Maxima n Minima

1.if f(x) = ax^2 + bx + c, how to find maximum, minimum .USE first derivation test.f'(x)= 2aX+b.find value of X for which , 2ax+b= 0.when a>0, at this value of X, f(x) is minimum.when a0, so at x=-2, f(x) will attain minimum value.f(-2) = 4-8+3 = -1.

if f(x) = -X^2 + 4X + 3f"(x) = -2x + 4X= 2.f(2) = -4+8+3 = 7 is the maximum value of f(x).

generalizing for f(x) = ax^2 + bx + c,x=-b/2a (2ax+c=o)gives maximum or minimum value of f(x) depending upon a>0 or a 3x/2 + 3x/2 + 5y/3 + 5y/3 + 5y/3 = 15.--------------1as I said, when sum of any quantities is constant, there product is maximum when they are equal.here sum is constant.so when 3x/2 = 5y/3. we get maximum value of x^2*y^3.taking 3x/2 = 5y/3 putting it in 1,=> 5(3x/2) = 15.=>x=2. and y = 9/5.answer is 2^2*(9/5)^3.

generalizing it, how to find maximum value of x^m*y^n where ax+by=P.a,b,x,y>0x^m*y^n is maximum whenax/m = by/ n = p/m+n

4. when the product of any quantity is constant, sum of the all the quantity is minimum, when they are equal.xy^3 = 64.find minimum value of x+12y.we need to adjust x+12y, accordingly.x+12y = x+ (12y/3)*3

now, x*(12y/3)^3= 64 *64 ( coz xy^3 = 64)-----------1

the product is constant. so the sum of the quantities will be minimum when quantities are equal.take x= 12y/3putting it in 1, we get x= 8=>12y/3 = 8, y = 2.minimum value of x+12y = 8+24 = 32.

generalizing it, how to find minimum value of ax+by where x^m*y^n=Pa,b,x,y>0ax+by is minimum whenax/m = by/n

1.if f(x) = ax^2 + bx + c, how to find maximum, minimum .USE first derivation test.f'(x)= 2aX+b.find value of X for which , 2ax+b= 0.when a>0, at this value of X, f(x) is minimum.when a0, so at x=-2, f(x) will attain minimum value.f(-2) = 4-8+3 = -1.

if f(x) = -X^2 + 4X + 3f"(x) = -2x + 4X= 2.f(2) = -4+8+3 = 7 is the maximum value of f(x).

generalizing for f(x) = ax^2 + bx + c,x=-b/2a (2ax+c=o)gives maximum or minimum value of f(x) depending upon a>0 or a 3x/2 + 3x/2 + 5y/3 + 5y/3 + 5y/3 = 15.--------------1as I said, when sum of any quantities is constant, there product is maximum when they are equal.here sum is constant.so when 3x/2 = 5y/3. we get maximum value of x^2*y^3.taking 3x/2 = 5y/3 putting it in 1,=> 5(3x/2) = 15.=>x=2. and y = 9/5.answer is 2^2*(9/5)^3.

generalizing it, how to find maximum value of x^m*y^n where ax+by=P.a,b,x,y>0x^m*y^n is maximum whenax/m = by/ n = p/m+n

4. when the product of any quantity is constant, sum of the all the quantity is minimum, when they are equal.xy^3 = 64.find minimum value of x+12y.we need to adjust x+12y, accordingly.x+12y = x+ (12y/3)*3

now, x*(12y/3)^3= 64 *64 ( coz xy^3 = 64)-----------1

the product is constant. so the sum of the quantities will be minimum when quantities are equal.take x= 12y/3putting it in 1, we get x= 8=>12y/3 = 8, y = 2.minimum value of x+12y = 8+24 = 32.

generalizing it, how to find minimum value of ax+by where x^m*y^n=Pa,b,x,y>0ax+by is minimum whenax/m = by/n

NO. OF SQUARES AND RECTANGLES IN A CHESSBOARD

in a chessboard,there are 8*8 squares.

in a 2*2 chessboard, there are 5 squares (4 small aquares, 1 big square).similarly in an n*n chessboard, there are 1^2 +2^2+....+n^2 squares.

so in a 8*8 chessboard, n=8=> no.of squares = 1^2 +2^2+3^2+.....+8^2= [n (n+1) (2n+1)]/6 (summation formula)= 204

rectangles :

in a 2*2 chessboard, there are 9 rectangles (4 1*1s,1 2*2,2 2*1s, 2 1*2s)

for an n*n chessboard, there are 1^3 +2^3+3^3+.....+n^3 rectangles.

so for an 8*8 chessboard, there are, rectangles = 1^3+2^3+....+8^3

=> no. of rectangles = [{n^2}{(n+1)^2}/4

n=8, we get no. of rectangles = 1296

**No. of rectangles that are not squares in an 8*8 chessboard=> 1296- 204= 1092

CIRCULAR MOTION

Consider a circle with circumference 200 sq metres and two people A and B moving in clockwise direction with speeds 5m/s and 9m/s

a) when do they meet at the starting point for the first time ?

b) after how much time will they meet for the first time ?

c) at how many distinct points they meet ?

d) if they move in opposite direction then in how many distinct points do they meet ?

ans a) time taken by A to reach the starting point for the first time is 200/5 = 40 sec,for the second time 80 secondsfor the third time 120 seconds and so on .......

similarly time taken by B to reach the starting place is 200/9for the second time 2 * 200/9for the third time 3 * 200/9

so they meet forfirst time at the starting point at the LCM of their time periodswhich are 40 and 200/9 in this case and hence they will meet at t= 200 for the first time .

ans b )this should be solved using the relative speed concept time taken for them to meet for the first time will berelative distance/ relative speedhere relative distance is 200 and relative speed is 9-5=4.so ans is 200/4= 50

ans c)when 2 bodies are moving incircular motion in same direction the number of distinct points where they meet is the difference of the speeds .here it is 9-5 = 4 distinct points.the lcm of the speeds must be 1

ans d ) when 2 bodies move in circular motion in opposite direction thenthe number of distict points they meet is the sum of the speedshere it will be 14 .the lcm of the speeds must be 1

e) find the number of distict points at which 2 bodies with speed 4 and 8 meet when they move in clockwise direcion and when they move in anticlock wise direction ?

here we first need to divide the HCF so we get 1:2

now using the formula stated above the number of distict point they meet when moving in same direction will be 2-1 =

the number of distict point they meet when moving in opposite direction is 2+1 = 3

f)consider three bodies a,b,c with speeds 5,9,13 respectively moving in clock-wise direction now number of distinct points at which all three meet ?

a-b = 4b-c = 4a-c = 8

so number of distinct points will be the hcf of the differences which is 4 ans.

g)consider three bodies a,b with speeds 5,9 respectively moving in clock-wise direction and c with speed 13 in anti-clockwise direction now number of distinct points at which all three meet ?

a-b=4a+c=18b+c=22

hcf is 2 so they all will meet at 2 distinct points.last 2 digits of a number]

divide the number by 100 then it will give the last 2 digits

if division is cumbersome u may use one of the following methods:

points to remember :

if last 2 digits are 25 then (abcdef25)^z where z is a natural number will always give the last 2 digit as 25.

if last 2 digits are 76 then (abcdef76)^z where z is a natural number will always give the last 2 digit as 76.

if the last digit is 1 say the number is (abcdefg1)^thgfds then the units digit will be 1 and the tens digit will be g*s

e.g (1231)^4563 last 2 digits will be 91.

(sdfdsf24)^ odd = last 2 digits 24(dfdsfd24)^even = last 2 digits 76

(dbfh26)^odd = last 2 digits 26(dfdsaf26)^even = last 2 digits 76

few examples :

last 2 digits for (71)^45 = 51

(2)^32 = (2^10)^3 * 2^2 = (1024)^3 *2^2 = asjahsj76 *4 = sdnsj04

(6)^76 = (3*2)^76 = (81)^19 * (1024)^7 * (2)^6 = aks21 * 24 * 64 = ss56

p.s. correct me if i any mistakes

Struggling between dreams and reality, dreaming in reality or living in dreams...

Following is another approach:

Last non-zero digit of 10! = 8For 20! = 8*8 = 4 [6 is dropped]This can continue for any number of 10s.

For example,

Last digit of 70! will be given by 8^7

For last digit of 8^7 = 2^21=( 2^10)^2 * 2= 76*2=2

Therefore, last digit of 70! is 2.

If you are asked for 73!, then just multiply the above 2 with 71*72*73or simply 2 with 1*2*3

Therefore, last non zero digit for 73! become 2*6 = 2 [ignoring 10s digit]Difference of Squares :

The concept is if there is any N = a^2 - b^2 and a and b are integers.So many ways a and b are chosen?If N = a^2 - b^2 = (a+b) (a-b), if you want to find integers a and b,then (a+b) and (a-b) should be both either odd or even

For example if we take N = 9 , in that case 9 = 9*1 = 3*3 (In thiscase both Odd)

if a+b = 9 and a-b = 1 therefore a = 5 and b = 4, thats why we have25-16 = 9and also if a+b = 3 and a-b = 3, a = 3 and b = 0 and thats why we 9-0= 9

so we see that only two ways we can represent the number 9 asDIFFERENCE OF SQUARESOF INTEGERS numbers.

