Quant… Basic Formulae

Consolidated some of the basic formula.
ALGEBRA :
1. Sum of first n natural numbers = n(n+1)/2
2. Sum of the squares of first n natural numbers = n(n+1)(2n+1)/6
3. Sum of the cubes of first n natural numbers = [n(n+1)/2]2
4. Sum of first n natural odd numbers = n2
5. Average = (Sum of items)/Number of items
Arithmetic Progression (A.P.):
An A.P. is of the form a, a+d, a+2d, a+3d, …
where a is called the ‘first term’ and d is called the ‘common difference’
1. nth term of an A.P. tn = a + (n-1)d
2. Sum of the first n terms of an A.P. Sn = n/2[2a+(n-1)d] or Sn = n/2(first term + last term)
Geometrical Progression (G.P.):
A G.P. is of the form a, ar, ar2, ar3, …
where a is called the ‘first term’ and r is called the ‘common ratio’.
1. nth term of a G.P. tn = arn-1
2. Sum of the first n terms in a G.P. Sn = a|1-rn|/|1-r|
Permutations and Combinations :
1. nPr = n!/(n-r)!
2. nPn = n!
3. nP1 = n
1. nCr = n!/(r! (n-r)!)
2. nC1 = n
3. nC0 = 1 = nCn
4. nCr = nCn-r
5. nCr = nPr/r!
Number of diagonals in a geometric figure of n sides = nC2-n

Tests of Divisibility :
1. A number is divisible by 2 if it is an even number.
2. A number is divisible by 3 if the sum of the digits is divisible by 3.
3. A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
4. A number is divisible by 5 if the units digit is either 5 or 0.
5. A number is divisible by 6 if the number is divisible by both 2 and 3.
6. A number is divisible by 8 if the number formed by the last three digits is divisible by 8.
7. A number is divisible by 9 if the sum of the digits is divisible by 9.
8. A number is divisible by 10 if the units digit is 0.
9. A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places, is divisible by 11.
H.C.F and L.C.M :
H.C.F stands for Highest Common Factor. The other names for H.C.F are Greatest Common Divisor (G.C.D) and Greatest Common Measure (G.C.M).
The H.C.F. of two or more numbers is the greatest number that divides each one of them exactly.
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
Two numbers are said to be co-prime if their H.C.F. is 1.
H.C.F. of fractions = H.C.F. of numerators/L.C.M of denominators
L.C.M. of fractions = G.C.D. of numerators/H.C.F of denominators
Product of two numbers = Product of their H.C.F. and L.C.M.

