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Hi all... After getting PG puyscar for most helpful in quants section this year , i really thought i should do something this year to atleast justify the puyscar.. i jus thought to start a thread and post the concepts i learned through PG.. m...

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kriti
@kritigup
273

can someone tell me how to solve questions like the one given below..i am not able to understand the concept involved here-

clerk has 5 boxes of different integral but unknown weights.the clerk weighted the boxes in pairs. he obtained the weights in kgs as 122,124,125,126,127,128,129,130,132,133. how much would the heaviest box weighs?

p.s. i dont know the answers..please explain

same type of questions comes for age also..

M not getting the correct answer by this method for 1st Feb 1984.

May b i m doing something wrong.:splat:...but kindly check!!!!

It is

- m is the month number. Months have to be counted speciall.... 22 Aug.

m is the month number. Months have to be counted specially for Zeller's Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year.) Because of this rule, January and February are always counted as the 11th and 12th months *of the previous year*. in here D is 83 not 84. Ans is 3 i.e wed.

One Year PGPM @ Great Lakes

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Know MoreFound it somewhere in PG only... thanks to a puy....posting it here so that it might help people who may not have seen it over there.

Zeller's Rule : With this technique named after its founder Zeller, you can solve any 'Dates and Calendars' problems.

Zellers rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.

Zeller's Rule Formula:

F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C

K = Date => for 25/3/2009, we take 25

In Zellers 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 => for 2009 = 09

C = The first two digits of century => for 2009 = 20

Example: 25/03/2009

F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 - (2 x 20)

= 25 + 12/5 + 09 + 09/4 + 20/4 - 2x20

=25+2+09+2+5-40

[ We will just consider the integral value and ignore the value after decimal]

= 43 - 40 =

Replace the number with the day using the information given below.

1 = Monday

2 = Tuesday

3 = Wednesday

4 = Thursday

5 = Friday

6 = Saturday

7 = Sunday

So it's Wednesday on 25th march, 2009.

If the number is more than 7, divide the no. by 7. The remainder will give you the day.

I hope you will find the above method very useful.

M not getting the correct answer by this method for 1st Feb 1984.

May b i m doing something wrong.:splat:...but kindly check!!!!

nice method...originally i used this method :

finding the last non-zero digit of N!The last non-zero digit of a number N is (4^n)*(2n)! ..... (for N = 10,20,30....)For nos other than multiples of 10, the result can be easily back tracked..

where n=N/10

for eg.. if u are asked for 36!, then find for 40! and then back tract to 36!..

But finding 30! and moving to 36! might be tricky...since it involves a 35 and will introduce an unnecessary 0..here N =40! and n=40/10 = 4

so the last digit is 4^n*(2n)! = 4^4*8! = 6*2 = 2

2 is the last non-zero digit of 40!,

now if u want to calculate for 36!,

then 36!*7*8*9*4 = 2 (i am writing all the last digit)

36!*6 = 2

so last non-zero digit of 36! = 2

Bro but with the approach above yours

Last Non.Zero digit for 40!

10!- 8

40! -8^4= 6(last non zero digit)

Now for 36!x7x8x9x4=6

36!*6=6(so wat wud be the the last non -zero digit nw???)

But you have written 36!*6=2 Hw???..tell me if m doing something wrong!!!!

Re: Rectangles Cut By a Diagonal Line

The general rule: If the lengths of sides (a x b) of the rectangle are

mutually prime, the number of squares cut is a+b-1

Thanks for this useful post but what do u mean by squares cut exactly...cud u pls elaborate??

Abhishek Jain
@jain4444
40,426

Pigeon hole theoram

if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons

Well this theorem, look apparently simple and trivial, but its extremely powerful. Lets take a test of it.

Example= Let A be any set of nineteen integers chosen from the arithmetic progression 1,4, . . . ,100. Prove that there must be two distinct integers in A whose sum is 104.

Now how do we go about this? remember n and n+1. The hint is to make n+1=19. Something clicked?

see we have 34 numbers of the form 3k+1, from 1 to 100. If we do not want a sum of 104 , we will break them in the sets of 2 integers whose sum is 104

{4,100},{7,97}..{49,55} and {1}, {52}. Clearly we have 16 two element sets and 2 one element set.