If we extend this concept.For odd Numbers,1) Any odd prime numbers can be represented in ONLY 1 way.For example : 3 = 3*1 ONLY so only 1 way2) Other than this any composite Odd numbers can be represented inMore than 1 way

For Even Numbers:1) Any number of the form 4K+2 CANNOT be represented at all, hence 0ways.For example Number 6 = 2*3 = 1*6 (So we don't get any combination foreither Both Even or Both Odd), hence Integer numbers IS NOT POSSIBLE.2) Any Prime factor Multiple of 4 is ALWAYS 1 way.For example : Number 8 = 2*4 = always 1 way.The reasoning behind is for prime number 2 when mutplied with 2 we canalways break up into factors of Even numbers. Hence 4*2 we can alwaysbreak up.And for odd prime numbers, we already have 2 twos in 4, so also we canbreak up into 1 ways into both even factors.

3) Any other even numbers more than 1 way.

Source : Byju's class Notes last year

Its great to keep all the concepts at a single place.

The general formula is (n-1)! * (111...n times) (Sum of the terms).

So for the 1st Case where the number of digits are always same thats is always a 5 digit number, then

4! * (11111) * 25 = 600 * 11111 = 6666600.

The Second case is when you take one digit, two digit etc.

For one digit.Sigma (Numbers ) = 25 = (1 + 3 + 5 + 7 + 9)

For two digit = 1! * 11 * (Sum of the numbers taking two at a time)= 1! * 11 * 4C1 * (1 + 3 + 5 + 7 + 9)

For three Digits :

= 2! * 11 * 4C2 * (1 + 3 + 5 + 7 + 9)

For four Digits :

= 3! * 11 * 4C3 * (1 + 3 + 5 + 7 + 9)

Adding all

(1 + 3 + 5 + 7 + 9) * (1 + 4C1 * 11 + 2! * 4C2* 111 + 3! * 4C3*1111 + 4! * 4C4 * 11111)

Hence General terms can be

Sigma (Terms) * (1 + 1! * (n-1)C(1) (11) + 2! * (n-1)C2 * 111 + ... (n-1)! * (n-1)C(n-1) * (1111...n times))

Good one even I learnt this conceptfind the sum of all the numbers greater than 10000 formed by digits 1,3,5,7,9 ? No repetition.

(1 + 3 + 5 + 7 + 9)*4!*11111 = 6666600

find the sum of all the numbers less than 10000 formed by digits 1,3,5,7,9 ? No repetition ---it means all single digit,double,triple,four digit nos.

(1 + 3 + 5 + 7 + 9)(1 + 11*4 + 111*4*3 + 1111*4*3*2) = 701025

There are some questions where a person A and person B starts at different time and meets at a specific time and then reaches the destination at same time from where the other has started.

This type of questions appear frequently in Mocks and other exams.

Lets say person A and B starts from points P and Q at different times and meets at point S and reaches the points Q and P at the same time.

P----------------------S----------------------Q------------------------------------------------Time taken by person A is say t1 and speed being s1.Time taken by person B is say t2 and speed being s2.

hence PS = t1*s1.SQ = t2 * s2.

As they reach the destination at the same time.

hence (t1 * s1)/(t2) = (t2*s2)/(t1)

hence s1/s2 = root(t2/t1)

This is the principle and depending upon the questions, use options to keep either the time ratio same or speed ratio same.

a)In a plane if there are n points of which no three are collinear, then The number of straight lines that can be formed by joining them isnC2. The number of triangles that can be formed by joining them isnC3. The number of polygons with k sides that can be formed by joining them isnCk.

(b)In a plane if there are n points out of which m points are collinear, then The number of straight lines that can be formed by joining them isnC2 mC2 + 1. The number of triangles that can be formed by joining them isnC3 mC3. The number of polygons with k sides that can be formed by joining themisnCk mCk. The answer should be (n-m)Ck + (n-m)C(k-1) *mC1 +(n-m)C(k-2) *mC2. [ for n-m >=k, otherwise based on the values of n,m,k expression will change.

(c)The number of diagonals of a n sided polygon arenC2 n = n (n 3)/2.

(d)The number of triangles that can be formed by joining the vertices of a n-sided polygon which has, Exactly one side common with that of the polygon aren (n 4). Exactly two sides common with that of the polygon aren. No side common with that of the polygon aren (n 4) (n 5)/6.Cauchy-Schwartz Equation:

If a , b , c , d are four real numbers, they always satisfy the relationship

(a^2+b^2)(c^2+d^2)>=(ac+bd)^2

This can be generalized to a large number of variables as(a1^2+a2^2+a3^2+.....)(b1^2+b2^2+b3^2+....)>=(a1b1 +a2b2+a3b3+....)^2

Questions:1>Find the least value of X^2+Y^2+z^2 if X+2y+3Z=14

Sol:-->(X^2+Y^2+z^2)(1^2+2^2+3^2)>=(X*1+Y*2+z*3)^2hence , min value= 14^2/14=14Few Concepts:

1> For any prime number p, (p-1)times same digit is repeated , then that number formed is exactly divisible by peg. 666666 is divisible by 7.

2>All perfect squares of the form3K+N where N=0,14K+N where N=0,18K+N where N=0,1,49K+N where N=0,1,4,7

3>For any two integers satisfying, 3 b^A

4>If three circles touch Each other in a row and they have two direct common tangents, then their radii are in GP series

5>Product of factors of a number = (Number )^(total no of factors/2)

6>All odd natural numbers can be represented as the difference of two perfect squares. All even numbers which are multiple of 4 can be also written as the difference of two perfect squares.

7>Suppose we get a question like(I) A and B start at same time towards each other, meet at a point after time T then reach their respective destinations after time T1 and T2, or (II)their start at different time ,meet each other at a time where A takes T1 to reach and B takes T2 to reach, and then reaches their destination after time T at the same time ,Then VA/VB=Root(T2/T1)And T = Root(T1*T2) where T is the same they take either to reach their meeting point in first or Destination in second case

8>If a Right Angled Triangle is rotated about its Axis to generate a cone, then the cone will have maximum volume when rotated about the smallest side as an axis and minimum when rotated about the side perpendicular to the smallest side as an axis.

9>If a n digit number is multiplied by an (n+1) digit number , the product has 2n or (2n+1) digits.

10> Difference between Principal Interest and Compound Interest for the second year=Pr^2And Difference between PI and CI for the third year=Pr^2(r+2)where r=(R/100)and P is the amount and R is the rate of interest.

11>If A and B takes K days when working together and X+K and Y+K days when working alone respectively to complete a work then (X/K)=(K/Y) or XY=K^2.

12>The number of times the sign of the coefficients of an equation changes, gives the number of roots of a quadratic or higher degree equation.

13>The number of ways of writing any number as a sum of two or more consecutive positive numbers = number of odd factors of that number-1. Any number that doesnt have an odd factor cant be expressed as a sum of two or more consecutive numbers.eg. 50 has two odd factors. So, it can be written in only one way as a sum of two or more consecutive positive numbers which is 11+12+13+14.POWER OF THREE METHOD

Q) You have 6561 balls. 1 is lighter than remaining. What is the minimum number of weighings required on a common balance to ensure that the odd ball is identified ?

(a) 8 (b) 7 (c) 6 (d)5

For such questions power of 3 is used.

Let there be 3 balls, 1 out of them is heavy, minimum number of weighings to find the heavy ball is = 1

First take any 2 balls, if they are in balance => the 3rd ball is heavy.if they are not in balance, we get the heavier ball.Thusminimumnumber of weighings = 1

For 9 such balls, where 1 is heavy , it would be = 2 .[ In the 1st weighing3 balls on each side of balance. if they are in balance, another weighing with remaining 3 balls is required.if they are not in balance, the heavier side having 3 balls is weighed again.]Thusminimumnumber of weighings = 2

Similarly for 10-27 balls,minimumnumber of weighings = 328 - 81 balls,minimumnumber of weighings = 482 - 243 balls,minimumnumber of weighings = 5244 - 729 balls,minimumnumber of weighings = 6730 - 2187 balls,minimumnumber of weighings= 72188 - 6561 balls,minimumnumber of weighings = 8

Therefore, answer would be (a)8.

Ex : For 425 balls,minimumnumber of weighings = 6 (Since 425 lies in the range 244 to 729.)

Basically, power of 3 can be summarized as :

3^n = Number of balls

where

n = minimum number of weighings on a common balance.

To Find out Pythagorean triplets

There is a formula

(N2-1)/2,( N2+1/2) of any number N.

For Example if the number is 3, the pythagorean triplets are (3)2-1/2 = 4and (3)2+1/2 = 5.

For any other number , i.e. !5, the triplets are (15)2-1/2 = 112 and (15)2+1/2 = 113.How to find the number of factor of form (3n + 0/1/2) of number N.

Its easier for the factors of kind 3n.Just find the number of factors of N/3 and that will be the answer.

