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Formulae and Shortcuts - making life easier!


Hermione

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couple more random facts

 

 

number of squares in a checkerboard = sum of squares of dimensions

eg squares in 4x4 board

= n(n+1)(2n+1)/6 == 4*5*9/6 = 30

 

number of rectangles in checkerboard.

= C(n+1, 2) * C(m+1, 2) where m X n dimensions

 

note: squares is the sum of squares and rectangles is the sum of cubes (easier to do it with combinations though)

 

 

number of cubes with various numbers of sides painted after broken to individual unit cubes

 

number with no paint = (n-2)^3

number with 1 side painted = 6(n-2)^2

number with 2 sides painted = 12(n-2)

number with 3 sides = 8 (always) all eight corners

 

derived nicely from the cubic equation n^3 = (n-2)^3 + 6(n-2)^2 + 12(n-2) + 8

 

Derangement combinatrics.

answers the question of number of ways things will be mismatched.

 

ex) there are 4 couples in a dance. how many ways can they be paired up such that no husband is with his wife?

4!( 1/2! - 1/3! + 1/4!) = 9

 

equation = n! ( alternating difference and sum of reciprocals until you reach n)

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Mixtures first....

 

1. when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing will be:

Vm = (v1.n1 + v2.n2) / (n1 + n2)

 

you can use this to find the final price of say two types of rice being mixed or final strength of acids of different concentration being mixed etc....

 

the ratio in which they have to be mixed in order to get a mean value of vm can be given as:

n1/n2 = (v2 - vm)/(vm - v1)

 

When three different ingredients are mixed then the ratio in which they have to be mixed in order to get a final strength of vm is:

n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (v2 - vm)(vm - v1)

 

2. If from a vessel containing M units of mixtures of A & B, x units of the mixture is taken out & replaced by an equal amount of B only .And If this process of taking out & replacement by B is repeated n times , then after n operations,

 

Amount of A left/ Amount of A originally present = (1-x/M)^n

 

3. If the vessel contains M units of A only and from this x units of A is taken out and replaced by x units of B. if this process is repeated n times, then:

 

Amount of A left = M [(1 - x/M)^n]

 

Ths formula can be applied to problem involving dilution of milk with water, etc...

 

(All these formulae have been taken from a test prep book issued by a private coaching center) ;)

 

is there any rule to assign smaller value to v1 and greter value to v2 and the greatest to v3??? :rolleyes:

 

Fahad

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For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.

 

For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)

 

Thanks in advance

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Guys, great job ... this thread is exactly what i was looking for and so far have found only here. many thanks to you all.

 

Got it from this blog: Divisibility Rules « TutorTeddy BLog

 

 

Let us take up an example. Consider the number 150. Let us find out if it is divisible by the other numbers:

 

150 is divisible by 2 because the last digit is 0.

150 is divisible by 3 because when we add the digits (1+5+0), we get 6 which is divisible by 3.

150 is NOT divisible by 4, since the last 2 numbers i.e., 5 and 0 are not divisible by 0.

150 is divisible by5 because the last digit is 0

150 is divisible by 6 because it is divisible by both 2 and 3

150 is not divisible by 7 because when we double 0 we get o again. If we subtract that from 15, we get 15, which is not divisible by 7.

150 is not divisible by 8 because the three digits 1, 5 and 0are not divisible by 8.

150 is NOT divisible by 9 because the sum of the digits 1+5+0=6, which is not divisible by 9.

150 is divisible by 10 because the last digit is 0.

 

 

Further Tips on Divisibility:

(i) A number divisible by 2 or 5

Any number ending in 0 or an even number is divisible by 2 e.g. 12, 256, 328, 2060. If the last digit of a number is 0 or 5, that number is divisible by 5 e.g. 150, 2025, 3175.

(ii) A number divisible by 4 or 25

Any number is divisible by 4 if the last two digits are divisible by 4, e.g. 132, 5276, 208. Similarly if the last two digits of a number are divisible by 25, that number is divisible by 25, e.g. 1375, 2500.

(iii) A number divisible by 8 or 125

For a number to be divisible by 8, its last three digits must be divisible by 8, e.g. 864, 1248, 3000. Similarly numbers ending with the last three digits divisible by 125 are divisible by 125, e.g. 4250, 12375, 12000.

(iv) A number divisible by 16 or 625

A number having its last four digits divisible by 16, will be divisible by 16, e.g. 31776, 28528. Similarly numbers having their last four digits divisible by 625 are divisible by 625, e.g. 83125, 125000.

(v) A number divisible by 3 or 9

If a number is divisible by 3, the sum of the digits is divisible by 3, e.g. 38451, 285612.

