Verdict!

As you may have read about my cat score and various interviews!

Here is the verdict!

IIM A: Waitlisted ( Waitlist didn't convert)


IIM C PGDM: Reject

IIM C  PGDCM: Convert

IIM L: Reject

IIM I: Convert

IIM K: Waitlisted ( Converted)

Joining: None

What's next: CAT 2009




Good Luck !

The Funny Side of Mathematics

During my interview for IIM Indore, which coincidentally was my first experience of  an "IIM Interview", I yelled a few times that I am good at mathematics and I should be asked questions on that. I also mentioned I would like to teach mathematics etc etc, lets leave the details.

The learned professor, who was supposed to ask the tech( acads maths etc) questions suddenly asked, " Can you prove 1=2?".

For a moment, I was like what? Then I told him that it is wrong thing, and all the proofs that you get in forwarded emails and fun sites are wrong and flawed.

He said, yes it is flawed, can you tell me the flaw?

I somehow ws not able to recollect this simple thing there, but I did it as soon as I was outside the room, interview blues I guess.

So the readers of this blog do not have such a problem, let me explain some ways to prove 1=2.

Method 1

x^2= x+x+x+x+x... ( x times)

differentiate both sides wrt x

then we get

2x=1+1+1+1+.. (x times)

2x=x

=>1=2

everything looks good, isn't it ? But it  is flawed. I leave it to the reader to figure out the flaw? if not, I will explain the flaw in one of teh comments. Kindly bear with me !

Method 2

1/4=1/4

(1/2)^2=(-1/2)^2

taking square root both sides

1/2=-1/2

add 3/2 both sides

2=1

This is much simpler and the flaw is also visible, i.e square roots are always positive .

Found it funny? Not yet?

One of my other friends was asked to tell when is pie day celebrated?

Innocent though he was, did not take a guess. You can always hazard a guess pie =3.14  so it has to be 14 march ( American calendar)

There is also a pie approximation day and yes you guessed it right, it is 22/7 that is 22 july ( english calendar).

Ramanujam's Number

A friend of mine, who as well proclaimed himself as math wiz in one of the IIM Interviews, was asked tell me about 1729?

And he was a wiz? Yes you guessed it right. He answered it well. 1729=10^3+9^3=1^3+12^3

It is the smallest number which can be written as the sum of two cubes in two different ways. It is a very famous number, so if you dare to proclaim maths as your hobby, you must know it.

Want to more? wikipedia, my friend! ( Indians must know about Ramanujam)

Thats all!

Cheers1

The New Look!

The Blog has gone a total change, old users will find the change quite sharp. Some of them are listed!

1. The theme has changed to a more classic and professional look.

2. We have moved from a three column theme to a two column theme.

3. The colour scheme has been changed.

4. Nothing more I can think of :)

Well, all of it has been done to signal the start of a new season. The users who will be taking CAT-09, you are in for a treat.  The blog will try to bring in the very best of quant and DI. There is a plan for some verbal content as well.

Best Regards

Rahul

XLRI Results Declared

XLRi declared its results today. It can be accessed here And not so surprisingly I was not selected. I had a poor Pi and an average GD. So the season does not start well

My CAT 2008 Score

Hi People,

I am sure quite a few of you who visit this blog took CAT 2008, as did I. And on 9th Jan 2009, came the results after about two months and the result was according to expectation at least for me, and I hope it was same for you.

My Scores

Quant     64 ( 16 Correct, 0 incorrect)  99.88%ile

Logic and DI  44 ( 12 correct, 4 incorrect) 98.43%ile

Verbal     76 ( 21 Correct, 8 Wrong)  99.49%ile

Overall  184  ( 49 Correct, 12 Wrong) 99.97%ile

Calls:

IIM Ahemdabad, Calcutta, Indore, Kozhikode, Lucknow!

Wish me luck for the next stage!!

Thanks a lot for being with me !

Best of Luck everyone!

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 !!

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)

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!

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 !!!

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.

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?

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/

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

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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

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