cracking the Math S.A.T

Math Tricks 1...squared numbers

2009-08-01T05:05:00.000-07:00

Welcome to a new world of advanced mathematical tricks that can save you alot of time in Math Exams. Okay I have shown you how to do most of the question types on the S.A.T. , it is time for me how to minimize your calculation time. Most times you made an error in simple calculation using a calculator and you haven't realise especially in the questions that have no multiple choices. Today i will show you how to calculate squared numbers close to 100 mentally...this is to stop relying on the calculator to often.

Final Chapter

2009-07-22T18:49:00.000-07:00

My fellow bloggers I have shown you all what I can on stratergies on the S.A.T. Now i have a new blog. http://elecmag-tech.blogspot.com. I have new futuristic ideas but let me not tell you it here, visit it. Thank you for reading, best regards...future C.E.O of Elecmag Technologies...M.Richards.

General

2009-01-15T06:59:00.000-08:00

For all my fellow readers I thank you for reading my blog. I know my last post has been almost a year now but times for me has been difficult for me. Now am back....o and Merry Christmas and Happy New Year ...I know it is late but better late than never

Arithmetic Progression :Part 2

2008-05-22T12:18:00.000-07:00

Welcome to part 2 of Arithmetic Progression. Today I am going to look at another formula called the () summation formula. This formula is useful for adding a series of numbers.

E.g. What is the sum of all the numbers between 1 & 20 inclusive?

There is the long, time consuming method of adding all the numbers between 1 & 20 OR use the summation formula. The summation formula is :

(n/2)( a + L ) where n is the number of numbers involved, a is the first number and L is the last number

OR

(n/2)( 2a + (n – 1)d ) where n and a is the same as above and d is the common difference.

These formulae give the same answer but they are used according to the question asked. In the example above the first and last number was given so the first formula was the most appropriate one to use. Using the formula:

n = 20, a = 1, L = 20

(20/2) (1 + 20) = 10(21) = 210

Note: you could have used the second formula also and in that case d, the common difference, was (+1).

e.g. What is the sum of the first 6 even numbers?

Note: in this case the last number is unknown so we must use the second formula!

a = 2, n = 6, d = +2

(6/2) (2(2) + (6 – 1) (+2) = 42

(you can check it out by yourself manually).

Now you are aware of how to use the formula, it is now time to look out for unnecessary working. Here is a famous example on the S.A.T

e.g. What is the sum of all the numbers between (-25) & 60?

There are two ways to do this.

Method 1. Find the sum of the all the numbers between 1 & 25 using the formula, then multiply the answer by (-1). This then gives you the value for the sum of the numbers between (-25) & (-1). Now find the value of the sum of the numbers between 1 & 60, and minus the first answer from the second answer.

Method 2. Since you know the sum of the numbers between (-25) and (25) will be zero (you do realize that the positive number will be cancelled by its negative counterpart right???) so you have to solve for the sum of the numbers between (26) & (60). You must be saying why not just do it like method one because it is much easier, that’s true but what if the question was find the sum of the number between 26 and 60? What would you have done? Add it manually? You will waste too much time; you do know the summation formula, so why not take advantage of that knowledge? First find the sum of the numbers between 1 & 25 then find the sum of the numbers between 1 & 60 and then minus the first answer from the last answer. SIMPLE!!

That concludes Arithmetic Progression Part 2.

Arithmetic Progression :Part 1

2008-05-14T04:49:00.000-07:00

So far we have done Geometric Progression, now we are going to look at Arithmetic Progression. Arithmetic Progression deals with reducing or increasing a start number by a fixed value or adding a series of consecutive numbers. There are two popular aspects on Arithmetic Progression but in this post we will look at only one.

First I am going to look at the reducing/increasing aspect of this topic. A.P. questions are well disguised in the S.A.T but it is very easy to identify it by remembering the principle form they appear by. As long as you can identify 3 items in the question, then you will know it is an A.P. question. Look for:

ü A start value.

ü The number of terms, ‘n’ (e.g. the number of years).

ü A fixed number which the start value is reducing/increasing by.

When you find all 3 items in a question it is revealed as an A.P. now let’s look at the A.P. formula. In the A.P. there are 2 useful formulae (also in the G.P. there are 3 but only on is useful in the S.A.T).

Now we are going to look at the first A.P. formula in this post. The first formula is called the ‘n’ term formula.

a + (n - 1) d

the ‘n’ term formula gives the value of the ‘nth term’. The ‘nth’ term means the value of the number you are asked to solve for. Let’s look at an example to clear all confusion.

E.g. A bag contains 1178 marbles. If 27 people, each having bags which can hold a maximum of 8 marbles, how many marbles remained in the bag?

Method 1:

8 x 27 = 216
1178 – 216 = 962

Therefore 962 marbles remained in the bag.

Second method:

a = 1178, n = 28, d = - 8

(NOTE: n equals 28 because n is referring to the number of bags involved and not the number of people involved and d equals – 8 because it is reducing by 8!)

Using the A.P. formula

1178 + (28 – 1)(- 8) = 962

you may have said that this question is easier to be done the first way, yes that’s true, but what if the same exact question came this way instead?

E.g. A bag contains 1178 marbles. It is to be shared in groups of 8 so that there are 962 marbles remaining in the bag. How many bags are needed?

Wouldn’t you use the formula in this case? The formula is useful in some cases; it is up to you to determine which way easier and less time is consumed for you. By manipulating the A.P. formula you can solve for ‘n’. That concludes A.P. part 1; in my other post I will show you the second useful A.P. formula.

Geometric Progression

2008-05-08T05:00:00.000-07:00

Today I am going to look at Geometric Progression, which is also known as G.P. Geometric Progression is a term used when a starting number, a, is reduced/increased by a fixed percentage or fraction. G.P. only involves the operation; division or multiplication. There is another topic similar to G.P. by increasing/reducing a start number, a, by a fixed value by means of operations; subtraction or addition. This is known as Arithmetic Progression also known as A.P. This topic will be taught in my next post.

G.P. is a similar form of gradual depreciation/appreciation and A.P. is a similar form of fixed depreciation/appreciation, but we will look at G.P. in this post.

As you have already known the depreciation formula from the previous post:

V x ((100 – d)/ 100)^n

The G.P. formula is almost exactly the same, it has the same principle but has different terms.

The formula for the G.P. is

ar^( n- 1)

Where ‘a’ is the beginning value (same as ‘V’ in the depreciation formula. Where ‘r’ is the common ratio (same as ((100- d)/100) in the depreciation formula) but there is a slight difference with the ‘n’.

