CAT 2008 Quant anlysis

The monster of all exams, IIM-CAT was held on 16th Nov. But, I must say the section 1, i.e quant, was not that monster which it was in 2007. But as we say when you expect a monster, even a mouse will appear as a rabbit.

The paper was well set, with a nice blend of sitters, easy questions and tough questions.

Any well prepared student should eb able to solve 12-14 questions with 90% accuracy, in around 40-45 mins

The geometry problems were so easy, it looked like, sitting and solving ncert book exercises of class 9.

The real gem where the function problems and the series involving roots.

The number theory problems can all be classified as mock type, as all of them have appeared in mocks

For a while, I was in the notion as if I was taking a mock which was a mix of cl ims and time. neither too tough or too easy, a blend..

My take on cutoff 35+-2

Good Luck !!

[polldaddy poll=1113052]

CAT 2008 Key and Solutions

Hi,

I will be posting my own solutions and keys for quant.

Any error on my part is totally unintended and I would not be liable to any damage claims

You are free to use these though

Rahul

Set 444

Q1.

The number of terms in (a+b+c)^20

it is same as number of solutions of a+b+c=20 which is 22c2=231 option 1

Q4)

Seed(n)=9 if the number is a multiple of 9

Hence 9,18, 495

55 numbers in total

5)

any two number is replaced by a+b-1

so basically we will end up with sum of all 40 numbers -39

40*41/2-39=781

Question 9

10/4=x/r

x=2.5r

A=2pir(r+h)=2pier(10-1.5r)

r=10/3

gives Area =100pie/3

Q10

obtuse angles are possible if

15 is the largest side or x is the largest side

take case 1

x+8>15 =>x>7

also x^2+8^2<15^2

x<13

so we get x=8,9,10,11,12

similarly we will get

18,19,20,21,22

hence total 10 values for x

Q11

m+(m+1)^2+(m+2)^3=9(m+1)^2

m^3-3m-2m^2=0

m=0,-2,3

Hence option 1

Q12

4 digit integers <=4000

first check upto 3999

we will get 3*5*5*5=375 and add 1

we get 376

Question 13

root(1+1/1^2+1/2^2)=3/2=2-1/2

hence we can write as 2008-1/2008 option 1

alternate

3/2=1+1/1.2

7/6=1+1/2.3

..

we add

1+(1-1/2)+1+(1/2-1/3)..+1+(1/2007-1/2008)=

2007+1-1/2008=2008-1/2008

Question 14

a/sinA=2r

17.5/3/9=2r

r=26.25

option 5

Question 15

f(x)f(y)=f(xy)

f(0)^2=f(0)

and f(1)^2=f(1)

consistent solution is f(1)=1

so f(2).f(1/2)=f(1)

4.f(1/2)=1

f(1/2)=1/4

Question 16)

7^2008

last 2 digits of 7 have a cycle of 07,49,43,01

Hence 2008=4k

hence last two digits is 01

Alternate is

7^2008=(7^4)^502=(2401)^502=(2400+1)^502=01

option 3

Question17)

[(2a)^2-2a^2/root(3)]/2a^2/root(3)=2root(3)-1

option 5

18)

let the roots be k-1, k, k+1

then 3k=a

k(k-1)+k^2-1+k(k+1)=b

3k^2-1=b

so min value of b is -1

Question 19 and 20

9a+3b+c=0

25a+5b+c=-3(4a+2b+c)

37a+3b+4c=0

gives a=b

hence

ax^2+ax+c=p(x-3)(x-q)

compare coefficients gives

q=-4

Hence 19 option  2

20 is option 5 cannot be determined

as we do not know a


we can just say a+b+c=2a+c=-10a

Question 25

first common term is 21


the common terms will form an AP of CD=lcm(4*5)=20

21+19.20=401 <417

21+20.20=421>417

Hence 20 terms !!

Mini Concept: Multinomial Theorem

Some Trickery of Multinomial Theorem

x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2

if we extend this to r variables then the formula becomes (n+r-1)C(r-1)

if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)

Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips

Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.

so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182


Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?

Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D

now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways

case 2) we leave 0 and pick two positive and one negative

then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2

case 3 it will be same as case 2
we get 50C2

hence total is 2.50C2+50=2500

Problem Of The Week 53

How many ways can a size k + 1 subset with maximum element m + 1 can be created from the given set S={1,2,3,.....,n+1} ?

Problem Of The Week 52

Suppose that and are positive numbers for which

,

what is the value of ?

Problem Of The Week 51

Let P(x)=ax^4 +bx^3+cx^2+dx+e be a polynomial with all integer coefficients and a=1. If√2+√5 is one of the roots of P(x)=0 , which of the following can be the value of |(b+c+d+e)|?

1) 103

2) 89

3) 63

4) 23

5) 5

Problem Of The Week 50

Find the number of unordered triplets (x,y,z) of positive integers such that x³+y³+z³=2008

1) 0

2) 1

3) 2

4) 3

5) None of these

Problem Of The Week 49

Find the maximum area that can be bound by four line segments of length 1, 2, 3 and 4. (you are not allowed to break a segment, you may join two :) )

1) 6√2

2) 2√6

3) 4√6

4) 3√2

5) None of these

Problem Of The Week 48

If square tiles are to be fitted on a square floor in such a way that the floor looks the same from all the sides. The tiles are availble in 6 different colors. In how many ways can the floor be made if it should have maximum possible colors?




1) 360

2) 60

3) 1296

4) 256

5) None of these

Problem Of The Week 47

The speed of the engine of a train is 150 km/hr. By connecting coaches to the engine its speed decreases. The decrease in speed is given by the function d=x^2+3x-2 (km/hr), where x is the number of coaches connected. Find the minimum speed ( in km/hr) with which the train can run?

1) 1

2 11

3) 22

4) 33

5) None of these

Problem Of The Week 46

Let be distinct positive reals. Determine the minimum value of and determine when that minimum occurs.

Source: ALtheman's problem column

MockaMania Oct 5 2008

Here are a few question from IMS simcat 11

Question 1) For what value of n>0 does the following pair of equations yield exactly three solutions for y?

|x|=|y|

x=y^2+n(n+3)-4

1) 0   2) 1   3) 4   5) no unique vale  6) none of these.




Question 2

The sum of 2k+1 consecutive natural numbers is 2n such that the sum of first k+1 natural numbers among these (2k+1) numbers is same as next k numbers. Which of the following cannot be the 7th least natural number among the 2k+1 natural numbers, if k>=3

1) 231   2) 175   3)  535  4) 325 5) none of these





Question 3

if x=1+1/(x+1/(1+1/(x+...))) ; then which of the following bet represents x

1) 1<x<2   2) 1.5<x<2  3) 1<x<1.5  4) 1<x<1.2    5) 0<x<1




Question 4

let a(n) be q sequence such that n is an inetger and n>=1


a(n)-a(n+1)=a(n+4)-a(n+5), find a(80)-a(84) if a(75)=103 and a(83)=205




1)- 51    2) -36 )  -17 4) 75  5) cannot be determined.

Question 5

Maya has six indentical pots, which she is planning to arrange in a straight line in her showcase. before that each of these pots is to be colored either red or yellow or green  or blue, such that at least one pot is coloured with each of the four colours. In how many different ways can she arrange the pots in the showcase so that now two pots of the same colour are adjacent?

