Download Tournament Funda There are 16 teams and they are divided into 2

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 solve
1.
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=5
So 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.
So guys whenever a similar questions appears you should not be taking more than 30 seconds.
Round Robin/League Games:
Suppose A,B,C,D,E,F play each other exactly once. Top 2 Qualify for Finals.
4 Standard Questions:
Q1. Find total no. of matches.
Q2. Min. no. of matches a team must win in order to have a chance of qualifying.
Q3. Max. no. of matches a team can win and still not qualify.
Q4. Min. no. of matches a team must win to reach next round undisputed/ guaranteed.
A1. Easiest of the lot.
A2. If top "k" teams are supposed to qualify for the next round then choose top k-1 teams and make
them win as many matches as possible. Distribute the remaining points equally/ as equal as possible.
A3. If top "k" teams are supposed to qualify then choose top k+1 teams and make them beat every other
team then you'll be left with matches between k+1 teams. Distribute as equally as possible.
A4. A3 + 1
For the example quoted.
A1. 6C2
A2.
A5B2C2D2E2F2
Answer is 2.
A3.
A4
B4
C4
Answer is 4
A4. Answer is 5.
UNDERSTANDING PnC, difference between identical and distinct
1. how many ways one can put
(a) 4 different boll in 4 different bag
(b) 4 identical ball in 4 different bag
(C)4 different boll in 4 identical bag
(d)4 identical boll in 4 identical bag
1) 4 identical balls in 4 identical bags: (This is easy one)
5 ways:
(4, 0, 0, 0), (3, 1, 0, 0), (2, 2, 0, 0), (2, 1, 1, 0) or (1, 1, 1, 1)
(2) 4 identical balls in 4 different bags: (This is also easy one)
It is like this:
a+b+c+d = 4
where a, b, c and d are number of balls in each bag.
So, 7C3 ways
(3) 4 different balls in 4 different bags: (This is also easy one)
Each ball can go to one of the 4 bags. So, it has 4 choices. So, it will be 4*4*4*4.. 4 choices for each.
So, 4^4.
(4) 4 different balls in 4 same bags: (This is calculation bsed..)
This is based on first case (1)
(4, 0, 0, 0) => 1 way only
(3, 1, 0, 0) => 4C3 ways
3 balls to be grouped can be chosen in 4C3 ways and remaining one in 1 way
(2, 2, 0, 0) => 2 balls can be chosen in 4C2 ways.
So, 6 ways. But, as we select 2, at the same time we make one more pair of remaing 2. So, at one time,
we take care of 2 cases. So, we can take 4C2/2 = 3 ways.
(2, 1, 1, 0) => 4C2
4C2 ways to chose 2 balls and then reaming will be 1 and 1.
(1, 1, 1, 1) => 1 way only
METHOD FOR OBTAINING SOLUTIONS TO LARGE EQUATIONS
let the eqn be 7x+4y = 41
to find out possible combinations , we are using remainder funda
divide by 4
so rem(7x/4) = rem(41/4)
hence x =3 thn find y= 5
its even better for much bigger values
e.g. 99x+2y=5481
here rem(2y/99)=36
y =18 n find x
now we can find a series of solutions to find next solution
just cross add n subtract othr , 7x+4y=41 we got 1st solution as x=3 y=5
nxt 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)
TO FIND THE SQUARE ROOT OF A NUMBER
to find sqrt
nearest square root +/- (depends u"ll see) (difference /2*nearest square root)
sqrt 23
nearest number = 25
sqrt 25 =5
so
diff=( 25-23)=2
5 - 2/(2*5)
5 - 1/5
4.8*4.8 = 23.04
ZELLER’S RULE
Zeller's rule is used to calculate the day on which any date falls for any year. With this technique you will
have the calendar for any given year available to you.
The rule is as follows
F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C
K = Date => for 25/3/2009, we take 25
In Zeller's rule months start from march.
M = Month no. => Starts from March.
March = 1, April = 2, May = 3
Nov. = 9, Dec = 10, Jan = 11
Feb. = 12
D = Last two digits of the year, using previous year for January and February. Thus for 2009 = 09
C = The first two digits of century => for 2009 = 20
Rules:
1.In Zeller's rule the year begins in March and ends in February. Hence, the month number from March
is 1, April is 2, May is 3 and so on up to January, which is 11, and February is 12.
2.January and February are counted as the 11th and 12th months of the previous year. Hence, if you are
calculating the day of any date on January 2026, the notation will be (month=11 and year= 2025) instead
of (month=1 and year=2026).
3.While calculating, we drop off every number after the decimal point.
4.Once we have found the answer we divide it by 7 and take the remainder. Remainder 0 corresponds
to Sunday; Remainder 1 corresponds to Monday ; Remainder 2 corresponds to Tuesday and so on....
