 / (1 + y) ===> (B)
Equation (B) can then be seen to be the negative of equation (A).
With the re-writing, we will be visually aided to see another form, a simpler one, of the simultaneous equations.
Moving forward with the solution...
12 - x = -(8 - x) = -8 + x
12 + 8 = 2x
x = 20 / 2 = 10 (ANSWER)
Putting x = 10 back into either equation (A) or (B),
We will get (12 - 10) (1 + y) = 15, if we select equation (A)
1 + y = 15 / 2 = 7.5
y = 7.5 - 1 = 6.5 (ANSWER)
The solution is not the issue in this post.
The key message here is the technique of "re-writing" the equations to reveal the simplicity of the question.
Maths is not that difficult if you look and think to make it easy.
Cheers!
.
Simple simultaneous equation problem comes as 2 straight forward mathematical expressions.
Example 1:
3x + y = 4
x + 2y = 3
But some may come in odd expressions (since life is always the case, which makes learning maths more exciting!)
Example 2:
(12 - x)(1 + y) = 15
(8 - x) (1 + y) = -15
Here you will notice that the unknowns are biased towards one side.
Approach 1:
Multiply the 2 factors to get something like example 1.
Using elimination method, remove one of the unknown.
Solve for the only one unknown left.
Using the result found, compute the other unknown.
Approach 2: (The focus of this post)
Re-write the expression to make it look simpler.
The example 2 can be re-written into below simpler form:
(12 - x) = 15 / (1 + y) ===> (A)
(8 - x) = -15 / (1 + y) ===> (B)
Equation (B) can then be seen to be the negative of equation (A).
With the re-writing, we will be visually aided to see another form, a simpler one, of the simultaneous equations.
Moving forward with the solution...
12 - x = -(8 - x) = -8 + x
12 + 8 = 2x
x = 20 / 2 = 10 (ANSWER)
Putting x = 10 back into either equation (A) or (B),
We will get (12 - 10) (1 + y) = 15, if we select equation (A)
1 + y = 15 / 2 = 7.5
y = 7.5 - 1 = 6.5 (ANSWER)
The solution is not the issue in this post.
The key message here is the technique of "re-writing" the equations to reveal the simplicity of the question.
Maths is not that difficult if you look and think to make it easy.
Cheers!
.
Tuesday, 30 March 2010
Using Equation To Create A Square Graphically
'
While studying maths, I have been exposed to equation that forms a circle.
We know that x^2 + y^2 = 1 creates a circle.
But I have been wondering what is an equation to form a square.
I had tried a few mathematical expressions till today.
And finally I found the interesting and mysteries equation.
It utilises the same concept as the circle except that hyperbolic trigonometry is applied.
Below is a graph plotted with that equation.
The corners are rounded though. Any one has any try with a more sharper corner?
Graph is a wonderful tool as it can present results visually with one view.
Appreciating maths and using it appropriately can reduce many complex problems.
Maths is interesting.
.
While studying maths, I have been exposed to equation that forms a circle.
We know that x^2 + y^2 = 1 creates a circle.
But I have been wondering what is an equation to form a square.
I had tried a few mathematical expressions till today.
And finally I found the interesting and mysteries equation.
It utilises the same concept as the circle except that hyperbolic trigonometry is applied.
Below is a graph plotted with that equation.
The corners are rounded though. Any one has any try with a more sharper corner?
Graph is a wonderful tool as it can present results visually with one view.
Appreciating maths and using it appropriately can reduce many complex problems.
Maths is interesting.
.
Saturday, 27 March 2010
Area Displacement Theory
.
Maths is not all about calculation.
There are always more to it than meet the eyes.
This is especially true when you are doing geometrical questions where you are involved with area, perimeter and so on.
Displacement theory or its equivalent is always done without the knowledge of many people.
What is this theory about?
Let's look at one example below.
In the diagram above, you will see a path (white coloured) going across a blue platform.
If you are asked to find the area of this path, what can you do to obtain this area?
If no data of dimension is given, it is definitely not possible.
Now if the width of the path and the vertical length of the blue platform is given, can you compute the answer?
Again , this need a bit of thinking.
Displacement theory kicks in here. Look at the diagram on the right.