1) 120    2) 84    3) 840   4) 600   5) 936

Question 6

The second rightmost digit of (102)^33 is

1) 8 2) 2  3) 4  4) 6  5) none of these




7) A number is said to be crazy number is the product of the digits is equal to products of the distinct primes in its prime factorisation. How many crazy numbers less than 100 have less than 3 distinct prime factors in their prime factorisation?




1) 4   2_ 5  3)8   4) 6  5) none  of these

More to follow

Problem Of The Week 45

Find the largest value for for pairs of real numbers which satisfy

Problem Of The Week 44

FInd the number of pairs (x,y) of integers 0<x<y such that sqrt(x)+sqrt(y)=sqrt(1984)

Problem Of The Week 43

Find the number of real roots (x+1)^x=x^(x+1)

Problem Of The Week 42

Let their be 7 special coins such that they are numbered from 1-7 on one side and blank on the other side. You toss all the 7 and let them fall and make your best guess as to which coin is 1, what is the probability that you are correct?

Problem Of The Week 41

Let s(p) be the sum of digits of a prime p. If p-s(p) is also a prime, then we call the prime a primal, how many primals below 100 are there ?

Problem Of The Week 40

When the mean, median and mode of the list 10,2,4,2,5,2,and x are arranged in increasing order, they form a non-constant arithmetic progression. FInd the sum of all such real x?

A)   3    B)  6  C) 9  D) 17   E) 20




Source : AMC 12

Problem Of The Week 39

A set of 2009 numbers from (1-2009) is written on the board. You are allowed to replace any two of these numbers by a new number which is either the sum or the absolute difference of these numbers, after 2008 such operations, Which of the following cannot be the last number left on the board?

A) 1  B) 3      C)  4     D) 5)    E) Cannot be determined

Problem Of The Week 38

My grandson, my son, and i share the same bday, this year all 3 of our ages have become prime numbers. i remarked to my son that For the last 17 years prior to this year, whenever your age was a prime number, so was mine and vice versa. "yes" he replied and said that 17 years from now, i will say to my son, for that last 18 years prior to this year, whenever your age was a prime number so was mine. How old are we?

Problem Of The Week 37

What is the sum of the smallest and the largest number of fridays the 13th that can occur in any year ?

Concept 6 Data Sufficiency

Concept 6: DATA SUFFICIENCY

"The ultimate goal of mathematics is to eliminate any need for intelligent thought."-A. N. Whitehead

Well, for starters try to think what's the meaning of the above quote, it has a nice and beautiful meaning. To make things so simple that an average mind is able to see it. So lets put this into practice, with another concept lesson.

Today, we will discuss DATA SUFFICIENCY, one of the very scoring problems in cat and other mba entrance exams. The good thing about DS is we get DS both in quant and DI, and it makes for 6-8 problems in almost every paper. As there are no theorems in DS, we will take things to note

Things To Note:

1) DS problems, we need to answer if it is sufficient information to answer, means, there should be one conclusive answer.  We do not need to find the answer, just if it can be found or not?

2) Some questions ask , is this true? So if we can find that the information available is enough to prove that it is not, we are still able to answer the question, that it is not true. Hence we are able to answer the question.

3) Check for all possibilties, that is using one statement, using second, then only combine the two.

Lets take up an example.

Instructions For DS Questions

Each question is followed by two statements, X and Y. Answer each question using the following instructions:
Mark (A) If the question can be answered by using the statement X alone but not by using the statement Y alone.
Mark (B) If the question can be answered by using the statement Y alone but not by using the statement X alone.
Mark (C) If the question can be answered by using either of the statements alone.
Mark (D) If the question can be answered by using both the statements together but not by either of the statements alone.
Mark (E) If the question cannot be answered on the basis of the two statements.


Example 6.1

In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(X) AB > AC

(Y) BD < DC

The problem is simple, but again see this, this question asks is M>N? The answer may be yes or no, but what we are concerned with is our ability to answer it and not the actual answer.

AB>AC gives us no idea, just imagine a few figures and you will know this.

Look at the other statement it says BD<DC, if BP and CQ are the perpendiculars then, BDP and CDQ are similar

so BD/DC=BP/CQ so we know the ratio and thus are able to answer and see this teh answer came NO.

so, we dont really want the answer, but the ability to give a unique answer.

So, answer is B)

This question was taken from QQAD practice test 4, here is the official solution.

Let D' be the midpoint of BC and let X' and Y' be the feet of the perpendicular from B and C to AD' respectively => X' = Y'. As D' shifts to right or left, we can know which of M or N is bigger. Thus, (Y) answers

(X) doesn't tell us anything

=> choice (B) is the right answer

Practice Problem 6.1

Is x=y?

X: (x+y)(1/x+1/y)=4?

Y: (x-50)^2=(y-50)^2

Practice Problem 6.2

What is the value of m and n?

X: n is an even number, m is an odd number, m>n

Y: mn=30

Note: Both the practice problems are old cat problems, enjoy!

Example 6.2

The distance of point P=(x,y,z) from the origin is sqrt(62) units, then find the coordinates of point P.

X: x+y+z=12

Y: x,y, and z are positive integers.

From original questions x^+y^2+z^2=62

From statement X

(x+y+z)^2=144

2(xy+yz+zx)=62

can't say for sure

From statement Y:

positive integers

but we can easily find two pairs (1,5,6), (2,3,7). can't find a unique solution

Option E

This question is taken from SIMCAT 9

Practice Problem 6.3

Rahim plans to draw a square JKLM with a point O on the side JK, but is not successful. Why is Rahim unable to draw the square?

X: The length of  OM is twice that of OL

Y: The length of OM is 4 cm

This problem is taken for CAT 07 paper.

Example 6.3Let p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?

(X) p(i) – p(j) is not a composite number

(Y) p(2i) + p(2j) is a composite number

One of my favourite problems !

p(i) – p(j) is not a composite number
=>i^2-j^2 is a prime as i ,j are positive integers and i >j, ( i^2-j^2) can't be 1
=>(i+j)(i-j)= prime
so i-j=1
let p be the prime so i=(p+1)/2
j=(p-1)/2
clearly p is not 2 hence all p is odd
p(i) + p(j)=80 +(p^2+1)/2

now p^2=6k+1 ( Refer my lession on prime numbers for this )
therefore
p(i) + p(j)=80 +(p^2+1)/2
becomes
80+(6k+2)/2=81+3k=3(27+k)
so not a prime => can be answered by using X

now, p(2i) + p(2j) is a composite number
4(i^2+j^2+20) is composite

now i and j can be anything
can't make any conclusions

=> choice (A) is the right answer

This is QQAD pratice test problem!

Thats all, do post your queries and suggestions !

Good Luck!

Problem Of The Week 36

Find the value of the expression ..........|||||||x-2|+2|-2|+2|..........

Problem Of The Week 35

How many 3-d igit numbers are such that one of the digits is the average of the other two?