Example:
Find the day on 26th June 1983
F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C
Here k=26, m=4, d=83, c =19
f= 26+(13*4-1/5)+83+83/4+19/4-2*19
= 105
105 divided by 7 leaves a remainder 0. Hence the day is a Sunday!
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)
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 - 8
Two faces - 12(L - 2)
One face - 6(L - 2)^2
Zero faces - (L - 2)^3
SEEDING IN TENNIS TOURNAMENTS
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.
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 resp
Time 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.
If ques: 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.
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.
REMAINDER CONCEPTS
1)If we need to form an AP with all primes, then there cant be more than 10 primes in that
AP. (The logic is something about the density of primes, google it if u like).
2)Any number repeated 3k times is perfectly divisible by 37.
3)Any number repeated 6k times is perfectly divisible by 11 & 13.
(2) and (3) remainders mein kaafi kaam aate h.
----------------------------------------------------------------Remainder funda.
Euler's Th.
Eg - N^n mod m.
Condition to apply Euler - N and m should be co-prime.
Euler of any number = Number*(1-1/p1)(1-1/p2).....
where, p1 etc are num of distinct prime factors.
E(m) = k(say)
the method is like this.
n mod k = p(say)
and N mod m = q (say)
so our prob will reduce to
q^p mod m.
Wilson's Th.
(p-1)! %p = -1
(p-2)! %p = 1 (this is wilson's corollary)
Fermats little Th
Any number repeated p-1 times is completely divisibile by p.
where p is a prime > 5
CHINESE REMAINDER THEOREM
This is nothing but here we break the denominator (generally very large) into its prime
factors (generally) or it can be non prime too. keeping in mind the numerator.
Then we use the Euler's to find the remainder.
Eg - x mod y
we break y into p,q and r.
now, x mod p = a
x mod q = b
x mod r = c
then our remainder wld be of the form.
pk1+a=qk2+b=rk3+c
now subsitute values for k.
jab ek common value aa jaye toh wo remainder.
NATURE OF ROOTS
If the coefficients of an equation are all positive then the equation has no positive root.
Hence, the equation 9x^5 + 6x^3 + 5x^2 + 3x + 1 = 0 has no positive root.
If the coefficients of even powers of x are all of one sign, and the coefficients of the odd
powers are all of opposite sign, then the equation has no negative root.
Hence, the equation 16x^4 - 12x^3 + 15x^2 - 12x + 1 = 0 has no negative root
If the equation contains only even powers of x and the coefficients are all of the same sign,
the equation has no real root.
Hence, the equation 4x^4 + 4x^2 + 2 = 0 has no real root.
If the equation contains only odd powers of x, and the coefficients are all of the same sign,
the equation has no real root except x = 0.
Hence, the equation 7x^5 + 6x^3 + x = 0 has only one real root at x = 0.
Descartes' Rule of Signs : An equation f(x) = 0 cannot have more positive roots than there
are changes of sign in f(x), and cannot have more negative roots than there changes of sign
in f( - x).
Thus the equation x^4 + 7x^3 - 4x^2 - x - 7 = 0 has one positive root because there is
only change in sign.
f( - x) = x^4 - 7x^3 - 4x^2 + x - 7 = 0 hence the number of negative real roots will be
either 1 or 3.
NUMBER OF SQUARES ON A CHESS BOARD
To find the number of axa squares on a chess board a formula of (9-a)^2 can be used.
For eg: the number of 1x1 squares will be (9-1)^2 = 64
the number of 2x2 squares will be (9-2)^2 = 49
and so on...
the total number of squares in a chess board will be n^2 upto 8 = 204
The total number of rectangles in a chess board will be
n^3 upto 8 = 1296
CONCEPT OF INEQUALITY
One concept of inequality from my side
n^(n+1) > (n+1)^n for all n>3.
Ex:- 1999^2000 > 2000^1999
99^100 > 100^ 99
(P.S if n = 3 , 2^3 < 3^2)
BASIC CONCEPT FUNDAS
1. If there are n number of straight lines , They intersect each other in nc2 ways
2. If there are m number of circles , They intersect each other in
2*(mc2) ways = m (m-1)= mp2 ways
3. When n straight lines and m circles intersect each other , they intersect in
at most nc2 + mp2 + 2.m.n
4. When n parallel lines intersect m straight lines , Then no. of parallelograms possible =
nc2 * mc2= mn (m-1) (n-1)/4
CONCEPT OF SUM OF N DIGIT NUMBERS FORMED BY A,B,C….
How to find the sum of all possible 4-digit numbers formed with 1,3,5,7?
sol. 6(1+3+5+7)(1+10+100+1000)= 106656
general formula for this is (n-1)!(sum of digits)(111....n times)
but when the digits are like 0,1,2,3,4 and we are asked to find the sum of all possible 5digit number in this case...then the answer to this question would be 24*10*11111 6*10*1111
CONCEPT OF MEETING POINTS IN CIRCULAR MOTION
A and B running around a circular path .Time to complete 1 lap for A-20mins ,B-8 mins
How many unique meeting points in the path??
