Formulae and Shortcuts - making life easier!

[priyankagre2005]

If SD of x1, x2, x3, ... xn is sigma then SD of x1+k , x2+k, x3+k ... Xn+k is also sigma

 

If SD of x1, x2, x3, ... xn is sigma then SD of x1*k , x2*k, x3*k ... Xn*k is k*sigma

 

Variance (kx) = k^2 Variance(x)

 

Binomial probability mass function: P(x) = nCx * p^x * q ^ (n-x)

where x is happening event, n is total number of event, p is probability of happening of event and q is probability of not happening.

[priyankagre2005]

z values that you need to memorize: PICTURE VIEW

 

- Area within .5 Standard Deviation above and below the mean is 38%

- Area within 1 Standard Deviation above and below the mean is 68%

- Area within 2 Standard Deviation above and below the mean is 95%

- Area within 3 Standard Deviation above and below the mean is 99.7

 

- Area below 1 standard deviation is 84%

- Area below 2 standard deviation is 97.7%

- Area above 1 standard deviation is 15.8%

- Area above 2 standard deviation is 2.27%

[sfahad_yunas]

Now thats excellent stuff

very helpful especially when the time is short for preparation.

good work

[Hermione]

Good job priyanka! :)

 

keep posting!

[priyankagre2005]

- If perimeters of a square and parallelogram are equal, then area of a square is always greater than area of a parallelogram.

- Similarly, if perimeters of a square and circle are same, then area of a circle is greater than area of a square.

[manwiththemission2005]

Awesome work priyanka. Keep going. Expectin you to come to WLT after your exam :) Cheers!

[nadia_dencheva]

there is mistake in info for geometric progression b^2=a.b it is not b^2=a+b

[priyankagre2005]

Nadia,

 

That has already been corrected. Check the complete post.

[Hermione]

Nadia,

 

That has already been corrected. Check the complete post.

Thanks for clarifying priyanka. :)

 

Yes. there are mistakes (mainly typos, which can not be avoided) in many formulae in this thread, including the very first formula on mixtures.

 

Post #1 - correction posted in #79

Post #7 - correction posted in #39

post #9 - clarifications posted in #44 and #45

post #61 - correction posted in #67

 

if any of you spot more mistakes feel free to post here and get it clarified. it will help us all.

[sfahad_yunas]

I have compiled all the formulae in a wordfile, thought if someone wanted to print it for themselves (for later reference). All the corrections and clarifications are included.

CopyRights Suja84_ibt and TMians ;)

[Hermione]

Thanks a lot sfahad_yunas! Thats great job :tup:

 

I had long been wanting to do this. But you know, I am quiet lazy! :blush:

[aarushi2001]

I have compiled all the formulae in a wordfile, thought if someone wanted to print it for themselves (for later reference). All the corrections and clarifications are included.

CopyRights Suja84_ibt and TMians ;)

 

thanks!! :tup: :)

[kronique]

I have compiled all the formulae in a wordfile, thought if someone wanted to print it for themselves (for later reference). All the corrections and clarifications are included.

CopyRights Suja84_ibt and TMians ;)

 

Awesome job budds.:tup:

[sfahad_yunas]

Not as good as the contributions made by suja_84_ibt and other TMians. This thread rocks!

[krishnakanthc]

A gold mine for us. Thanks A lot

We can put it for sale on ebay ;) what do you think ?

[bscout]

Posted by sreenivas in this thread.

 

ONE VARIABLE INCREASED/ DECREASED PROBLEMS:

 

PRICE INCREASED AND REDUCTION OF THE CONSUMPTION:

 

1. Price of sugar is increased 25%. How much percent must a house hold must reduce his consumption of sugar so as not to increase his expenditure?

 

how much time u require to this problem? just try this short cut less then 5 sec u will get the answer

 

 

% REDUCTION= (INCREASE/100+INCREASE)* 100

lets try this with short cut

 

increase = 25% so reduction = (25/ 100+25 ) * 100

= (25/125) * 100

= 1/5* 100

= 20 %

so house hold have to decrease 20% of their consuption to keep constant .

 

PRICE DECREASED INCREASE IN CONSUPTION:

 

PROBLEM : 2

Certain familyhave fixed budget for ice cream purchase for year . but,Ice cream price decreased by 20% due to winter season. find by how much % a consumer must increase his consumpion of ice creame so as not to decrease his expenditure.

 

Here the short cut

 

%INCREASE IN CONSUMPTION = (REDUCTION/ 100- REDUCTION) *100

 

 

just as mentioned above

 

 

decrease icecreame price= 20

 

= 20/(100-20) *100

=20/80 * 100

=1/4 *100

= 25%

 

BOTH VARIABLES INCREASED/ DECREASED PROBLEMS

TYPE3:

 

Petrol tax is increased by 20% and the costumer comsumption also increased by 20%.Find the % increase or decrease in the expenditure

OR

Water tax is increased by 20% and consumption also increased by 20% find what is the net effect in change?