So if we make a set of 19 integers, we will have to pick both the integers from atleast one of the two element sets, which will give us a sum of 104.

We are done here.

If you still have doubts, let me explain again, suppose you are four friends ( boys) and there are three girls. And each one of you like a girl out of the three. So at least one of the girls will be liked by two boys.

Lets solved a more involved example, wherein we need not prove a thing, but find a thing. Some people may be feeling cat does not want us to prove but find. Here is how we do that.

Example = Let there be n balls with Ram. he decides to colour one ball with colour 1, two balls with colour 2 and so on upto, fifty balls with colour 50. At the end of it , all n balls are used, and no ball is coloured twice. Ram then draws balls from the lot at random, without replacement. What is is the minimum number of balls that he must draw in order to gurantee drawing 10 balls of the same color?

What the hell is his problem. Why coloring and then taking out. Stupid chap. Let us help him with the math now.

see if he picks all the balls with colors which are less than 10 it will come upto (1+2+3..+9)=45.

Now for the worst case he will pick 9 balls each from rest of the balls, which is 41*9

so total is 41*9+45=41*10+4=414. ( avoid multiplying, be watchful)

now if he picks one more ball, atleast one of the set will be of 10. so we are done

he needs to draw 414+1=415 balls.

Practice Problem 1 = A circular table has exactly 60 chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest value of N?

Practice Problem 2 = We call a set "Sum-free" if no two elements of the set add upto a third element of the set. What is the maximum size of the "sum-free" subset of {1,2,3...2n-1}?

if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons

Well this theorem, look apparently simple and trivial, but its extremely powerful. Lets take a test of it.

Example= Let A be any set of nineteen integers chosen from the arithmetic progression 1,4, . . . ,100. Prove that there must be two distinct integers in A whose sum is 104.

Now how do we go about this? remember n and n+1. The hint is to make n+1=19. Something clicked?

see we have 34 numbers of the form 3k+1, from 1 to 100. If we do not want a sum of 104 , we will break them in the sets of 2 integers whose sum is 104

{4,100},{7,97}..{49,55} and {1}, {52}. Clearly we have 16 two element sets and 2 one element set.

So if we make a set of 19 integers, we will have to pick both the integers from atleast one of the two element sets, which will give us a sum of 104.

We are done here.

If you still have doubts, let me explain again, suppose you are four friends ( boys) and there are three girls. And each one of you like a girl out of the three. So at least one of the girls will be liked by two boys.

Lets solved a more involved example, wherein we need not prove a thing, but find a thing. Some people may be feeling cat does not want us to prove but find. Here is how we do that.

Example = Let there be n balls with Ram. he decides to colour one ball with colour 1, two balls with colour 2 and so on upto, fifty balls with colour 50. At the end of it , all n balls are used, and no ball is coloured twice. Ram then draws balls from the lot at random, without replacement. What is is the minimum number of balls that he must draw in order to gurantee drawing 10 balls of the same color?

What the hell is his problem. Why coloring and then taking out. Stupid chap. Let us help him with the math now.

see if he picks all the balls with colors which are less than 10 it will come upto (1+2+3..+9)=45.

Now for the worst case he will pick 9 balls each from rest of the balls, which is 41*9

so total is 41*9+45=41*10+4=414. ( avoid multiplying, be watchful)

now if he picks one more ball, atleast one of the set will be of 10. so we are done

he needs to draw 414+1=415 balls.

Practice Problem 1 = A circular table has exactly 60 chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest value of N?

Practice Problem 2 = We call a set "Sum-free" if no two elements of the set add upto a third element of the set. What is the maximum size of the "sum-free" subset of {1,2,3...2n-1}?

- 12 Likes

Abhishek Jain
@jain4444
40,426

Type # 1.

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

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

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

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

i.e. 504 + 4 = 508

Type # 2

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

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

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

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

Type # 3

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

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

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

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

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

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

208%11 = 10

210k%11 = k

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

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

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

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

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

Type # 4

What if there is no relation between divisors n remainders?

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

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

in such cases...take 1 case n target another case...

e.g. i take the case 7 with 13...and target 6 with 11.