(A)   96   (B) 112   (C) 120   (D) 104   (E) 256

Problem Of The Week 34

Rakesh has the habit of always pouring his tea from the cup into the saucer before drinking it. He fills both the cup and the saucer to only 90% of their capacity (subject to the availability of tea). He also does not drink any tea which is below the 15% mark in the saucer. If he has to pour the tea from the cup into the saucer at least three times before emptying the cup (each time drinking from the saucer till it reaches the minimum level), then what is the maximum possible ratio of the volume of the cup to that of the saucer respectively? (1) 3 : 1 (2) 9 : 4 (3) 5 : 2 (4) 8 : 3  (5) None of these.

Problem Of The Week 33

The sum of base-10 logarithms of divisors of 10^n is 792. what is n?

(A) 11  (B) 12  (C) 10  (D) 13 (E) 14

Problem Of The Week 32

Four Equilateral triangles are formed taking one of their sides as the sides of the square, the third vertices of equilateral triangles being inside the square. The ratio of the area of fig formed by the third vertices of the triangles to that of the square is nearly

Problem Of The Week 31

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

Problem Of The Week 30

Let b be an odd square and n be any natural number. Then b=n^3+2n^2. Find b?

Problem Of The Week 29

Let x,y,z be distinct positive integers such that x+y+z=11. Find the maximum value of (xyz+xy+yz+zx)?

Problem Of The Week 28

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let p be the probability that at least two of the three had been sitting next to each other. If p is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?

Problem Of The Week 27

A golden rectangle is a rectangle in which the ratio of the width to length is the same as that of the length to the sum of the length and width. Which of the following is also true about a golden rectangle?
I. The ratio of the length to width is the same as the ratio of the width to the difference of the length and width.
II. The product of the length and width is equal to the product of the sum of the two sides and the difference of the two sides.
III. The length has to be greater than two times the width.

(1) Only I and II (2) Only II and III
(3) Only I and III (4) All the three statements
(5) None of these

Problem Of The Week 26

A three digit number is such that the sum of the digit in the hundred’s place and the ten’s place is 1 more than the digit in the unit’s place. It is also given that the digit in the ten’s place exceeds the square of the digit in the hundred’s place by 1, and that the square of the digit in the units place diminished by 7 is the same as the sum of the squares of the other two digits. What is the number?

(1) 346                 (2) 256                 (3) 458                 (4) 526       (5) None of These.

Concept 5 Pigeon- Hole Principle

I am delighted with the response, my concept lessons are being well received. Thank you.

Today, we will be taking something called "Combinatorics". It is generally the nemesis of many students, especially the ones who do not understand why do we need to arrange something, and that too in some weird way. Well I have sympathy for you, but no matter how chaotic our lives be, we still like to maintain some order, and therein comes the concept of ordering, arranging, partitioning and so on. And all of it come together to make a branch of mathematics called Combinatorics.

It will be injustice to combinatorics, if I write just once lesson, so I will try to write a few more, for the moment, I will pick up one of the darling principles of Combinatorics, known as Pigeon-Hole Principle.

Theorem 5.1if 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 5.1 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 5.2 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 5.1 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 5.2 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}?

Thats all in this lesson, this is more than enough, if you need more, look into the mock papers, you will find something.

I would like to thank Mr. David Santos, as I have used his book on number system extensively to make this lesson, but I have tried to add my own flavor to it.

Problem Of The Week 25

Let X and Y be distinct 3 digit palindromes such that X>Y. How many pairs (X,Y) exist such that X-Y is also a 3 digit palindrome ?

Problem Of The Week 24

Let x,y be positive integers such that 19x+90+8y=1998. Let a be the max value of x and b  the max of y.

Find a+b.

Problem Of The week 23

Let f(n) be the number of 1s in all numbers from 1to n. then f(1)=1, f(8 )=1 and f(10)=2 and f(11)=4. Find f(10^100)?

A) 10^99+1 B) 10^100+1 C) 10^101+1 D) 10^102+1 E) None of these.

Problem Of The Week 22

Find the sum of all prime numbers below 100 which divide 3^32-2^32?

A)110          B)  115                   C)127                    D) 132 E) None of these

Problem Of The Week 21

if ||x-2|-1|<6 and ||y-1|-2|<9 where x and y are integers, then find the range of x-2y?

Tipster: When the equations look too complex, option hunting is a good method

Learn By Example I

Today we will discuss a problem, I am not giving it as an exercise as I want to discuss it, rather than discuss a whole new concept.

How many digits cannot be the unit's digit of the product of three 3-digit numbers whose sum is 989

When we read the problem, it looks like what the hell is it asking? But, it is not that difficult a problem if we go by a method

let the three  3 digit numbers be xyz, abc and pqr

then what the question essentially says is z+c+r=9 or 19

now we have to check if d is the last digit of the product of xyz, abc and pqr, then what are the values d cannot take.

it can be easily checked that it works for 0 and 2 so d can be o or 2

for d=1, (z,c,r)=(9,3,3) or(9,9,1) and others

9+9+1=19 satisfies hence d can be 1

d=3, (z,c,r)=(3,1,1),(3,7,3),(9,7,1) but none of which sum to 9 or 19

hence d cant be 3

d=4 (1,1,4) (1,2,2) (7*3*4),(4*4*4),(8*8*1)(9*8*2) and others

d=4 works

d=5 is obvious (5*1*3)

d=6 is (1*1*6) ,(1*2*3) (4*4*1)

which works

hence d=6 works too

d=7  works for 7*1*1

d=8 (4*2*1) (6*8*1) (6*2*4)(2*2*2) (2*3*8)

d=8 does not work

d=9 will work for (9*7*3)

so d=3,8 does not work

Hence 2 digits cant be unit digits !

Problem Of The Week 20

A classroom has 4 tables, each of which has 4 chairs. There are 16 students in the class, 4 of whom are friends. If the teacher assigns the seats randomly to the students, what is the probability that the 4 friends will be sitting together at a table?

Tipster: To every probability question, there are two parts, one to find the total possible roster and the other, to find the ones which suit our case( or the ones which not). We can't just find one and leave the other.

Problem Of The Week 19

New Problem: A fresh one, I created it this morning, trying to find the most adequate data!

In a triangle ABC, altitude AD=6 is drawn to cut BC at D. From D, altitude DE=3 is drawn to cut AC at E. If it is know that AB =12. Find the ratio of the area of ABC to area of DEC?

A) 4:1 B) 16:1 C) 25:1 D) 5:1 E) Cannot be determined

Concept 4 Prime Numbers

I was recently reading a blog by one MIT student, wherein he quotes one of his professors, " Mathematicians are annoyingly Precise" and he further adds, " think deeply of simple things". So thats what we will do, be precise and rigorous in our deep thinking of simple things, that is exactly what CAT wants us to do. Lets roll!

Theorem 4.1 Prime numbers are odd, except for 2, and they have exactly two factors, the number and 1 itself.

You would be wondering, why I have started with this, but whole prime number theory is based on this only.

From here, I will formulate something, which I use excessively in problem solving !. But first lets solve an example. This question is taken from My quant problem set III, the link of which can be found," free material for cat".

Example 4.1 Let a,b,c,d be distinct prime numbers satisfying :

2a+3b+5c+7d=162

11a+7b+5c+4d=162

Given that abcd=k. Find the number of distinct values of k?

A)  0                   B) 1                  C)  2                   D) 3                 e) 4

How we go about this? We were told in school, that n variables need n equations, but we have n-2 here. A road-block? No, a call to think deeply. Just see how we can reduce variables or increase equations :) .