Shortcut - Take speed ratio of A and B to smallest coprimes ---> 8:20=2:5
If A and B running in same direction no. of unique meeting pt.s will be difference of the
smallest ratio. So here 5-2=3 unique pts.
A and B running in opp. dir. no. of unique meeting pt.s will be sum of the smallest ratio. So
here 5+2=7 unique pts.
Basically we need to see before 1st meeting at starting pt (after start of the race) the faster
person has taken lead of how many full laps.That num is the required unique num of
meeting pts.
TRIANGLES: SCALENE & OBTUSE
Triangles: Scalene, Acute, Obtuse, Right angled: How to find the 3rd side if 2 sides are
given
Suppose a,b & c (suppose largest of the 3) are the 3 sides of the triangle then one basic
rule which always follow is (a+b)>c.
The triangle is right angled if c^2=(a^2 +b^2)
The triangle is acute angled if c^2<(a^2 +b^2)
The triangle is obtused angled if c^2>(a^2 +b^2)
The triangle is scalene if all 3 sides are of unequal length.
LAST NON ZERO DIGIT OF A FACTORIAL
Lets say D(N) denotes the last non zero digit of factorial, then the algo says
D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]
D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest
Integer Function
Find the last non zero digit of 26!*33!.
Solution Scheme and Approach
D(26)=6*D[26/5]*D(6)=6*D(5)*D(6)=6*2*2=4[D(5) means last non zero digit of 5!=120
which is 2, same for D(6)]
D(33)=4*D[33/5]*D(3)=4*D(6)*D(3)=4*2*6=8
Hence last non aero digit of 26!*33!=4*8=2
Remember
D(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
GENERALIZED RULE TO FIND NUMBER OF POINTS THE DIAGONAL PASSES THROUGH IN
ANY RECTANGLE
The generalized rule to find the no of points the diagonal passes through is (sum of length &
breadth) - HCF (length & breadth).
EG. for example in case of a rectangle of sides 16 n 17 n having square of area 1 , the
diagonal will pass through (16+17)-1 i.e 32 squares as 16 is co prime to 17
SUCCESSIVE CHANGES
Two successive change of a % and b % is equivalent to
a + b + ab/100
The sign matters. Increase/Profit gives +ve sign. Decrease/Loss gives –ve sign.
COIN PROBLEM
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 coins
Minimum : 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. )
ESCALATOR PROBLEMS
Problem #1
A 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= 2/3
Hence no. steps is 26+30(2/3) = 46.
Problem #2
An escalator is descending at constant speed. A walks down and
takes 50 steps to reach the bottom. B runs down and takes 90 steps
in the same time as A takes 10 steps. How many steps are visible
when the escalator is not operating?
Sol: Lets suppose that A walks down 1 step / min and escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom. In the same time escalator would
have covered 50n steps. So total steps on escalator is 50+50n.
Again it is given that B takes 90 steps to reach the bottom and time taken by him for this is
equal to time taken by A to cover 10 steps i.e 10 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 = 1
Hence total no. of steps on escalator is 100.
Problem #3
There 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/min
Given that A takes 50 steps and hence in the same time escalator will
take 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.
FACTS, INFERENCES AND JUDGEMENTS
FACTS:
1.They can be verified or discovered
2.Applicable to all Universal truths.
3.Anything that is seen, heard or read is a fact.
4.Anyone elses opinion stated by the author is a fact. (The personal opinion will become a
judgement)
5.Figures, Statistics or other data without an opinion attached to it is a fact. ( with an
opinion attached becomes an inference).
INFERENCES:
1.A logical conclusion based on a set of facts. (Logical conclusion meaning, anyone seeing
those set of facts should be able to arrive at that.)
2.Change the facts and the conclusion will no longer be valid.
3.Inferences are self-explanatory. It tells you why. (If you have a statement which looks
like an inference and you form a why question on it, you should be able to get an answer. If
you dont, it will become a Judgement.)
4.An inference is an opinion that doesnt change according to perspective.
5.If the set of sentences given to you are connected, you can only do UPLINKING to decide
that a statement as an inference. Meaning You should have the facts preceeding this
statement. If the facts are provided as the next statements it cannot be an inference. It will
be a judgement.
Ex:
A.X is a good student when it comes to attendance.
B.X attends college regularly.
In this example, statement A is a judgement, statement B is a fact.
If you change the order like below
A.X attends college regularly.
B.X is a good student when it comes to attendance.
In this case A is a fact and B is an inference.
6.One fact can lead to multiple inferences.
7.Inference will have verbal bridges with facts ( words like leads to, points to, indicates,
shows, explains, expresses)
8.A cause and effect relation ship can be termed as an inference.
9.A statement backed up by an example in the same sentence will be an inference.
JUDGEMENTS:
1.Any statement that could lead to an approval or disapproval.
2.It is the authors personal opinion.