It is the displaced or closed up portion of the blue platform that does the trick.
Here you will notice the dashed line forming a white rectangluar area on the right-most side of the white blue platform.
Are you able to find the area of this white rectangular piece?
The width of this rectangle piece is ACTUAL the width of the white path!
You should now be able to calculate the area of this rectangular piece since the path width and length of the rectangular block is known or deduced now.
How this is possibe is through the "hidden" clue or step of closing up the path revealing the simpler rectangular area that any decent maths student can calculate.
Hence, maths is wonderful in that it tests you not only about applcations of maths tools, but your other "intelligence".
Having known displacement theory here, I believe you are really for the Math Challenge 23.
Go there and answer the question, and be quick before others grap the position one ...
:-D
.
Maths is not all about calculation.
There are always more to it than meet the eyes.
This is especially true when you are doing geometrical questions where you are involved with area, perimeter and so on.
Displacement theory or its equivalent is always done without the knowledge of many people.
What is this theory about?
Let's look at one example below.
In the diagram above, you will see a path (white coloured) going across a blue platform.
If you are asked to find the area of this path, what can you do to obtain this area?
If no data of dimension is given, it is definitely not possible.
Now if the width of the path and the vertical length of the blue platform is given, can you compute the answer?
Again , this need a bit of thinking.
Displacement theory kicks in here. Look at the diagram on the right.
It is the displaced or closed up portion of the blue platform that does the trick.
Here you will notice the dashed line forming a white rectangluar area on the right-most side of the white blue platform.
Are you able to find the area of this white rectangular piece?
The width of this rectangle piece is ACTUAL the width of the white path!
You should now be able to calculate the area of this rectangular piece since the path width and length of the rectangular block is known or deduced now.
How this is possibe is through the "hidden" clue or step of closing up the path revealing the simpler rectangular area that any decent maths student can calculate.
Hence, maths is wonderful in that it tests you not only about applcations of maths tools, but your other "intelligence".
Having known displacement theory here, I believe you are really for the Math Challenge 23.
Go there and answer the question, and be quick before others grap the position one ...
:-D
.
Wednesday, 24 March 2010
Percentage Increase in Area
.
Is there any formula or maths expression showing the ncrease in area when its length and its breadth are increase by m% ?
If you cannot find one, it does not matter. You can easily derive one!
Let us work on this and show the others how simple maths can help us solve daily issue.
Let the length be x and breadth be y.
If x and y increase by m%,
length becomes x + x(m/100), and breadth becomes y + y(m/100).
Area is length x breadth.
Thus new area becomes [ x + x(m/100)] [ y + y(m/100)]
This gives us an area of xy + (m/100)xy + (m/100)xy + (m/100)(m/100)xy.
From the above maths expression, we can deduce that increase in area is:
2 x m% + (m% x m%)/100
Example with numbers will convince readers better, therefore ......
Example
If the increase in perimeter is 10%, what is the increase in area?
Answer is 2 (10%) + (10% x 10%) /100 = 20% + 1% = 21%
Easy isn't it?
For other post related to this concept in percentage increase, see the post Percentage Increase in Perimeter.
Maths is interesting.
;-D
.
Is there any formula or maths expression showing the ncrease in area when its length and its breadth are increase by m% ?
If you cannot find one, it does not matter. You can easily derive one!
Let us work on this and show the others how simple maths can help us solve daily issue.
Let the length be x and breadth be y.
If x and y increase by m%,
length becomes x + x(m/100), and breadth becomes y + y(m/100).
Area is length x breadth.
Thus new area becomes [ x + x(m/100)] [ y + y(m/100)]
This gives us an area of xy + (m/100)xy + (m/100)xy + (m/100)(m/100)xy.
From the above maths expression, we can deduce that increase in area is:
2 x m% + (m% x m%)/100
Example with numbers will convince readers better, therefore ......
Example
If the increase in perimeter is 10%, what is the increase in area?
Answer is 2 (10%) + (10% x 10%) /100 = 20% + 1% = 21%
Easy isn't it?
For other post related to this concept in percentage increase, see the post Percentage Increase in Perimeter.
Maths is interesting.
;-D
.
