Formulae and Shortcuts - making life easier!

[Hermione]

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) ;)

[yasseramin]

thanxxxxxxxxx

 

 

[clap] [clap]

[Hermione]

Lets make this thread an archive of formulae. (I dont know if there already exists one in the forum. I havent gone through all previous mails)

 

Everybody make an attempt to post any formula or short cut that you have got. Be it a standard formula that everybody knows or a rare one. coz there are people in this forum who are from non-math background and they might find it useful. (even if you are not sure how a formula works, as long as you know the formula does work for all cases, post it). So, lets get the ball rolling...

 

P.S. for any clarifications regarding how a formula works, start a new thread and let this thread be exclusively for listing formulae. that way we can reduce the length of this subforum and enable those, who will probably read this as a last minute revision, to spend lesser time...

[manasi4gre]

Thanks Suja.:)

[rkhanna]

I agree with you , we should have separate thread on formuls for the last minute glance before the real gre.

[manwiththemission2005]

Thanks a lot Suja. I hope this thread goes ROCKING jus like the PROBABILITY thread which I started b4. :)

[Hermione]

More formulae:

 

PROGRESSION:

 

Sum of first n natural numbers: 1 +2 +3 + .... + n = [n(n+1)]/2

Sum of first n odd numbers: 1 + 3 + 5 + .... upto n terms = n^2

Sum of first n even numbers: 2 + 4 + 6 + ... upto n terms = n(n+1)

ARITHMETIC PROGRESSION

 

nth term of an Arithmetic progression = a + (n-1)d

Sum of n terms in an AP = s = n/2 [2a + (n-1)d]

where, a is the first term and d is the common differnce.

 

If a, b and c are any three consequtive terms in an AP, then 2b = a + c

GEOMETRIC PROGRESSION

 

nth term of a GP is = a[r^(n-1)]

sum of n terms of a GP:

s = a [(r^n - 1)/(r-1)] if r > 1

s = a [(1 - r^n)/(r-1)] if r

 

sum of an infinite number of terms of a GP is

s(approx.) = a/ (1-r) if r

 

If a, b and c are any three consequtive terms in a GP, then b^2 = a + c

 

PS: Everybody!!! please contribute or this thread will never grow.

[gre1600]

suja great job :tup:

[manwiththemission2005]

Problems on trains

 

 

 

 

 

 

 

a km/hr = (a* (5/18)) m/s

 

a m/s = (a* (18/5)) km/hr

 

 

 

 

Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

 

 

 

 

Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l+b) metres.

 

 

 

 

Suppose 2 trains or 2 bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relative speed = (u-v) m/s

 

 

 

 

Suppose 2 trains or 2 bodies are moving in the opposite direction at u m/s and v m/s, where u>v, then their relative speed = (u+v) m/s

 

 

 

 

If 2 trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then the time taken by the trains to cross each other = (a+b)/(u+v) sec

 

 

 

 

If 2 trains of length a metres and b metres are moving in same directions at u m/s and v m/s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec

 

 

 

 

If 2 trains(or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then

 

(A's speed):(B's speed) = (root(B):root(A))

 

[Hermione]

Good job manwithmissions!!!

 

TM...wake up. post more formulae...

[Hermione]

This one is a nice formula for finding the number of unique divisors for any number and also the sum of those divisors.... such questions are there in powerprep and so you might also get it in your real GRE.

 

If N is a number such that

N = (a^m) (b^n) (c^p)....

where, a, b, c, ... are prime numbers, then the number of divisors of N, including N itself is equal to:

(m+1) (n+1) (p+1)....

and the sum of the divisors of N is given by:

S = [(a^m+1) - 1]/[a - 1] * [(b^n+1) - 1]/[b - 1] * [(c^p+1) - 1]/[c- 1].....

Example:

for say N = 90, on factorizing you get 90 = 3*3*5*2= (3^2)*(5^1)*(2^1)

then the number of divisors of 90 are (2+1)(1+1)(1+1) = 12

the 12 divisors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

And the sum of the divisors would be

[(3^3) - 1]/[3 - 1] * [(5^2) - 1]/[5 - 1] * [(2^2) - 1]/[2 - 1]

= (26/2) (24/4) (3/1)

= 234

 

Though this method looks more complicated than listing the factors and adding them, once you get used to this formula, it saves lot of time..