G.P. really gives the same answer as the depreciation formula, but it is not the same ‘n’ as in the depreciation formula.

The ‘n’ in the depreciation formula tells you the amount of years as in (Dec 31^st) not as in (Jan 1^st). The ‘n’ in the G.P. formula tells you the amount of years as in (Jan 1^st) hence it is (n – 1) in the G.P. formula so that it works out to be the exact value as in the depreciation formula.

The G.P doesn’t also workout percentages (depreciation/appreciation) it can also be used in different cases.

E.g. An amoeba (single cell) doubles every 30 minutes. How many cells are in the container after 3 hours?

This can be done manually………..>

3 hours = 6 x 30 minutes…..thus it doubles 6 times!

2^6 = 64

Start amount | time elapsed |

1 | 30 minutes |

2 | 1 hour |

4 | 3 x 30 minutes |

8 | 4 x 30 minutes |

16 | 5 x 30 minutes |

32 | 6 x 30 minutes |

64

So you started at 1 and ended with 64, hence there are 7 terms. (the ‘n’ in the G.P. formula gives the number of terms!)

Using the G.P. formula

ar^( n- 1)

a = 1 r = 2 and n = 7, we will get 64.

Questions like these are disguised in the S.A.T but with practice you will able to identify and use the G.P. formula efficiently. Well, that concludes G.P. ( you can try the depreciation question from the previous post using the G.P. method). The next topic is on Arithmetic Progression.

Depreciation/ Appreciation

2008-05-04T08:49:00.000-07:00

Today’s topic will feature a combination of a variety of topics which includes percentages, depreciation/ appreciation and Geometric Progression. I have not taught Geometric Progression (G.P) but in this very chain of topics I will teach it.

I am going to teach depreciation/ appreciation today, it is the main topic but it involves percentages and (G.P) is a method which depreciation/ appreciation can be done! Let’s look at 2 types of depreciation questions! First we will look at the gradual depreciation type.

E.g. A house of value $500,000 depreciates annually by 2%. What will be the new value of the house 5 years from now, starting January 1^st, 2008? This is the very exact question from 3 posts ago!

Solution: Method 1

Year | value at start | depreciation | end year value

1 | $500,000 | $500,000 x 2% =$10,000 | $490,000

2 | $490,000 | $490,000 x 2% =$ 9,800 | $480,200

3 | $480,200 | $480,200 x 2% =$ 9,064 | $470,596

4 | $470,596 | $470,596 x 2% =$9,411.92| $461,184.08

Therefore the value in the 5^th year is $461,184.08

Note: the end year value for the 4^th year is the value at start on the 5^th year and 5 years from now means after for years. The S.A.T questions are all designed this way, so the trick in the question is the grammar and not the method how to solve it.

Solution: Method 2 (easier method)

Since the depreciation is 2% annually, the value after each year is 98% of the start.

After 1 year the value is $500,000 x .98= $490,000

After 2 years the value is $500,000 x .98 x.98= $480,200

After 3 years the value is $500,000 x .98 x.98 x.98=$470,596
After 4 years the value is $500,000 x (.98)^ 4 =$461,184.08

If you know the original value V, the rate of depreciation d% per annum, and the number of years n, the depreciated value is given by

V x ( (100 – d)/100)^ n
This is known as the depreciation/ appreciation formula.

Note: To do appreciation questions just change the minus sign to a plus sign!

Now let’s look at the other type of depreciation known as fixed appreciation.

E.g. A company vehicle bought for $125,000 in 2004 depreciates by a fixed value by 15% of the vehicle. What would be the value of the vehicle given by the company (book value) after 3 years?

Solution: Method

Value of the vehicle is $125,000.
Depreciation for the first year is $125,000 x 15%=$18,750
Depreciation for 3 years is $18,750 x 3=$56,250
Book value after 3 years is $125,000 - $56,250 =$68,750

This is the simplest for of depreciation. The next post I will show you how to use the (G.P) formula to do these questions.

Percentages Part C

2008-04-27T07:24:00.000-07:00

Hello and welcome to the last part in percentages. Percentages are really straight forward but the wording of the problem can be a little confusing at times. Some problems you can do mentally, some questions you can solve in a few seconds by the use of a calculator and some questions require you to think more than the previous questions.

Now I will look at the last question posted in ‘Percentages Part A’. The last question is actually easier than the question about the house, but there is a reason why this question is one of the last questions you will most likely encounter in the exam. The reason is a result of the answer choices and the requirements for this question. For average test takers, when they reach this question, usually they have a few seconds remaining. With my guidance on how to approach the S.A.T you will have at least two minutes remaining! Let’s look at the question;

60% of x is 30% of y. if w = 4x, what percentage is (x + y)/w?

a) 15%
b) 25%
c) 60%
d) 74%
e) cannot be determined from the information above

When you are running low on time students tend to build up an anxiety fear and their brains are out of focus and are easily distracted by eye catchers. When time is low some students take the risk and guess answers. Let’s look at the answer choices.

Why choice a) 15%? This is to distract you as ½ of 60% is 30% hence ½ of 30% is 15%. (THIS IS INCORRECT!)

Why choice b) 45%? This is to distract you as ½ of (60% + 30%) = 45%. NOTE: X and Y are not percentages, they are unknown numbers!

Why choice c) 60% ? This is a distracter! This is because, under pressure, 90% of students guess the answer c.

Why choice d) 75%? This is the actual answer but this is a well placed answer for this question. This is a trap for students that who underestimate and over think in Mathematics. Advance students in Mathematics scan each choice, like what I have just done, to choose an appropriate answer when time is low. (x + y)/ 4x can be misinterpreted for (30% + 60%)/ 120% (THIS IS INCORRECT!). Some Math students know that this is wrong but they think of how the average person will attempt it and you must do that also so that you won’t waste time looking at ridiculous choices hence saving you time!

Why choice e)? This is because there are no values given. (NOTE: YOU WILL ENCOUNTER THIS ANSWER A LOT ON THE S.A.T.)

Those are the reasons why this question is hard. It’s because of the time you will most likely have to do this question that really brings out the difficulty. This question is a dependant question. Y and W depends on X, because of this, no matter what number you choose for X, you will always get the correct answer as long as you do it properly. You can choose any value for X but when working with percentages it is much more economical, in reference with time, to use the number 100. No matter what value for X you choose you will get 75% as the answer. This concludes straight forward percentages. There are also more questions involved in percentages but they are a combination of other topics such as Geometric Progression (G.P.)! The question about the house involves G.P. and that will be the topic of my next post. Thank you for reading and I wish you all the luck in your future exams!

