My Test Audio Pod Cast- Episode# 1

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Boeing 747 Take Off

Number System

Number Systems

1. Real Numbers
a. Rational - That can be expressed in the form p/q where q <> 0 and p, q are integers
b. Irrational Numbers - That cannot be expressed in the form p/q where q <> 0 and p, q are integers
c. Natural - All consecutive numbers from 1 onwards...
d. Whole - The set if all natural numbers including 0
e. Even - Divisible by 2
f. Odd - Not Divisible by 2
g. Integers - Positive and and negative
h. Prime numbers - The numbers that have no factor other than itself and unity- Ex 2, 3, 5, 7
2. Composite Numbers - The numbers that have factors other than itself and unity- Ex 9, 6
Complex Numbers - Numbers that are expressed in the form a+bi, where i = (-1)^(1/2)

3. Properties:

1. 1 is neither a prime number nor a composite
2. 0 is an even number
3. 2 is the only even prime

4. Operations on even \ odd numbers

2. Even (+ -) Even = Even
3. Odd (+ -) Odd = Even
4. Even * Even = Even
5. Odd * Odd = Odd
6. Odd + - Even = Odd
7. Odd * Even = Even

5. For any two numbers N1 and N2, their product is equal to the product of their LCM and HCF ->N1xN2 = LCM(N1,N2) x HCF(N1,N2)

6. If N1 and N2 are co prime, then HCF(N1,N2) = 1, so N1xN2 = LCM(N1,N2)

7. Sum of the first n natural numbers - [n(n+1)]/2
8. Sum of the squares of first n natural numbers - [n(n+1)(2n+1)]/6
9. Sum of the cubes of first n natural numbers - [n(n+1)/2]^2
10.Sum of even numbers - n(n+1)
11.Sum of odd numbers = n^2

12.If a^p x b^q x c^r are factors of a number N, then the total of factors of N are (p+1)(q+1)(r+1), where a , b and c are prime factors

EX- The unit's digit in the expansion of 2 raised to power 51 is :

2 has a cyclicity of 4 and 51(mod 4) = 3 , so the unit's digit will be same as in 2^3 = 8


EX- How many 0s are there in the expansion 100!

To solve this you need to find the highest power of 10 that divides 100!

Now 10 = 5X2

100/2 = 50
100/2^2 = 25
100/2^3 = 12
100/2^4 = 6
100/2^5 = 3

So the max power of 2 in 100! Is 96

Similarly – We can find the highest power of 5 in 100!

100/5 = 20
100/5^2 = 4
100/5^3 = NA

i.e. = 20+4 = 24


So we have 100!/10 = 100!/(2^96X5^24) = 100!/(10^24 X 2^72)

So the number of 0s in 100! Is 24

EX - The difference between the biggest and the smallest number formed by using 0, 1, 2, 3, 4 exactly once is :

(1) 28000 (2) 32591 (3) 27143 (4) 41976

Soln - 43210 - 01234 = 41976

13.You can perform all the arithmetic operations with remainders that you can with the numbers. For example if the two numbers N1 and N2 on division by “a“, leave remainders R1 and R2, then the remainder when (N1+N2)%a = R1+R2 or (R1+R2)(% a) if r1+r2 > a. Similarly (N1 – N2)%a = (R1 – R2)%a and also (N1*N2)%a = (R1*R2)%a. (read % as remainder from divisor).

14.Fermat's Little Theorem:

a. If p is a prime number and a any integer, then (a^p-a)% p = 0 i.e. a^p- a is completely divisible by 'p'.

b. If a and p are co prime then a^(p-1)%p = 1

15.Euler's Generalization
If 'a' is any number co-prime to 'n' then - a^e(n) % n = 1, where e(n) = euler's quotient

16.For any positive number n, 2<=(n+1/n)^n <=3

17.(n!)^2 > n^n

18.If n is an odd number, then n(n^2-1) is divisible by 24.

19.If n is an odd number greater than 3, then (n^2-1) is divisible by 24.

20.If the sum of 2 number x and y is constant x+y = k, then their product is MAX when x = y = k/2 and this product is = k^2/4
21.On the same lines if the product of two numbers is constant i.e. xy = k, then their sum is MIN when x = y = root(k) and the value is = 2root(k)

Note: This concept can be extended to more than 2 numbers.

Modular Mathematics
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This one is for ignorant buddies who find remainder problems and theorems related to it hard to understand.

