Some math questions are simple and can be answered easily.
See this math challenge 20 and its answer in the comment.
But a twist of the questioning will and can make it more challenging without changing the math expression.
Below is one:
Base on math challenge 20, if all unknowns CANNOT be repeated, what are they?
Again they are integers and below 10.
Enjoy the math thrill answering this.
Merry Christmas!
^.^
Friday, 25 December 2009
Saturday, 19 December 2009
Math Challenge 20
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Can anyone come out the answers for A, B, C and D in the below math expression?
A2 + B2 + 2C2 = D2
The rule is that the unknowns are all integers and below 10.
Happy trying, and
don't forget that maths is interesting!
:-)
Can anyone come out the answers for A, B, C and D in the below math expression?
A2 + B2 + 2C2 = D2
The rule is that the unknowns are all integers and below 10.
Happy trying, and
don't forget that maths is interesting!
:-)
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!
.
Sunday, 29 November 2009
Square Root | An Exciting Outcome
*
Everyone knows what a square root is for and what it does to a number.
But have you tried multiple square rooting ?
What do I mean?
Let's take an example.
Start with a number, say, 3.
i) Square root this 3.
ii) You will get a number after step one.
iii) Square root this new number again.
iv) Continue the steps above and look closely at the number.
Findings:
You will notice that the number after multiple square rootings, will give you closer and closer to or approaching the number "1".
Now let's start with another number. This time, let's choose, 0.4
After doing the same procedures, you will again notice that the answer is getting and approaching the number "1"!
Amazing isn't it?
What other wonders can you find from this Square Root maths operator?
Share with me in the comment section.
Maths is interesting.........
:D
Everyone knows what a square root is for and what it does to a number.
But have you tried multiple square rooting ?
What do I mean?
Let's take an example.
Start with a number, say, 3.
i) Square root this 3.
ii) You will get a number after step one.
iii) Square root this new number again.
iv) Continue the steps above and look closely at the number.
Findings:
You will notice that the number after multiple square rootings, will give you closer and closer to or approaching the number "1".
Now let's start with another number. This time, let's choose, 0.4
After doing the same procedures, you will again notice that the answer is getting and approaching the number "1"!
Amazing isn't it?
What other wonders can you find from this Square Root maths operator?
Share with me in the comment section.
Maths is interesting.........
:D
Monday, 23 November 2009
Logarithm Operation Explained (2)
'
Logarithm operator exists as a mathematical tool to allow us to convert a number to another form in terms of base and power.
Can you make 0.5 in terms of 10y or 5y ?
The above is a maths question that can use logarithm to solve.
The application requires the Power rule that states that logkDm is equivalent to mlogkD.
With that knowledge, to convert the above question of 0.5 to various base, we simply log the 0.5 to its respecive base.
Let's go for the base 10.
Original: 0.5 = 10y
Performing "log" on both sides:
log10 0.5 = log1010y
-0.301 = y log1010
-0.301 = y
or another way to put the outcome is 0.5 = 10-0.301
We have managed to convert the original number of 0.5 to one with base 10 and a power (index) of "-0.301".
We can also similarly do the same for a base of 5. => 0.5 = 5y
Here we just "log" to base 5.
log5 (0.5) = log5 5y = y
-0.431 = y
Thus 0.5 = 5-0.431
From the 2 examples above, you can see the wonders of having logarithm as a conversion tool.
Once you understand this principles, you will appreciate logarithm.
Cheers!
:D
Logarithm operator exists as a mathematical tool to allow us to convert a number to another form in terms of base and power.
Can you make 0.5 in terms of 10y or 5y ?
The above is a maths question that can use logarithm to solve.
The application requires the Power rule that states that logkDm is equivalent to mlogkD.
With that knowledge, to convert the above question of 0.5 to various base, we simply log the 0.5 to its respecive base.
Let's go for the base 10.
Original: 0.5 = 10y
Performing "log" on both sides:
log10 0.5 = log1010y
-0.301 = y log1010
-0.301 = y
or another way to put the outcome is 0.5 = 10-0.301
We have managed to convert the original number of 0.5 to one with base 10 and a power (index) of "-0.301".
We can also similarly do the same for a base of 5. => 0.5 = 5y
Here we just "log" to base 5.
log5 (0.5) = log5 5y = y
-0.431 = y
Thus 0.5 = 5-0.431
From the 2 examples above, you can see the wonders of having logarithm as a conversion tool.
Once you understand this principles, you will appreciate logarithm.
Cheers!
:D
Friday, 20 November 2009
Logarithm Operation Explained
.
Everyone knows what happen when we add a number to another.
We also know what happen when we do subtraction.
There is also no problem with multplication or division of numbers.
These are all simple mathematical operations that are basic.
What about doing a logarithmic operation in math?
This is a slightly complex but interesting question.
When we do a logarithmic operation on a number, what we actually do want out of the mathematical process is the power or index with reference to a base number.
105 has a base number of 10 and index of 5.
Doing a "log" of the above will reveal an answer of 5.
This is the power after doing a "log" operation.
Thus, when anyone does a logarithmic operation on a number (or expression), he is trying to find the index with respect to a base reference.
log 105 = 5.
Note:
Full written log expression is "logkP". When the "k" is left out, it implies that k = 10.
And log 10 = 1.
I hope the above can explain why we do logarithmic operation and its significant.
There must be a reason for each math operation, otherwise we will be learning and doing some insane process on earth!
Maths Is Interesting!
:-D
Everyone knows what happen when we add a number to another.
We also know what happen when we do subtraction.
There is also no problem with multplication or division of numbers.
These are all simple mathematical operations that are basic.
What about doing a logarithmic operation in math?
This is a slightly complex but interesting question.
When we do a logarithmic operation on a number, what we actually do want out of the mathematical process is the power or index with reference to a base number.
105 has a base number of 10 and index of 5.
Doing a "log" of the above will reveal an answer of 5.
This is the power after doing a "log" operation.
Thus, when anyone does a logarithmic operation on a number (or expression), he is trying to find the index with respect to a base reference.
log 105 = 5.
Note:
Full written log expression is "logkP". When the "k" is left out, it implies that k = 10.
And log 10 = 1.
I hope the above can explain why we do logarithmic operation and its significant.
There must be a reason for each math operation, otherwise we will be learning and doing some insane process on earth!
Maths Is Interesting!
:-D
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