Tuesday, 23 March 2010
Percentage Increase in Perimeter
.
Percentage is a nice and mystery word in maths.
Why do I say that?
Look at the example below:
If the perimeter has increased by 30%, does the length also increases by the same amount?
The answer is obviously YES.
Next,
If the perimeter is increased by 30%, does the area covered by it also increases by the same amount?
??? The answer needs some pondering, right?
Answer to this:
If the perimeter is increased by 30%, the length and width will both increase by 30%.
This makes the area increase by 2(30%) + (30% x 30%) = ?
(I will explain this maths calculation in a later post.)
For now, let's concentrate on the maths operation.
What do you get from 30% x 30%?
30% = 0.3
Thus 30% x 30% = 0.3 x 0.3 = 0.09 = 9%
This is a potential mathematical mistake.
Error: 30% x 30% = 900% !
So, increase in area becomes 60% + 9% = 69%
Interesting how the mind works.
If the mind is not clear when doing maths, common mistakes do occur.
With more practice, however, this form of mistakes will be lesser.
Hence, be careful when dealing with parameter such as perimeter, length and AREA.
Know their relation and be aware of the "catch" when this type of maths question is being asked.
Do not fall for the maths trick.
:-)
.
Percentage is a nice and mystery word in maths.
Why do I say that?
Look at the example below:
If the perimeter has increased by 30%, does the length also increases by the same amount?
The answer is obviously YES.
Next,
If the perimeter is increased by 30%, does the area covered by it also increases by the same amount?
??? The answer needs some pondering, right?
Answer to this:
If the perimeter is increased by 30%, the length and width will both increase by 30%.
This makes the area increase by 2(30%) + (30% x 30%) = ?
(I will explain this maths calculation in a later post.)
For now, let's concentrate on the maths operation.
What do you get from 30% x 30%?
30% = 0.3
Thus 30% x 30% = 0.3 x 0.3 = 0.09 = 9%
This is a potential mathematical mistake.
Error: 30% x 30% = 900% !
So, increase in area becomes 60% + 9% = 69%
Interesting how the mind works.
If the mind is not clear when doing maths, common mistakes do occur.
With more practice, however, this form of mistakes will be lesser.
Hence, be careful when dealing with parameter such as perimeter, length and AREA.
Know their relation and be aware of the "catch" when this type of maths question is being asked.
Do not fall for the maths trick.
:-)
.
Wednesday, 17 March 2010
Hidden Clues in Maths Questions
There are different levels in any educational system.
This goes with the learning of mathematics too.
At various level of learning, you will be presented with different level of complexity.
At the elementary stage, you will be shown maths questions that are real straight forward type.
At intermediate, a bit of mind twisting has to be done to resolve any challenge.
At the highest level, the questions come embedded with hidden clues to be discovered by learners and used to continue with the solving process.
But hidden clues are now becoming the norm among intermediate level due to its benefits to prevent pure memorising of mathematical technique.
A example of this interesting "hidden clue" can be seen in my Math Challenge 23.
There anyone taking up the challenge needs another step in order to "see" through the simple trick of solving the issue.
(Note: The challenge requires only one step to calculate the area of the path).
Multi-discipline is thus needed for merit of helping get the answer.
Knowledge in utilising maths tools and technique are not sufficient these days.
Maths students have to know some basic theory of motional replacement to understand Math Challenge 23.
Hence, to master mathematics, it will be good to read more, especially, topics outside maths.
This enlarge your understanding of real-life cases roped into maths questions.
Maths is interesting in this manner since it involves not only one learning discipline but encompasses more.
Enjoy maths. It widens your perspective of the world.
:-)
This goes with the learning of mathematics too.
At various level of learning, you will be presented with different level of complexity.
At the elementary stage, you will be shown maths questions that are real straight forward type.
At intermediate, a bit of mind twisting has to be done to resolve any challenge.
At the highest level, the questions come embedded with hidden clues to be discovered by learners and used to continue with the solving process.
But hidden clues are now becoming the norm among intermediate level due to its benefits to prevent pure memorising of mathematical technique.
A example of this interesting "hidden clue" can be seen in my Math Challenge 23.
There anyone taking up the challenge needs another step in order to "see" through the simple trick of solving the issue.