I believe that algebraic number theory is one of the most intriguing topics of mathematics. Numbers have enchanted mathematicians for centuries and still there is much to be explored. Here is a set of theorems that can be used to solve almost all the problems related with finding remainders. Before starting lets prove that number of primes are infinite. Suppose I have found first few primes as p(1) = 2; p(2) = 3 and so on up to p(n). Product p(1)*p(2)*…*p(n) is divisible by all the primes we have found so far.
That means p(1)*p(2)*…*p(n) + 1 is not divisible by any of the known primes. That means either this number is a prime or it is a composite number having prime factors different from that are known (Caveat: -We have no right to say that this number is prime unless we prove it. A number of coaching institutes say this number is a prime but this is not necessarily true). Hence if we have first n primes we can find at least one more prime and that proves infinitude of primes. Wasn’t it a simple way of proving something of quite an elusive nature? Well that’s the beauty of mathematics.

Why does ISO insist upon standardizing the methods and approaches of running a business? Because when you follow same process again and again, you become proficient at using the process that brings the efficiency and quality in products and services. So if we follow a prescribed set of rules, we will become proficient at solving even difficult remainder problems.

It is necessary to know how a remainder is formally defined. Suppose we have two positive numbers a, b where a>b. a = b*q + R is true for infinite sets of integers (q, R).
Lets say that R is the residue (because we have not defined remainder as yet). We define remainder as value of R that is numerically less than the divisor b and either negative or positive. Or ½r½ < b. A sample statement “ When a number is divided by 29 leaves a remainder 5.” should be interpreted as N = 29x + 5. I won’t explain the trivial cases that are given in almost all the materials, as when a number is divided by two different divisors, remainders are same then remainder when LCM divides the number gives the same remainder etc. Here are the theorems you must know.

First thing we should be clear about is that whatever operation (addition, subtraction or multiplication) happens on the numbers, same operation happens on the remainders of the numbers from a given divisor. Suppose 2 numbers x and y when divided by divisor ‘a’ leave remainders r and R i.e. x = am + r and y = an + R.
(x + y) = a(m + n) + (r + R) hence remainder when (x + y) is divided by a is either (r + R) or remainder when (r + R) is divided by a ( in case r + R is greater than a).
Similarly (x – y)%a = (r – R)%a and also (x*y)%a = (r*R)%a. (read % as remainder from divisor).

I presume everyone knows the application of binomial theorem in finding remainders so won’t explain here.

RULE 1. Chinese remainder theorem: - Do not panic by looking at the name. It is just another buddy that helps in solving remainder problems. I will explain it using a sample problem.
Q. A number when divided by 7 leaves remainder 5 and when divided by 11 leaves remainder 3, find a general solution. There may be easier methods available for solving these questions but motive here is to understand the approach that more or less remains the same even when we move towards questions of higher level. From given statement we can say that number N = 7x + 5 = 11y + 3.

Or x = (11y – 2)/7
Lets say that y when divided by 7 leaves a remainder r i.e. y = 7a + r. 11 leaves either 4 or –3 as remainder from 7 (at times negative remainder makes the calculations easier). Since x is an integer, it gives remainder 0 when divided by 7 therefore 4r – 2 is divisible by 7 since r is the remainder from 7, you’ll have to try values of r up to 6. At r = 4, 4r – 2 becomes divisible by 7. So we can say here that y = 7a + 4. Putting the value of y in N=11y + 3 = 11(7a + 4) + 3 = 77a + 47. Putting a = 0, 47 is the first number of this type and 77a + 47 is the general series of such numbers.

I hope you can understand the process. Now lets say that in the same question another condition is given that when divided by 13, number leaves a remainder 6.
That means N = 13b + 6 along with the previous conditions. We have already solve it for two conditions and found the combined form as 77a + 47. We just need to solve this with the new form 13b + 6.
77a + 47 = 13b + 6
Or b = (77a + 41)/13.
Again using the same process.
77%13 = -1 (Usefulness of negative remainders)
41%13 = 2 (Positives are equally useful)
That clearly means that a%13 = 2
(-1)*(2) + (2) = 0{I have done the same operation on remainders as done on numbs}
Therefore a = 13c + 2; and putting this value in 77a + 47 = 77(13c + 2) + 47
= 1001c + 154 + 47 = 1001c + 201.
Whatever other remainder conditions are given, we can solve in same fashion.