For factors of form (3n + 1) or (3n + 2)We know that,(3n + 1)(3n + 1) = (3k + 1)(3n + 1)(3n + 2) = (3k + 2)(3n + 2)(3n + 2) = (3k + 1)

Lets consider the example of N = 3136 = (2^6)*(7^2)i) (3n + 1)We know that 2 is of form (3k + 2), but 4 is of form (3k + 1).7 is also of form (3k + 1).

So all the factors of form (3k + 1) are expressed as (2^2k)*(7^n)k van take 4 values(0, 1, 2, 3) and n can take 3 values (0, 1, 2)=> Number of such factors = 4*3 = 12

ii) (3n + 2)We know 2 is of form (3n + 2)

So, all factors of form (3n + 2) can be expressed as (2^(2k + 1))*(7^n)k can take 3 values (0, 1, 2) and b can take 2 values=> Number of such factors = 3*3 = 9

Similarly, we can go for (4n + 1) or (4n + 3) Or some other form

Equations:

1) Quadratic Equation ax^2+bx+c = 0, has maxima = -D/4a at x = -b/2a [D = b^2-4ac]2) All polynomials of degree 1 would be a straight line. hence if a area under the curve is asked, solve for x = 0 and y =0 and we can get the two points and draw straight lines. This helps in elimininating some options.3) if the Polynomial is given as ax^n + bx^(n-1) +...+ z = 0

then sum of the roots (taken one at a time) = -b/asum of roots taken two at a time = +c/asimilarly it continues with alternate -/+ signs.

4) For Quadratic equations, if a>0, then the graph would be a parabola with opens upward and if a (209) = 209*(1 - 1/11)(1 - 1/19) = 180

=> Since 5 and 209 are coprime, we can say that5^180 = 1 (mod 209)

Fermat's Little Theorem

It is an extension of Euler's Theorem.If in Euler's theorem n is a prime number, thena^(n - 1) - 1 will be divisible by n is a and n are coprime.

We can see that (n), where n is a prime no will be:-(n) = n(1 - 1/n) = n - 1

That why when n is prime we can write a^(n - 1) - 1 will be divisible by n.

An application of Euler's Theorem

Suppose we have number N having digit 'a' written 'k' times, where k is Euler's totient of 'n'. We have to find the remainder when N is divided by 'n' and it is also given that n and 10 are coprime

N = aaa...a (k times) = (a/9)(10^k - 1)

Now, since 10 and n are coprime, we can say that (10^k - 1) is divisible by n as k is Euler's totient of 'n'. Now since n and 9 are also coprime, we can say that

(a/9)(10^k - 1) will be divisible by n.

That meansany digit written (n) times will be divisible by n if n is coprime to 9 and 10

For ex:-Find the remainder when 99999....9(54 times) is divided by 19.

Now, we know that (19) = 18 and 54 = 3*18 and also 19 is coprime to 10 and 9. Hence we can say that:-

99999....9(54 times) is divisible by 19

Number Systems (Concept 1)

Lets brush up some painted cube funda

We assume the cube is divided into n^3 small cubes.

no. of small cubes with ONLY 3 sides painted: 8( all the corner cubes )

no. of small cubes with ONLY 2 sides painted:

A cube is painted on 2 sides means, it is on the edge of the bigger cube ,and we have 12 edges, each having n cubes. but since the corner cubes are painted on 3 sides, we need to neglect them. so in effect, for each side we will have (n-2) small cubes with only 2 sides painted.thus, then number is, 12 * (n-2)

no of small cubes with ONLY 1 side painted:

for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with only one side painted. and we have 6 faces in total.so th number is, 6*(n-2)^2

no of small cubes with NO sides painted:

if we remove the top layer of small cubes from the big cube we will end up a chunk of small cubes with no sides painted.this number will be equal to, (n-2)^3.

Also, remember for Cuboids with all different sizes, the following are the results:

a x b x c (All lengths different)

Three faces - 8 (all the corner small cubes of the cuboid)

Two faces - There are two (a-2) units of small cubes on one face of the cuboid and there is a pair of such faces. Hence, number of such small cubes corresponding dimension a of the cuboid = 4(a-2).

Similarly, for others.

So, total with two faces painted = 4(a - 2) + 4(b - 2) + 4(c - 2)

One face - Since each face of the cuboid is a combination two different dimensions, hence for the face which is a combination of a and b dimensions, the number of small cubes is 2* (a-2)(b-2) [multipliesd by 2 because there are 2 such faces for the combination]

Similarly, for others.

So, total with one face painted =2(a - 2)(b - 2) + 2(a - 2)(c - 2) + 2(b - 2)(c - 2)

Zero faces - The entire volume of small cubes except for two cubes in each of the rows and columns will not be painted at all. hence this is the simplest ...

(a - 2)(b - 2)(c - 2)

You can put different integer values for number of small cubes producing different edge lengths of cuboid to get varied results.

To verify for a cube, put a=b=c=L, you get

Three faces - 8Two faces - 12(L - 2)One face - 6(L - 2)^2Zero faces - (L - 2)^3

Type # 1.

find smallest no. other than k, that leaves remainder k when divided by w,x,y...

to solve such questions, take lcm of w,x,y...and add k to it.

e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...

take lcm of 6,7,8,9 and add 4

i.e. 504 + 4 = 508

Type # 2

find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.

unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it

here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23

Type # 3

Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,

we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208

now we have one more condition...remainder 1 with 11.

concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.

i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.

so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1

208%11 = 10

210k%11 = k

therefore, 10 + k shud leave remainder 1 when divided by 11.

hence, k = 2. and the no. is 208 + 210 x 2 = 628

e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7

for first 3 conditions....no. is 120 + 2 = 122

hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362

Type # 4

What if there is no relation between divisors n remainders?

e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.

in such cases...take 1 case n target another case...e.g. i take the case 7 with 13...and target 6 with 11.

which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.

a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72

now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.

to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...

hence the no. is of the form 72 + 143 k.

72 + 143k % 7 = 2 + 3k

now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..

a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.

hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.

now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.

For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.

there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...

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 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)

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.

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

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

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)

Consider the two equations

a1x+b1y=c1a2x+b2y=c2

Then,If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.If a1/a2 = b1/b2 c1/c2, then we have no solution.If a1/a2 b1/b2, then we have a unique solution. Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1

|a| + |b| = |a + b| if a*b>=0else, |a| + |b| >= |a + b| 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

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.

If for two numbers x*y=k (a constant), then their SUM is MINIMUM ifx=y (=root(k)). The minimum sum is then 2*root (k). 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.

For any 2 numbers a, b where a>b

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

(GM)^2 = AM * HM For three positive numbers a, b, c

(a + b + c) * (1/a + 1/b + 1/c)>=9 For any positive integer n

2= 4abcd (Equality arises when a=b=c=d=1)

(n!)2 > nn

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

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

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)

e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinityNote: 2 < e < 3 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]

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

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.

Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)If a number N is represented as a^x * b^y * c^z where {a, b, c, } are prime numbers, thenthe total number of factors is (x+1)(y+1)(z+1) ....

the total number of relatively prime numbers less than the number isN * (1-1/a) * (1-1/b) * (1-1/c)....

the sum of relatively prime numbers less than the number isN/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....

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

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

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

. 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.

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)! * 11111 (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+* (5-1)! * 11111 (since 5 digits are there)= 25 * 24 * 11111=6666600

WINE and WATER formulaLet 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 Pascals Triangle for computing Compound Interest (CI)

The traditional formula for computing CI isCI = P*(1+R/100)^N P

Using Pascals Triangle,

Number of Years (N)-------------------1 12 1 2 13 1 3 3 14 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.1210CI = 2 * 100 + 1* 10 = Rs.210 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 % = 1,000Then, 100 % = 20,000.Hence, C.P = 20,000& S.P = 20,000+ 1000= 21,000 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.

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

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

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.

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.

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

If A can finish a work in X time and B in Y time and A, B & C together in S time thenC can finish that work alone in (XYS)/ (XY-SX-SY)B+C can finish in (SX)/(X-S); andA+C can finish in (SY)/(Y-S) In case n faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5

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

When an unbiased coin is tossed even no. (2n) of times, then,P (no. of heads=no. of tails) = 1-(2nCn/22n) 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

If the two bodies are moving in opposite direction.

# They will meet first time after covering a total distance of d# For subsequent meetings they will meet after covering a total distance of 2d.

So we can generalize it to be d(2n+1)

Similarly, If the two bodies are moving in same direction.

# They will meet first time after covering a total distance of 2d# For subsequent meetings they will meet after covering a total distance of 2d.