We subtract the two equations and get 9a+4b=3d => 4b=3(d-3a)

RHS is divisible by 3, so should be LHS and therefore b=3

put this in the initial equations, and we are sure the max value of a can be =7 ( i leave it to u to figure out how, a hint: all prime numbers are distinct, and we have used 3, we are left with the two smallest as 2 and 5 :) ).

Back again 3a=d-4=>d=3a+4 gives us (a,d)=(5,19),(11,37)..  but clealry the second set wont work, very large values. We found the set, just by using the constraint, all are distinct primes and 3 has been used.

so we have b=3,a=5 d=19, there is no further need to go as we need the no of values of k which will obviously be 1. But for the sake of completeness we can check c=2 :)

Seems like a marathon, but no its a 3-4 minute problem, once you start doing what I want you to !

Now, if you have understood this concept, you should be able to get the practice problem, which is taken from one of the simcats.

Practice Problem 4.1 A boy spends Rs. 81 in buying some pens and pencils. If a pen costs Rs.7 and a pencil Rs 3, What is the ratio of pens to pencils when the maximum number of pens are purchased such that no extra money is given to the shopkeeper?

A) 3:2    B) 2:1   C) 5:4   D) 7:2   E) none of these1

The next concept which I am going to take up is Prime squares:)

Theorem 4.2 : All prime squares ( p>3) are of the form 6k+1, i.e , p^2=6k+1, for all primes p>3.

Lets try to prove this,  any three numbers  (p-1)p(p+1) will be divisible by 6. but as p is a prime greater than 3, it would neither be divisible by 2 nor 3, hence p^2-1=6k so p^2=6k+1.

Some purists will say, that as p is a prime greater than 3, then, p^2-1=24k+1, yupp I agree, but 24k+1 becomes cumbersome to handle sometimes. The proof is simple again, p is odd so both will be divisible by 2 and one by 4. also one of them by 3. hence p^2-1=24k so p^2=24k+1

But, I have always used 6k+1, may be just used to it. You may pick the one that suites you.

Kindly note, this is a necessary condition not a sufficient one, means all prime square will be of form 6k+1, but all no of 6k+1 cant be prime square :)

Lets handle our next example based on this.

Example 4.2 : Find the number of primes p, such that p^2+3p-1 is also a prime?

A quick check will tell 2 does not satisfy and 3 does.

now we check for higher primes

p^2+3p-1=6k+1+3p-1=3(2k+p) hence divisible by 3, not  a prime

So, only one prime p=3 . We are done here !

More to follow, do tell us how you like this, press the rating button on the left :) !

Miscellaneous Problem Package I

1) If |2-3/x|<=1/3 and |3-x/y|<=1/6 and ,x,y are positive reals. Which of the follwoing is a possible value of

x^2/(2x-y) ?

A) sqrt(3)-1  B) sqrt(2)  C) sqrt(1/2)   D) sqrt(3)/2   E)  sqrt(3/2)

2)There are two numbers A and B. A can be expressed as a product of 13 and a two-digit
prime number and B can be expressed as a product of 17 and a two-digit prime number. If
the unit’s digit of the product of A and B is 7, then how many distinct products of A and B
are possible?
A . 48              B. 55               C. 80                    D. 110            E. 120

3) 10 beads and two similar diamond pendants are required to form a diamond necklace. If the beads are same, how many different diamond necklaces can be formed?

A)   1    B   5   C)   6  D) 8  E 10

4) Assume the given sum of the series, 7.5 + 15.5 + 10.5 + 13.0 + 13.5 + 10.5....+ x – y + z is
20634, (x, y, z > 0) then what is the value of the expression (2x + 3y + 5z)?

A) 1084     B) 1284     C) 1464   D) 1684  E) None of These.

5) Let P be the product of all natural numbers between 45 and 293 that have an odd number of
factors. Find the highest power of 12 in P.

A)  6   B)  8  C) 10   D) 12  E) none of these

6)Assume r, s and t be three distinct integers between 0 and 10. if 1/r+1/s-3/t=2/(5r)
then find the
number of distinct values of (r + s – t).

A) 1  B)  2  C)   4   D) 3  E) none of these

7 How many integers less than 300 are relatively prime to either 10 or 18?
A)  140     B) 141    C) 142                   D) 139              E) 138

Source: OLD mocks( 2007 and before)

Problem Of The Week 18

Let p,q be prime numbers and n be a natural numbers. Find the number of ordered triplets (p,q,n) such that

1/p+1/q+1/(pq)=1/n




A) 0       B)   1   C)  2  D)  4   E) none of These

Tipster: All prime( >3) squares are of the form  6k+1 where k is a positive integer. Note this is a necessary condition not a sufficent one

Problem Of The Week 17

Two points are selected randomly on the surface of a sphere of radius R. What is the expected distance between them, along the surface of the sphere?

Problem is simple if you think logically, else you will have issues. This question is taken from teh CAT quant Blog by Suresh, the link of which you can see on the right!

Tipster:  The sum of the two legs of a right angled triangle is equal to sum of diameters of incircle and circumcircle :)

Concept 3 Circle and Triangles ( Part 1)

Geometry as a section wa spretty popular in CAT till 2004, consiting of 1/3rd of the problems and people used to think if they could handle it well they are clear with quant cutoff. And really that used to be the case. Cat has changed the trend reducing geometry every year since and last year in CAT 2007, there was not a single problem from geometry.  But, we can for sure be ready for nice geometry problems, so that if they come, we are up and ready for it.

Geometry has some major theorems. One should be clear about them, the ones on similarity of triangles, congruency of triangles, pythagoras, area and volume formula. Kindly refer to a text book for revising such concepts, I would recommend Quantum Cat By Arihant Publishers. Lets roll then !

The major theorems which we always need are :

Theorem 3.1 Pythagoras Theorem : a^2+b^2=c^2 where a,b,c are sides of a right angled triangle.

Clearly, C is the largest side, we call it hypotenuse.

The triplets of real numbers (a,b,c) which satisfy the above theorem is called pythagorean triplets. They are of real interest in all kinds of work.

Example 3.1 The length of one of the legs of a right triangle exceeds the length of the other leg by 10 cm but is smaller than that of the hypotenuse by 10 cm. Find the hypotenuse.

The obvious solution is  (a-10)^2+a^2=(a+10)^2 ( I have jumped a step)

solving we have a^2-20a=20a =>a=40 ( a can't be zero, its side of a triangle)

hypo is a+10=50

P.s : we have avoided the cumbersome assumption of sides as a,a+10 and a+20

Tipster clue: See this , the smallest integer pythagorean triplet is (3,4,5) so all numbers of the form (3k,4k,5k) will be pythagorean!

Practice Problem 3.1 FInd the sum of the lengths of the sides of a right angled triangle if the Circumradius=15 and inradius=6

Theorem 3.2  Sin law

a/Sin A=b/SinB =c/SIn C=2R   where a,b,c are sides opposite <A , <B and <C respectively and R is circumradius of Triangle ABC.

Very useful theorem, though we have entered the domain of trigonometry, but trigonometry, plane geometry and coordinate geometry are very important for each other to co exist.