PERCENTAGES :
1. If A is R% more than B, then B is less than A by R / (100+R) * 100
2. If A is R% less than B, then B is more than A by R / (100-R) * 100
3. If the price of a commodity increases by R%, then reduction in consumption, not to increase the expenditure is : R/(100+R)*100
4. If the price of a commodity decreases by R%, then the increase in consumption, not to decrease the expenditure is : R/(100-R)*100
PROFIT & LOSS :
1. Gain = Selling Price(S.P.) – Cost Price(C.P)
2. Loss = C.P. – S.P.
3. Gain % = Gain * 100 / C.P.
4. Loss % = Loss * 100 / C.P.
5. S.P. = (100+Gain%)/100*C.P.
6. S.P. = (100-Loss%)/100*C.P.
Short cut Methods:
1. By selling an article for Rs. X, a man loses l%. At what price should he sell it to gain y%? (or)
A man lost l% by selling an article for Rs. X. What percent shall he gain or lose by selling it for Rs. Y?
(100 – loss%) : 1st S.P. = (100 + gain%) : 2nd S.P.
2. A man sold two articles for Rs. X each. On one he gains y% while on the other he loses y%. How much does he gain or lose in the whole transaction?
In such a question, there is always a lose. The selling price is immaterial.
Formula: Loss % =
3. A discount dealer professes to sell his goods at cost price but uses a weight of 960 gms. For a kg weight. Find his gain percent.
Formula: Gain % =
RATIO & PROPORTIONS:
1. The ratio a : b represents a fraction a/b. a is called antecedent and b is called consequent.
2. The equality of two different ratios is called proportion.
3. If a : b = c : d then a, b, c, d are in proportion. This is represented by a : b :: c : d.
4. In a : b = c : d, then we have a* d = b * c.
5. If a/b = c/d then ( a + b ) / ( a – b ) = ( d + c ) / ( d – c ).
TIME & WORK :
1. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n
2. If A and B work together for n days, then (A+B)’s 1 days’s work = 1/n
3. If A is twice as good workman as B, then ratio of work done by A and B = 2:1
PIPES & CISTERNS :
1. If a pipe can fill a tank in x hours, then part of tank filled in one hour = 1/x
2. If a pipe can empty a full tank in y hours, then part emptied in one hour = 1/y
3. If a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours, then on opening both the pipes,
the net part filled in 1 hour = (1/x-1/y) if y>x
the net part emptied in 1 hour = (1/y-1/x) if x>y
TIME & DISTANCE :
1. Distance = Speed * Time
2. 1 km/hr = 5/18 m/sec
3. 1 m/sec = 18/5 km/hr
4. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is 2xy/(x+y) kmph.
PROBLEMS ON TRAINS :
1. Time taken by a train x metres long in passing a signal post or a pole or a standing man is equal to the time taken by the train to cover x metres.
2. Time taken by a train x metres long in passing a stationary object of length y metres is equal to the time taken by the train to cover x+y metres.
3. Suppose two trains are moving in the same direction at u kmph and v kmph such that u>v, then their relative speed = u-v kmph.
4. If two trains of length x km and y km are moving in the same direction at u kmph and v kmph, where u>v, then time taken by the faster train to cross the slower train = (x+y)/(u-v) hours.
5. Suppose two trains are moving in opposite directions at u kmph and v kmph. Then, their relative speed = (u+v) kmph.
6. If two trains of length x km and y km are moving in the opposite directions at u kmph and v kmph, then time taken by the trains to cross each other = (x+y)/(u+v)hours.
7. If two trains start at the same time from two points A and B towards each other and after crossing they take a and b hours in reaching B and A respectively, then A’s speed : B’s speed = (√b : √
SIMPLE & COMPOUND INTERESTS :
Let P be the principal, R be the interest rate percent per annum, and N be the time period.
1. Simple Interest = (P*N*R)/100
2. Compound Interest = P(1 + R/100)N – P
3. Amount = Principal + Interest
LOGORITHMS :
If am = x , then m = logax.
Properties :
1. log xx = 1
2. log x1 = 0
3. log a(xy) = log ax + log ay
4. log a(x/y) = log ax – log ay
5. log ax = 1/log xa
6. log a(xp) = p(log ax)
7. log ax = log bx/log ba
Note : Logarithms for base 1 does not exist.
AREA & PERIMETER :
Shape Area Perimeter
Circle ∏ (Radius)2 2∏(Radius)
Square (side)2 4(side)
Rectangle length*breadth 2(length+breadth)
1. Area of a triangle = 1/2*Base*Height or
2. Area of a triangle = √ (s(s-(s-b)(s-c)) where a,b,c are the lengths of the sides and s = (a+b+c)/2
3. Area of a parallelogram = Base * Height
4. Area of a rhombus = 1/2(Product of diagonals)
5. Area of a trapezium = 1/2(Sum of parallel sides)(distance between the parallel sides)
6. Area of a quadrilateral = 1/2(diagonal)(Sum of sides)
7. Area of a regular hexagon = 6(√3/4)(side)2
8. Area of a ring = ∏(R2-r2) where R and r are the outer and inner radii of the ring.
VOLUME & SURFACE AREA :
Cube :
Let a be the length of each edge. Then,
1. Volume of the cube = a3 cubic units
2. Surface Area = 6a2 square units
3. Diagonal = √ 3 a units
Cuboid :
Let l be the length, b be the breadth and h be the height of a cuboid. Then
1. Volume = lbh cu units
2. Surface Area = 2(lb+bh+lh) sq units
3. Diagonal = √ (l2+b2+h2)
Cylinder :
Let radius of the base be r and height of the cylinder be h. Then,
1. Volume = ∏r2h cu units
2. Curved Surface Area = 2∏rh sq units
3. Total Surface Area = 2∏rh + 2∏r2 sq units
Cone :
Let r be the radius of base, h be the height, and l be the slant height of the cone. Then,
1. l2 = h2 + r2
2. Volume = 1/3(∏r2h) cu units
3. Curved Surface Area = ∏rl sq units
4. Total Surface Area = ∏rl + ∏r2 sq units
Sphere :
Let r be the radius of the sphere. Then,
1. Volume = (4/3)∏r3 cu units
2. Surface Area = 4∏r2 sq units
Hemi-sphere :
Let r be the radius of the hemi-sphere. Then,
1. Volume = (2/3)∏r3 cu units
2. Curved Surface Area = 2∏r2 sq units
3. Total Surface Area = 3∏r2 sq units
Prism :
Volume = (Area of base)(Height

Some useful fractions to learn

we have to calculate 1/x when x is
x    1/x 
2     .5
3     .{3}
4     .25
5     .20
6     .1{6}
7     .{142857}
8     .125
9     .{1}
10   .1
11    .{09}

Find all sets of consecutive integers that add up to 1000.