In the above numbers, sum of digits 3+8+4+5+1 = 21 and 2+8+5+6+1+2 = 24, which are all divisible by 3. Hence, the numbers are divisible by 3.

In a similar manner, a number is divisible by 9, if the sum of all its digits is divisible by 9, e.g. 1548, 653229.

In the above numbers, the sum of digits = 1+5+4+8 = 18, 6+5+3+2+2+9 = 27. Hence, the numbers are divisible by 9

(vi) A number divisible by 11

A number is divisible by 11 if the difference between the sum of the digits in the even places and the sum of the digits in the odd places is either 0 or a number divisible by 11, e.g. 65274, 538472.

In 65274, difference in the sum of numbers in the odd places and the sum of numbers in the even places = (6+2+4) – (5+7) = 0

In 538472, difference in the sum of numbers in the odd places and the sum of numbers in the even places = (5+8+7) – (3+4+2) = 20 – 9 = 11, a multiple of 11.

Hence the two numbers are divisible by 11

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I have organized all the formulas from this threat into the OneNote format, where it is easy to navigate by topic. Unfortunately, I can't attach it here for some reason, so if you need it..write me a msg with your email. i'll send it you. it's not perfect, but i like it more than MS Word format

 

wishes

Kseniya

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I have organized all the formulas from this threat into the OneNote format, where it is easy to navigate by topic. Unfortunately, I can't attach it here for some reason, so if you need it..write me a msg with your email. i'll send it you. it's not perfect, but i like it more than MS Word format

 

wishes

Kseniya

 

Can you email me at CONTACT DETAILS DELETED BY MODERATOR. Please use Test Magic messaging system.

Edited by wasleys
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If “a” men or “b” women can do a piece of work in x days, then “m” men and “n” women can together finish the work in (abx)/(an+bm) days. But what will be the formula when there will be three persons,suppose men,women & boy . I am giving a Question

 

if 1 man or 2 women or 3 boys can do a piece of work in 44days,then the same piece of work will be done by 1man , 1 woman and 1 boy

TIME AND WORK

 

 

 

  • If A can do a piece of work in x days, then A’s one day’s work=1/x
  • If the ratio of time taken by A and B in doing a work is x:y, then, ratio of work done is 1/x :1/y=y:x. And the ratio in which the wages is to be distributed is y:x
  • If A can do a work in x days and B can do the same work in y days, then A and B can together do the work in (xy)/(x+y) days
  • If “a” men or “b” women can do a piece of work in x days, then “m” men and “n” women can together finish the work in (abx)/(an+bm) days
  • If A is x times efficient than B, and working together, they finish the work in y days, then Time taken by A=y(x+1)/(x), Time taken by B=y(x+1)
  • If A and B can finish a work in “x” and “ax” days respectively, that is if A is “a” times efficient than B, then working together, they can finish the work in (ax)/(a+1) days
  • If A and B working together can complete a work in x days, whereas B working alone can do the same work in y days, ten, A alone will complete the work in (xy)/(y-x) days.
  • Pipe A can fill a tank in x hrs and B can empty a tank in y hrs.If both pipes are opened together, the tank will be filled in (xy)/(y-x) hrs
  • A pipe can fill a cistern in x hrs but due to leakage in the bottom, it is filled in y hrs, then the time taken by the leak to empty the cistern is (xy)/(y-x) hrs

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If “a” men or “b” women can do a piece of work in x days, then “m” men and “n” women can together finish the work in (abx)/(an+bm) days. But what will be the formula when there will be three persons,suppose men,women & boy . I am giving a Question

 

if 1 man or 2 women or 3 boys can do a piece of work in 44days,then the same piece of work will be done by 1man , 1 woman and 1 boy

 

I'm not a fan of formulas, but if you are going to use them then its sometimes important to know how these formulas are derived. The way the original formula was derived was considering that 'a' men do 1/x work in a day. So each man does 1/(a*x) work in a day. Similarly, each woman does 1/(b*x) work in a day. Therefore, 'm' men do m/(a*x) work per day and 'n' women do n/(b*x) work per day. It will therefore take:

1 work / ( m/(a*x) work/day + n/(b*x) work/day) = abx / (an + bm) days to complete 1 piece of work. Notice that I skipped the simplification.

 

Extending this with boys, lets assume 'c' boys do a piece of work in x days also. So 'c' boys do 1/x work in a day. And each boy does 1/(c*x) work in a day. Therefore, r boys do r/(c*x) work per day. It will therefore take:

1 work / ( m/(a*x) work/day + n/(b*x) work/day + r/(c*x) work/day). Again skipping the simplification, this equates to:

a*b*c*x / (a*b*r + a*c*n + b*c*m) days

 

Without using the * for multiplication (just to make it look neater) its: abcx/( abr + acn + bcm )

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