So we can generalize it to be 2dn.Cars start in opposite direction with speeds 60km/h and 40km/hRelative speed = 100km/h

Distance b/w them (d)= 120 km

Time taken to cover (d) = 1.2hrs

their meeting will go by the general formula ( 2n+1)d, so their next meeting will be when both of them cover distance of 3d,5d,7d.....so on

Time taken to cover (2d) = 2.4hrs

No of meeting in 20 hrs = [(20 - 1.2)/2.4 ] + 1 = 8 meetings.first meeting at d ie 120mts ..time taken 1.2 hourssecond meeting 3d ie 360 mts... time taken (360-120) = 240 = 2.4 hoursthird meeting 5d ie 600 mts ..time taken ( 600-360) = 240 = 2.4 hours...

thus, 1.2+2.4+2.4+2.4.... n times< 201.2 + 2.4 ( 7) = 1.2 + 16.8 = 18 , 20///thus 1+7 = 8 times????Tournament Funda

There are 16 teams and they are divided into 2 pools of 8 each. Each team in a group plays against one another on a round-robin basis. Draws in the competition are not allowed. The top four teams from each group will qualify for the next round i.e round 2. In case of teams having the same number of wins, the team with better run-rate would be ranked ahead.1. Minimum number of wins required to qualify for the next round _____?2. Minimum number of wins required to guarantee qualification in the next round _____?

Now, i don't know how many of you are aware of the following method. But 1 thing I mention in advance that this should take only 30 seconds to solve1.1 group is consisting of 8 teams. So each team will play 7 match each. Suppose each of the 8 teams were seeded and we consider the case where a higher seeded team will always win.So the number of wins for the 8 teams would be 7,6,5,4,3,2,1,0 with highest seeded team winning all and lowest seeded team losing all.For minimum number of wins we allow 3 teams to win maximum number of matches. Of the remaining 5 teams just find out the mean of their number of wins.In this case it would be (4+3+2+1+0)/5=2.So 5 teams can end up with 2 wins each and a team with better run rate will qualify with 2 wins.

2.In this case consider the mean of first 5 higher seeded teams(7+6+5+4+3)/5=5So it may be the case that 5 teams can end up having 5 wins each. And hence 1 team will miss the second round birth. So minimum number of wins to guarantee a place would be 6.

let the eqn be 7x+4y = 41to find out possible combinations , we are using remainder fundadivide by 4so rem(7x/4) = rem(41/4)hence x =3 thn find y= 5

its even better for much bigger valuese.g. 99x+2y=5481here rem(2y/99)=36

y =18 n find x

now we can find a series of solutions to find next solutionjust cross add n subtract othr , 7x+4y=41 we got 1st solution as x=3 y=5nxt solution will be (3+4,5-7)=>(7,-2) nxt will be (11,-9)we can reverse the operation n get solution in othr way means (4-3,7+5)

SHORTCUT METHOD TO FIND RANK OF A GIVEN WORDThis shortcut method is used when the lettors of the given word are not repeated.Given word isMASTERThe letters of the are M,A,S,T,E,R.Write the alphabetical order of the letters of the given word ' MASTER ' as A,E,M,R,S,TNow strike off the first letter M.A,E,M,R,S,T.Then count the no.of letters before M, and it is equal to 2,which is the coefficient of 5!.Again strike off the first letter A.A,E,M,R,S,TThen count the no.of letters before A and it is equal to 0 which is coefficient of 4!Again strike off the first letter S.A,E,M,R,S,TThen count the no.of letters before S and it is equal to 2 which is coeffcient of 3!Again strike off the first letter T.A,E,M,R, S, TThen count the no.of letters before T and it is equal to 2 which is coeffcient of 2!Again strike off the first letter E.A,E, M,R, S,TThen count the no.of letters before E and it is equal to 0 which is coeffcient of 1!Finally add 1 to the above values to get the rank of the word MASTER as follows:2(5!) + 0(4!) +2(3!)+2(2!)+0(1!)+1=257

Re: Rectangles Cut By a Diagonal Line

The general rule: If the lengths of sides (a x b) of the rectangle aremutually prime, the number of squares cut is a+b-1

Thus, your example: (3 x 5) gives 3+5-1 = 7

Other examples: (8 x 5) gives 8+5-1 = 12(9 x 4) gives 9+4-1 = 12(9 x 5) gives 9+5-1 = 13

BUT (9 x 6) DOES NOT give 9+6-1 = 14. Instead you must proceed asfollows:

First divide (9 x 6) through by common factor 3 to get (3 x 2)Then apply the rule to (3 x 2) to give 3+2-1 = 4Now multiply by the factor 3 again to get 12 (which is correct).

Let's do a square figure, say (5 x 5). We divide through by 5. Thisgives (1 x 1). Applying the rule gives 1+1-1 = 1. Now multiply up bythe factor 5 again to get 5. We know this is correct because in ANYsquare figure the number of squares that are cut will be equal to theside of the square.

The general procedure for a rectangle (a x b) is as follows:

If a and b are relatively prime the answer is a+b-1

If a and b have a common factor c, first divide through by c to get(a/c x b/c).Then apply the rule to get a/c + b/c - 1.Finally, multiply through again by c to get (a+b-c).

For example, with (9 x 6) the correct answer is 9+6-3 = 12.Type 1: Cube problems

A cube is given with an edge of unit N. It is painted on all faces. It is cut into smaller cubes of edge of unit n. How many cubes will have x faces painted?

In these types of questions, the first thing that we need to figure out is the number of smaller cubes. For this, we look at one particular edge of the big cube and figure out how many smaller cubes can fit into this. It will be N/n. So, the number of smaller cubes will be (N/n)3

A cube has 6 faces and none of the smaller cubes will have all faces painted. As a matter of fact, none of the smaller cubes will have even 5 or 4 faces painted. The maximum number of faces, which will be painted on a smaller cube, will be 3. This will happen only in the case of the smaller cubes that emerge from the corners of the big cube.

So, number of smaller cubes with 3 faces painted = 8 (Always)

For 2 faces to be painted, we will have to consider the smaller cubes that emerge from the edges of the big cube (leaving out the corners). So, the smaller cubes on every edge will be (N-2n)/n. There are 12 edges in a cube.

So, number of smaller cubes with 2 faces painted = 12 * (N-2n)/n

For 1 face to be painted, we will have to consider the smaller cubes that emerge from the face of the big cube (leaving out the corners and the edges). So, the smaller cubes on every face will be [(N-2n)/n]2. There are 6 faces in a cube.

So, number of smaller cubes with 1 face painted = 6 x [(N-2n)/n]2

For no face to be painted, we will have to consider the smaller cubes that emerge from the inside of the big cube (leaving out the outer surface which was painted). Imagine this as taking a knife and cutting a slice of width n from every face of the cube. You will be left with a smaller cube with an edge of N-2n. Number of smaller cubes that you can make from the resulting cube is [(N-2n)/n]3

So, number of smaller cubes with 0 face painted = [(N-2n)/n]3

Let us take an example to elucidate this type of problem.

Example,

A painted cube is given with an edge of 15 cm. Smaller cubes are cut out from it with an edge of 3 cm each. How many cubes will have 3 faces painted, 2 faces painted, 1 face painted and no face painted.

Solution,

Total number of smaller cubes = (15/5)3= 125

3 faces painted= 8 cubes.

2 faces painted: Consider an edge of size 15 cm. We have removed the corners that take away 3 cm from each corner of the edge. Now our edge is of 9 cm. 3 cubes of 3 cm each can come from it. There are 12 edges. So, there will be 3 * 12 = 36 cubes.

1 face painted: Consider a face. If we have removed 3 cm from each edge of the face, we will be left with a square of side 9 cm or area 81 sq cm. There can be 9 smaller squares that can be formed on that face with an area of 9 sq cm each. These 9 will be the cubes which will have 1 face painted. There are 6 faces. So, there will be 9 * 6 = 54 cubes.

No face painted:Cut slices of 3 cm each from each face of the cube. We will be left with a smaller cube of edge 9 cm. Number of smaller cubes that can be formed from it is (9/3)3= 27. So, 27 cubes will have no faces painted.

You can use this to verify the formulas above and also note that 8 + 36 + 54 + 27 = 125. This means that there is no need to find out all four using the formula, just find any three of them and the other would emerge by using the total. In an exam, this might save you some valuable time.

Type 2: Matchstick Game

You are playing a matchstick game with Mr Bond. There are n matchsticks on a table. On a players turn, he can pick any number of matchsticks upto p (p is typically quite smaller than n). Whosoever picks the last matchstick loses the game. It is your turn first. How many matchsticks should you pick (assuming that you are smart and will play to win) that you will always win?

First remove 1 matchstick from consideration, as that would be the matchstick that Mr Bond will pick and lose the game.

Find out Remainder [ (n-1) / (p+1) ] = q

You should pick q matchsticks in the first turn.

After that if Mr Bond picks r sticks, you should pick p+1-r sticks and you will win the game.

Let us see an example.

Example,

There are 105 matchsticks on a table and a player can pick any number of matchsticks from 1 to 10. The person who picks the last matchstick loses the game. You are playing the game against Mr Bond and it is your turn first. How many matchsticks should you pick in the first turn such that you always win the game?

Solution: You should pick Remainder [(105 - 1)/(10+1)] = 5 matchsticks to win the game.