3.It is open to challenge.
4.It doesnt tell you why. ( see point 3 of inference for clarity)
5.It changes according to perspective.
A fact or an inference or a judgement will be as per the above mentioned rules but it is not
mandatory that each satisfies all the conditions.
Some examples:
-Rahul says that he likes federers back hand. FACT (Someone elses opinon)
-I like federers back hand JUDGEMENT (Authors opinon)
-Galloping real estate prices and hardening interest rates mean difficult times for the great
Indian middle class. INFERENCE ( Notice the verbal bridge mean)
-The maharastra government seems to have developed a sweet tooth having pumped in
hundreds of crores to sugarcane, the cash starved government is now set to help mango.
INFERENCE ( maharastra government has developed sweet tooth.. why? You get an
answer)
-The current season is likely to see the output drop to less than 3 lakh tonnes. JUDGEMENT
( the phrase is likely means the authors opinon. And doesnt tell why the output is going to
drop)
-The Clinton camp feels that the attack was largely an act of nervousness on the part of
senator obamas camp because senator Clinton was ahead in the opinon polls to win the
democratic nomination to run the president. FACT (the phrase the Clinton cam feels we can
verify whether Clinton camp feels so or not)
A- The rise of Indian currency will impact exporters adversely FACT (rise of currency will
always impact exporters adversely universal truth)
B- Exporters will need to consider how to protect their operating margins- INFERENCE (by
uplinking)
BASE SYSTEMS
Let's start with having a look into the most familiar number system - the one with base 10.
We know that (5678 )10 = 5*10^3 + 6*10^2 + 7*10 + 8
The points to remember here are:
1.The unit digit of this number system represents the remainder when the number (e.g.
5678 ) is divided by base, i.e. 10.
2.The last two digits give us the remainder when we divide the number by 100, i.e. 10^2.
3.There can be no digit more than 9 (i.e. base-1) in a system of numbers with base 10.
Let us analyze the same set of rules for base-2 number system, another one of those very
frequently used.
For this, we will analyze (1011000101110)2, which is binary equivalent of (5678 )10.
1.The unit digit of this number system represents the remainder when the number is
divided by base 2, i.e. (10)2.
2.The last two digits give us the remainder when we divide the number by 2^2, i.e. 102.
3.There can be no digit more than 1 (i.e. base-1) in a system of numbers with base 2.
4.(1011000101110)0 > (5678 )10, i.e. value in 2 base is numerically more than the value
in 10 base.
The above few points will solve each and every question based on bases if little generalized
for a system of base n.
1.The unit digit of any number system represents the remainder when the number is
divided by base n, i.e. (10)n.
2.The last two digits give us the remainder when we divide the number by n2, i.e. 10^2.
3.There can be no digit more than n-1 (i.e. base-1).
4.The numerical value of a particular number in base n is always greater than the numerical
value of the same number in any base greater than n.
For egg: since base 2 is less than base10, (1011000101110)10 > (5678 )10.
And this brings us to one minor rule,
5.Two numbers can only be compared if they are in the same base.
Q. If Z is a set of three digit numbers of base 10 which are divisible by 3 and end with 1
when written in base 5 and base 7 notations, then how many elements does Z have?
Since the numbers when represented in bases 5 and 7 leave 1 at the unit's place, we
deduce that when the numbers are divided by 5 or 7, remainder is 1. (point 1)
Or, the number is of the form 5k'+1 and 7k''+1, i.e. 35k+1.
Since number is a 3 digit number divisible by 3, we have 35k+1 = 3m, where k>2
Solving this, we get values of k as : 4, 7, 10 28, i.e. total 9 terms.
Once we are thorough with the number system concepts, we can put it to use in many
ways. Following are few such applications:
1.When a number is getting skipped every time.
A faulty speedometer skips at 6 and 7. Ram didn't know that and he quotes : "Our trip
yesterday was 1234 km long as per the speedometer ratings". What was the actual length
of the trip?
Since 6 and 7 are getting skipped every time, we can conclude that there are only 8 digits
that can be displayed on speedometer. Or speedometer functions in base-8 number system.
Therefore, the actual length of the journey = (1234)8 = 1*83 + 2*82 + 3*8 + 4 = 512 +
128 + 24 + 4 = 668 km.
2.Notorious King problem:
A king has 1000 bottles of wine out of which 1 is poisoned. If king, being ruthless and
clever, wants to test his wines on his prisoners, find the minimum number of prisoners that
will be required to carry out this experiment.
You can notice that a bottle will be either poisonous or non-poisonous - 2 possibilities.
So, all we have to do is to represent 1000 in base 2. Highest power of 2 that is greater than
1000 will be the number of prisoners.
Since 210 > 1000, Minimum number of prisoners required = 10.
3.Case of negative weights:
We have to measure all the weights from 1 to 728. What is the minimum number of unique
weights required to achieve this?