Percentages Part B

2008-04-24T18:42:00.000-07:00

Welcome to part B of percentages. In this post I am going to show you. Now let’s look at the first example in the previous post.
e.g. 30% of x is 27, what is x?

x = 100%
>>>>>>>>>>>>>>30% of x = 27
(divide by three)>> 10% of x =9
(multiply by ten)>> 100% of x = 90

x = 100%, so first solve for a factor of 100; you could have solved for 1% which gives you 0.9%, or 5% which gives you 4.5 but in this case finding for 10% was most appropriate as 30% and 27 are easily divisible by 3 which results to a whole number rather than a decimal number. This question can be done mentally with a few practice ( in some cases you will have to find for 1% then multiply by 100 to get the value but the S.A.T gives you easy numbers that are easily divisible to produce 100% such as 60%,15%,30% etc. They would not give you numbers such as 81% or 29%)

Let’s look at the second question.

60% of y is 120, what is 75% of y?

60% of y = 120
divide by 60>>>> 1% = 2
multiply by 75 >> 75% = 150

It is the same as before, find for 1% and then find for the percentage asked for, but if you realized….you could have done this mentally and this is what I want you to get into. I am teaching your mind how to solve patterns rather than use pencil and paper. Your mind has to be rapidly clicking. Its mental time!!!

What do the numbers 60 and 75 have in common? The answer is the number 15!

In your mind you have to visualize 15 is ¼ of 60 hence ¼ of 120 is 30. Also you must realize 60 + 15 = 75, so just add 15% to 60% hence add 30 to 120 to get 150. This is how I solved it and I want you to be able to do it.

The third question is a bit tricky but it can be done two ways. I will recommend only one method, the geometric progression method, but I will show both methods. Because this question involves another topic I have not covered in this course I will not solve it in this post. (Look in future post please for the solution of this question for it is very important!!!!)

This concludes the end of this post. Refer to percentages Part C for the conclusion of percentages! Bye.

Percentages Part A

2008-04-24T17:43:00.000-07:00

First of all I have to apologize to my readers for my late post. This will not happen again, at least I hope so. Today I am going to lecture on percentages. Percentages are a very common topic in Mathematics and especially on the S.A.T. In the S.A.T, percentages come in some of the forms I have provided below.

The first form is given by this example. 30% of x is 27, what is x? This is the simplest form, and it usually comes in the first ten questions of a Mathematics section. A similar form is; if 60%y is 120, what is 75% of y?

Another form comes in the form of a worded problem. These are a little tricky especially the wording of problems referring to depreciation/appreciation. Now I will give you a depreciation question but I will not show you how to solve it in this post. When I am doing appreciation/depreciation, which is my next post, I will refer to this very question! This is the question.

A house of value $500,000 depreciates annually by 2%. What will be the new value of the house 5 years from now, starting January 1st, 2008? This is a medium level question and it occurs between questions 15 and 20. Do not underestimate the power of this question, it between questions 15 and 20 for a specific reason.

The last example is very common on the S.A.T and is the most difficult questions pertaining to percentages and is found in the last five questions of the Mathematics sections. They come in the form of only letters and percentages, but then again it will be easy for you, so do not be intimidated by these questions!

60% of x is 30% of y. if w = 4x, what percentage is (x + y)/w?

a) 15%
b) 25%
c) 60%
d) 74%
e) cannot be determined from the information above

In my next post I am going to show you how to solve these questions and also my mind method of solving these questions of solving these questions.

Advance Class Part C: Product and Sum of the Roots!

2008-04-13T09:41:00.000-07:00

Today I am going to show you a cool trick how to check to make sure your answers for the solution of a quadratic equation is correct without plugging the values back into the equation. Have you ever heard the term “Sum of the Roots and the Product of the Roots”? Well if you have not, today I am going to teach you it. The symbol I will use for alpha is (K*) and for beta is (V*). Where (K*) is one solution and (V*) is the other solution.

The sum of the roots is alpha plus beta (K* +V*) and the product of the roots is (K* x V*). This method works like this, (K* + V*) must be equal to (–b/a) and (K* x V*) must be equal to (c/a). Are you still confused? Okay, let’s look a question I had previously done.

14x^2 – 13x – 27 = 0
In this case a = 14, b = - 13 and c = -27

We have already solved the equation to produce solutions of x = 27/14 and x = - 1

The Visual method showed you how to solved it, but what if you forgot the visual method? (The conditions are in red above)

Let’s apply the sum of the roots! In the equation (- b/a) equals to (- (-13/14)) which is equal to (13/14).
The sum of the roots (K* + V*) equals to ( 27/14 + (-1)) = ( 13/14)
Therefore the first condition is satisfied!

Let’s apply the product of the roots! In the equation (c/a)=(-27/14).
The product of the roots (K* x V*) = ( (27/14) x (-1)) = (-27/14)
Therefore the second condition is satisfied!

If these two conditions are satisfied, then your solutions are correct and there will be no need to plug in your values to double check you answer. This method applies for all quadratic equations.

Advance Class Part B: Visual Method

2008-04-09T09:53:00.000-07:00

Welcome to Part B of this post. First I am going to show you my visual method how to solve for x. This technique is very useful on the day of the S.A.T. It is not advisable to use this technique in school because by skipping steps you can lose valuable points on an exam. The S.A.T requires no working only answers, so working doesn’t count. Now this visual technique can only be used on certain quadratic equations. The entire purpose of this technique is to save you about 15 seconds, which is a lot in the exam for one question. For student who already did the S.A.T and are going to do over the exam should know how important 15 seconds can be. Now let’s get started. Consider the equation…..

14x^2 - 13x -27 = 0

by just watching the question I can tell the answers are x = -1 and x = 27/14 . There is a special pattern to it. Ok let’s solve this first.

Perform the number splitting method.

14x^2 + 14x – 27x – 27 = 0

Factorize

14x(x +1) – 27(x +1) = 0
(14x – 27)(x +1) = 0
14x = 27…………and x = -1
x = 27/14 ………..and x = -1

Now you must be thinking I solved this question first and wrote the answer…..well I will teach you how to do it.

Remember the number splitting technique, look at that line, do you realize anything about the co efficient. This visual method can only work if and only if the first and last numbers are used in the number splitting. If this condition is met, then the two answers are according to the conditions of the equation.

If the equation is in the form…..:ax^2 –bx + c = 0
The solutions are x = c/a and x = 1

If the equation is in the form…..:ax^2 + bx + c = 0
The solutions are x = -c/a and x = -1

If the equation is in the form……:ax^2 - bx - c = 0
The solutions are x = c/a and x = -1

Now you must be pondering how to remember this? OK simple, the only thing you should be remembering is that you ensure the first and last number is used in the number splitting technique.