 

the short cut: [(A+B) + AB]/100

 

increase A: 20

increase B: 20

= (20+20) + (20*20)/100

= 40+ 400/100

= 40 +4

= 44 % net increase

 

ONE INCREASED ONTHER DECREASED PROBLEMS:

 

Shop keeper decreased the price of a article by 20% and then increased the artical by 30% what is the net effect of the artical is it increased or decreased?

 

SHORT CUT IS SAME AS ABOVE

 

so first decreased the price so we have to take as negative value for A

decrease A : -20%

increase A : 30%

= (-20 + 20)+ (-20)*20/100

= (0) +(-400)/100

= -4%

so net effect is 4% loss to the shop keeper.

[sfahad_yunas]

Sorry, I left out some explaination and formulae. This is the updated alpha version. Report bugs if any one finds :d

[sfahad_yunas]

formulae

[simba]

2 cents from me too............... got it from some other forum dedicated for some other thing.

 

Quantitative Ability – POINTS TO REMEMBER

1. If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.

Eg: x3+3x2+2x+6=0 has no positive roots

2. 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.

3. 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)

4. Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.

5.

ü For a cubic equation ax3+bx2+cx+d=o

· Sum of the roots = - b/a

· Sum of the product of the roots taken two at a time = c/a

· Product of the roots = -d/a

ü For a bi-quadratic equation ax4+bx3+cx2+dx+e = 0

· Sum of the roots = - b/a

· Sum of the product of the roots taken three at a time = c/a

· Sum of the product of the roots taken two at a time = -d/a

· Product of the roots = e/a

6. If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)

7. Consider the two equations

a1x+b1y=c1

a2x+b2y=c2

Then,

ü If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.

ü If a1/a2 = b1/b2 c1/c2, then we have no solution.

ü If a1/a2 b1/b2, then we have a unique solution.

8. Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1

9. |a| + |b| = |a + b| if a*b>=0

else, |a| + |b| >= |a + b|

10. The equation ax2+bx+c=0 will have max. value when a0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a

11.

ü If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.

ü If for two numbers x*y=k (a constant), then their SUM is MINIMUM if

x=y (=root(k)). The minimum sum is then 2*root (k).

12. Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.

13. For any 2 numbers a, b where a>b

ü a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)

ü (GM)^2 = AM * HM

14. For three positive numbers a, b, c

ü (a + b + c) * (1/a + 1/b + 1/c)>=9

15. For any positive integer n

ü 2

16. a2 + b2 + c2 >= ab + bc + ca

If a=b=c, then the case of equality holds good.

17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1)

18. (n!)2 > nn

19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s

20. If n is even, n(n+1)(n+2) is divisible by 24

21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)

22. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity

Note: 2

23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]

24. (m + n)! is divisible by m! * n!

25. When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.

26. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)

27.

ü The sum of first n natural numbers = n(n+1)/2

ü The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

ü The sum of cubes of first n natural numbers is (n(n+1)/2)2/4

ü The sum of first n even numbers= n (n+1)

ü The sum of first n odd numbers= n2

28. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then

ü the total number of factors is (x+1)(y+1)(z+1) ....

ü the total number of relatively prime numbers less than the number is

N * (1-1/a) * (1-1/b) * (1-1/c)....

ü the sum of relatively prime numbers less than the number is

N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....

ü the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)

29.

ü Total no. of prime numbers between 1 and 50 is 15

ü Total no. of prime numbers between 51 and 100 is 10

ü Total no. of prime numbers between 101 and 200 is 21

30.

ü The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6

ü The number of rectangles in n*m board is given by n+1C2 * m+1C2

31. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.

32. Certain nos. to be remembered

ü 210 = 45 = 322 = 1024

ü 38 = 94 = 812 = 6561

ü 7 * 11 * 13 = 1001

ü 11 * 13 * 17 = 2431

ü 13 * 17 * 19 = 4199

ü 19 * 21 * 23 = 9177

ü 19 * 23 * 29 = 12673

33. Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.

34. If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.

35. To find out the sum of 3-digit nos. formed with a set of given digits

This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)

Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.

Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)

= 25 * 24 * 11111

=6666600

36. Consider the equation x^n + y^n = z^n

As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.

37. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then,

N^(p-1) – 1 is always divisible by p.

38. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.

145 = 1! + 4! + 5!

39.

ü Where a no. is of the form a^n – b^n, then,

· The no. is always divisible by a - b

· Further, the no. is divisible by a + b when n is even and not divisible by

a + b when n is odd

ü Where a no. is of the form a^n + b^n, then,

· The no. is usually not divisible by a - b

· However, the no. is divisible by a + b when n is odd and not divisible by

a + b when n is even

40. The relationship between base 10 and base ‘e’ in log is given by

log10N = 0.434 logeN

41. WINE and WATER formula

Let Q - volume of a vessel, q - qty of a mixture of water and wine be removed each time from a mixture, n - number of times this operation is done and A - final qty of wine in the mixture, then,

A/Q = (1-q / Q)^n

42. Pascal’s Triangle for computing Compound Interest (CI)

The traditional formula for computing CI is

CI = P*(1+R/100)^N – P

Using Pascal’s Triangle,

Number of Years (N)

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

1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

… 1 .... .... ... ... ..1

Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1:

Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331

The coefficients - 1,3,3,1 are lifted from the Pascal's triangle above.