[Hermione]

Let the coordinates of P1 be (x1,y1) and of P2 be (x2,y2)

 

- The distance from P1 to P2 is:

d = sqrt[(x1-x2)2+ (y1-y2)2]

- The coordinates of the point dividing the line segment P1P2 in the ratio r/s are:

([r x2+s x1]/[r+s], [r y2+s y1]/[r+s])

- As a special case, when r = s, the midpoint of the line segment has coordinates:

([x2+x1]/2,[y2+y1]/2)

- The slope m of a non-vertical line passing through the points P1 and P2:

slope = m = (y2 -y1)/(x2 -x1)

• Two (non-vertical) lines are parallel if their slopes are equal.
  
• Two (non-vertical) lines are perpendicular if the product of their slopes = -1.
  
• Slope of a perpendicular line is the negative reciprocal of the slope of the given line.
  

[Dimas]

I am impressed.

[apprentice]

Good job suja..This will really help all TMians.. even to just revise before exam..

[coolmavs]

Great job Suja, this thread is a real boon.....especially just before the exam....

[Anand Iyer]

Great work Sujatha! Kudos to you...

 

I posted these formulas as a page at my blog so that it could be useful for future test taker (including me) ;)

[greenhorn]

The population of a town decreases by 'x%' during the first year, decreases by 'y%' during the second year and again decreases by 'z%' during the third year. If the present population of the town is 'P', then the population of the town three years ago was::

 

P*100*100*100

-----------------------

(100-x)(100-y)(100-z).

 

 

The population of a town is 'P'.It decreases by 'x%' during the first year, decreases by 'y%' during the second year and again deceases by 'z%' during the third year. The population after 3 years will be:

P*(100-x)(100-y)(100-z)

--------------------------

100*100*100.

 

If 'X' litres of oil was poured into a tank and it was still 'x%' empty, then the quantity of oil that must be poured into the tank in order to fill it to the brim is:

X*x

------- litres.

100 - x

 

 

 

If 'X' liters of oil was poured into a tank and it was still 'x%' empty, then the capacity of the tank is:

X*100

---------- litres.

100 - x

 

A candidate scoring x% in an examination fails by "a" marks, while another candidate who scores y% marks gets "b" marks more than the minimum required pass marks. Then the maximum marks for that exam =

100(a+b)

----------

y-x .

 

 

 

 

The pass marks in an examination is x%. If a candidate who secures y marks fails by z marks, then the maximum marks is given by

 

100(y+z)

-----------

x.

[Hermione]

1. If one operation can be performed in m ways and another operation in n ways, then the two operations in succesion can be done in m*n ways

 

2. The linear permutation of n distinct objects (that is, the number of ways in which these n objects can be arranged is n! and the circular permutation of n distinct objects is (n-1)! But if the clockwise and anticlockwise directions are indistinguishable then the circular permutations of n different things taken at a time is (n-1)!/2

 

3. But out of these n objects, if there are n1 objects of a certain type, n2 of another type and n3 of another, and so on, Then the number of arrangements (linear permutations) possible is n!/n1!n2!...nz!

 

4. The total number of ways of arranging r things from n things is given by nPr = n!/(n-r)!

 

5. The number of ways to select r things out of n things is given by nCr = n!/(r!*(n-r)!)

 

6. nPr = r! * nCr

[coolgal]

COUNTING

 

SUM OF FIRST “n” NATURAL NUMBERS = n(n+1)/2

Sum of first “n” ODD integers = n*n

Sum of first “n” EVEN integers = n(n+1)

Sum of the squares of the first n integers = n(n+1)(2n+1)/6

Sum of the cubes of first n integers =(n(n+1)/2)^2

 

IF n is even, then

No. of odd no.s from 1 to n is n/2

No. of even no.s from 1 to n is n/2

 

If n is odd then,

No. of odd no.s from 1 to n is (n+1)/2

No. of even no.s from 1 to n is (n-1)/2

 

 

 

POWERS AND INDICES

 

To find the unit digit of p^n

 

If there is an odd no. in the unit place of p eg 741,843 etc

Multiply the unit digit by itself until u get 1.

 

 

Example:

If u need to find the unit digit of (743)^38:

 

Multiply 3 four times to get 81.