(Note: The challenge requires only one step to calculate the area of the path).
Multi-discipline is thus needed for merit of helping get the answer.
Knowledge in utilising maths tools and technique are not sufficient these days.
Maths students have to know some basic theory of motional replacement to understand Math Challenge 23.
Hence, to master mathematics, it will be good to read more, especially, topics outside maths.
This enlarge your understanding of real-life cases roped into maths questions.
Maths is interesting in this manner since it involves not only one learning discipline but encompasses more.
Enjoy maths. It widens your perspective of the world.
:-)
Monday, 7 December 2009
Percentage | Common Mistake
*
Percentage problems can be tricky at times when you are careless.
Let's us look at one maths word problem related to it.
Question:
There are 3 persons, John, Mary and Jane.
John is richer than Mary by 10%, and Mary is richer than Jane by 20%.
Is John richer than Jane by (10 + 20)% = 30% ?
Most students, upon quick thinking, will acknowledge that 30% is the correct answer.
Is it so?
To verify the answer, let us assume that Jane has $1000.
As such, Mary will have (100+20)% of $1000 = 1.2 x $1000 = $1200.
John is then 1.1 x $1200 = $1320 richer than Jane ==> By 32%.
If 30% is correct, we should get 1.3 X $1000 = $1300.
The latter number (dollar) is not the same as the first worked out solution.
Why?
Mistake in understanding what is percentage:
To assume that John is 10% + 20% richer than Jane is incorrect.
This is due to the fact that percentage has to take a common reference for this to be correct.
In the word problem, the percentages of comparison are not to a common reference.
The first one is to Mary, while the next is to Jane.
These made the denominator of the ratio different.
Thus adding the percentage up is a mistake, and an easy one too!
.
Percentage problems can be tricky at times when you are careless.
Let's us look at one maths word problem related to it.
Question:
There are 3 persons, John, Mary and Jane.
John is richer than Mary by 10%, and Mary is richer than Jane by 20%.
Is John richer than Jane by (10 + 20)% = 30% ?
Most students, upon quick thinking, will acknowledge that 30% is the correct answer.
Is it so?
To verify the answer, let us assume that Jane has $1000.
As such, Mary will have (100+20)% of $1000 = 1.2 x $1000 = $1200.
John is then 1.1 x $1200 = $1320 richer than Jane ==> By 32%.
If 30% is correct, we should get 1.3 X $1000 = $1300.
The latter number (dollar) is not the same as the first worked out solution.
Why?
Mistake in understanding what is percentage:
To assume that John is 10% + 20% richer than Jane is incorrect.
This is due to the fact that percentage has to take a common reference for this to be correct.
In the word problem, the percentages of comparison are not to a common reference.
The first one is to Mary, while the next is to Jane.
These made the denominator of the ratio different.
Thus adding the percentage up is a mistake, and an easy one too!
.
Wednesday, 21 October 2009
Tricky Angles | Be Aware!
*
Geometry in maths can means dealing with angles from a square or a rectangle.
Normally the question is to determine an unknown angle given some shape and angles.
However, mistakes can happen when basic knowledge of relationship between angles and shapes are not proper understood.
Here, I will stress on the square and rectangular matters. This is basic but can pose a tricky problem to the unwarys. Poor thing.....
Let's look at the diagram below.
Here, if sides M and N are the same, that is, if the box is a square, angle A will be 45 degree.
This is so since the corner where angle A lies is 90 degree divided EQUALLY by half due to the diagonal lines reaching to the opposite side. (symmetrical sides).
However, if the side M and N are not equal in length, then angle A WILL NOT be 45 degree. It will depends on the ratio of side M and N.
Note this message and unnecessary mistake can be avoided.
Sometime it is to test the logical thinkng through maths, by not telling you angle A is 45 degree but stating that the box is a square.
This type of maths problem will require you to calculate another angle but using angle A which is not given.
It is tricky but good to have. Your brain will be stretched to make it "flexible" for future use.
Maths is good in this sense as it twists our mind and makes our life interesting!
Work hard as well as smart.
For more examples on avoiding unnecessary mistakes, visit this time calculation post.
.
Geometry in maths can means dealing with angles from a square or a rectangle.