Now you know CHINESE REMAINDER THEOREM.
You need practice to master it. Do you require questions on this? I don’t think so. You can make as many as you wish to.

RULE 2: FERMAT’S LITTLE THEOREM

Using the properties of infinite arithmetic progressions, Fermat proved the theorem that for a prime number ‘p’ that is co prime with another number ‘a’; when a to the power
(p-1) is divided by p, remainder is 1. Or a^(p-1) % = 1. Don’t forget that this theorem is defined only for prime divisors. E.g. 2^4%5 =1; 3^10%11 = 1. Or in more general form
a^{n(p-1)}%p = 1. So if the question is 2^1000%11, remainder will be 1 because power of 2 is a multiple of (11-1).

What if divisor is not a prime? Then we have eular’s generalization of Fermat’s rule.
That generalization says that a^e(n)%n = 1. Where e(n) is eular’s number for divisor n. Also a is co prime with the divisor n. Eular’s number e(n) for divisor n is defined as number of natural numbers less than ‘n’ and co prime with it. How to find e(n)? Suppose I want to find eular’s number of 1001. Prime factors of 1001 are 7,11,13.
e(1001) = 1001(1-1/7)(1-1/11)(1-1/13). In general for a number n having prime factors p1, p2, p3…. e(n) = n (1-1/p1)(1-1/p2)(1-1/p3)…
This knowledge about eular’s theorem is sufficient. Now that we know fermat’s rule and eular’s rule, lets try a sample question.

What are last two digits of 3^96? Last two digits of this number can be obtained by finding remainder when this number is divided by 100. 100 = (2^2)*(5^2). Understand the things clearly here. For 2^2 i.e. 4, eular’s number is 4(1-1/2) = 2 and for 5^2 (25), eular’s number is 25(1-1/5) = 20. Using eular’s theorem, 3^2 %2^2 = 1 and
3^20 %5^2 = 1. if we take LCM of the powers of 3 in two cases(LCM is 20) we find that 3^20 %100 = 1 or last two digits of 3^20 are 01. It can be proved for any other number co prime with hundred. That explains why cyclicity of last two places is generally 20 for all numbers co prime with 100. Last 2 places of 3^96 will therefore be same as last two places of 3^16. (Because 3^96 = 3^80 * 3^16 and 3^80 ends in 01) . How to find last two digits of 3^16? 3^20 = 3^16 * 3^4 lets say last two digits of 3^16 are xy.
Therefore xy * 81 = 01 (considering only last two digits) that’s true for xy = 21.

Another question. What is the remainder when 3^1001 is divided by 1001?

1001 = 7*11*13. 3^6 %7 = 1; 3^10 %11 = 1 and 3^12 %13 = 1. That implies
3^{LCM (6,10,12)} %1001 = 1 or 3^60 %1001 = 1. Using this we can reduce the previous problem into a simpler one. 3^1000 = 3^960 * 3^41, from 3^960, remainder is 1 therefore reminder when 3^1001 is divided by 1001 is same as remainder when 3^41 is divided by 1001. 3^41 %7 = 3^5 %7 = 5 {I presume it should be clear by now}
3^41 %11 = 3
3^40 %13 = 3^5 %13 = 9. It is a number which when divided by 7,11 and 13 gives remainders 5,3 and 11 respectively. N = 7x + 5 = 11y + 3 = 13z + 9. Further we know how to solve it (HINDI CHEENI BHAI BHAI).

I am attaching a list of questions few of which are collected from various threads and a few are made by me. Lets solve all those using the standard methods we have learnt. Solving the problems will make concepts clearer. Remember; try to standardize your approach.




1.What is the remainder when 17^19 + 13^19 is divided by 25?
2.Find the remainder when 104^303 is divided by 101.
3.Find the last two digits in the expansion of 2^ 999.
4.Find the remainder when 2 ^ 1990 is divided by 1990.
5.Remainder when (128 )^500 is divided by 153.
6.Find last 3 digits of 3^1994.
7.What is the remainder when 2^2001 is divided by 2001?
8.What is the remainder when 2^2002 is divided by 2002?
9.What is the remainder when 13^2404 is divided by 2310?
10.What is the remainder when 2^10013 is divided by 3125?
11. Find last 4 digits of (2319)^{10^12 + 2}.
And a lengthy one here.
12. What is the remainder when 17^28820 is divided by 30030?