 

(743)^38=(743)^36 X (743)^2

 

 

36 is a multiple of 4, and 3 when multiplied 4 times gives 1 in the unit digit.Therefore, when multiplied 9 x 4 times, it will still give 1 in the unit digit.

the unit digit of (743)^38, hence will be 1 x 9 =9

 

In short

(…..1)^n =(…..1)

(….3)^4n=(…..1)

(….7)^4n=(…..1)

(….9)^2n=(….1)

 

If the unit digit of p is even and u need to find the unit digit of (p)^n

 

Multiply the unit digit of p by itself until a 6 is in the unit place

(…2)^4n=(….6)

(….4)^2n=(….6)

(….6)^n=(….6)

(….8)^4n=(….6)

 

 

 

For numbers ending with 1,5,6, after any times of multiplication, you get only 1, 5, 6 respectively.

 

 

 

Number of numbers divisible by a certain integer:

How many numbers upto 100 are divisible by 6?

Soln:

Divide 100 by 6, the resulting quotient is the required answer

Here,

100/6 = 16x6+4 16 is the quotient and 6 is the remainder.

Therefore, there are 16 numbers within 100 which are divisible by 6.

 

PERCENTAGES

 

• If the value of a number is first increased by x% and later decreased by x%, the net change is always A DECREASE= (x^2)/100
  
• if the value of a number is first increased by x% and then decreased by y%, then there is (x-y-(xy/100))% increase if positive , and decrease if negative
  
• If the order of increase or decrease is changed, THE RESULT IS UNAFFECTED
  
• If the value is increased successively by x% and y%,then final increase is given by x+y+(xy/100) %
  

[manwiththemission2005]

You r pretty cool :-) as your name says so

[Hermione]

Great formulae Coolgal and greenhorn! keep posting :tup:

[bscout]

Suppose you have a name with n letters, and there are k1 of one letter, k2 of another letter, and so

on, up to kz. For example, in ELLEN,

 

n = 5, k1 = 2 [two E's], k2 = 2 [two L's], k3 = 1 [one N]).

 

Then the number of rearrangements is n!/k1!k2!...kz!

 

Taking from http://mathforum.org/library/drmath/view/56109.html.

 

This formula was provided by mauveÄster in another thread.

[Hermione]

1. Area of a triangle with base b and height h = (1/2)*b*h

2. The area of an equilateral triangle with side a is [sqrt(3)/4]*a^2

3. The area of any triangle given the length of its 3 sides a, b and c:is sqrt[s(s-a)(s-b)(s-c)] where s= (a+b+c)/2

[puzoo]

 

Hi,

 

This is taken from one of the GMAT websites.

if you click on the link it will actually bring you to the website 4gmat.com

 

 

 

Math-o-pedia: Did you know? Number Theory

 

The product of any three consecutive integers is divisible by 6.

 

Similarly, the product of any four consecutive integers is divisible by 24.

 

 

Math-o-pedia: Did You Know? : Permutation and Combination

 

When n dice (n > 1) are rolled simultaneously, the number of outcomes in which all n dice show the same number is 6, irrespective of the value of n.

 

Similarly, when n fair coins (n > 1) are tossed simultaneously, the number of outcomes in which all n coins turn up as heads or as tails is 1, irrespective of the value of n.

 

 

Math-o-pedia: Did you know? : Speed Time

 

When an object travels the first x hours at p km/hr and the next x hours at q km/hr, the average speed of travel is the arithmetic mean of p and q.

 

However, when the object travels the first x kms at p km/hr and the next x kms at q km/hr, the average speed is the harmonic mean of p and q.

 

 

Math-o-pedia: Did you know? : Number Theory

 

Any perfect square has an odd number of factors including 1 and the number itself and a composite number has an even number of factors including 1 and itself.

 

Any perfect square can be expressed in the form 4n or 4n+1.

 

 

Math-o-pedia: Did you know?: Profit Loss

 

If the selling price of 2 articles are equal and 1 of them is sold at a profit of p% and the other at a loss of p%, then the 2 trades will result in a cumulative loss of ((p^2)/100)%.

 

If the cost of price of 2 articles are equal and 1 of them is sold at a profit of p% and the other at a loss of p%, then the 2 trades will result in no profit or loss.

 

 

Math-o-pedia: Did you know? : Progressions

 

Arithmetic mean of 'n' numbers will always be greater than the geometric mean of those 'n' numbers which will be greater than the Harmonic mean of those 'n' numbers.

 

Arithmetic mean of 2 numbers = geometric mean of '2' numbers = harmonic mean of '2' numbers if both the numbers are equal.

 

[veena]

Great job guys.I am sure all TMians will find this very helpful.

 

Thanks.