Here a particular weight can be used in 3 ways - positive, negative or not used at all.
Therefore we need to find n where 3n >728
=> n = 6
Therefore the unique weights are 1,3,9,27,81,243. Using these weights, you can now
measure anything from 1 to 728.
LIAR PROBLEMS
See, Few Basic Fundas Evolve Whenever We Do Such Kinda Problems. Few Things You have
to keep in mind, And Problems are solved in less than 5 minutes.
First Condition: A Truth Teller Never Lies. It Means A Truth Teller, Will Never Say Like this "I
am A Liar"...He Will Always Say - "I Speak Truth" Or "He Lies" and so on
Second Condition: The Best part of these questions is A Liar also never says that "I am a
Liar"...He Will Always Say...He is A Liar Or He Speaks Truth...He Will Never Talk Of Himself
If He is Lying.
Only These Two Conditions And Problem is Solved.
Have, An Example:
Three Persons, A,B,C are either truth teller, liar or alternator.
A says: I am A Liar. C is A Truth Teller.
Then What Is B ?
Ans. Let Us Assume, A Is Truth Teller. But he can't be, his both statements contradict. So
he is for sure not a truth teller.
Assume, A is Liar. But Liar Never says, he lies...So he can't be Liar Also
So, We conclude, A is Alternator. Hence His Statements Must be, one false, and one true.
Now
A's First Statement: I am A Liar - False. He Is An Alternator
A's Second Statement: C is Truth Teller - Has To Be True.
So, B is A Liar.
So, remember these two commandments. And your life would be easy.
LAST MINUTE READS – A NAGA25FRENCH COMPILATION
Sourced from
Pagalguy.com
http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t43729/p-1642787/r-1742160
1. THE BASE SYSTEM
Concept of a digital system. It will be clear to everybody that it was an arbitrary
choice to think of the digits as being multiplied by the various powers of 10. Here we
can replace the number 10 by any other whole number greater than 1. But we must
be consistent and multiply the various digits by the powers of the same number. This
number is called the base, while the method of writing numbers with the aid of such
a base (hence the positional system) is called a digital system. If 10 is its base, we
speak of the decimal system. It is obvious that 10 was chosen for the base because
the fingers are a preeminently suitable aid in counting (and also adding). When
someone has used all fingers of both hands, then he has to start the counting anew,
remembering that he has reached one" decade."
Counting on the fingers might alternatively have led to the base 5; this happens not
to be the case, probably because the base 5 is somewhat small. In counting on the
fingers, the base 5 would have the advantage that we could count the units on one
hand, indicating every multiple of 5 by a finger of the other hand. In this way, we
could count up to 25 on both hands together. In the digital system with the base 5
we have only the digits 0,1,2,3,4. In this system, the numerals are written down
according to altogether the same method, and with as much ease, as in the decimal
system, namely in the following way: 1-2-3-4-10-11-12-13-14-20-21-22-23-24-3031-32-33-34-40-41-42-43-44-100-101-102-103-104-110-111-112-113-114-120121-122-123-124-130 etc.
As you see, numbers having many digits arise earlier than in the decimal system,
and this has to be considered as a drawback of the system with the base 5. From the
foregoing it appears that the base of the system is the smallest number that is
written with two digits, that is, as 10. The second (third) power of the base is the
smallest number that is written with 3 (4) digits, that is, as 100 (1000).The writing
of the numerals occurs according to the same method in any digital system. If the
base is larger than 10, then we must devise new symbols, as for ten and for eleven
in the duodecimal system(BASE 12). If we take 't' and 'e' as these symbols, the
initial letters of the corresponding words, then the numerals will be written as
follows:
1-2-3-4-5-6-7-8-9-t-e-10-11-12-13-14-15-16-17-18-19-1t-1e-20-21-22......29-2t2e-30-31.......3t--3e-40-41-.......-4e-50-
CONCEPT
Say a person has to measure all the weights from 1 kg to 127 kg and only 1 side of a
balance weight can be used to put the weights. What is the minimum number of weights
that he needs?
The highest weight that has to be measured is 127. The least power of 2 greater than 127 is
7(say n). Now the minimum number of weights required is n=7.
These weights are in powers of 2. In this case the weights are 2^0=1, 2^1=2, 2^2=4,
2^3=8, 2^4=16,2^5=32, 2^6=64.
Now let us that weights can be put in both sides of a balance weight. In this case the
weights are in powers of 3.
3^0=1, 3^1=3, 3^2=9, 3^3=27, 3^4=81 and 3^5= 243.
Hence the minimum number of weights required would be 6.
In case, only one side is used for putting weights, if we assume that we have n number of
weights, then each of these n weights can have only 2 states, either they are being used on
the one usable side, or they are not. That implies a binary state. Its like a yes or no switch,
the coefficients are either 0 or 1, for each of these n weights.
So, we use a binary system, or as put forward by mipul, err, kid, is nothing but using
exponents of 2.