How to remember the solutions for x? First write x = c/a and x = 1 (leave a little space to insert any negative signs).To know the sign of x = c/a, watch the sign of b, the sign of c/a is always opposite to b in the equation.To know the sign of x = 1, watch the signs of b and c in the equation. If the signs of b and c are the same the sign of x = 1 is always negative. Now this technique requires lots and lots of practice. I don’t expect you to master this soon but if you do, it will be a great asset to you in the S.A.T.

Advance Class Part A: Completing the Square

2008-04-06T17:03:00.000-07:00

Today we will look at 2 approaches on how to solve quadratic equations. One is a brand new method and the other one is a visual method which I developed through years of practice and experience. Also I am going to show you a very advance method on how to check your answers. I didn’t show you this method before because I wanted you to learn the correct method first and them my shortcuts later on; when you are comfortable with any quadratic equation.

Firstly let’s learn this new method; it is called completing the square. This method is not advisable to use on the S.A.T, but I am teaching people who are also going to school and have problems in mathematics. Some questions in math involve completing the square. Normally these quadratic equations cannot factorize. Let’s look at a question; 2x^2 – 5x + 1 = 0

to begin completing the square you must divide the entire equation by the co efficient of x^2; in this case its 2, so divide throughout by 2.

Dividing by two gives: ………>>> x^2 – (5/2)x + (1/2) = 0

Now have only terms of ‘x’ on the L.H.S.

x^2 – (5/2)x = -(1/2)

Now you must add half the co efficient of ‘x’ and then square that answer to both sides of the equation.
In this case the co efficient of ‘x’ is (5/2), half of that is (5/4) and then that answer squared is (5/4) ^2.

this gives..>> x^2 – (5/2)x + (5/4) ^2 = - (1/2) + (5/4)^2

By doing this step above you have made the L.H.S a perfect square and it can be written in the form (x + (b/2)) ^2. b in this case is – (5/2), so the L.H.S can be written as (x – (5/4) ) ^2.

The equation is now….>> (x – (5/4) ) ^2 = - (1/2) + (5/4)^2
……………………….>> (x – (5/4) ) ^2 = (17/16)
Square root both sides..>> x – (5/4) = +/- sqrt (17/16)
………………………..>> x – (5/4) = +/- (4.123)/4
………………………..>> x = (5/4) +/- (4.123)/4
………………………..>> x = 2.28 and x = 0.22

In my next post I will teach you the cool trick to check your answers and the visual method I perfected!

Simultaneous Equations Part C: Linear and Quadratic Solving!

2008-03-30T09:14:00.000-07:00

Today I am going to do simultaneous equations again. So you know how to do the ‘elimination method’ and the ‘substitution method’ and you also know how to solve simultaneous equations when you are given a pair of linear equations but what about if you were given to solve a pair of simultaneous equations which has a linear equation and a quadratic equation?? What will you do?? Simple the way to solve it is exactly the same as how to solve a pair of linear equations simultaneously! The only catch is that we only use the substitution method. We can also use the ‘elimination method’ but it is much more feasible to use the ‘substitution method’. Let’s look at an example.

2x^2 + 3y^2 = 21……….eq’n 1
3x + 2y = 7………….…..eq’n 2

First step: Use the linear equation (eq’n 2) and make any unknown (x or y) the subject of the formula. Ok let’s do x first!

3x + 2y = 7
3x = 7 – 2y
x = (7 – 2y)/3

Ok let’s do y

3x + 2y = 7
2y = 7 – 3x
y = (7 – 3x)/2

For this question I will substitute y = (7 – 3x)/2 …..this is because you should always use the denominator with the lowest value.

Second step: sub. y = (7 – 3x)/2 into eq’n 1

2x^2 + 3y^2 = 21
2x^2 + 3( (7 – 3x)/2)^2 = 21
Do you remember how to expand (a - b)^2 ?? Well if you do expand it just like (7 – 3x)^2..but we are squaring (7 – 3x)/2……so jus square the top first to get …..49 – 42x + 9x^2 and now square the bottom……2^2 = 4.

Third step: re-write the question with the substitution of the unknown so that there is only one unknown in the quadratic equation.

2x^2 + 3( (49 – 42x + 9x^2)/4)) = 21

Now this can be solved just like a normal quadratic equation. REMEMBER YOU MUST GET 2 ANSWERS FOR X BECAUSE IT IS A QUADRATIC EQUATION!!!

By solving you get x = 3/5 and x = 3. Now you must put each value you got for x into the linear equation to solve for y.

When you do the substitution, you get y = 13/5 when x = 3/5 and y = -1 when x = 3.

NOTE: YOU GET 4 ANSWERS FOR A LINEAR AND A QUADRATIC AND 2 ANSWERS FOR A LINEAR AND A LINEAR

Believe me, the S.A.T questions are so different, but to do it you still the need to have the knowledge to do the questions. On my next post I will show you how the S.A.T brings their questions. They either give you it in written form which will be a long question to read or a visual form…we will look into more details in this, for now good luck learning simultaneous equations because that’s all to it.

Simultaneous Equations Part B: Elimination Method

2008-03-27T10:34:00.000-07:00

Hi again, last time we talked about solving linear equations simultaneously by the method of substitution. Now I am going to teach you the other method: elimination. As I said before any method can be used to solve any simultaneous equation so I am going to use the same question to demonstrate the “elimination method”. The trick to it is knowing which method is most suitable to use, usually it is the elimination method, but the safer method to use is the “substitution method”. This is because the “substitution method” is less liable to encounter errors. Let’s look at the same question as before:

y = -x + 8…………………eq’n 1
4y = 3x – 24………………eq’n 2

Elimination means getting rid of one term, either the x or y. The elimination process must occur only when the same term exist in both equations. What do I mean by this?? Let’s take a look. Let’s eliminate the y-term.

First step: multiply equation y = -x + 8 by 4. By doing this all terms in this equation must increase by a factor of 4. The equation now becomes 4y = -4x + 32…..let’s call it equation 3. Note that the equation has not been altered by its numerical value.

Second step: Minus equation 2 from equation 3. This is where a lot of people encounter difficulties, but those with difficulties at this point, I will teach you a nice technique to solve your problems. All you have to do is change all the signs of all the terms of the subtractor.