Round IDMr. Bond PicksSticks LeftYou PickSticks Left

Round 15100 5 = 9510 + 1 5 = 695 6 = 89

Round 2889 8 = 8110 + 1 8 = 381 3 = 78

Round 3778 7 = 7110 + 1 7 = 471 4 = 67

Round 4467 4 = 6310 + 1 4 = 763 7 = 56

Round 51056 10 = 4610 + 1 10 = 146 1 = 45

Round 6845 8 = 3710 + 1 8 = 337 3 = 34

Round 7134 1 = 3310 + 1 1 = 1033 10 = 23

Round 8223 2 =2110 + 1 2 = 921 9 = 12

Round 9912 9 = 310 + 1 9 = 23 2 = 1

As only 1 stick is left, Mr. Bond will have to pick it and lose the game. I recommend, that you try out such scenarios with a friend. Nothing validates a concept more than a real-life implementation, especially if it is on a bet. :)

For those who are still wondering what did just happen (as I did when I first read this concept), I suggest you pick up Hitchhikers Guide to the Galaxy and read about how God vanished in apuff of logic.

Questions on Seedings in a tennis tournamentThese questions can be easily found in some or the other MBA exam, some might feel these are easy marks but for some these are leave alone type.So I would try to explain the concept a bit which might help you guys in tackling these questions.Starting from the very beginning for those who are unfamiliar to such questions.The question will be like, there are 64 players in a knock out tournament and every player is ranked (seeded) from 1 - 64. And now the matches are played in such a manner that in round one the 1st seeded player plays with the 64th, 2nd with the 63rd and so on.The players who win move on to the next round whereas others are out of the competition.In second round the winner of match 1 will play winner of the last match (which was between seed 32 and seed 33), and winner of match 2 will meet the winner of second last match in round 1 and so forth.Thus after N number of rounds winner is declared.In these Questions: the term UPSET means when a lower seeded player beats the higher seed.Now the Questions: I will start with basic and gradually move up:Q1: Which seeds will play Match no 4 and Match no 9 in Round 1 of a 32 player tournament?Ans: Easy: 4 Vs 29 & 9 Vs 24 respTime for trick: Notice one thing the sum of the seedings in every match will be equal to total players + 1.i.e. 1 + 32 = 33, 2 + 31 = 33.In round of 64, sum of seeds will be 65, and in round of 16, sum of seeds will be 17. And so forth. (This will be useful in solving complicated questions)This way we can easily calculate the opponents in any round.For eg: If ques is: In a tournament of 128 players who will play 36 in round II if there are no upsets?No need to do any back calculations: Just see in Round 2 there will be 64 players. So the opponent of 36 will be = 65-36 = 29. Similarly u can calculate for anyone.Now this was without upsets, lets take a ques on upsets:Q: Who will meet Seed 68 in the Quaterfinals of a 128 format tournament, if seed 5 lost in the prequarters and there was no other upset?Now the ques seems complicated but its not if we go step by step using the above method: try once to solve the ques urself and then read further:So We know Seed 68 is in Quarters that means he has defeated Seed 129 - 68 = 61 in round 1. Now 61 would have played 65 - 61 = 4 in round 2 (Which now 68 will be playing). Now, 68 has defeated 4 as well as he is in quarters.Now look at the quarters opponent of 4 (68 will be playing with him) - its 9-4 = 5. We know 5 has lost in pre quarters where his opponent was 17 - 5 = 12.So opponent of 68 will be 12.Once you are through with the concept these questions will be a cakewalk.Funda Credit: Trueindian of NCR DT'09

Adding One point which i was neglecting when i was solving this question first time so may be other can face the same :

In UPSET when a lower seeded player beats the higher seed it captures the opponent Rank.

Like if 64 rank player beats top player means rank 1 player.Now next round 64 will be top player.(dont change the rank while solving the question just understand the logic of upset in game)concepts of RACES:-(Winner's time)/(loser's time) = (loser's time)/(winner's distance) = (beat time + start time)/(beat distance + start disatnce)when 3 runners and same length (L):-(L - X12)*X23 = L*(X13 - X12)L = length of trackX12 = 1st beat 2ndX13 = 1st beat 3rdX23 = 2nd beat 3rdwhen length of race change in each of case:-1st beat 2nd by X12 metres in a race of L1 metres1st beat 3rd by X13 metres in a race of L3 metres2nd beat 3rd by X23 metres in a race of L2 metresthen X12 , X13 , X23 are to be converted on a desired length of race say , L metersX'12 = (X12/L1)*LX'13 = (X12/L3)*LX'23 = (X23/L2)*Lthe using the formula :(L - x'12)*X'23 = L*(X'13 - X'12)

We know, that an hour hand covers 30 degrees for every hour; a minute hands covers 6 degrees for every minute and an hour hand covers 1/2 degree for every minute.

Now make a small table as given in attachment. (where you want to find the angle between hands at 'x'hours:'y'min)The angle hence would be: {6y - (30x + y/2)} or {(30x + y/2) - 6y}

Example:For 5hours, 50 min. See the table (attachment): difference = 125For 3hours, 40min. again table: difference = 130.

Pretty fast for any clock question. Try it out for a few examples. You would definitely get speed.

Problem #1A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. How many steps are there ??Sol:Let's suppose that escalator moves n steps/sec.It is given that if he walks he takes 30 sec and covers 26 steps.So in that 30 sec escalator would have covered 30n steps.Hence the total number of steps on the escalator is 26 + 30n----(1)

Similarly when he runs he takes 18 sec and covers 34 steps.So in 18 sec escalator covers 18n steps.Hence total steps on the escalator must be 34 + 18n-------(2)

Equating (1) & (2) 26 +30n = 34 + 18n we get n= 2/3Hence no. steps is 26+30(2/3) = 46.

Problem #2An escalator is descending at constant speed. A walks down andtakes 50 steps to reach the bottom. B runs down and takes 90 stepsin the same time as A takes 10 steps. How many steps are visiblewhen the escalator is not operating?

Sol: Lets suppose that A walks down 1 step / min andescalator moves n steps/ minIt is given that A takes 50 steps to reach the bottomIn the same time escalator would have covered 50n stepsSo total steps on escalator is 50+50n.

Again it is given that B takes 90 steps to reach the bottom and timetaken by him for this is equal to time taken by A to cover 10 steps i.e10 minutes. So in this 10 min escalator would have covered 10n steps.So total steps on escalatro is 90 + 10n

Again equating 50 + 50n = 90 +10n we get n = 1Hence total no. of steps on escalator is 100.

Problem #3There is a escalator and 2 persons move down it. A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps, find the no. of steps in the escalator while it is staionary.

Sol: Let A take 1 step/min.Hence B takes 3 steps/min. Let the escalator take n steps/minGiven that A takes 50 steps and hence in the same time escalator willtake 50n steps. So total no. of steps must be 50 + 50n

It is given that B takes 75 steps (which means he takes 25 min)So in the same time escalator will cover 25n steps.Hence no. of steps must be 75 +25n

Equating 50+50n = 75+25n we get n = 1.Hence total no. of steps must be 100.

P.S.: For keeping thread clean, concise and to the point, let us try not to post our queries too much in the thread. What we can do is, add a funda or two with each query you put.. or when your query is done.. edit that very post and put a funda over there.. it can save our own time searching the whole thread.. If it is possible... just a thought

PPS: I have tried many problems with this funda and have solved them successfully. I have not yet come accross any question that defies the above method but still, I am not saying it won't fail. But again, it is better than having nothing.

The 1m/s thing is a guess only (for those who asked about it in the later post as this section is added later).et A has m elements Set B has n elements.

(1)If m>n One to One func =0If n>m One to One func = nPm

(2)Number of funcs = n^mNumber of relns = 2^(mn)

(3)No of Many to One functions = Number of func- Number of One to One=n^m -nPm

(4)If n>m Number of Onto func =0If m>n Number of Onto func=1. Method for finding the last 2 digits of any number with base ending in 24(eg: 1024 ... (12345678924)^123456789

----> For any number ending with (24)^even , last 2 digits will always be 76.

----> For any number ending with (24)^odd, last 2 digits will always be odd.

The Concept will be useful in finding the last 2 digits of numbers with base 2 raised to any power as well since (2)^10 =1024 ( which ends with 24 )

eg: Find Last 2 digits of (2)^2003 ..

=> (2)^3 * ^200=> 8*(1024)^200 (which is even)=> 8*76 = 08 which is the answer.2. ______76)^ any power : Last 2 digits would always be76..

Last 2 digits of a number ending with 5

=> For even bases of 5 : eg: Numbers ending with 05,25,45,65,85 : Last 2 digits will always be25

=> For odd bases of 5 : eg: Numbers ending with 15,35,55,75,95 . There are 2 cases . Below are the 2 cases

1. Foreven powers. last 2 digits would be25. eg : (1115)^21445898 last 2 digits would be 25.

2. Forodd powerslast 2 digits would be75. eg : (1115)^21334547 last 2 digits for it would be 75.

One obvious assumption in the above case is that the numbers ending with 5 have powers > 13. Let me assume an example. Let us assume that there are 2 people A and B playing a game and they have some 35 coins to play with. The last person to pick up the coin would be declared as the winner. Minimum 1 and a maximum of 3 coins can be put by any player on the table at a time. We need to find the best winning strategy.