Solve these first and I’ll post more related with the topic.


Concepts - Cubes
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We usually find questions invovling cubes in CAT. like, a big cube painted red is cut into 64 smaller cubes and find the no of smaller cubes with no sides painted etc. etc.

lets assume the cube is devided into n^3 small cubes.
then,

no. of small cubes with ONLY 3 sides painted : 8( all the corner cubes )

no. of small cubes with ONLY 2 sides painted :

A cube is painted on 2 sides means, it is on the edge of the bigger cube ,and we have 12 edges, each having n cubes. but since the corner cubes are painted on 3 sides, we need to neglect them. so in effect, for each side we will have (n-2) small cubes with only 2 sides painted.
thus, then number is, 12 * (n-2)

no of small cubes with ONLY 1 side painted :

for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with only one side painted. and we have 6 faces in total.
so th number is, 6*(n-2)^2

no of small cubes with NO sides painted :

if we remove the top layer of small cubes from the big cube we will end up a chunk of small cubes with no sides painted.
this number will be equal to, (n-2)^3.

IBM in Software services - Key Points

Key points to remember about the entry and growth of IBM in Software services:

1. In 2002 IBM acquired PricewaterHouseCoopers. This gave it access to PWC's 30000 consultants and not it could bid as aggressively for consulting projects as an Accenture could. Before this IBM was absent from the high end consulting business.

2. As part of its overall strategy, IBM also shed its commodity PC business by selling it to Chinese PC maker Lenovo in early 2005.

3. IBM lacked a low cost base for executing services work.The big India push of the past three years — which includes the takeover of the BPO company Daksh eServices for $160 million and ramping up its delivery centres in Bangalore — is meant to remove the second weakness.

A small point here — the India strategy is part of IBM’s overall plan to create four delivery hubs in the BRIC (Brazil, Russia, India, China) countries to bring down cost of operations. However, as we will see, India is way, way ahead of the other three.

4. Brazil - To leverage the latin american markets and as well as a nearshore option for US customers.
Russia - Nearshoring destination for European clients. Its big advantage in Russia is a large per capita pool of engineers — 3,500 for every 1 million people.
Then there’s China. Compared to India, IBM has just 7,000 people in China doing offshore, but the country has a huge domestic market over which IBM has established a stronghold over a 15-year period.

5. In 2005, the BRIC countries collectively ploughed in $3.8 billion in revenues and employed close to 60,000 people.

6. IBM bet on three key trends taking off — Linux, on-demand computing and, finally, emerging markets not just as a low-cost base, but also as a market

Questions Marks that remain on whether IBM will achieve the target:

1. The PWC acquisition has not really brought the kind of consulting business that IBM was expecting.
2. IBM still has a lot of flab in its european and american establishents. It need to cut atleast half of the 260,000 american workforce, but the question is how fast can its India division take the load of a 130000 head count.

A couple of influential IBM watchers feel that for India to make a significant difference to IBM’s bottomline and topline, Big Blue needs to make at least one more spectacular acquisition, perhaps of one of the top five Indian services firms. That remains to be seen in the future!!

Star TV - The new star of broadcasting

Star TVs presence in India and vital stats:


What star would do next:

1. Promote regional content. From being a primarily HIndi channel it needs to create more regional content.

2. The second is launching new channels. Star One was launched in 2004 with the intention of keeping the audience for high-end Hindi entertainment within the Star fold. The same year, Star Utsav went on air to get people to convert to cable TV in the interiors

3. The third is the international market where Star, strangely, has been left behind by Zee. In March 2005, Zee got a juicy chunk of its income from distributing its channels in Europe and the US.

4. And the fourth: Star is doing all sorts of things to squeeze more from advertising. There is the bet on new categories such as retail, real estate and branded jewellery. The other, says Kevin Vaz, executive vice-president (ad sales), is to expand time bands.Star Plus has already extended the definition of prime time from 9-10 p.m. to about 8-12 p.m. now. Also, Star One has plugged the leakage of revenues from small advertisers. “It was launched for the advertiser who can’t afford a high TRP product (Star Plus),” says Vaz.