4y = -4x + 32……………..eq’n 3
(-) = (-) + (-)
4y = 3x – 24………………eq’n 2

4y - 4y = 0, - 4x - 3x = -7x and 32 – (-24)= 32 + 24 = 56…….by subtracting both equations I eliminated the y-term.
4y = -4x + 32……………..eq’n 3
(-) = (-) + (-)
4y = 3x – 24………………eq’n 2
0 = -7x +56
7x = 56
x = 8

Now substitute x = 8 into the equation y = -x + 8 which results to y = 0.

What if you were given a pair of simultaneous equations like the one below:

2y = 3x +7………….eq’n 1
5y = 15x + 10………eq’n 2

To do elimination we must get the same term in both equations. So multiply equation 1 by a factor of 5 and multiply equation 2 by a factor of 2.

2y = 3x +7………….eq’n 1……..x 5
5y = 15x + 10………eq’n 2……..x 2

10y = 15x + 35……..eq’n 3
10y = 30x + 20……..eq’n 4

Now we have the same y-term in both equations now repeat the same process as above.

Simultaneous Equations Part A: Substitution Method

2008-03-25T09:16:00.000-07:00

Today I am going to talk about solving simultaneous equations. Simultaneous equations can appear in two forms: linear and linear or linear and quadratic. Luckily for you, the S.A.T only gives linear and linear, however quadratic and linear may also come but the method of solving them on the S.A.T is based on a graph, so it is a lot easier than solving it the proper way. There are two methods of solving simultaneous equations: substitution or elimination. First I shall demonstrate the substitution method.

Consider the linear equations

y = -x + 8…………………eq’n 1
4y = 3x – 24………………eq’n 2

ok, in equation 1 , y = -x + 8, and in eq’n (equation) 2 there exist a term 4y. So substitute y = -x +8 into the term 4y in eq’n 2.

y = -x + 8
4y = 3x – 24

substitute , y = -x + 8 into eq’n 2

4(-x + 8) = 3x – 24…………………now expand this
-4x + 32 = 3x – 24………………….now put like terms on one side
24 + 32 = 3x + 4x…………………...now simplify
56 = 7x……………………………...now solve for x
8 = x………………………………...now sub x into eq’n 1 to get y
y = -(8) + 8
y =0

Therefore the answer for the simultaneous equations are x = 8 and y = 0. But what this really means??? It means that if these two lines were drawn on graph paper, the answer gives the point of intersection. For these two linear equations, the point (8,0) is their point of intersection. This is how the basic simultaneous equations are solved, but the S.A.T doesn’t give a question based only on a quadratic equation. Take a look at this actual question from a S.A.T book.

Line q is given by the equation y = -x + 8, and line r is given by 4y = 3x – 24. If the line r intersects the y-axis at point A, line q intersects the y-axis at point B, and both lines intersect the x-axis at point C, what is the area of the triangle ABC?

Luckily for you it is the last question in the section, and this is as hard as it comes, but to tell you the truth this question is so easy with my technique. Unfortunately this topic covers 3 areas in mathematics, simultaneous equations, area and coordinate geometry. The technique I possess is in the coordinate geometry aspect so I will refer to this exact question when I am doing coordinate geometry.

Solving this question above only gives you point C. when I am doing coordinate geometry I will solve this question. Ok now it is time to look at the other method of solving simultaneous equations: elimination.

Factorization

2008-03-22T06:46:00.000-07:00

Factorization is the most foundational topic in algebra. Every Quadratic Equation requires factorization in it. Today I am going to show you how to factorize and the deadly error most people do when factorizing. Consider the linear equation

2acx – 6acy
To factorize all you have to do is take out all the common terms existing in both terms (at the same time) and bracket the remainder of the sum.

2ac(x – 3y)………………that’s the final answer.

Consider the equation……6qp^2 +12pq^2

What are the common terms? 6 occurs in both, only one q occurs in both and only one p occurs in both. Hence the factors are 6,p and q.

6pq(p + 2q)………………..that’s the final answer.

Consider the equation………bc^3 – b^2c^2
what are the common terms? b occurs in both and not c, but c^2 occurs in both. Hence the common factors are b and c^2. You must take out the greatest factor occurring in the terms of the Equation.

bc^2( c – b)………………….that’s the final answer.

That’s about it when factorizing, now I am going to tell you what error to avoid. Consider the equation…….2x^2 – 16x = 0

…………………2x^2 – 16x = 0

divide by 2……... x^2 – 8x = 0
divide by x……… x – 8 = 0

solve for x……….x = 8

This is the biggest error you can make in factorizing. Did you notice my error? Did you notice it was a quadratic equation? If you did realize, where are my 2 solutions for x???

This is the correct and only way to solve the same equation above.

………………2x^2 – 16x = 0

Factorizing…… 2x(x – 8) = 0

Solve for x……. x = 0 and x = 8

Never divide an equation by a letter term, you can only divide by a number term provided only and only if the equation is equal to zero. Never in any point in time divide by x. You will lose a term x = 0

In the S.A.T the exam does not give you factorizing questions, instead they give you questions that involve a little factorizing such as solve for x in the equation…….x^2 + 5x – 6 = 0

This concludes factorization, I hope my viewers understand every step!

Quadratic Equations part C ( Perfect Squared Quadratic Equations)

2008-03-19T08:42:00.000-07:00

In quadratic equations you must get two values for x, this does not necessarily mean that you must get two different values for x. There is only one exception. It is true that you must get two different values for x but this does not apply to Perfect Squared Quadratic Equations. In Perfect Squared Quadratic Equations (p.s.q.e. for short) you get 2 values but 1 unique solution. What do I mean by this?

Consider the example taken from Quadratic Equations part A (2 posts down)>>>>> 3x^2 -7x -6 =0. This is an ordinary quadratic equation. The solution gives 2 values of x and two unique solutions: x = -2/3 and x = 3.

Now consider the (s.p.q.e) >>>>> x^2 + 4x + 4 =0.
by implementing the split number technique, you will get a result of >>>>>>x^2 + 2x +2x +4 = 0
Factorize> x(x + 2) + 2(x + 2) = 0
Simplify> (x + 2)(x + 2) = 0
Solve>>>> x = -2 x = -2…(2 solutions for x,but 1 unique answer)

The only obstacle in your way is, how can you tell which equations are (p.s.q.e). Consider the form of a quadratic equation:

ax^2 + bx + c = 0

when a = 1 ….as long as (b^2)/4 = c………the quadratic equation is a (p.s.q.e). if a =/= 1 all you have to is divide every term by a!
Consider the equation:

3x^2 - 15x + 18.75 = 0………….a = 3 …then divide by 3
You get: x^2 – 5x + 6.25 = 0……then
use the (b^2)/4 = c, condition. 25/4 = 6.25 ….BINGO!