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

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

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

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

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

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

72 + 143k % 7 = 2 + 3k

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

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

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

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

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

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

ORIGINALLY POSTED BY MAXIMUS

- 6 Likes

Abhishek Jain
@jain4444
40,426

The general rule: If the lengths of sides (a x b) of the rectangle are

mutually prime, the number of squares cut is a+b-1

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

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

(9 x 4) gives 9+4-1 = 12

(9 x 5) gives 9+5-1 = 13

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

follows:

First divide (9 x 6) through by common factor 3 to get (3 x 2)

Then apply the rule to (3 x 2) to give 3+2-1 = 4

Now multiply by the factor 3 again to get 12 (which is correct).

Let's do a square figure, say (5 x 5). We divide through by 5. This

gives (1 x 1). Applying the rule gives 1+1-1 = 1. Now multiply up by

the factor 5 again to get 5. We know this is correct because in ANY

square figure the number of squares that are cut will be equal to the

side of the square.

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

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

If a and b have a common factor c, first divide through by c to get

(a/c x b/c).

Then apply the rule to get a/c + b/c - 1.

Finally, multiply through again by c to get (a+b-c).

For example, with (9 x 6) the correct answer is 9+6-3 = 12.

:splat::splat:

- 14 Likes

Attached JPEG

Can anyone pour how to factorize a quadratic equation with high value of C

in Ax2 + Bx + C = 0????

Kindly refer JPEG

In this particular question,by looking at the signs of B and C,you can say that the factorization would be of type (x-a)(x-b).Now you can check C and try some divisibility rules over it.Like 4293 is divisible by 9(sum of 4293 is 18].Similarly,we will see that 81 divides 4293. 81*53= 4293. Also 81+ 53 =134,which is B.I hope someone comes up with a particular technique rather hit and trial method.

- 3 Likes

Can anyone pour how to factorize a quadratic equation with high value of C

in Ax2 + Bx + C = 0????

Kindly refer JPEG

Questions regarding finding remainders have appeared a lot of times in CAT. Here are few ways by which u can find the remainders quickly

1)EULER'S THEOREM:

If M and N are two co-prime nos. and N=a^P*b^q*c^r. where (n)=N*(1-1/a)*(1-1/b)*(1-1/c).

Then remainder M^(n)/N=1

For example:

Find the remainder when 5^37 is divided by 63

5 and 63 are co primes,

63= 3^2*7

So, (63)= 63*(1-1/3)*(1-1/7)

=36

Therefore , remainder( 5^37/63) = remainder (5^36*5/63)

=1*5

=5

It's a very easy concept, apply it on other nos and try to find the remainder

If P is a prime number and N is prime to P, then (N^P-N) is divisible by P

e.g: what is the remainder when (N^7-N) is divided by 7

ans. 0

If P is a prime no. then remainder (P-1)!+1/P=0, which means that remainder when (P-1)! Is divided by P is P-1

For example: remainder when 40! Is divided by 41

= 41-1

= 40

- 10 Likes

Z(n)= last two digit

Z(n1)= 4, if the tens digit is odd

6, if the tens digit is even

Z(n)= Z(n1)*Z(n/5)!*z( factorial of units digit)

e.g: find the last non zero digit of 36!

Z(36)= 4*Z(36/5)!*z(6)!

=4*7!*6!

=4*(last non zero digit of 7!)(last non zero digit of 6!)

=4*4*2

=2 ans

2). Number of numbers less than and prime to a given number:

If N is a natural number such that N= a^p*b^q*c^r, where a, b, c are different prime nos. and p,q, r are different positive integers then, the number of positive integers less than and prime to N = N(1-1/a)(1-1/b)(1-1/c)

e.g: how many first 1200 nos. are not divisible by 2,3 and 5??

1200= 2^4*3^1*5^2

Therefore number of numbers not divisible by 2, 3 and 5 will be equal to the numbers prime to 1200

=1200(1-1/2)(1-1/3)(1-1/5)

=1200*1/2*2/3*4/5

=320 ans

Thus there are 320 such nos which are not divisible by 2, 3 and 5

- 13 Likes

Hey,

can some one just tell me or paste a link or upload a pdf file as the tips, concepts, fundas in geometry and mensuration??

This is one chapter which I am not able to do not matter how much or how hard I am practising.. plz help

Thanks

Alok Biyani, Kolkata

- 1 Like

Great thread Puys!!!!

- 1 Like

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