Saturday, August 30, 2008

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 !

Friday, August 29, 2008

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 × 5x –5|x|+1> 1 is:

1] x  >  –1  and  x <  5 / 3
2] x  >  1
3] 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 × 5x –5|x|+1> 1. Now, to make it little more friendly, let me change it to   5 × 5x –5|x|+1> 1 (If it has to work for 3 × 5x it must work for 5 × 5x. Please note that this cannot be used everywhere). Now, the question changes to

5 × 5x –5|x|+1> 1
= 5x+1 –5|x|+1> 1

The above cannot have a solution because at best 5x+1 will be equal to 5|x|+1 when x is positive. When x is negative 5|x|+1 will be greater than 5x+1. Since 3 × 5x is less than 5 × 5x, 3 × 5x – 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










































20911 X 19

30811 X 28

40711 X 37

50611 X 46

60511 X 55

70411 X 6464 is a Perfect Square!

80311 X 73

90211 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!

Thursday, August 28, 2008

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/

Wednesday, August 27, 2008

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 !

Tuesday, August 26, 2008

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

Monday, August 25, 2008

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 !