It is a (p.s.q.e), perform the number split technique and you got your answer. Most people would not recognize it as a (p.s.q.e), which is where practice comes into play. You need to build up your mental to recognize this. If you are now beginning to learn quadratic equations I advise you to do the quadratic formula to solve this one.

Wait! There is an easier way to solve (p.s.q.e). when you have confirm that a quadratic equation is a (p.s.q.e) by ensuring the condition (b^2)/4 works you can solve it in that same line. If you are doing an exam for school you must show the steps to earn marks but in the S.A.T, they don’t watch your steps, they just want the answer, so let us give them the answer. This is to save time in the exam. When you have a (p.s.q.e) the answer for x is always (-b/2). Yes, it is that simple. The only way to master the S.A.T is to master time saving techniques. The Math S.A.T is easy, it is metal stamina is your real test! That concludes Perfect Squared Quadratic Equations

Quadratic Equations part B ( the formula)

2008-03-16T09:48:00.000-07:00

Welcome to quadratics again. Now I am going to talk about the rare case I know where the split numbers technique cannot be used. Consider the quadratic equation:

3x^2 - 7x + 1 = 0……….we do the product and sum condition as in the previous topic below.

product…..condition….(a x c) sum …condition…..(a + c) = b

1. (+1) x (+3) = +3…………………..(+1) + (+3) = +4

This is the only factor of 3, thus we must find another way to solve it as the sum condition did not satisfy the number splitting condition. The only way to solve this is by using the quadratic formula:

x = ( -b (+/-) sqrt (b^2 -4ac) )/2a

b = -7 a = 3 c = 1 ……..just substitute the values in the equation and you will get x = 2.18 and x = .15. You must be wondering how I got two values for x! In the formula above, do you see a (+) above a (-)? To get two values work out x using one sign at a time. x = 2.18 when I use the (+) sign and x = 0.15 when I used the (-) sign.

This formula can be used to solve ANY quadratic equation. The only problem is that it is time consuming. It is advisable to use the number splitting technique if it is possible. When working with huge numbers such as a = 6, b = - 31 & c = 18, you should use the quadratic equation because although it may be able to number split you would have probably spent too much time trying to find the condition line such that a + c = b.

NEVER USE THE FORMULA FOR PERFECT SQUARE QUADRATIC EQUATIONS. PERFECT SQUARES CAN BE SOLVED IN MERE SECONDS!

Quadratic Equations part A

2008-03-15T09:29:00.000-07:00

Quadratic equations can be puzzling at first, but it is so easy. The only really ‘difficulty’, if you want to believe that there is actually a difficulty, is finding the split numbers. Split numbers is a term I use, so don’t be scratching your head thinking what split numbers are. Well first you have to identify what a quadratic equation is. In “Linear Equation”, the topic below this one, I demonstrated how to identify a quadratic equation. Now let’s look at the typical quadratic equation:

1) 3x^2 - 7x - 6 = 0
To begin any quadratic equation you must have 4 terms in the equation. All quadratic equations you see come with 3 terms like the question above. You have to create another term by doing my split number method. The split number only involves altering the x term (e.g. -7x, as in the question above.)

ax^2 + bx +c = 0

1. the first step is to multiply the coefficient of x^2, a, (in this question, the coefficient is 3) by the constant, c, (in this question, the constant is -6). Ok I am going to put it in letter form so you will not get confuse.
the first step result is ………..a x c₌ac

2. The second step is to number split. You must get the correct numbers so that it satisfies this condition…..a + c = b. so let’s take a look of all the factors of (a x c)
a x c = 3 x (-6) = -18……now let’s look at the factors of -18

product …condition... (a x c) sum…..condition... (b)

1. (+18) x (-1) = - 18..............................(+18) + (-1) = 17
2. (-18) x (+1) = -18……………….............(-18) + (+1) = -17
3. (+9) x (-2) = -18…………………...........(+9) + (-2) = +7
4. (-9) x (+2) = -18…………………...........(-9) + (+2) = -7
5. (+6) x (-3) = -18…………………...........(+6) + (-3) = +3

6. (-6) x (+3) = -18…………………...........(-6) + (+3) = -3

These are all the factors of -18. Now we must look for the line that satisfies the condition …. a x c = b. there is only one line where this condition meets, in any quadratic equation. There is only one rare case know of, which involves the number 6 and in my next session I will talk about it in more details. In this case the only line which satisfies the condition and its line 4. In line 4, the product is -18 and the sum is -7, exactly what we need.

3) *Using the sum condition in line 4, replace -7x with (-9 +2)x, solve for x ( you must get two values). Now we have completed the number splitting, now rewrite the equation. NOTE: the equation has not been altered in terms of its numeric value, it is the exact equation. The equation before was

3x^2 -7x -6 =0………now it is…..3x^2 + (-9 + 2)x - 6 = 0
…………………………………............3x^2 – 9x +2x – 6 = 0

Now factorize…………………........3x(x – 3) + 2(x – 3) = 0

Now group…………………….........(3x + 2)(x – 3) = 0
Now solve individually…………...(3x + 2) = 0 (x – 3) = 0
(solve it like a linear equation)……………x = -2/3……x = 3

You now know how to solve quadratic equations. Now you need practice. For those who don’t know how to factorize look for future topics on my site, so you can learn, then you can fully understand this topic.

*DO NOT ADD BACK (-9 +2) TO GET BACK (-7) AS YOU WILL DEFEAT THE PURPOSE OF NUMBER SPLITTING, YOU SPLIT (-7) TO GET (-9 +2) SO THAT YOU HAVE FOUR TERMS IN THE QUADRATIC EQUATION.

Linear Equations

2008-03-12T12:01:00.000-07:00

Today’s topic deals with how to solve linear and quadratic equations. Before we begin this topic, I will first teach you how to identify a linear and a quadratic equation. A linear equation comes in the form, ax + b = c ……where a, b & c are any value. The main idea of identifying a linear equation is that, the highest power of x existing in the equation must be 1. A quadratic equation comes in the form ax^2 +bx + c = d ……where a, b, c & d are any value. The main idea of identifying a quadratic equation is that, the highest power of x existing in the equation must be 2.

Now let’s learn how to solve linear equations. This is easy when compared to solving quadratic or even cubic equations. Luckily for you, the S.A.T doesn’t give you any cubic equations to solve, but the quadratic equations are popular.

1) 2x +3 = 9……..what is x?