Now, for the above question , let us get some facts clear :

Given: Total: 35 coinsMinimum : 1 coin and maximum 3 coins.

Now , for A to win he should remove the remainder of 35/(3+1) which is 3 coins first and then he should also make sure that whatever move B makes he adds it upto 4 (1+3 ) . This concept is very simple and easy to comprehend. We can use it even in real life to check the results.

If A removes 3 coins and then its B's turn and B removes 3 coins (lets say), then in the next move A would remove 1 coin . In this way the sum would be become 4 and number of coins left with both of them would be 35 - (3+3+1) = 28.

B will never be able to win in such a scenario.

However , if in the above question insatead of 35 coins, there are 36 coins, then A would never be able to win. Because whatever the number of coins A takes out B would make sure that he makes the sum equal to 4 this time so that in the end its his turn to pick up the last coin (Note that 36 being a multiple of min+max (1+3) hence this situation has arrived, otherwise the first situation would hold valid. )

We are obviously assuming that both players are playing to win and are equally intelligent.

I hope this solves some doubts related to the coin problems...4. 1) X is the side of largest equilateral triangle that can be drawn inside the square of side 'a'a< 1.1(a)

2)If A can do some work in 'a'hr more than a&b; togather& B can do same work in 'b'hr more than a&b; togather,then both can complete work in sqrt(ab)hrs

3)Co-Efficient of a^p * b^q * c^r in (a+b+c)^n = n! /(p! * q! * r!) (n>p,q,r)

4) Area of x + p +|y +q| = a is same as area of |x|+y = a

5) Maxima & Minima of f(x)-> find f'(x)-> find value of x, taking f'(x) =0-> find f''(x) and put value of x (that we found in step-2)-> if f''(x) = +ve, it is minima at x= -ve, it is maxima at x6) Circular Race :A runs at a m/s and B runs at b m/s->Running in same direction- time for First meet = l/(a-b) (a>b, l = length of track)- time for 1st meeting at starting pt = LCM (l/a , l/b)-> Running on opp direction- time for First meet = l/(a+b) ( l = length of track)- time for 1st meeting at starting pt = LCM (l/a , l/b)-> 3 people are running (C at speed of c m/s)- time for First meet = LCM(l/a-b , l/a-c) (a>b,a>c, l = length of track)- time for 1st meeting at starting pt = LCM (l/a , l/b,l/c)Tournament Model Containing League Matches And Rounds

As we have come across many sets In the History of CAT...Tournament Model Sets Play A Vital Role...You to bear a few Strategies to Crack these kind of Questions in the Exam...

Concept

Let The Number of Teams Be 'n' And Each team Plays with Each other..Much of Like IPL and EPL...Then The Number of Matches Which Are to Be Played Are: nC2...For Example There Are 8 Teams..Then The Number of Matches = 8C2 = 28

Number of wins necessary for a team to ensure its place in next round is: +1..For Example..There Are 8 teams playing...So the Number of wins necessary for a team to ensure its place in next round is Greatest Integer Function(root(8c2)) + 1 = 5+1 = 6

If there is Football/Hockey tournament going on..And Partial Data is provided...(Remember A Hockey Set in CAT 2004)...Always Look For Key Facts:Number Of Wins = Number Of LossesSum Of Draws In A Tournament Will Always Be An Even NumberSigma(GF - GA) = sigma(GD) = 0...AlwaysIf point system like...2 for win,1 for tie to each, 0 for loss is given...Always look out for Total Number Of Points Awarded In A Tournament...Because Number of Points Awarded In A Match will always be 2..(One Team would Win and Other Will Lose or Both of them Would Have A Draw)...Total Points Awarded In a Match would be = Number Of Matches * Total Points Given Per MatchIf a variable data is given like...3 for win..And 1 each for draw...Use Equations..Here We Know that..Points Awarded In A match when it results in A win/loss = 3..And Points Awarded In A match ending in A draw = 2....Then Use Equations..Let x matches ended In Draw, and y match had definite result(either win/loss)..Then...2x+3y = Total Points Awarded...And x+y = Total Number Matches...From here..It would be Easy to find which of the matches Ended in A draw or definite result.Consider A Score Line of this Type:

Team_____GF____GA___GD___India_______ 4______4_____0____Pakistan____ 4______5_____-1___Australia____ -______-_____-1___Spain_______6______4______2___England_____ -______3_____2____Brazil_______ -______6_____-____

In here..We don't know GF by Australia, England, Brazil...GA by Australia...GD of Brazil.....But We Know that....Summation of All Goal Difference = 0So...GD for Brazil = Let say xx+0+(-1)+(-1)+2+2 = 0So x = -2Which Gives GF = 4 by Brazil...

Some Cents On Cricket Sets Too

Cricket Sets Like Football/Hockey Sets Are Similar. Instead of GF,GD,GA...we have Net Run Rate...Which is Calculated by X-Y

Where X = Total Runs Scored By A Team Till That Match/Total Numbers of Overs PlayedAnd Y = Total Runs Scored Against That Team till that Match/Total Number Of Overs Played By All the Opponents

In Simple Words...Let there Be 3 teams..India, Pakistan, Australia...Match 1 = India(301-6 in 50 overs) beat Pakistan(256-10 in 45 overs) by 45 runsMatch 2 = India(245-7 in 42 overs) beat Australia(241-5 in 50 Overs) by 3 wicketsMatch 3 = Australia(276-10 in 48 Overs) beat Pakistan(250-5 in 50 Overs) by 26 runs

Here Runs Scored By India = 546 in 92 oversRuns Scored Against India = 497 in 95 oversSo NRR = 546/92 - 497/95 = +0.7(approx)

Similarly We can calculate NRR for Australia As Well As Pakistan ...Which Comes out to be -0.11 and -0.58 respectively.....Here Also We can See that Roughly Summation of NRR of All teams = 0

ProblemsConsider A tournament having 4 teams A,B,C,D...The Table given below Shows the proceedings of the tournament. The Table shows NRR of the team after that match....

Team_____First Match_____Second Match____Third MatchA__________+0.8____________A2_____________+0.6____B__________B1______________-0.4___________+0.066__C__________+1______________+0.25__________C3_____D__________D1______________-0.5____________D3____

A team Plays its second match, once every team has played its first match..Similarly, it plays 3rd match only after every team has played its 2nd match.If A team was allout within the stipulated 50 overs, it was deemed to have batted 50 overs.All matches ended in a victory..except 1..which was a tie...In these victories, all the teams that batted first were the winners in all the matches except one which happened to be the last match played by C.

Q1. Which Teams were Involved In a tie ?1. A and D2. A and B3. B and D4. CBD

Q2. What is Value of A2 ?1. +1.32. +1.053. +0.654. +0.525

Q3. By how many runs A won the match against C1. 502. 903. 454. 25

Plz post some previous useful posts if u have come across.... It will b helpful for all of us....Together we can conquer anything...So keep pouring ideas..5. To find the number of ways in which a number say 'n' can be written as the sum consecutive natural numbers.

1.find the number of odd factors of 'n'say 'm' is the number of odd factors of n2.no.of ways will be=m-1 0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6! = 720; 7! = 5040.

Funda 1: Rightmost non-zero digit of n! or R(n!)

R(n!) = Last Digit of [ 2ax R(a!) x R(b!) ]

where n = 5a + b

Example: What is the rightmost non-zero digit of 37! ?

? R (37!) = Last Digit of [ 27x R (7!) x R (2!) ]

? R (37!) = Last Digit of [ 8 x 4 x 2 ] = 4

Example: What is the rightmost non-zero digit of 134! ?

? R (134!) = Last Digit of [ 226x R (26!) x R (4!) ]

? R (134!) = Last Digit of [ 4 x R (26!) x 4 ]

We need to find out R (26!) = Last Digit of [ 25x R (5!) x R (1!) ] = Last digit of [ 2 x 2 x 1 ] = 4

? R (134!) = Last Digit of [ 4 x 4 x 4 ] = 4

Funda 2: Power of a prime p in a factorial (n!)

The biggest power of a prime p that divides n! (or in other words, the power of prime p in n!) is given by the sum of quotients obtained by successive division of n by p.

Example: What is the highest power of 7 that divides 1342!

? [1342 / 7] = 191

? [191 / 7] = 27

? [27 / 7] = 3

? Power of 7 = 191 + 27 + 3 = 221

Example: What is the highest power of 6 that divides 134! ?

As 6 is not a prime number, we will divide it into its prime factors. 3 is the bigger prime, so its power will be the limiting factor. Hence, we need to find out the power of 3 in 134!

? [134/3] = 44

? [44/3] = 14

? [14/3] = 4

? [4/3] = 1

? Power of 3 in 134! = 44 + 14 + 4 + 1 = 63

Example: What is the highest power of 9 that divides 134! ?

As 9 is not a prime number, we will divide it into its prime factors. 9 is actually 32. The number of 3s available is 63, so the number of 9s available will be [63/2] = 31.

Highest power of 9 that divides 134! is 31.

Highest power of 18 and 36 will also be 31. Highest power of 27 will be [63/3] = 21.