5. Last but not the least Star has its cash reserves and, therefore, its ability to outspend competition

Source - BusinessWorld

Why should you not set up Chip Fabrication Unit in India

1.Already there is huge global overcapacity in chip manufacturing. There is no way that we can compete with China. Besides, you can simply pick up what you want from Japan, South Korea and China. Setting up a fab means an investment of $2-3 billion. It is a highly capital intensive business.
2. The cost advantage that we have is in the cheap labor costs. But due to poor infra structure, the transportation and shipment costs are very high in India so that advantage is lost on us.
3. Each chip needs a customer. So, logically, you need to tie-up with the Nokias and IBMs of the world before building a fab. It’s like buying a Boeing 747-400. Unless you are full up on travellers, you will incur losses.
4. A quote from Vinod Dham - Better known as Father of the Pentium processor-

"When I was with Intel at Albuquerque in 1989 in a fab plant the size of a football field that was set up with an investment of $1 billion. We were working on the 486 chip, but it was not ready. Each day we lost $25 million. So, to ensure that the operators got practice, we used memory chips. All that was used to make key chains and earrings. We simply cannot afford something like that in India."

Well you see these are reasons enough not to set up a Chip fabrication plan in India.

Source - BusinessWorld

The Mukesh Ambani Group's SEZ Gambit

Mukesh Ambani group has set its sights on creating world class business and residential cities in the form of 4 intergrated SEZ. One of the flagship SEZ is being set up in Navi Mumbai - Maha Mumbai region which when complete will be comparabe with China’s Shenzhen. The investment in this project could well be more than the Rs 73,000-crore annual revenues of his flagship petrochemicals company, if one goes by the Rs 25,000-crore investment, both in the form of debt and equity, planned over a decade in just Navi and Maha Mumbai.

Salient Features of the SEZ plan:

No of Hubs - 4
Size - > 25000 acres each
Investment By Reliance - USD 50 bn in Navi and Maha Miumbai
Outside Investment (from business enterprises, the hotel industry, retail, healthcare etc ) - USD 50 bn
Deadline - 10 yrs for Navi and Maha Mumbai
Targeted users will include manufacturing industries as well as services like warehousing, BPO and biotechnology.

The group's estimate that the investemt of $5 bn will attract 10 times investment in the form of FDI is considered conservative since similar Chinese SEZs attracted FDI to the tume of 40 times the investment.

The proposed Sewri trans-harbour sea link is vital to fuel the growth of the Mukesh Ambani group’s Navi Mumbai SEZ — it will cut down the commute time from South Mumbai to Sewri from a couple of hours to just around 40 minutes

Roadblocks and Pit falls:

1. Unlike in China, the land acquisition in India is slow and lengthy process and land ownership is fragmented. In China all land is owned by the state.
2. How quickly the SEZ projects take off also depends on the government’s ability to execute infrastructure upgrade and attract foreign investment. A key driver of growth of Suzhou, another Chinese SEZ, has been its road and rail infrastructure, apart from access to airports and seaports along the southern coast. But, in India, much remains to be done in this regard.
With Ambani's prior experience in developing the 7400 acres Jamnagar refinery and the 140 acre Dhirubhai Ambani knowledge city, it seems that this venture in going to be another feather in Ambani group's cap.

Source - Businessworld

The TCS billion dollar roadmap - Key Points

Here I have summarized an article from Business World about TCS future growtn intiatives:

1. Develop Global delivery capability in Offshore, Onshore and Nearshore. TCS plans to achieve this by augmenting its own existing centres especially the near shore ones (post the $260 million ABN Amro deal). It already has 20 — 10 in North America, one in the UK, and three each in Europe, Latin America and the Asia-Pacific. The other 22 are in India.

2. Acquisitions and greenfield investments. Last years buyout of BPO company Comicrom in Chile brought 1,200 people on board, while the $847-million Pearl Group contract gave it a 950-man BPO delivery capability in the UK.

3. Equal the competition - Given that the competition is already formidable — IBM has 25 such centres and Accenture has 40 — TCS’ bid to scale up fast is not without reason.

4. De-centralization of dcision making. As TCS plans to go global decision making cannot be restricted to a central command led bt Ramadorai. So the global centres will function as independent units with more localization of authority.Decentralisation has become all the more critical because of the rapid expansion in TCS’ headcount – it closed fiscal 2005-2006 with almost 63,000 people on its rolls, of which 6.5 per cent are non-Indian

5. Taget 2007 - To cross USD 4 bn due to the growth initiatives taken so far.

The "Big Daddy of" all airliners

Its the biggest.