This can also be extended to cases like, if we have N bags containing fixed amount of
money, and the total money in all these bags combined is 100 rupees. Now, what can be
the minimum possible total number of bags, such that one is able to pay any amount
ranging from 1 rupee to 100 rupees, the constraint being that when u pick up a particular
bag, all the money in the bag has to be transferred, that is, partial transfers aren't allowed.
In the second case, where both the sides can be used for measurement, the number of
possible states is 3. Either it is not used, or it is on side 1, or on side 2. the coefficients, can
be thus, +1, 0, or -1. Thus, we use a trinary system, or powers of 3. Those who have
studied logic circuits, should be at more relative ease of grasping this topic.
This can be extended to cases where, both the parties dealing with the transactions have
moneybags, and not just one like i mentioned above.
Now attempt these questions:
1. Sammael had 3000 bananas with him and he had to take them to a market 1000 km
away on his only camel, which can carry a maximum of 1000 bananas at a time. The camel
eats 1 banana for every 1km, or a part thereof, that it travels while carrying the
bananas.What is the maximum number of bananas that Sammael can take to the market?
2. There are 1000 bottles containing water. One of the bottle contains poison. There are
1000 prisoners available who are willing to drink the water in order to find out the bottle
containing poison. Each person can drink the water only once from any bottle or any
combination of bottles.
(a).What is the minimum number of prisoners who must drink the water in order to
determine the bottle containing poison?
(b). Now minimum number of people are selected to drink the water. All those prisoners
who survive after drinking water are set free. What is the minimum and maximum number
of prisoners that will be set free?
Answers:
1. 833
Sammael, the Camel wala, has given a nice explanation for this. So i advise all to refer to
his post.
Now if anyone has no idea about how to solve this one, then try trial and error method. But
remember that u must not allow the camel to cover more than 500 km at a stretch. That
would result in loss of bananas. So when doing by trial and error method always first think
of 1/3 or 1/2 of the distance.
In this case we first cover 333 km, put 667 bananas over there and repeat this 3 times. So
now we are having 2001 bananas at a distance of 667 km from the market. Now we move
another 500 km forward and put 500 bananas over there. We go back to get the rest of
bananas and travel 500 km ahead and get those 500 bananas. We leave that 1 banana. (
But Sammael being a greedy man, would go back to get that 1 banana forgetting that the
camel his also his baap). Now we are left with 167 km to cover. So maximum bananas that
can be carried to the market is 1000-167=833.
2. Minimum 10 people would be required. 2^10=1024. Since there are 1000 bottles to be
tested, minimum 10 people would be required.
Another approach to this could be that of a truth table. Look at the following table for 4
variables
ABCD
1 1 1 1 ( Bottle 1)
1 1 1 0 ( Bottle 2)
1101.
1100.
1011.
1010.
1001.
1000.
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
.
.
.
.
.
.
.
( Bottle 16)
This is the case for 16 bottles. Here 4 people would be required to find out the bottle
containing poison. If bottle 1 contains poison, all would die. If bottle 2 contains poison, A,B,
and C would die but D will survive. Hence there would be a unique case for every different
bottle containing poison.
2. DESTRUCTION OF A PARAJUMBLE
Follow these steps religiously:
Identify a Dangler: A dangler is a sentence which is not connected with the rest. It would
definitely talk about the same subject, but you will not be able to link it with others. A
Dangler has to be an opening line or concluding line.
Ways to identify danglers:
By Pronouns: If there is a pronoun present which cannot be traced to any of the nouns in
the other sentence, then it has to be a dangler. Probably its a concluding line of some para,
e.g. and should be the first sentence.
By Subject Matter: Danglers can be spotted by looking at the subject, if author is quoting
someone, there should be a reference of it in following line. If there isn't one, Its a opening
line of a new para. e.g. .... Place it in last
Linking. Linking two sentences is a common practice and everyone knows about it, so just
mentioning the methods for linking, if you want would explain each one..
By Pronouns
By Structure words, very very important. Genral idea followed by example or specific idea
De-Linking... Its to separate different sentences. We have to pick the tone of each sentence,
it can be:
Positive (P)
However followed by positive (HP)
Negative (N)
However followed by Negative (HN)
Neutral (E)
Now to switch between the tones, from P to N or N to P, there must be a HP or HN.
e.g. following can be Structure of a para
P P P P HN
P P P HN N
P P HN N N
P HN N N N
N N N N HP
N N N HP P
N N HP P P
N HP P P P
The source of these PJs are very reliable so rest assured that author will not randomly
change the tone. So there can never ever be a pair like N P or N P.
3. FUNCTIONS AND GRAPHS
Let us take a function y=f(x) and we know what it graph looks like.
(Kindly,look at at some well known graphs like x^2 , x^3 , 1/x , 1/(x^2) , ln x , e^x ,sin x
,cos x , x| etc.)