Note: To find out the highest power of a composite number, always try and find out which number (or prime number) will become the limiting factor. Use that to calculate your answer. In most cases you can just look at a number and say that which one of its prime factors will be the limiting factor. If it is not obvious, then you may need to find it out for two of the prime factors. The above method can be used for doing the same.

Funda 3: Number of ending zeroes in a factorial (n!)

Number of zeroes is given by the sum of the quotients obtained by successive division of n by 5.

This is actually an extension of Funda 1. Number of ending zeroes is nothing else but the number of times n! is divisible by 10 or in other words, the highest power of 10 that divides n!. 10 is not a prime number and its prime factors are 2 and 5. 5 becomes the limiting factor and leads to the above-mentioned idea.

Example: What is the number of ending zeroes in 134! ?

? [134/5] = 26

? [26/5] = 5

? [5/5] = 1

? Number of ending zeroes = 26 + 5 + 1 = 32The Race is on

In a race of length L units, if

1) A beats B by a units and B beats C by b units, then A beats C by a+b-ab/L units

2) A beats B by a units and A beats C by b units, then B beats C by L(b-a)/L-a units

How to find different pythagorus triplets.

The Formula is a mixed fraction n

hence N can take values from 1,2.....

hence the first mixed fraction becomes 1 = 4/3One side of the pythagorus theorem is denominator which is 3 and other side is numerator which is 4, the hypotenuse is 4+1 = 5

Hence the triplet is 3,4,5.Hope this helpsslight error in the formula.. it is n + n/(2n+1)o Find out Pythagorean triplets

There is a formula

(N^2-1)/2,( N^2+1)/2 of any odd number N>1.

For Example if the number is 3, the pythagorean triplets are (3)^2-1/2 = 4and (3)^2+1/2 = 5.

For any other number , say 5, the triplets are (5)^2-1/2 = 12 and (5)^+1/2 = 13.

All other triplets are multiple of these...Last Non Zero Digit Of a factorial

This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task

Concept

Lets say D(N) denotes the last non zero digit of factorial, then the algo saysD(N)=4*D*D(Unit digit of N)D(N)=6*D*D(Unit digit of N); Where is greatest Integer Function

Find the last non zero digit of 26!*33!.Solution Scheme and ApproachD(26)=6*D*D(6)=6*D(5)*D(6)=6*2*2=4D(33)=4*D*D(3)=4*D(6)*D(3)=4*2*6=8

Hence last non aero digit of 26!*33!=4*8=2

RememberD(1)=1 D(2)=2 D(3)=6 D(4)=4 D(5)=2 D(6)=2 D(7)=4 D(8 )=2 D(9)=8

Now solve:(1)Last Non Zero Digit of 99!*26!.(2)How many zeros are at the end of 13!+14!+15!+16!+17!+18!

source quantomaniaThe generalized rule to find the no of points the diagonal passes through is (sum of length & breadth) - HCF (length & breadth).

In ur example, we have subtracted 1, which is the HCF(16,17)In this case the outer square is not given, so we will start from pi (as the rule says 4-pi-2,and the outer sqaure is not present).

Now pi=pi*r*r....equating area so that we can find the radius.

r=1

therefore circumference=2*pi*1=2pi

For the inner square the area is a^2=1..

therefore a=1

perimeter=4*a

therefore ratio=2*pi/4=pi/2Plz mohit if you could solve the below prblm based on your trick which appeared in CAT99

The figure below shows two concentric circle with centre O. PQRS a square, inscribed in the outer circle. It also circumscribes the inner circle, touching it at points B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

solution is pi/2want to know the solution thru your trick as it will be an easier approachWe assume the cube is divided into n^3 small cubes.

no. of small cubes with ONLY 3 sides painted: 8( all the corner cubes )

no. of small cubes with ONLY 2 sides painted:

A cube is painted on 2 sides means, it is on the edge of the bigger cube ,and we have 12 edges, each having n cubes. but since the corner cubes are painted on 3 sides, we need to neglect them. so in effect, for each side we will have (n-2) small cubes with only 2 sides painted.thus, then number is, 12 * (n-2)

no of small cubes with ONLY 1 side painted:

for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with only one side painted. and we have 6 faces in total.so th number is, 6*(n-2)^2

no of small cubes with NO sides painted:

if we remove the top layer of small cubes from the big cube we will end up a chunk of small cubes with no sides painted.this number will be equal to, (n-2)^3.

Also, remember for Cuboids with all different sizes, the following are the results:

a x b x c (All lengths different)

Three faces - 8 (all the corner small cubes of the cuboid)Two faces - There are two (a-2) units of small cubes on one face of the cuboid and there is a pair of such faces. Hence, number of such small cubes corresponding dimension a of the cuboid = 4(a-2).Similarly, for others.So, total with two faces painted = 4(a - 2) + 4(b - 2) + 4(c - 2)One face - Since each face of the cuboid is a combination two different dimensions, hence for the face which is a combination of a and b dimensions, the number of small cubes is 2* (a-2)(b-2)Similarly, for others.So, total with one face painted =2(a - 2)(b - 2) + 2(a - 2)(c - 2) + 2(b - 2)(c - 2)Zero faces - The entire volume of small cubes except for two cubes in each of the rows and columns will not be painted at all. hence this is the simplest ...(a - 2)(b - 2)(c - 2)

You can put different integer values for number of small cubes producing different edge lengths of cuboid to get varied results.

To verify for a cube, put a=b=c=L, you get

Three faces - 8Two faces - 12(L - 2)One face - 6(L - 2)^2Zero faces - (L - 2)^3In Logical Reasoning, very often we encounter problems based on games or tournaments. The first thing that as a CAT taker you need to realize is that the tournament-based format of questions offers the examiners a multitude of options to test you about. So, there cannot be a one-size-fits-all way of solving such kind of questions. However, if you looked at the CAT papers of the past few years you can spot some patterns. Let us discuss a couple of them.

Type 1:These questions are typically in the form of a set where the data will be either in the standard tabular format or a format which you would never find on Cricinfo or for that matter any other ESPN-like website. The different for the sake of being different format essentially tests a candidate's ability to infer data presented in unusual ways.

Example,

Each diagram communicates the number of runs scored by the three top scorers from India, where K, R, S, V, and Y represent Kaif, Rahul, Saurav, Virender, and Yuvraj respectively. The percentage number in each diagram denotes the percentage of total score that was scored by the top three Indian scorers in that game.

You can get the complete set of questionsfrom here.

I will not get into the details of solving this particular set. Once you interpret the information, the questions are really simple. The catch in this question (this type of questions) is to interpret the given data.

Let us view this same information in a different way

PakistanSouth AfricaAustralia

Runs by Kaif2851

Runs by Rahul4955

Runs by Saurav7550

Runs by Virender130

Runs by Yuvraj4087

Runs by Top 3

Top 3 as a %age of total90%70%80%

Total runs

As you can see, the triangular format is just a twisted way of representing simple tabular data. It is just a little more intimidating to see the triangles in a pressure situation and that intimidation is exactly what you should avoid. With some very simple addition and calculation operations, you would be able to solve this problem set.Bottom line:Even if it takes a couple of extra minutes, it is best to represent information using a format that you are more comfortable with.

Type 2: For some reason, Tennis appears to be a favorite among CAT exam setters. Actually Tennis does offer some very interesting possibilities such asseeds,unconventional scoringand the knockout feature. Knockouts are inherent feature of the sport of Tennis and hence used frequently by exam setters.

Note: In a knockout tournament,

No of matches = No of players 1

Let us look at few ideas related to questions on seeded players. Lets say in a tournament there are n players and they are seeded (ranked) from 1 to n. Typically this n is a power of 2 such as 32, 64 or 128.

In the first round the highest seeded player plays the lowest seeded player, the second highest seeded player plays the second lowest seeded player and so on. To put it into perspective,

Round 1 Match 1 Seed 1 Vs Seed n

Round 1 Match 2 Seed 2 Vs Seed (n-1)

Round 1 Match 3 Seed 3 Vs Seed (n-2)

.

.

.

Round 1 Match n/2 Seed n/2 Vs Seed (n/2) + 1

In the second round, the winner of Match 1 plays the winner of Match n/2; winner of Match 2 plays the winner of Match n/2 1 and so on.

In this kind of a question, an upset happens when a lower seeded player beats a higher seeded player.

The questions are typically of the format,

Q: Who will play match 36 in Round 1?

A:It will be played between the 36thhighest seed and the 36thlowest seed.

The 36thlowest seed can be sometimes difficult to figure out but you can figure it out easily by calculating (n+1) 36.

Note:The rthmatch in Round 1 will be played between Seed r and Seed n+1-r

Q:If there are no upsets, then in Round 2 who will play the 5thmatch?

A:One way of solving this question would be to figure out the winners of Round 1 and then figuring out the 5thfrom the top and the bottom.

If there is no upset, then seed 5 will be there. The other player would be (n/2+1 5)

Note:If there are no upsets, then the rthmatch in the pthround will be played between,

Seed 'r' and Seed (n/2p - 1) + 1 - r

Q:Who will meet Seed 37 in the Quarter-finals of a tournament in which 64 players are taking part? Other than Seed 37s matches, there were no other upsets.