Normally the question is to determine an unknown angle given some shape and angles.
However, mistakes can happen when basic knowledge of relationship between angles and shapes are not proper understood.
Here, I will stress on the square and rectangular matters. This is basic but can pose a tricky problem to the unwarys. Poor thing.....
Let's look at the diagram below.
Here, if sides M and N are the same, that is, if the box is a square, angle A will be 45 degree. However, if the side M and N are not equal in length, then angle A WILL NOT be 45 degree. It will depends on the ratio of side M and N.
Note this message and unnecessary mistake can be avoided.
Sometime it is to test the logical thinkng through maths, by not telling you angle A is 45 degree but stating that the box is a square.
This type of maths problem will require you to calculate another angle but using angle A which is not given.
It is tricky but good to have. Your brain will be stretched to make it "flexible" for future use.
Maths is good in this sense as it twists our mind and makes our life interesting!
Work hard as well as smart.
For more examples on avoiding unnecessary mistakes, visit this time calculation post.
.
Saturday, 17 October 2009
Solving Maths Visually
.
Maths is interesting!
There are many exciting concepts and techniques one can apply.
Read on ...
There are many ways to solve a maths question.
Normally it involves taking many steps with related sequence.
However, there are also simple ways to handle a maths problem.
One such solving method is simply through visual steps.
This has no working at all.
What do I mean?
Let's look at an example.
Example:
Determine the angle B from the diagram below. Angle A is 40 degree.
Angle B is same as angle A = 40 degree.
There is no working at all. Just the visual determination.
Concept of this trigonometrical question in geometry:
When 2 straight lines cross and meet at an angle to each other, the angle opposite to any one is the same.
As such, angle C is also equal to angle D.
This is visual maths.
Interesting?
:-D
Maths is interesting!
There are many exciting concepts and techniques one can apply.
Read on ...
There are many ways to solve a maths question.
Normally it involves taking many steps with related sequence.
However, there are also simple ways to handle a maths problem.
One such solving method is simply through visual steps.
This has no working at all.
What do I mean?
Let's look at an example.
Example:
Determine the angle B from the diagram below. Angle A is 40 degree.
Solution:
There is no working at all. Just the visual determination.
Concept of this trigonometrical question in geometry:
When 2 straight lines cross and meet at an angle to each other, the angle opposite to any one is the same.
As such, angle C is also equal to angle D.
This is visual maths.
Interesting?
:-D
Tuesday, 6 October 2009
Time Mathematics | Tips On Avoiding Mistakes
Dealing with time and its calculation can be a bit tricky sometimes.
There are the hours, the minutes and the seconds to handle.
We can add or subtract them.
We do not operate in the hundreds, or ones, like any simple arithmetric.
We are calculating in terms of sixties.
1 hour = 60 minutes
1 minute = 60 seconds
Hence when given a maths question on subtracting two times, how do we go about to avoid mistakes?
Let's take an example to illustrate.
Question:
John start his journey to the market at 09:15am. If he arrived at the market at 10:05am, how long did he travelled?
Approach 1:
We can use the conventional method of carrying back 60 to the minutes, since the ending minute is smaller than the starting minute. And reducing the 10 to 9 (hour).
Next, we can then subtract with the new numbers.
That is 9:65 - 9:15 = 50 minutes.
This answer is fine.
Approach 2:
Change all the times to mintes.
10:05 ==> (10 x 60) + 5 = 605
09:15 ==> ( 9 x 60) + 15 = 555
Subtracting the two new numbers gives 50 minutes directly.
And that is the answer!
Remarks:
Comparing approach 1 with approach 2, you would notice that the latter seems to be simpler.
Why so?
This is because we have simplified the mixture of hours and minutes to only one dimension, that is, the minutes.
Thus, subtracting the newly created numbers involved only the minutes, avoiding the distraction of handling the hours.
To do maths, clear the mind of the unnecessary.
In the example above, we have removed the "hours" factor to focus purely on the "minutes".
Maths is interesting in that it is up to us to "play" with the techniques.
We can work with it or go against it.
The choice is up to us to select.
Hope you pick up some tips to make maths interesting.
Cheers!