We can get the modified graph using the following methods:1) y=f(x+a) : Translate the graph of y=f(x) along X-axis by a units in the direction opposite
to the sign of 'a' .
2) y=f(x)+a : Translate the graph y=f(x) along Y-axis by a units in the direction same as
the sign of 'a' .
3) y= -f(x) : Reflect the graph of y=f(x) about X-axis.
4) y=f(-x) : Reflect the graph of y=f(x) about Y-axis.
For Even functions,graphs for y=f(x) and y=f(-x) are same.
5) y=f(x) : Reflect that part of the graph y=f(x) which is below the
X-axis about X-axis.
(Note:There won't be any graph in this case below the X-axis
i.e. for -ve values of 'y' because R.H.S. f(x) is always +ve
hence 'y' can't be -ve. )
6) y=f(|x) : Just consider the part of the graph y=f(x) for x>0 i.e the part of the graph to
the right of Y-axis and omit the rest. Now reflect this new graph about Y-axis to obtain the
final graph.
I hope that this post might be of great help to few and of some help to many.
Try some graphs to check out what actually happens.
4. TOURNAMENT MODEL CONTAINING LEAGUE ROUNDS AND MATCHES
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 Losses
Sum Of Draws In A Tournament Will Always Be An Even Number
Sigma(GF - GA) = sigma(GD) = 0...Always
If 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 Match
If 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 = 0
So...GD for Brazil = Let say x
x+0+(-1)+(-1)+2+2 = 0
So x = -2
Which 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 Played
And 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 runs
Match 2 = India(245-7 in 42 overs) beat Australia(241-5 in 50 Overs) by 3 wickets
Match 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 overs
Runs Scored Against India = 497 in 95 overs
So 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 RoughlySummation of NRR of All
teams = 0
5. COINS PROBLEMS
Let us say Vipul and Shashank are playing a game. In the game they decide to put coins in
a bag. They put the coins in the bag in turns. Now the coins can be put in the bag only in
numbers of 10 or 20 i.e Vipul or Shashank can put either 10 coins or 20 coins in the bag.
Any other number of coins is not allowed to put in the bag. The one who makes the number
of coins in the bag=100 wins the game. Say at an instance the number of coins in the bag
are 80 and then Shashank puts 20 coins in the bag, thus making the number of coins=100.
Then Shashank wins.
Now the one who starts by putting 10 coins in the bag will always win. And the one starting
with putting 20 coins in the bag will lose. Why??
Let the lowest integer number of coins that is not possible with a single attempt be t.
In this case the t= 30.
Target value(also called Vipul's value) = desired value -t= 100-30= 70.
So me must try to put the other member on this figure called target value. When on 70, a
player can put either 10 or 20. In both cases he would lose.
Now if we keep on subtracting t from target value, we would have a similar situation.
Now 70-30=40. So if a player is on the figure 40, there is no way he can win the game.
Let a player be on 40 coins. He can put either 20 or 10, then other would put 10 or 20
respectively. Thus getting him on the target value.
Extending this further, 40-30=10. So a player when on 10 can never win the game.
So whoever starts the game with 10 coins is always going to win.
Now let us say the coins can be put in the bag in numbers of 10, 20 and 30.
Here t=40.
Target value= 100-40=60.
60-40=20.
So the one starting with 20 coins in this case will never lose. And if one starts with 10 coins,
he will lose.
I think this much understanding should be enough to answer the following questions:
Two friends Shashank and Vipul are playing a game. They start with an unlimited supply of
coins, with denominations 10,20,40 and 50. They take turns placing a single coin into the
pot, which is initially empty. They decide a target amount and the amount in the pot should
not exceed this target amount.
6. CONCEPT OF INCLUSION EXCLUSION
Five balls of different colours are to be placed in three boxes of different sizes.Each box can
hold all five balls, The no of ways in which we can place the balls in the boxes so that no
box remains empty??
The question mean that each box shld get atleast one.
We do it using Principle of inclusion-exclusion.we include favorable cases and exclude
unfavorable cases.
Favorable cases are Total number of ways distributing 5 different balls to 3 different boxes
is 3^5.
Unfavorable cases are when two boxes not getting any ball.we have to exclude the
distributions in this case,this can be done like this
select any one box from 3 in 3c1 ways and distribute the balls to the remaining balls in 2^5
ways, hence total ways which we need to exclude is 3c1*2^5
and one more case is when one box not getting any ball, select two and distribute the balls
to remaining box in 1^5 ways.
hence it is 3^5 - 3c1*2^5 + 3c2*1^5 = 150 ways.
Note:It is similar to finding number of ONTO functions.
here's a general formula:
n^m - nC1*(n-1)^m + nC2*(n-2)^m - nC3*(n-3)^m + ... +(-1)^(n-1)*nC(n-1)
conventional approach 1 :
Given 5 balls can be placed in 3 boxes as follows:
i. 1 ball can be placed in each of two boxes and 3 balls are placed in the third box.
ii. Two balls are placed in each of the boxes and 1 ball is placed in the third box.
