Math Tricks 1...squared numbers

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

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

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

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

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

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

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.