:-D
There are the hours, the minutes and the seconds to handle.
We can add or subtract them.
We do not operate in the hundreds, or ones, like any simple arithmetric.
We are calculating in terms of sixties.
1 hour = 60 minutes
1 minute = 60 seconds
Hence when given a maths question on subtracting two times, how do we go about to avoid mistakes?
Let's take an example to illustrate.
Question:
John start his journey to the market at 09:15am. If he arrived at the market at 10:05am, how long did he travelled?
Approach 1:
We can use the conventional method of carrying back 60 to the minutes, since the ending minute is smaller than the starting minute. And reducing the 10 to 9 (hour).
Next, we can then subtract with the new numbers.
That is 9:65 - 9:15 = 50 minutes.
This answer is fine.
Approach 2:
Change all the times to mintes.
10:05 ==> (10 x 60) + 5 = 605
09:15 ==> ( 9 x 60) + 15 = 555
Subtracting the two new numbers gives 50 minutes directly.
And that is the answer!
Remarks:
Comparing approach 1 with approach 2, you would notice that the latter seems to be simpler.
Why so?
This is because we have simplified the mixture of hours and minutes to only one dimension, that is, the minutes.
Thus, subtracting the newly created numbers involved only the minutes, avoiding the distraction of handling the hours.
To do maths, clear the mind of the unnecessary.
In the example above, we have removed the "hours" factor to focus purely on the "minutes".
Maths is interesting in that it is up to us to "play" with the techniques.
We can work with it or go against it.
The choice is up to us to select.
Hope you pick up some tips to make maths interesting.
Cheers!
:-D
Sunday, 30 August 2009
Misconception of Percentage
Maths is interesting!
It catches you nicely when you are not aware of its full implication.
Many of its techniques and concepts has wide boundaries.
Any maths learners has to understand its fundamentals and basic concepts in order to "escape" being caught with wrong usage.
Percentage is one area I wish to mention in this post.
You may find that percentage or percent is a simple term in maths.
Besides being a ratio, it is a comparative element.
It tells how big the target is to the original, or how small it is.
Misconception of Percentage:
The numerator is always smaller than the denominator (which is the total).
Is it true?
A definite NO!
Reason:
If a ratio has its numerator less than the denominator ==> The numerator is relatively smaller in size than the total.
If the numerator is larger than the denominator ==> The numerator is bigger in size than the total.
An example can illustrate the concept.
Example:
If a costume is now priced at 90% of its original, $100, it means that the price is not only $90.
It is lesser than the original, since the ratio is 90 / 100 or 0.9.
If the costume is newly priced at 120% of its original, $100, it means that the price is now at $120!
A price value more than the original.
It is a practical real-life number.
Interesting? You bet.
:) Happy maths learning.
.
If a ratio has its numerator less than the denominator ==> The numerator is relatively smaller in size than the total.
If the numerator is larger than the denominator ==> The numerator is bigger in size than the total.
An example can illustrate the concept.
Example:
If a costume is now priced at 90% of its original, $100, it means that the price is not only $90.
It is lesser than the original, since the ratio is 90 / 100 or 0.9.
If the costume is newly priced at 120% of its original, $100, it means that the price is now at $120!
A price value more than the original.
It is a practical real-life number.
- Percentage can be more than 100%.
- The numerator can be more than the denominator.
Interesting? You bet.
:) Happy maths learning.
.
Wednesday, 26 August 2009
Usage of Percentage
'
In maths learning, every learners will come across the terms "percent" or "percentage".
Percentage is a ratio.
It is a ratio between two numbers.
The denominator is normally the total of an item.
The unit used is %.
The above may be common knowledge for anyone doing percent maths problems.
However, what is the presentation of the solution that is appropriate?
Let's us do an example.
Example:
Class A has 30 students. If 40% of them are girls, how many are boys?
Solution:
Number of boys in the class = 30 - 40%
Now, is this "30 - 40%" correct?
The idea may be there, but someting is amiss.
What is this "40%"?
Can we just write 40% as it is?
The answer is NO!
Percentage is a ratio of the total (the class size of 30).
We cannot simply write 40% if we want to know the actual number.
The correct way is to present the step or working as: (40/100) x 30 = 12 students
We cannot simply write : 30 - 40%.