That question is simple. All you have to do is carry all the numbers on one side and keep x on the other side. So it look like
2x = 9 – 3 …….2x = 6………x = 3
but the questions in the S.A.T have a little addition.
1) 2x +3 = 9……..what is 4x – 3 ?.......you can solve for x like above and then substitute it into 4x – 3 to get 9, but why solve for x? They asked for 4x – 3. Let me show you my technique. As the above, when you got 2x = 6, multiply both sides by 2 so that you get, 4x =12….. and then minus 3 and you get 9. As I told you before a large percentage of the Math in the S.A.T doesn’t require you to solve for x. You have to manipulate the question for you to gain time in the exam. Linear equations appear in the first 5 questions of the exam.

2) 6 + 5(2x + 3) = 31…..another linear equation.

First you must expand the brackets and then solve it just like above……6 + 10x + 15 = 31
10x = 31 – 15 – 6
10x = 10

x = 1

That’s about it for linear equations. Wasn’t that easy? Now we are going to take a look at handling quadratic equations.

Welcome to my first lesson on Mathematics tutoring.Algebra Expansion part B

2008-03-09T07:28:00.000-07:00

Manipulating algebra, especially in your mind, can be a little demanding at first but eventually you will get so good at it that you may be even able to help your colleagues and siblings. For this topic all you have to do is memorize only two lines:

(x – y)(x + y) = x^2 – y^2…………that’s one, the other is

(x – y)^2 =/= x^2 – y^2………….(=/=) means not equal to

I wrote a topic on the “ 6 Deadly Sins to Avoid on the day of the S.A.T.” , if you take a look at example 4 , there is a question;

x^2 – y^2 = 77 and x + y = 11. What is the value of x?

You can do the long way of substituting y= 11 – x, then substituting it into the other equation so that it becomes

x^2 - (11 – x) ^2 = 77, then solve for x,

x^2 – 121 + 22x – x^2 = 77

22x = 198…………thus x =9

This was the time consuming way. Now instead of doing it this way, we are going to implement the technique of “the difference of two squares” method.

x^2 – y^2 = 77…but we know that x^2 – y^2= (x +y)(x – y)

x +y = 11 (it was given)

77 = (x + y)(x – y)……77 = 11(x – y) , thus (x – y) = 7

Now add (x + y) and (x – y) …you get 2x = 18, thus x = 9

This can be done mentally with a lot of practice, took me 20 seconds. Now let’s move on to the final topic for today, learning to save time by plugging sensible values. The question is: If half of a number is equal to 4 more than twice the number, what is the number?

a) -3

b) -8/3

c) 0

d) 7/2

e) 4

Let’s call the unknown number x. So half of the number is .5x. 4 more means 4 +, so for more twice the number means 4 + 2x. So the question in algebraic forms means

.5x = 4 + 2x ……..by just looking at the question you can tell that x is a negative value. So c), d), and e) are out. Now plug in a) – 3 and you will notice it is not the answer, hence b) is the answer. So without working out the question, it was solved by just plugging in one value, a sensible value.

This is a useful strategy when you are running low on time, but by all means take advantage of this technique to gain time. You don’t have to solve all the questions in the S.A.T.; you can use the plug in method, especially when large numbers are involved. Most people start with choice c) for plugging method, but this is not always sensible, just look at the question and choose the most obvious one to you for the easy ones and the most unobvious one for hard ones. The “plug in method” is an effective way to save time, but it is up to you to know when this method is feasible. This concludes my first lesson on algebra.

Welcome to my first lesson on Mathematics tutoring.Algebra Expansion part A

2008-03-09T07:19:00.000-07:00

Welcome to my first lesson on mathematics tutoring. Today I am going to talk about ways in which you can handle algebra without struggling.

1. Learning how to expand “squared brackets” > (a +/- b) easily.

2. Learning how to manipulate squared brackets without expansion by using the concept of the “difference of two

squares”- (a + b)(a – b) and “squared brackets” > (a +/- b)^2

3. Learning how to save time by plugging sensible values.

Let us learn how to expand “squared brackets” easily. First of all you must be able to identify a “squared bracket”. A “square bracket appears in the form …..(a – b)^2 or (a + b)^2. To expand this easily…without doing it the long tiring way like:

(a – b)(a – b) = a^2 – ab –ba + b^2

= a^2 – 2ab + b^2

You may not think it is long to do but in the S.A.T. they use large numbers. Sometimes students make mistakes with the signs especially the one that comes before b^2.

To avoid this learn my technique I used. Some students may very well know this already but I am also here to help students who are weak in algebra.

When handling “squared brackets” always remember this format. Firstly square the first number, (a) becomes (a^2). Secondly multiply the first number (a) and last number (b) then multiply the resulting product by (- 2) if the sum is (a – b)^2 or by (2) if the sum is (a + b)^2. Lastly square the last number, (b) becomes (b^2). After all this is done, add all three steps together so that:

(a – b)^2 = a^2 – 2ab + b^2 and

(a + b)^2 = a^2 + 2ab + b^2

Wasn’t that simple? It looks complicated at first but with a few practice you can store it in long term memory. Okay, this was the first topic on my lesson. Let’s move on to topic 2.

6 Deadly Sins to Avoid on the day of the S.A.T

2008-03-06T14:00:00.001-08:00

Today I am going to tell you what my intentions are. My intentions are to help you in specific areas in the Mathematics section. I will show you all the techniques I possess so that it can be an asset to you on the day of the exam. Time is very essential in this exam and you need as much time you can get to check over any silly errors you made or if you skipped out a question.

Before I begin to show you techniques on the Mathematics, I am going to help you, by showing you, ways in which you can save a lot of time on the exam. Here is a list of the 6 deadly sins to avoid on the day of the S.A.T

1. NEVER read the instructions at the top of the Mathematics section. The rules are exactly the same you read when doing the practice books at home. Reading those instructions can cost you valuable time in the exam, as much as 45 seconds. You may laugh and say what is 45 seconds……..trust me you would not want to waste as much as 20 seconds.

2. NEVER learn the formulas on the day of the exam. Do not even waste time flipping back pages to look for formulas. You must enter the exam room with that knowledge already, especially the special triangles formulae.

3. NEVER hesitate to write all over the S.A.T exam paper. Be free to circle questions you have encountered difficulties with. Questions that you are not sure about, that you think have 2 possible answers and you are not sure which one it is; put an asterix (*) next to both answers on the S.A.T booklet. The reason for this is that, you will save some precious seconds by not trying to remember your two possible answers.

4. NEVER be so hasty to use a calculator. By actually using a calculator, it can cost you some precious time on the S.A.T, this is because 45 %( just an estimation, it can be higher or lower) of the questions can be solved mentally by just looking at it and manipulating the sum in your mind.

e.g. x^2 – y^2 = 77 and x + y = 11. What is the value of x?