The number of ways in both cases = 20 + 30 = 50
There are exactly 3 such arrangements.
Hence, required number of ways = (50)(3) = 150
conventional approach 2 :
no boxes are empty = maximum prob - boxes are empty
= 243 - { 6*(10 + 5) + 3) } = 150 ways
7. CALENDER BASICS
Many people do disasters when they face questions related to calculation of dates. So I am
today posting some concepts on Calendars which can help u guys in calculating any day of
any date or year.
Firstly let me brush up some basics :
(a) : A Leap year is a year which is divisible by 4.
(b) : A Leap Century is a century which is divisible by 400.
(c) : A century is a multiple of 100 i.e. 100,200,500,800 etc
(d) : ODD DAYS : remainder obtained when divided by 7.
(e) : Every Leap year has 2 odd days and Every ordinary year has 1 odd day.
ODD days in a year or a no of years are the left over days which are extras and cant be
filled within a week i.e. the days left when all the weeks are counted and is calculated as a
remainder by dividing by 7.
Calculation of ODD days
Suppose we want to find the no of odd days within a century i.e. 100 years.
-> the first step is to divide the century into no of leap years and ordinary years.
-> 100 / 4 = 25
-> now in 25 years the 100th year is not a leap year becoz as i told a century shud be
divisible by 400 to be a leap year. so subtract 1 from 25 i.e. 24 is the no of Leap years and
(100-24) i.e. 76 is the no of ordinary years.
-> 100 = 24 LY + 76 OY
-> As i told every Leap year has 2 odd days and Every ordinary year has 1 odd day. So 100
= 24(2) + 76(1) = 124 odd days
-> divide 124 odd days into weeks by dividing by 7 and we get 5 left odd days.
Therefore we found a Century has 5 odd days.
It is benificial to remember :
200 years = 3 odd days. (found using same method but subtract 2 dis time coz neither
100th year nor 200th year are leap years)
300 years = 1 odd day.
400 years = 0 odd day.
and Every Multiple of 400 = 0 odd day or we can say a year repeats after 400 years
(f) : Every year repeats after 400 years
Now Some Examples :
(a) Direct Method
Q. 21st Jan 1932 was my birthday, which day was I born ?
Solution: Remember, always find odd days till a year less and then calculate further for the
year required. In dis case we calculate till 1931.
1931 can be written as 1600 + 300 + 31 years
Now check odd days in all these years
- we know 1600 years will have 0 odd days being a multiple of 400
- 300 years will have 1 odd day as i mentioned before
- Now Calculate no of odd days in 31 years by the method i explained earlier.
31 years = 7 LY + 24 OY = 7(2) + 24(1) = 38 odd days = 3 odd days left.
Total odd days = 0+1+3 = 4 odd days
Now check year 1932.
we require days will 21st jan 1932 = 21 days = 0 odd day
Grand total of odd days = 4+0 = 4 odd days.
Taking 0 odd day as sunday
We see that 4 will be Thursday.
dis means 21st Jan 1932 will be thursday.
PS : This method is lengthy but will clear some basics CAT requires a method in which u
may be knowing some day beforehand.
(b) My Method
Q1. My Dad was Born in 14th Nov 1984 and dat was Wednesday. I am his Son and
scheduled to be born in 15 Dec 2020. Can u tell me the day
Solution : According to my method we can get the day in the same date of the required
year i.e. 14th Nov 2020
->
->
->
->
2020 - 1984 = 36 years
find odd days in 36 years by the same method i explained before.
36 years = 9 LY + 27 OY = 9(2) + 27 = 45 odd days = 3 odd days left.
Now add 3 more days to wednesday which gives us Saturday.
-> So, 14th Nov 2020 will be a Saturday.
But we Require 15th Dec 2020.
So calculate days left upto 15th dec 2020 from 14th Nov 2020.
Nov = 16 days
Dec = 15 days
Total = 16+15 = 31 days = 3 odd days left.
Now Finally add 3 more days to Saturday which gives us Tuesday.
Therefore 15th Dec 2020 will be a Tuesday.
Q2. In the above Question If I was to be born in 15th Sept 2020 then wat shall be the day
??
Solution. We found 14th Nov 2020 will be a Saturday.
We have to think that after 15th Sept 2020 we added some odd days to dat day and
reached 14th Nov 2020 and found dat it was a Saturday.
-> So let the day of 15th Sept 2020 be x. (x is the day like 0 is a sunday,1 is monday and
so on)
-> Sept = 15 days left, Oct = 31 days left, Nov = 14 days upto 14th Nov
-> Total days = 60 = 4 odd days
-> this means (x + 4)%7 = Saturday = 6 (becoz 6 is a saturday )
-> this means x-2 is divisible by 7 which means x shud be 2 (becoz 0 is divisible by 7)
Now x=2 which means x is a Tuesday