Theorem 3.3 Cosine law

a^2=b^2+c^2-2bcCos A ( the notations remain the same as Theorem 3.2). The theorem can be similarly used for other angles too.

Practice Problem 3.2 Find the angle between the diagonal of a rectangle with perimeter 2p and area (3/16)p^2

Example 3.2 Find the length of the base of an isosceles triangle with area S and vertical angle A.

How do we start with this, we can offcourse going to need some basic geometry knowledge. let me tell you all of it. First the vertical angle of an isosceles triangle is the angle between the two equal sides( unless otherwise mentioned). The Perpendicular dropped on the unequal side from the opposite vertex, bisects the vertical angle as well as bisects the side. It means if we have a triangle ABC with AC=AB and AD perpendicular to BC then BD=BC and <BAD=<CAD.

The last thing we need is that area of a triangle is (1/2)bcsinA or (1/2)b^2Sin A for an isosceles triangle as b=c

now given (1/2) b^2sin A=S.......(1)

Now as AD bisects the vertical angle and then use BD=bsin(A/2)

hence BC=2BD=2bSin(A/2)

we can put the value of b from (1) and we are done !

Practice Problem 3.3  Find the largest angle of a triangle in which the altitude and the median drawn from the same vertex divide the angle at the vertex into three equal parts

Lets do a more involved example. This came in IMS SimCat 9. Nice and easy problem, but it might scare you for a moment if you look at the figure they drew. So I am not giving it :)

Example 3.3 In Triangle ABC, AD,BE and CF are the medians which intersect at G. ABCH is trapezium with AH=5units , and BC=10units  and Area( Tr BHC)=35 Sq units. Find the ratio of Area( BDFG): Area( ABCH). ( note we have  H and C on same side of B :) )

Here we again need to know this. The three medians divide the triangle into three triangle of equal area . Also they divide it into three quadrilaterals of equal area. So Area( BDFG)= (1/3)Area(ABC)

Next comes, the traingles drawn on the same base and between same parallel lines have equal area. Hence Area(ABC)=Area(BHC)=35 as we know the base BC, we know the altitude D= 2Area/base=70/10=7

Area of trapezium =(1/2)altitude( sum of paralle sides)= (1/2)7(10+5)

so our ratio is (35/3)/(15.7/2)=2/9

We are done :)

Problems Package ( Number Theory I)

1) If set A ={ 26, 29, 34, ...., n^2+25,...}
set B = {13501, 13504, ...., n^2+13500,....}
how many common elements do the two sets have?

2) What is the smallest positive integer with 6 positive odd integer divisors and 12 positive even integer divisors?

3) 4. Compute the number of squares between 4^9 and 9^4.

4) Given any positive integer n, Fidn the sum of all possible last digits of n^5-n?

MockaMania : Mocks on 14th Sept

Ims Simcat 9( Some good Problems from Quant)

The paper had cat2007 pattern, only difference was it was +2, and -0.5

1) If a^k has k^4 divisors, where k is a natural number, then which of the following is true?

I a=k=1   II a^k>=210   III a^k>=2^k; k>=2

A) Only I     B)   only II    C)  Only III    D) I or II E) I or III

2) f(n)=2g(n)+f(-n); for all non-zero integers n

g(n)= n*g(n-1) for all n>0 and g(0)=1

then Find g(-10)+g(-9)+....+g(0)+...g(9)+ g(10)

A) 10!+1    B)2*10!+1   C) 2*10!    D) 1 E) None of these

3) Distinct two digit numbers are written one after the other to form a six-digit number. How  many six-digit numbers thus formed have four consecutive 1s in them?

A) 90    B) 64    C)  65    D) 56  E )72

4) x=3m-1 and y=5n-1 where m,n,x,y are natural numbers less than 16. Find the number of pairs (m,n) satisfying the equations x^2=2y^2-7

A) 0   B) 1  C) 2   D)  3  E ) 4

5) FInd the number of real solutions of the system of equations

y=|x-1|+|x-2| and y+1=x(3-x)


A) 0   B)   1 C) 2   D ) 3   E) 4

6) A natural number ( greater than one) is called squareful  number if its prime factorisation contains at least one square. How many squareful numbers below 101 are there

A) 63  B) 61  C) 39  D) 67 E) 41

7) How many pair of consecutive natural numbers less than 51 are squareful numbers as defined above

A) 2   B)  3 C)  4   D)  5 E ) 6

8) A binary number is called tri-one if it has exactly three 1s. If all tri-ones are arranged in ascending order, what is the rank of the least 8 digit tri-one number?

A) 35 B) 32 C) 34 D) 40 E) 36

9) How many tri-ones as defined above , less than  110 in decimal, when converted to decimal is divsible by 5 in base 10?

A) 5  B)  6 C) 7   D) 8 E )10

updated!

Problem Of The Week 16

Points M and N are taken on the hypotenuse of a right triangle ABC so that BC=BM and AC=AN.. Find <MCN

A relatively easy problem, still try to do it. You will benefit from it

Problem Of The Week 15

a, b, c, and d are the solutions of x^4+7x^2-3x+2=0. Find a polynomial with integer coefficients whose roots are (a+b+c)/d^2, (a+b+d)/c^2, (a+c+d)/b^2, and (b+c+d)/a^2. Find the sum of roots of this equation

A very simple problem, one should take about 1-1.5 mins to solve this. If you take more, revise equations please!!

Problem Of The Week 14

Let X=[x] +{x} where [x] and {x} are respectively the integer and fractional parts of a real number x. Find the number of real number which exist such that they satisfy the following conditions:

a) 0<=x<=100

b) {x^2}={x}




A) 9000                  B) 9900                     C) 9990                    D) 9901                 E) 9991

Problem Of The Week 13

Suppose we have 3 real numbers x,y and z such that x^3=3y-2, y^3=3z-2 and z^3=3x-2 . Find the number of triplets of (x,y,z) which exist?




A) 0     B) 1                         C)  2          D) 3                E) None of these

Problem Of The Week 12

Let us have a series of numbers 16,1156,111556, .... . The series is formed by k ones, (k-1) fives and one 6 for k>0. Then each number of the series is a

A) Prime

B) Perfect Square

C) Prime Square

D) Atleast one of the foregoing

E) None of the foregoing

Conundrum 2 : Middle Three Digits

A five-digit perfect square in the form of 5abc6 has a thousands digit a, hundreds digit b, and tens digit c.  If a is less than or equal to b and b is less than or equal to c, what is the sum of a + b + c?

Problem Of The Week 11

Triangle ABC is right angled at C. m<ABC=60 and AB=10. Let P be a point randomly chosen inside triangle ABC, such that BP extended meets CA at D. What is the probability that BD>5root(2)?

