Misc Funda

8/10/2019 Misc Funda

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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.


2/37

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.

A 5 B 2 C 2 D 2 E 2 F 2

Answer is 2.

A3.

A 4

B 4

C 4

Answer is 4

A4. Answer is 5.


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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


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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


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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)


6/37

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


7/37

ZELLERS 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!


8/37

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 sidepainted. 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.


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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


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11/37

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


12/37

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.


13/37

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 oddpowers 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.


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15/37

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 5-

digit 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 fasterperson 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 scalene if all 3 sides are of unequal length.


16/37

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.


17/37

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. )


18/37

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 #2An 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 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.


19/37

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. Ifyou 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 belowA.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.


20/37

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)


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22/37

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 digitsthat 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 than1000 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 nowmeasure anything from 1 to 728.


23/37

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 "Iam 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.


24/37

LAST MINUTE READS A NAGA25FRENCH COMPILATION

Sourced from

Pagalguy.com

http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-

43729/p-1642787/r-1742160
http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-43729/p-1642787/r-1742160http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-43729/p-1642787/r-1742160http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-43729/p-1642787/r-1742160http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-43729/p-1642787/r-1742160http://www.pagalguy.com/forums/cat-and-related-bschools/pg-underdogs-team-09-t-43729/p-1642787/r-1742160


25/37


26/37

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 ofweights, 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.


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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 ofbananas 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

A B C D

1 1 1 1 ( Bottle 1)1 1 1 0 ( Bottle 2)

1 1 0 1 .

1 1 0 0 .

1 0 1 1 .

1 0 1 0 .

1 0 0 1 .

1 0 0 0 .


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0 1 1 1 .

0 1 1 0 .

0 1 0 1 .

0 1 0 0 .

0 0 1 1 .

0 0 1 0 .0 0 0 1 .

0 0 0 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.


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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 woulddefinitely 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 openingline 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 NN 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.


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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.


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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 andEPL...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 toensure 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 outfor 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 = 0So...GD for Brazil = Let say x

x+0+(-1)+(-1)+2+2 = 0

So x = -2

Which Gives GF = 4 by Brazil...


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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 numberof 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.


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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 = 50There 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


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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


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-> 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

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