The correct way has to be 30 - (40/100) x 30 = 30 - 12 = 28 students.
2 mistakes, in concept, were made:
1) Number cannot subtract a percentage. Their units are different.
2) Percentage is a ratio, an indication of the proportion of a piece to the total. It is a relative term, not an absolute number. Thus, an absolute number cannot operate with a relative term.
Understanding this concept of "percent" and "percentage" will be handy and avoid the unnecessary trouble and anxiety of doing maths.
Tips:
Thinks of a slice of cake when doing "Percent". It has meaning only when compared to the whole cake.
Remember: Maths Is Interesting! And it WILL be get more interesting.
:D
In maths learning, every learners will come across the terms "percent" or "percentage".
Percentage is a ratio.
It is a ratio between two numbers.
The denominator is normally the total of an item.
The unit used is %.
The above may be common knowledge for anyone doing percent maths problems.
However, what is the presentation of the solution that is appropriate?
Let's us do an example.
Example:
Class A has 30 students. If 40% of them are girls, how many are boys?
Solution:
Number of boys in the class = 30 - 40%
Now, is this "30 - 40%" correct?
The idea may be there, but someting is amiss.
What is this "40%"?
Can we just write 40% as it is?
The answer is NO!
Percentage is a ratio of the total (the class size of 30).
We cannot simply write 40% if we want to know the actual number.
The correct way is to present the step or working as: (40/100) x 30 = 12 students
We cannot simply write : 30 - 40%.
The correct way has to be 30 - (40/100) x 30 = 30 - 12 = 28 students.
2 mistakes, in concept, were made:
1) Number cannot subtract a percentage. Their units are different.
2) Percentage is a ratio, an indication of the proportion of a piece to the total. It is a relative term, not an absolute number. Thus, an absolute number cannot operate with a relative term.
Understanding this concept of "percent" and "percentage" will be handy and avoid the unnecessary trouble and anxiety of doing maths.
Tips:
Thinks of a slice of cake when doing "Percent". It has meaning only when compared to the whole cake.
Remember: Maths Is Interesting! And it WILL be get more interesting.
:D
Saturday, 25 July 2009
Meaning of "Average"
'
I managed to see a word problem shown below:
" Mary obtained an average of 80 marks for 2 tests. What marks has she to get for an average of 80 marks for 3 tests?"
If a maths student understand the meaning of "average", she can do this without even working out the mathematical steps.
The first statement in the question stated that Mary scored on both tests an average of 80 marks. This implied that for one test, she obtained 80 marks. Same as for the second test.
For the next test (the third one), to get an average of again 80 marks, it meant that all the test she has to get 80 marks each. This is the power of "average". That is to say, all the test can be concluded as the same score.
The actual differences between the individual marks can be offset from each other to achieve a final "average" of same marks (here, 80 marks).
The maths problem is to test the understanding of the word "average" in maths, and its concept.
Thus, knowing concepts do help in solving maths questions.
It does not mean working out the mechanical computation steps to get answers.
With strong maths concept, anyone can solve simple problems without doing working.
But that is maths also (in the brain, the step-less way).
Interesting. Maths is so. Fear not.
:-)
I managed to see a word problem shown below:
" Mary obtained an average of 80 marks for 2 tests. What marks has she to get for an average of 80 marks for 3 tests?"
If a maths student understand the meaning of "average", she can do this without even working out the mathematical steps.
The first statement in the question stated that Mary scored on both tests an average of 80 marks. This implied that for one test, she obtained 80 marks. Same as for the second test.
For the next test (the third one), to get an average of again 80 marks, it meant that all the test she has to get 80 marks each. This is the power of "average". That is to say, all the test can be concluded as the same score.
The actual differences between the individual marks can be offset from each other to achieve a final "average" of same marks (here, 80 marks).
The maths problem is to test the understanding of the word "average" in maths, and its concept.
Thus, knowing concepts do help in solving maths questions.
It does not mean working out the mechanical computation steps to get answers.
With strong maths concept, anyone can solve simple problems without doing working.
But that is maths also (in the brain, the step-less way).
Interesting. Maths is so. Fear not.
:-)
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