Problem Of The Week 10

Find the length of the shortest path from (0,0) to (12,16) in the x-y plane which does not go inside the curve x^2-12x+y^2-16y+75=0

Problem Of The Week 9

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

Solutions to Power Play 1

The Answer key is as

1) C  2)  E  3) B  4) A 5) E  6) D 7) B 8) E 9) C  10) E

Question 1 is easy just use a-1/a= 1 and 1/a-a=1 will give two roots sum them

Question 2  10^j-1o^i=10^i(10^(j-1)-1)

1001=7.11.13=10^3+1

so clearly 10^3+1 divides 10^6-1 and therefore 10^6k-1

hence j-i=6k, k is a natural number

applying other constraints we get option e

Question 3) 1+2+3..30=30.31/2=31.15=465

[465/2]=232 now suppose a subset A of S doe not have sum more than 232 then A' must have sum more than 232 hence 1/2 of the subsets of S will have sum more than 232

so 2^30/2=2^29

question 4 use the concept of reflection we will get min distance sum as 5root(2)

Question 5) Function is not correctly defined as 0 is not a natural number

so option e)

Question 6)

let the radius be r
and the point of tangency be P and Q and triangle be ABC. P lies on AB and Q on BC

let AP = m and BQ = n

m^2 = 15^2-x^2
n^2 = 20^2-x^2
m = 9 and n = 16 x =12 arcPQ = 6pi

question 7) see n numbers product is n and sum is zero

if n is odd then sum can't be zero

similalry check for other cases it will easily come out n =4k

question 8 toughest problem of the test

let g(n)=p(n)-n

then g(17)=-7=g(24)

let a,b be integers such that p(n)=n+3

then a-17 divides g(a)-g(17)=3-(-7)=10

similarly for 24

hence we find a-17 and a-24 both divide 10 this means k=a-17 and k+7 both divide 10

this means k=-5 or-2

a=19 b=22

hence ab=418

question 9 can be easily done

question 10) tricky enough problem

look for a series and solve it you will get E

Blog On Vacation

I will be on vacation for about a week, will come online for very short periods of time, but I think there is enough material to keep you engaged. If you want more, kindly click the links listed in blogroll to the right and enjoy some interesting problem for CAT.

Laters !

Quantologic!

Edit: Blog is back !!!

Power Play I ( 30 August 2008)

We will start with power plays now. It shall consist of 10 Problems and You are expected to solve them in 25-30 Mins. Each correct answer carries 5 marks. The first two wrong answers carry (-1) marks each, the next two (-2) marks each and so on. Maximum time is 30 mins, but you are expected to solve in 25 mins!

Good Luck !

Please hit the comment button and post  your keys, I will post the official keys and solution in 7 days time!

And We will have a better mechanism hopefully, by the next powerplay!

Power Play I

1) Two different positive numbers a and b differ from their reciprocals by 1. Find a+b

A)   1                   B)  √6            c) √5              D)  3          E) None of these.




2) How many positive integer multiples of 1001 can be expressed in the form 10^j-10^i where i and j are integers such that 0<=i<j<99?

A)  15                    B) 90                     C) 64                     D) 720                     E) 784




3) How many subsets of the set {1,2,3...,30} have the property that the sum of all elements is more than 232?

A) 2^30                   B)    2^29                      C) 2^29  -1            D)  2^29+1           E ) None of these




4) Let the point P be (4,3). Choose a point Q on y=x line and another point R on the line y=0 such that sum of lengths  PQ+QR+PR is minimum. Find this minimum length.

A) 5√2                B)  3+3√2               C) 5                   D) 4                  E) 10




5) Let f be a function defined on odd natural numbers which return natural numbers such that f(n+2)=f(n)+n

and f(1)=0 . Then f(201)?

A) 10000                 B)20000                         C) 40000                  D) 2500                E) None of these




6) A semicircle  is inscribed in a right triangle so that its diameter lies on the hypotenuse and the centre divides the hypotenuse into segments  15 cm and 20 cm long. FInd the length of the arc of the semicricle included between its points of tangency with the legs.

A)  2π                B) 3π                      C)  4π                D)  6π                  E) none of these




7) The product of n numbers is n and their sum is 0. Then n is always divisible by

A)  3                       B) 4                               C)  5                         D)  2       E) None of the foregoing




8 ) Let P(x) be a polynomial with integer coefficients such that P(17)=10 and P(24)=17 . It is further know that P(n)=n+3 has two distinct integer solutions a and b. Find a.b

A) 10                     B) 220                      C)  190                       D)  48                       E) 418




9) Let A and B be two points on the plane. Let S be the set of points P such that PA^2+PB^2 is at most 10. Find the area of S

A)  2π                B) π                      C)  4π                D)  6π                  E) none of these




10) Let T={9^k: k is an ineteger 0<=k<=4000}. Given that 9^4000 has its 3817 digits and its leftmost digit is 9. Find the number of elements in T which also have leftmost digit as 9.

A) 92                          B) 93                    C) 183                          D) 184                       E) 185




Good Luck !

Problem Of The Week 8

Let x and y be real nos such that

[x] + 3{y}=3.9 and

{x} + 3[y]=3.4

[z] and {z} have their usual meanings.

Find the number of ordered pairs (x,y).

a) 0                    b) 1                   c) 2                     d)  3                        e) none of these

Gejo's Way Of Approaching The Quants Section

This is a useful article, by one of IMS Faculty. I thought of sharing with you guys. Have a look
http://www.imsindia.com/myims/index.php?option=com_content&task=view&id=112&ac=0&Itemid=59

Gejo's way of approaching the Quants Section!!!

The quant section of the CAT is feared by many – just the look of a question can cause worry in the mind of the test taker. The best way to tackle a CAT Quant question is to solve them logically than mathematically. Of course, you need to have your basic fundas in place to solve any quant question using logic.

The quant section of the CAT is feared by many – just the look of a question can cause worry in the mind of the test taker. The best way to tackle a CAT Quant question is to solve them logically than mathematically. Of course, you need to have your basic fundas in place to solve any quant question using logic.

Let us look at some examples – these are a few questions which are based on actual CAT questions.

1. Let n be the total number of different 5-digit numbers with all distinct digits, formed using 2, 3, 4, 5 and 6 and divisible by 4. What is the value of n?

1] 44 2] 32 3] 36 4] 38 5] 40

Permutations & Combinations is probably the most ‘hated’ topic. However, if you understand the basics and use logic, then it is the most fun-filled topic of all. Let us get to this question. As mentioned, we need to find out the 5 digit numbers divisible by 4 formed by the digits given in the question. To be divisible by 4, the last 2 digits should be divisible by 4. So to arrive at the answer, the first step is to find out combinations of the 2 digits from 2,3,4,5 & 6 that are divisible by 4 – eg: 24. Then, for each such combination, the last 2 digits are fixed. The remaining 3 digits can be arranged in 3! Ways = 6 ways. So the answer would be - 6 multiplied by the total no. of combinations of 2 digits divisible by 4. The answer necessarily should be a multiple of 6 and therefore the answer is 36 – option [3]. We just got lucky here, by the way, since there is just one option that is divisible by 6!