Step 2:

CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)

If N =2, we would have had,

Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210

CI = 2 * 100 + 1* 10 = Rs.210

43. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:

Final Difference% = X - Y - XY/100

Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?

Applying the formula,

Final difference% = 40 – 25 - (40*25/100) = 5 %.

So if 5 % = 1,000

Then, 100 % = 20,000.

Hence, C.P = 20,000

& S.P = 20,000+ 1000= 21,000

44. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.

45.

ü Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)2

ü The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003

46.

ü If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.

ü If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.

ü If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time

ü If A can finish a work in X time and B in Y time and A, B & C together in S time then

· C can finish that work alone in (XYS)/ (XY-SX-SY)

· B+C can finish in (SX)/(X-S); and

· A+C can finish in (SY)/(Y-S)

47. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5

48.

ü When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0

ü When an unbiased coin is tossed even no. (2n) of times, then,

P (no. of heads=no. of tails) = 1-(2nCn/22n)

49. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!

Eg)1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?

Here n=10, m=3 (i.e. A, B, C)

Hence, P (A>B>C) = 1/3!

= 1/6

Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?

P (M>S) = 1/2!

= 1/2

50. CALENDAR

ü Calendar repeats after every 400 years.

ü Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.

ü Century has 5 odd days and leap century has 6 odd days.

ü In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.

ü January 1, 1901 was a Tuesday.

51.

ü For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)

ü For any regular polygon, the sum of interior angles =(n-2)*180 degrees

So measure of one angle is (n-2)/n *180

ü If any parallelogram can be inscribed in a circle, it must be a rectangle.

ü If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).

52. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)

53.

ü For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is

0.5*d1*d2, where d1, d2 are the length of the diagonals.

ü For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where

s=(a + b + c + d)/2

Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.

ü Area of a Rhombus = Product of Diagonals/2

54. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for

[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]

55. Area of a triangle

ü 1/2*base*altitude

ü 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B

ü root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2

ü a*b*c/(4*R) where R is the circumradius of the triangle

ü r*s ,where r is the inradius of the triangle

56. In any triangle

ü a=b*cos C + c*cos B

ü b=c*cos A + a*cos C

ü c=a*cos B + b*cos A

ü a/sin A=b/sin B=c/sin C=2R, where R is the circumradius

ü cos C = (a^2 + b^2 - c^2)/2ab

ü sin 2A = 2 sin A * cos A

ü cos 2A = cos^2 (A) - sin^2 (A)

57. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1

58. Appollonius Theorem

In a triangle ABC, if AD is the median to side BC, then

AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2)

59.

ü In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

ü In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.

60. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.

61. Let W be any point inside a rectangle ABCD, then,

WD2 + WB2 = WC2 + WA2

62. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))

63.

ü Distance between a point (x1, y1) and a line represented by the equation

ax + by + c=0 is given by |ax1+by1+c|/Sq(a2+b2)

ü Distance between 2 points (x1, y1) and (x2, y2) is given by

Sq((x1-x2)2+ (y1-y2)2)

64. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.

[Illusionz]

Excellent!

[jyothsnams]

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)../images/smilies/frown.gifB's speed) = (root(B):root(A))

 

can u explain the reason behind it

[devisai85]

this is very good job to improve the math

[Hermione]

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))

 

can u explain the reason behind it

I am sorry I missed this earlier.

 

My solution is this:

 

Lets say the initial distance between A and B was (x+y) kms. A and B cross each other after traveling a distance of x and y km respectively.

 

------------*-------------------------

A---- x------------ y---------------- B

 

Speed = distance travelled / time taken

 

So, we can form the following equations:

 

1. x/ A’s speed = y/ B’s speed, as they take the same time to cross that distance.

 

2. A takes 'a' seconds to travel the distance y after crossing B. Hence, y/A’s speed = 'a' secs => a * A’s speed = y

 

Similarly, x/B’ speed = 'b' secs and b * B’ speed = x

 

Now substituting these for x and y in eqn. 1

 

b * B’s speed/ A’s speed = a * A’s speed/ B’s speed

 

That is, A’s speed squared/ B’s speed squared = b/ a

 

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

[bscout]

I came up with this formula and I think it hasn't been posted before.

 

If we have to find: "How much of a v1 something should be melted into n2 of a v2 something to produce an alloy which is R something?" we can use this formula:

n1=n2(R-v2)/(v1-r)

 

As example:

 

How much of a 75% copper alloy should be melted into 62 kg of a 35%

copper alloy to produce an alloy which is 50% copper?

 

n1=62*(0.5-0.35)/(0.75-0.5)= 37.2.

 

I hope this help.

[kvreddy]

Good job manwithmissions!!

 

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

 

To be frank, even though it is OK to learn these instant formulas, you really need to know how you arrive at those, so that you can apply them properly, because most of the GRE CAT questons are not straight forward application of these formulas. They twist it a lot, especially if you are shooting for high scores