You can do the long way of substituting y= 11 – x, then substituting it into the other equation so that it becomes

x^2 - (11 – x) ^2 = 77, then solve for x,

x^2 – 121 + 22x – x^2 = 77

22x = 198…………thus x =9

You probably would have used a calculator and about 2 minutes to do this and 2 minutes is too much to spend on one question.

In my next post I will show you my technique I used, how it required no calculator; just mind manipulation and it took me 15 seconds.

5. NEVER spend more than a minute on any of the first 18 questions. This is because they are relatively easy when compared to the last 5 questions. The last five questions have a lot more reading than the others. Remember all questions carry the same point, so try to get at least the first 22 correct. This is what you need if you want a high score. If you get 6 wrong in the entire Mathematics section, you can get a score such as 700. It is a fair score but if you intend to go to top universities you need score like 760 and above. The higher your scores are the better your chances are in acquiring a scholarship. Students, especially who have difficulties with financial aid, should aim for high scores. Students who need financial help can visit here www.snagem19.collegehg.hop.clickbank.net for more details. Always remember time is your friend in this exam.

6. NEVER forget to check over your work even if you have a few seconds on the clock remaining. That few seconds of checking over can save you some points if you can spot the error in time.

This concludes the deadly sins of the S.A.T, now let's begin to learn my techniques.

Elements to consider when doing the SAT

2008-03-04T13:32:00.000-08:00

The SAT exam can be intimidating but it is easy as you think it is hard. The SAT is just an ordinary exam so why should you be afraid of it; yes of course the SAT scores determine what future university you plan to attend but in order to accomplish this feat you have to master the elements of the SAT. In this section I would show you how to master the math aspect of the SAT exam. These are the elements I mastered for me to master the SAT Math section.

1. Basic knowledge

2. Practice

3. Distractions

4. Analyzing

5. Over confidence

Are you familiar with this list? I am sure you are acquainted with this list but are you acquainted with all? The SAT Math comprises of all the basic knowledge you covered all your life; Area, Geometry, Ratios etc., but the SAT has a sneaky way of turning all your knowledge you possess to all the knowledge you think you don’t possess. To prevent this you have to practice not only on the Math itself but also on the time. Actually the time was too much for me exactly....I completed the Math sections with a lot of time to spare. Your mind has to be very determined to accomplish this. You must be thinking it’s impossible but I will show you eventually how I did it. You think its best to study in a quiet environment to avoid distractions, well you are so wrong. You should get accustomed in studying in an environment with mild noise. This is because you are distracted easier when you study in a too quiet because you get accustomed to the quiet environment. Studying with music, especially music you like, will and is the most distracting factor. You may not realize but it is because your mind is actually consuming the song rather than the Math. Your brain triggers your memory function and you sing the song in your mind rather than do the Math…am I right??? Yes I am because it happened to my colleagues and I. In the future I will show you how to master the Math SAT. The SAT questions are very simple but you wonder to yourself, how you got such a low score. Easy answer, you didn’t solve for what the SAT asked for. An example of a question is…..how many square numbers between 1 and 26?

a) 3

b) 4

c) 5

d) 6

e) 7

If you chose c)5 ,then you are wrong. The correct answer is b)4 . This is because the SAT asked you for the number of square numbers between 1 and 26…which are 4, 9, 16 & 25. The fact that they did not end the question with the word “inclusive” omits the square number 1. Proper analysis of the SAT questions can earn you a higher score.

Over confidence is an emotion that 80% of test takers fail to realize. You go with a mentality that you know everything, but somewhere, somehow you obtain a score you didn’t deserve. This is a battle you have to fight on your own.

Well this concludes the end of my introduction. I hope you blogger readers can find my articles interesting and worth your while. Now let’s begin to master the SAT Math…

an introduction of the format of the SAT

2008-03-03T17:01:00.000-08:00

Hi again, before we start I am going to share some basic information about the SAT. This information can be useful especially to students who are new to the SAT. The SAT test is given seven times a year in the United States and around the world, but the exam is one day in the month and on that scheduled day everyone does the exam at the same time. Yes it’s true. This is because the SAT wants everyone to have an equal chance, meaning that no one from another country can help someone from another country. There are two ways to register for the SAT test, online or by mail. For student who don’t reside in the United States, it is advisable that your register online. Online registration is completed on the website of the College Board. Go to www.collegeboard.com for more details.

The SAT can be one of the longest exams you have ever taken in your life at a student’s level. It is on average about 3 hours long. That’s exactly what I said, 3 hours long with two 10 minute breaks and they are 2 hours apart. Well most of you are aware of the format of the SAT. The essay part of the writing section is always first on the test, and the multiple choice part of the writing section is always last on the test. The critical reading section of the SAT test consists of two 25 minute sections and one 20 minute section. The mathematics section of the SAT test consists of two 25 minute sections and one 20 minute section. The writing section of the SAT test contains a 35 minute multiple choice section and a 25 minute essay section.

Now the SAT is not just an ordinary test unlike your normal examinations in school when pertaining to the rules. The rules of the SAT are very strict. Not conforming to these rules can lead you from disqualification not only from that test but future tests. Obey all rules if you are serious about the SAT.

an intoduction to my readers

2008-03-03T15:49:00.000-08:00

Hi students, my name is Mark Richards. I live in the Caribbean on an island call Trinidad. I am 19 years and I am now beginning to start my future career of becoming an electrical and computer engineer in university. In my spare time I like to read the newspapers, play video games (mostly role playing games), playing chess, lime with my friends and teach people like you. I like to help students who are struggling on academics especially when it involves Mathematics.

Mathematics is my passion. Some people may think I sound like a geek with hobbies like chess and playing games, I do it for a reason. Chess is a mental game. It requires a lot of concentration and it builds up your mind stamina. I got so involved that I was Trinidad and Tobago’s Under 17 champion for 3 years. For all chess players who are reading this article, playing chess will be a great asset for your mind, especially when you are planning to do the SAT.

I did the SAT 1 and SAT 2. My score on the SAT 1 was 1980.

I got 800 in Math, 590 in Critical Reading and 590 in Writing. On the SAT 2, I got 2000. I got 720 in Math1, 640 in Math2 and 630 in Physics. The SAT was the most difficult and time consuming examination I ever took, so far in my life.

I am here to help you in the Math section. A lot of students get a lot of problems in the Math aspect of the SAT and I am here to share all my techniques I possess, to help my fellow students out. Some of you are very new to the SAT so I am just going to give you a little introduction on the SAT first.