Of course, you have to be strong in the concepts so that you can solve this question in less than 10 seconds. I must remind you that short cuts happen when your concepts are strong.
2. The set of all integer values of x such that 3 × 5^x –5^|x|+1> 1 is:

1] x  >  –1  and  x <  5 / 3
2] x  >  1
3] x  ε I
4] x  < 5 / 3
4] No solution

Here is a question that looks scary! To solve this question, I am going to do a little manipulation. The equation given in the question is 3 × 5^x –5^|x|+1> 1. Now, to make it little more friendly, let me change it to   5 × 5^x –5^|x|+1> 1 (If it has to work for 3 × 5^x it must work for 5 × 5^x. Please note that this cannot be used everywhere). Now, the question changes to

5 × 5^x –5^|x|+1> 1
= 5^x+1 –5^|x|+1> 1

The above cannot have a solution because at best 5^x+1 will be equal to 5^|x|+1 when x is positive. When x is negative 5^|x|+1 will be greater than 5^x+1. Since 3 × 5^x is less than 5 × 5^x, 3 × 5^x – 5^|x|+1 > 1 will have no solution. Hence option [4]

Sometimes, we could change the question a bit without changing the answer outcome.
Most of the time test takers avoid dangerous looking question, which is a bad idea. You must read every question and give it a fair shot. You leave a question only after this. Be prudent not to waste a question just to save time! But also remember, not to get stuck on a question for long.

3. The capacity of tank B is 1.5 times the capacity of tank A. One tap fills tank A in 9 hrs and other tap fills tank B in 11 hrs. Both the taps are started at the same time initially. After 7 hrs, both of them are closed. Then remaining part of tank B is filled with the water taken from tank A. After this, how much time will it take to fill tank A with its tap?

1]  9.4 hrs       2]  2.1 hrs      3]  5.8 hrs      4]  6.9 hrs      5]  1.7 hrs
A – In less than 30 secs, if you apply logic, you can eliminate all options but 4. To explain this, it will take lot of words. But let me try.

Here is the story. There are two tanks, tank A & tank B and two taps, I will call them tap A & tap B [you will know why]. Given the capacity of tank B is 1.5 times the capacity of tank A. Also given tap A takes 9 hours to fill tank A and tap B takes 11 hrs to fill tank B. Both the tanks have been filled for 7 hrs. At this point, tank A needs 2 more hrs of water from tap A and tank B needs 4 more hrs of water from tap B. Now, the tank B is filled using water in tank A. We need to find out how much time will it take for tap A to fill tank A.

The answer will be 2 hrs + tap A time equivalent to 4 hr of tap B.

If the above statement seems confusing, let me explain it a bit. If the water was not transferred to tank B, then the tap A would have taken 2 hours (9hrs – 7hrs). That is the 2 hrs. Now the second term – tap B needed 4 hours of water from tap B (11hrs – 7 hrs). This is filled by tank A. 4 hrs of tap B water is filled by the tank A, therefore, tap A would need to fill water equivalent to 4 hr of tap B. [Confusing? Read it again, slowly!]

Let us now eliminate some options – Option 1 is out since tap A will take only 9 hours to fill tank A. Option 5 is also definitely out.  Option 2 seems to be out, at this moment, let us not eliminate it.

The fight is between 6.9 hrs, 5.8 hrs & 2.1 hrs. Look at this – tank B is 1.5 times bigger than tank A. While tap B takes 11 hrs to fill tank B & tap A takes 9 hrs to fill tank A. If the flow of tap A & B were same then tap A : tap B should have been  1: 1.5. However, it is 9: 11. You can see that tap B is faster than tap A. So 4 hrs of tap B > 4 hrs of tap A. Therefore, the tap A time equivalent of 4 hr of tap B > 4 hrs. So the answer has to be greater than 6. Hence Option [4]

The shorter way to solve this question seems so long, that is only because I am explaining the logic to a 3rd person. While reading this question, it is quite natural that you would straight away want to apply work, pipe cistern formula. In this case, the question can be solved using ratios.
Capacity Ratio : 1: 1.5
Tap ratio 9: 11
So for every 1 hr of tap B = (1.5 X 9)11 hr for tap A = 1.23 hr for tap A
4 hrs of tap for tap B = 1.23 X 4 = 4.9 hrs.
Therefore, the answer is 2 + 4.9 = 6.9 hrs.
After reading a question, take a moment to think and understand the question thoroughly before solving it

4. Consider two different cloth-cutting processes. In the first one, n circular pieces are cut from a square piece of side a in the following steps. The original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is:

1] 1  : 1     2]  √2  :  1     3]  n( 4 - π )  :  4n - π     4] 4n - π  :  n(4 - π)

For ease of calculation, let  4x = a

For process 2,



Radius = a/2 = 2x

The area is   4πx²

Scrap clot area  =  a² - 4πx²






For Process 1, assume n = 4



Radius of 1 circle = x

The area of each circle =  πx²

Total area of the 4 circles =  4πx²

Scrap clot area =  a2 -  4πx²



The answer has to be 1 : 1
I do not know how many would even try reading the question just because it is long question. Once you read the question and understand what needs to be done, then a little logic would help you reach the solution in no time.

5. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect square?
1]  4      2]  0      3]  1       4]  4      5]  2
Here, we need to find a perfect square which looks like aabb [a & b are digits]. Now, we need to use a little logic to arrive at the fact that aabb = 11 X a0b [for eg. 11 X 102 = 1122, 11 X 304 = 3344]
For aabb to be a perfect square a0b should be of the form 11x² so that aabb = 11² X  x² . The first task is to list down all possibilities of a0b being divisible by 11. a+b should be equal to 11 or 0 (this is the divisibility rule for 11 : sum of odd digits – sum of even digits should be either 0 or 11). a+b cannot be equal to 0 (for this, a & b both have to be equal to 0). a+b = 11.

The possibilities for a0b that are divisible by 11 are

209	11 X 19
308	11 X 28
407	11 X 37
506	11 X 46
605	11 X 55
704	11 X 64	64 is a Perfect Square!
803	11 X 73
902	11 X 82

There is one solution – 704 X 11 = 7744 = 88²
Ans: 3



Boom! One line question but not necessarily a one line answer. Many make this mistake of thinking that the level of difficulty of a question is directly proportional to the length of the question. There is NO such relation. You must solve each question on its merit. [not by the length or the look!]. This question needs you to first crack aabb = 11 X a0b. It may not come directly. If you crack this one step, you get the answer. This question becomes time consuming depending on the logic you use.





6.



It is an application of a basic funda. Again, the question looks scary and many would miss it.
During the analysis of the SimCATs, I would suggest that before looking at the explanatory answers, solve each question yourself. Then you look at the explanatory answers and see if you can find alternative methods to solve every question. This will help you build on your ‘logical cells’. You still have good 3 months for the CAT and you can crack it – just ensure that you use logic & common sense more.

Problem Of The Week 7

Two positive integers differ by  60.The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

A) 100                 B) 156                   C)  392                   D) 452                       E) none of these

P.S: If You are not able to solve this, Kindly refer to the concept 1 Perfect squares. Any student who has read that lesson and practised the problems should be able to solve this, if not reread the lesson!

Conundrum I : Help Varun!

Varun and Rahul sit together, Rahul asks varun find the sum of the sum of the digits of all natural numbers upto 10. Varun quickly answers 46. Rahul again asks what if we find upto 50, varun takes some time and answers 330. Rahul stands up and says, Can you find it for 2008, and do it before I count upto 20?

Your task is to help Varun, How will you do that?

Problem Of The Week 6

Let f(x) be as shown above. How many solutions does f(f(x))=6 have?

a) 2                b)  3                     c) 4                         d) 5                           e) 6

Problem Of The Week 5

Let a,b,c be real numbers such that for all real numbers x, such that |x|<=1, we have |ax^2+bx+c|<=100. Determine the maximum value of |a|+|b|+|c|

Top 50 WordPress Blog

Rarely, does it happen that a 2 day old blog appears in the Blog Of The Day, but this blog has achieved the remarkable feat. Quantologic was ranked 50th on 28 August, by WordPress.com. Thanks for all the support, I hope to bring more quality stuff.