It weighs 277,000 kg when empty and 560,000 kg @full load.
Wing span 79.8m (261ft 10in), length 72,75m (238ft 8in).
Height 24,08 m (79ft).
Capacity - Flightcrew of two. Standard seating for 555 passengers on two decks in a three class arrangement.

Who is this? The Airbus A380 - The "Big Daddy of" all airliners. The most ambitious civil aviation aircraft designed till date.

Formerly known as the A3XX, Airbus' double-decker passenger jet, the A380, will be the largest airliner ever built. Lengthwise, it would nearly stretch from goal line to goal line of a football field while its wing tips would hang well beyond the sidelines. Three full decks will run along the entire length of the plane. Upper and main decks will serve as passenger areas, and will be connected by a grand staircase near the front of the plane and by another smaller staircase at the back. Although the lower deck will be reserved primarily for cargo, it could be outfitted for special passenger uses such as sleeper cabins, business centers or even child care service. In a one-class configuration, the A380 could accommodate as many as 840 passengers. The more likely three-class configuration will still offer an unprecedented 555 passenger seats. Either way, the A380 would offer 30% - 50% more seating than its direct competition, the Boeing 747-400.
Although the A380 will be able to fly a distance of over 10,000 miles, the plane's usefulness will not be limited to long-haul flights. For instance, many flights within Japan are among the highest in passenger capacity and would be well suited for A380 service, despite their short distances. Whatever the flight distance, a new breed of engines will be required to lift the plane's 1.2 million pounds into the air. Rolls Royce and GE/Pratt & Whitney are both working on engines to provide thrust that will max out at 75,000 pounds. By comparison, the first American jet airliner in service, the Boeing 707, was powered by only 10,000 pounds of thrust.

As amazing as it will be for this behemoth to take off into the air, the A380 faces significant challenges on the ground as well. To integrate into existing airports, the A380 must fit the standard airport-docking plan. The plane's nearly 262-foot wingspan meets this requirement by about 18 inches. Its outer-most engines, however, would hang just beyond the standard 150-foot runway width, requiring upgrades at many airports. The plane's weight will be distributed to 20 landing gear wheels, actually producing less weight per wheel than the 747. The cockpit location, between the main and upper decks, is designed to give pilots a vantage point on the runway similar to that of current airliners.

Due to recent technological advances, Airbus claims the A380 will be a more efficient plane than its rival, the 747. Airbus states the A380 will use 20% less fuel and will fly quieter, cheaper and more environmentally friendly than the 747. Airlines seem to be impressed. So far, ten carriers have declared their interest in the plane, placing options to order a total of 66 planes.

The Big Daddy of all planes - Pictures

I am back!!

Yes! I am back, after a long winter hibernation that lasted longer than the Polar Bear's. In these 5 months a lot many things have happened. I lost some and won some. As the B-School results trickle out I can see what I have got. NM, FMS and XLRI ll have the same answer for me - No. I am not crestfallen as you would assume. In 3-4 odd months of preparation, along with a demanding job, managing 98.85 percentile in VA and an overall 86%ile in XAT 2005 is a not an un - welcome sign. If you have attempted XAT 2005 you would know what I am talking about.
So here I am setting the mark again, getting the aim right, getting back on the tracks with mud on my face. CAT 2006 here I come

Biz News -I

Flying start- Go Air

The fourth low-cost carrier in India, Go Air, took to the air on 4 November. Go Air, promoted by the Wadia group, will offer services to Goa, Ahmedabad, and Coimbatore from Mumbai. Initially, the airline will commence service with one A320 aircraft with a single class, 180-seat configuration, and plans to expand its fleet to 36 aircraft in three years.

VodaFone-Airtel

A little more than two years after it pulled the plug on India, the world's largest mobile company, Vodafone plc, is back. While Air Touch Vodafone had a 21 per cent stake in RPG Cellular (Madhya Pradesh) earlier, this time round, it has picked up a 10 per cent stake in Bharti Tele-Ventures, India's largest mobile operator, for Rs 6,700 crore. It's by far the biggest deal ever in the Indian telecom industry. (See 'Recent Telecom Deals')The £34-billion, UK-based Vodafone, which has 165 million subscribers in 27 countries, has paid Rs 47,652 ($1,083) per Bharti subscriber. That's a lot higher than the Rs 16,296 per subscriber paid by Essar to pick up BPL and the Rs 17,325 per subscriber to pick up Max's 3.16 per cent holding in Hutchison Essar.