There are total 8 such series:
1.
Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000. (-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000
2.
Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202. (-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000
3.
Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70. (-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000
4.
Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000
5.
Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52. 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000
6.
Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70. 55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000
7.
Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202. 198 + 199 + 200 +201 + 202 = 1000.
8.
Sum of 1 number starting from 1000. 1000 = 1000

The smallest no. such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original no.

Ques Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.
Answer The answer is 285714. If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.

Sum of last digit = sum of rest digits

Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5. How many such 3-digit numbers are there?
Ans There are 45 different 3-digit numbers.

Take the combinations
The last digit can not be 0.
If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)
If the last digit is 2, the possible numbers are 202 and 112.
If the last digit is 3, the possible numbers are 303, 213 and 123.
If the last digit is 4, the possible numbers are 404, 314, 224 and 134.
If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.

Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....
Thus, total numbers are 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.

Divisibility tests

A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).

A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.

A number is divisible by 4, if last 2 numbers are divisible by 4. Likewise the number is divisible by 8, if its last 3 digits are divisible by 8. So this holds for powers of 2.

A number is divisible by 5 if the last digit is 5 or 0.

Most people know (only) those 3 rules. Here are the rules for divisibility by the PRIMES up to 50. Why only primes and not also composite numbers? A number is divisible by a composite if it is also divisible by all the prime factors (e.g. is divisible by 21 if divisible by 3 AND by 7). Small numbers are used in these worked examples, so you could have used a pocket calculator. But my rules apply to any number of digits, whereas you cannot test a 30 or more digit number on your pocket calculator otherwise.
Lets assume L is the last digit and A is the remaining truncating number....

Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.
There are similar rules for the remaining primes under 40, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.
(A-2L) / 7


Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
(A-L) / 11
     Another approach is to sum up numbers in 2 parts - 1st part containing all odd positioned numbers and other part containing all numbers positioned at  even positions. Find the difference and if the difference is divided by 11, i.e. if it is 11 or 0 it means number is divided by 11. eg. 19151 ,
odd position sum = 3, even position sum = 14; Difference = 11, that means no. is divisible by 11.

Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

(A+4L) / 13

Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
(A-5L) / 17




  Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
(A+2L) / 19

I think what I have above written is sufficient..but if you want more rules please carry on... :)

Test for divisibility by 23. Add seven times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.
Example: 17043-->1704+7*3=1725-->172+7*5=207-->20+7*7=69 which is 3*23, so 17043 is also divisible by 23.
(A+7L) / 23




Test for divisibility by 29. Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.
Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.
(A+3L) / 29


Test for divisibility by 31. Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.
(A-3L) / 31

Test for divisibility by 37. This is (slightly) more difficult, since it perforce uses a double-digit multiplier, namely eleven. People can usually do single digit multiples of 11, so we can use the same technique still. Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.
Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.
(A-11L) / 37


Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.
Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.
(A+2L) / 41


Test for divisibility by 43. Now it starts to get really difficult for most people, because the multiplier to be used is 13, and most people cannot recognise even single digit multiples of 13 at sight. You may want to make a little list of 13*N first. Nevertheless, for the sake of completeness, we will use the same method. Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.
Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.
(A+13L) / 43


Test for divisibility by 47. This too is difficult for most people, because the multiplier to be used is 14, and most people cannot recognise even single digit multiples of 14 at sight. You may want to make a little list of 14*N first. Nevertheless, for the sake of completeness, we will use the same method. Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.
Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.
(A+2L) / 19


I've stopped here at the last prime below 50, for arbitrary but pragmatic reasons as explained above.
Lets summarize A+mL divisibilities :

Number	Coeff of L (m)
7	-2
11	-1
13	4
17	-5
19	2
23	7
29	3
31	-3
37	-11
41	2
43	13
47	-14

Other blogreaders (sadly even people from .edu domains, who should be able to do the elementary algebra themselves) have asked why I sometimes say ADD and for other primes say SUBTRACT, and ask where the apparently arbitrary factors come from. So let us do some algebra to show the method in my madness.
We have displayed the recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, if it's a 1 we are going to SUBTRACT later. Then we will use the leading digit(s) of the multiple as our multiplier M.
Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also. Q.E.D.

Special numbers

A number of 9 digits has the following properties:

• The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.

• Each digit in the number is different i.e. no digits are repeated.

• The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

Answer
The answer is 381654729 One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.
At odd points we have  3 1 5 7 9
At even points we have 8 6 4 2

Armstrong number
Armstrong numbers are the sum of their own digits to the power of the number of digits. example:153 = 1³ + 5³ + 3³[as sum of the digits are 3], 1⁴ + 6⁴ + 3⁴ + 4⁴ = 1634.