We first need to analyze which round would be the quarterfinal,

Round 1 (32 matches), Round 2 (16 matches), Round 3 (8 matches pre-quarter), Round 4 (4 matches quarterfinals).

In Round 1, Seed 37 must have defeated 64 + 1 37 = 28

In Round 2, Seed 37 played the match that Seed 28 would have played. Seed 28 would have played against Seed 32 + 1 28 = Seed 5

In Round 3 (pre-quarters), Seed 37 played the match that Seed 5 would have played. Seed 5 would have played against Seed 16 + 1 5 = Seed 12 and won it.

In Round 4 (quarterfinals), Seed 37 would meet the player that Seed 5 would have met. Seed 5 would have met 8 + 1 - 5 = 4.

Hence, Seed 37 will meet Seed 4 in the quarterfinals.

As a matter of fact, even the above solution is not the most optimal one. Because once you realize that Seed 37 defeated Seed 5, he would keep meeting the opponents that Seed 5 would have met.

I think I have taken enough of your time with this lengthy post but this time would be well spent if such a question appears in CAT or some other management entrance exam.Question:Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament was conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprised several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup.

The tournament rules such that each match results in a winner and a loser with no possibility of a tie. In the first stage, a team earns one point for each win and no points for a loss. At the end of the first stage, teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams form each group advanced to the next stage.

Now questions were asked on: Total number of matches, minimum number of wins required for a team to guarantee advance (or possible advance) to next stage, maximum number of matches that a team can win in the first stage without advancing, etc.

In first stage, teams are divided into two groups of 8 teams each. There they play a match against everyone exactly one ie8C2matches in every group. So 2 *8C2= 56 matches for the first stage.

In second stage, there are 8 teams in a knockout stage. There will be one winner, so 8 1 = 7.

So,total number of matchesis 56 + 7 = 63

For a team to advance to the second stage, it should be among the top 4 in its group. Total points on stake in a group is the same as the total number of matches which is8C2= 28. To guarantee advance, it can have 3 teams with the same or more points. There can be 5 teams with 5 wins or 5 points. So, 5 wins is not good enough to ensure a birth in round 2. However,6 wins will guaranteeits advance. This also tells us that a team might have5 wins but still not advance.

To figure out the minimum wins required to possibly advance, let us look at the method for n teams.

(n/2) 1 teams should win maximum Number of matches.

(n/2) + 1 teams should have exactly the same number of wins.

So in this question, the top 3 teams can have a maximum of 7 + 6 + 5 = 18 points.

All other teams (5) have a combined score of 28 18 = 10 points. Their individual score is 2 points each and one of these five teams will advance to second stage.

So,minimum wins required to advance is 2.

Let us look at another type of question in which we are given a table and we have to fill it. Given below is a random table at the end of hockey tournament. For each win two points were awarded and for a draw one point was given. We also know that the South Africa Spain match was a draw. No two teams have the exact same count for Win/Draw/Loss and Australia has won more matches than Spain. Figure out the result of every match from the table given below,

Team NamePlayedWonDrawLostPoints

India06

Pakistan6

Australia3

Spain3

South Africa

Each team has played 4 matches.

A team can get a score of 6 in two ways,

3 Wins and 1 loss or 2 wins and 2 draws.

India did not lose, so it will have 2 wins and 2 draws whereas on the other hand Pakistan will have 3 wins.

Number of matches played will be5C2= 10

The total Number of points at stake is 20. South Africa has the leftover points which is 2.

We also know that the South Africa Spain match was a draw.

Team NamePlayedWonDrawLostPoints

India42206

Pakistan43016

Australia43

Spain4Min(1)3

South Africa4Min(1)2

2 Points can be achieved by 1 Win, 0 Draw, 3 Loss OR 2 Draw, 2 Loss.

We know that both Spain & South Africa have at least 1 Draw. This means that South Africas 2 points are by 2 Draw, 2 Loss.

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CAT 2011 Logical Reasoning: Scoring goals in questions based on Football & Hockey Games & TournamentsRavi Handa |18 October 2011| Tags:MBA test prep,Logical reasoning,Games and Tournaments

In myprevious post, we discussed two common types of games and tournament questions. So, if you are looking for questions on new types of data representation or questions based on seeding in a tennis tournament, do read that post. However, there is another frequently appearing class of questions in the games and tournaments category Football/Hockey tournament questions in which we have to find out Goals scores, winners, ties, etc.

In such tournaments, all competitors play a fixed number of matches. Points are awarded for wins/draws/losses. Then an overall ranking is calculated using the total points or average points per match. Sometimes other factors such as goals scored/goals faced also come into the picture to resolves ties in ranking.

Let us look at a question from CAT 2000 (full set of questions here).

Question:Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament was conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprised several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup.

The tournament rules such that each match results in a winner and a loser with no possibility of a tie. In the first stage, a team earns one point for each win and no points for a loss. At the end of the first stage, teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams form each group advanced to the next stage.

Now questions were asked on: Total number of matches, minimum number of wins required for a team to guarantee advance (or possible advance) to next stage, maximum number of matches that a team can win in the first stage without advancing, etc.

In first stage, teams are divided into two groups of 8 teams each. There they play a match against everyone exactly one ie8C2matches in every group. So 2 *8C2= 56 matches for the first stage.

In second stage, there are 8 teams in a knockout stage. There will be one winner, so 8 1 = 7.

So,total number of matchesis 56 + 7 = 63

For a team to advance to the second stage, it should be among the top 4 in its group. Total points on stake in a group is the same as the total number of matches which is8C2= 28. To guarantee advance, it can have 3 teams with the same or more points. There can be 5 teams with 5 wins or 5 points. So, 5 wins is not good enough to ensure a birth in round 2. However,6 wins will guaranteeits advance. This also tells us that a team might have5 wins but still not advance.

To figure out the minimum wins required to possibly advance, let us look at the method for n teams.

(n/2) 1 teams should win maximum Number of matches.

(n/2) + 1 teams should have exactly the same number of wins.

So in this question, the top 3 teams can have a maximum of 7 + 6 + 5 = 18 points.

All other teams (5) have a combined score of 28 18 = 10 points. Their individual score is 2 points each and one of these five teams will advance to second stage.

So,minimum wins required to advance is 2.

Let us look at another type of question in which we are given a table and we have to fill it. Given below is a random table at the end of hockey tournament. For each win two points were awarded and for a draw one point was given. We also know that the South Africa Spain match was a draw. No two teams have the exact same count for Win/Draw/Loss and Australia has won more matches than Spain. Figure out the result of every match from the table given below,

Team NamePlayedWonDrawLostPoints

India06

Pakistan6

Australia3

Spain3

South Africa

Each team has played 4 matches.

A team can get a score of 6 in two ways,

3 Wins and 1 loss or 2 wins and 2 draws.

India did not lose, so it will have 2 wins and 2 draws whereas on the other hand Pakistan will have 3 wins.

Number of matches played will be5C2= 10

The total Number of points at stake is 20. South Africa has the leftover points which is 2.

We also know that the South Africa Spain match was a draw.

So, now our table looks like this,

Team NamePlayedWonDrawLostPoints

India42206

Pakistan43016

Australia43

Spain4Min(1)3

South Africa4Min(1)2

2 Points can be achieved by 1 Win, 0 Draw, 3 Loss OR 2 Draw, 2 Loss.

We know that both Spain & South Africa have at least 1 Draw. This means that South Africas 2 points are by 2 Draw, 2 Loss.

Team NamePlayedWonDrawLostPoints

India42206

Pakistan43016

Australia43

Spain4Min(1)3

South Africa40222

3 Points can be achieved by 1 Win, 1 Draw, 2 Loss OR 3 Draw, 1 Loss.

As no two teams have the same Win/Draw/Loss count, one of the above applies to Australia whereas the other one applies to Spain. As Australia has won more matches, it will get the 1 Win, 1 Draw, 2 Loss.

Our final table will look like this,

Team NamePlayedWonDrawLostPoints

India42206

Pakistan43016

Australia41123

Spain40313

South Africa40222

Now, let us try and analyze the match results for the 10 matches (in no special order).

Pakistan has won 3 and lost 1. Pakistan cannot win against India as India did not lose a match. So,

Match 1 India beat Pakistan

Match 2 Pakistan beat Australia

Match 3 Pakistan beat Spain

Match 4 Pakistan beat South Africa

We also know the result of South Africa Vs Spain,

Match 5 South Africa & Spain drew the match.

Spain has lost against Pakistan and it needs to draw all other matches.

Match 6 India & Spain drew the match

Match 7 Australia & Spain drew the match

Australia cannot draw another match as it has only 1 draw. It cannot win against India as India has no losses. So, it must have lost against India and the win must have come against the remaining team ie South Africa.

Match 8 India beat Australia

Match 9 Australia beat South Africa

The only match remaining between India & South Africa must have been a draw as India scored wins against Pakistan and Australia.

Match 10 India & South Africa drew the match

Phew!! I hope that you lasted this long without actually playing the game of Banging head against the wall.