Regards

QuantoLogic

http://botd.wordpress.com/2008/08/28/growing-blogs-853/

Problem Of The Week 4

Let a,b be integers then find the no of pairs of (x,y) such that
[x]+2y=a and
[y]+2x=b  where  [z] means greatest integer <=z

a) 0          b) 1                 c) 2              d) 4                e) Cannot be determined

Problem Of The Week 3

Find the sum of all positive integers n such that x,y,z are positive factors of n-1, x>y>z and x+y+z=n

a) 13               b)  31                  c) 44                d)   105                   e) none of these

Problem Of The Week 2

Let us define [x] as the greatest integer which is less than or equal to x and logp(q) is logarithm of q to the base p. Find a natural n such that

[log2(1)]+[log2(2)]+[log2(3)]+....+[log2(n)]=2008

a) 312                   b) 313                 c) 314                    d) 315            e) none of these

Problems based on concept 2

1)Two real non negative numbers satisfy that ab>=a^3+b^3, find the maximum value of a+b

a) 1/2  b)  1 c) 3/2 d) 2 e) none of these

2) Let x(n) be a sequence of real numbers such that x(1)=2 and x(n+1)=2x(n)/3+1/(3x(n))

then  for all n>1 which is always true

a) x(n) >1  b) x(n) <2 c) 1<x(n) <3/2 d) 1<x(n)<2  e) 3/2<x(n) <2

3) if p  and q are real positive numbers such that p+q=1 then fidn the minimum value of (p+1/p)^2+(q+1/q)^2

a) 5/2         b) 25/2           c) 15/2          d) 6 e) none of these

Concept 2 Inequalities I

Concept 2 Inequalities

Lets move on to our next concept, i.e Inequalities. Inequalities are generally present in cat and similar MBA papers, the question can be direct or indirect.

Concept 2.1 AM-GM Inequality

It means that AM( arithemetic mean) of  a set of positive numbers is always greater than or equal to the GM( geometric mean). The equality holds when the numbers are equal

(a+b+c)/3 >=(a+b+c)^(1/3)..........( 2.1)

Example 2.1 If a,b,c are positive numbers prove that (a+b)(b+c)(c+a)>=8abc

what we will do is  use AM-GM multiple times

(a+b)/2 >=sqrt(ab)

=>(a+b)>=2sqrt(ab)

similarly for others

(b+c)>=2sqrt(bc)

(c+a)>=2sqrt(ac)

then multiplying these three inequalities we get the desired result!

Practice Problem 2.1show that (n^n)[(n+1)/2]^(2n)>(n!)^3

Practice Problem 2.2 if x,y,z be the lengths of the sides of a triangle then prove that (x+y+z)^3>=27(x+y-z)(y+z-x)(z+x-y)

Practice Problem 2.3 show that for any natural number n, (n+1)^n>2.4.6....2n

Example 2.2 Show that for any natural number n 2^n>=1 +n.2^[(n-1)/2]

Lets see how we do this

2^n>=1+n.2^[(n-1)/2]

2^n-1>=n.2^[(n-1)/2] ( can you recognise the form?)

its the sum of a GP

we need to use AM-GM on the sum of GP

[1+2+2^2...+2^(n-1)]/n>(1.2.2^2...2^(n-1))^(1/n)

(2^n-1)/n> ( 2^(1+2+3..+n-1))^(1/n)=(2^[n(n-1)/2])^(1/n)=2^((n-1)/2)

so

2^n-1>2^((n-1)/2)

so we are done !!

Concept 2.2 Cauchy- Schwartz Inequality

If a,b,c and x,y,z be real numbers ( positive, negative or zero) then

(ax+by+cz)^2<=(a^2+b^2+c^2)(x^2+y^2+z^2)

Equality holds iff  a:b:c::x:y:z

Example 2.3 if x^4+y^4+z^4 =27 find min value of x^6+y^6+z^6

use cauchy on x^3,y^3,z^3 and  x,y,z

then (x^6+y^6+z^6)(x^2+y^2+z^2)>=(x^4+y^4+z^4)^2....(1)

use cauchy on the numbers x^2,y^2,z^2 and 1,1,1

then (x^4+y^4+z^4)(1+1+1)>=(x^2+y^2+z^2)^2

3(x^4+y^4+z^4)>=(x^2+y^2+z^2)^2...(2)

squaring both sides of 1 and using 2 we get

(x^4+y^4+z^4)^4<=3[(x^6+y^6+z^6)^2](x^4+y^4+z^4)

putting x^4+y^4+z^4=27 and taking positive square root we get

x^6+y^6+z^6>=81

Practice Problem2.4  if a,b,c be positive numbers such that a+b+c=4 find minimum value of a^3+b^3+c^3

Practice Problem 2.5 Find the min value of 2x+y if xy=8 and x,y are positive numbers

For any queries, post your doubts here itself !

Problem Of The week 1

If we define P(x)=1+x+x^2+..+x^6. Then what will be the remainder when P(x^7) is divided by P(x)?

a) 0        b)  1      c)   7     d) 49         e) none of these

Problems Based On Concept 1

254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is
a) prime
b) composite
c) prime square
d) a ,b,c
e) a,c

255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such m's

256) Find the number of pairs of positive numbers (x,y) such that x^3-y^3=100 and (x-y) and xy are integers.

a) 0          b) 1                 c) 2                d) 4                e) none of these

Concept 1 Perfect Squares

Concept I Perfect Squares

There has been a huge surge in the number of questions about perfect squares, in almost all mocks. The basic trick to any such question is assuming the number as a perfect square of an integer k and then using techniques of completion of square and then the formula of (a^2-b^2) and solving using divisibility theory

Example I Find all natural n such that n(n+16) is a perfect square

step 1 n(n+16)=k^2

step 2 (n^2+2.8.n+8^2)-8^2=k^2

step3 (n+8+k)(n+8-k)=64

see now lhs and rhs both are integers then both of (n+8-k) and (n+8+k) are divisors of 64. But note that we add the two equations we will get 2n+16, so teh sum of two divisors should be even hence both divisors even or both odd

so n+8+k=32,16,8,4,2 and n+8-k=2,4,8,16,32

but see this n is positive hence k is positive, thus n+8+k>n+8-k
so only two options

and solving we get 2n+16=34,20
so n=9,2

Note : The source of this problem is Pomona Wisconsin mathematics talent search exam!

Practice problem!!
Find the sum of all such positive integers m's such that m^2+25m+19 is a perfect square

Now we will extend the method to other kinds of problems
Basically what we used in the above problem is difference of square method

lets take an example
x^6=y^2+127, find the no of pairs of postive integers (x,y)

first step in this problem is recognising that 127 is a prime
then we move to
(x^3+y)(x^3-y)=127
so clearly 2x^3=128 x=4 and y=63

so one pair (4,63)

Welcome to QuantoLogic

A new blog for Quant lovers, especially those taking cat, gmat or other MBA entrances, we will be covering various concepts and other material from time to time

enjoy !