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INFERENCE OR JUDGEMENT
Introduction
Any statement can belong to one of the three types. Fact, inference or judgment. In this category of questions in CAT, you are expected to figure out which type it is.
Fact, Inference, Judgment.
Facts
• Can always be verified
• Facts are events which have occurred or results which have been observed
• Since they need to occur in order to become a fact, facts can not be based
in the future.
• May contain consequences, but even these can be verified. See example 1 below.
Examples of a Fact:
1.The government has been supplying free drugs since 2004, and 35 000 have benefited up to now- though the size of the affected population is 150 times this number.
2.Only about 13 million children in the age group of 6 to 14 years are out of school.
3.The truth is that we have more red tape- we take eighty- nine days to start a small business, Australians take two.
4.The economies of the industrialized western world derive 20% of their income from the sale of all kinds of arms.
Inference and judgment unlike facts are not real happenings or situations. They are some kind of a conclusion that we arrive at based on the facts. What then, is the difference between an inference and a judgment?
Inference or Judgment?
Judgments are arguable and contestable. Inferences are rock solid. Although both judgments and inferences are based on facts, in the latter the conclusion is so unquestionable that it becomes fact itself.
Inferences
• Are conclusions drawn based on facts.
• Are logical consequences concluded from evidence.
• They always appear along with facts. They are not baseless statements and are not questionable
Examples of Inference
1. The recent initiatives of networks and companies like AIDScare Network, Emcure, Reliance-Cipla-CII, would lead to availability of much-needed drugs to a larger number of affected people.
2. According to all statistical indications, the Sarva Shiksha Abhiyan has managed to keep pace with its ambitious goals.
3. Every red tape procedure is a point of contact with an official, and such contacts have the potential to become opportunities for money to change hands.
4. Even without war, we know that conflicts continue to trouble us- they only change in color.
Judgments
• Are opinions, suggestions and recommendations
• Include a lot of quantities that cannot be measured, such as happiness, beauty, joy etc.
• Many a times, judgments are not accompanied by facts at all but are only opinion statements. When there is no fact involved, the statement can only be a judgment statement.
• A judgment is an honest attempt to make reasonable observations about the given facts but they do not conclusively prove anything.
Examples of Judgments:
1. So much of our day-to-day focus seems to be on getting thins done, trudging our way through the tasks of living- it can feel like a treadmill that gets you nowhere; where is the childlike joy?
2. We are not doing things that make us happy; that which brings us joy; the things that we cannot wait to do because we enjoy them so much.
3. This is the stuff that joyful living is made of- identifying your calling and committing yourself wholeheartedly to it.
4. When this happens, each moment becomes a celebration of you; there is a rush of energy that comes with feeling completely immersed in doing what you love most.
Question Format
The question consists of four statements.
Each is either a fact or an inference or a judgment.
You are expected to determine which is which and match the order with the four options that follow.
Sample Question
E.g. 1
(1) In its 15th report, the Parliamentary Standing Committee on Law and Justice has recommended creation of an all India judicial service (AIJS) on the pattern of the All India Civil Services and directed the Law Ministry to take immediate steps for setting up such a service.
(2) As of now, while most government departments have all India service recruits, selected after an all India competitive examination conducted by the Union Public Service Commission every year, the judiciary is the only set-up that does not have an all India selection process.
(3) In this scheme of things, the measure of uniformity in the standards for selection will improve the quality of personnel in different High Courts.
(4) The quality of dispensation of justice will also improve considerably right from the bottom to the top.
(1) FIJI (2) FFJJ (3) FIIJ (4) IFJI (5) JIIJ
Answer: 2
E.g. 2
1. Recently nicotine, the main chemical additive in cigarettes, was declared addictive by the Food and Drug Administration.
2. This explains why smokers continue to use cigarettes even though smokers are aware of the constantly warned about health dangers in cigarettes.
3. Studies show that this year alone cigarettes will kill over 420,000 Americans, and many more will suffer from cancers, and circulatory and respiratory system diseases.
4. Thus, it’s high time that cigarettes be banned in the country.
(1) FIJI (2) FFJJ (3) FIIJ (4) FIFJ
Answer: 4
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DATA SUFFICIENCY
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What is Data Sufficiency?
In DS questions you are required to find out whether a given set of information or parts of it are sufficient to lead you to an answer to a given question.
Cognitive skills tested through DS:
Data sufficiency questions test your ability to reason quantitatively, unlike problem solving section, where the focus lies in testing your abilities with numbers.
Format
Every Data Sufficiency problem consists of a question followed by two statements. You have to decide not what the answer is, but whether the question can be answered based on the information given in the two statements.
Given below is a simple example to give a feel of the format type:
E.g. 0
What is x?
A. x + y = 17
B. 4x + 4 = 18
Options:
1. If the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone.
2. If the question can be answered by using either statement alone.
3. If the question can be answered by using both the statements together, but cannot be answered by using either statement alone.
4. If the question cannot be answered even by using both statements together.
Answer: 1
Solution: 4x+4=18
4x=14
x=3.5
Thus, B gives the value of x. However, A alone can not lead us to the value of x as it is one equation in two unknowns.
Solving a DS question
Treat each of the two statements and the question posed as an independent question. This simplifies things a lot, and takes care that you do not use inferences from a previous statement in analyzing the next statement.
Sample Question
E.g. 1
What is the value of modulus X?
Statement 1: x = – (modulus x)
Statement 2: x ^2 = 4.
Options:
1. If the question can be answered by one of the statements alone, but cannot be answered by using the other statement alone.
2. If the question can be answered by using either statement alone.
3. If the question can be answered by using both the statements together, but cannot be answered by using either statement alone.
4. If the question cannot be answered even by using both statements together.
Solution:
We would be tempted here, to use both the statements to arrive at a value of x=-2. But then that is not what we need.
The question is asking for modulus X. Statement 1 is not sufficient because it only tells you that x <=0. However, statement 2, implies x = 2 or -2. Regardless, modulus x = 2. Hence, the answer is 1.
CAT Galaxy – Questions asked in past CATs
Options:
1. If the question can be answered with the help of one statement alone
2. If the question can be answered with the help of any one statement independently
3. If the question can be answered with the help of both the statements together
4. If the question can not be answered even with the help of both the statements together
E.g.2. After what time will the two persons Tez and Gati meet while
moving around a circular track? Both of then start from the same point
and at the same time.
1. Tez moves at a constant speed of 5m/s while Gati starts at a speed of 2m/s and increases his speed by 0.5m/s at the end of every second there after.
2. Gati can complete one entire lap in exactly 10 seconds.
Solution: D.
The statement 1 alone is sufficient to calculate the time when two persons will meet around the circular track. But time calculated will be different in case both are moving in the same direction from the time calculated when they are moving in opposite direction. Since no unique solution is achieved, the answer is 1.
E.g.3. Is the number completely divisible by 99?
1. The number is divisible by 9 and 11 simultaneously.
2. If the digits of the number are reversed, the number is divisible by 9 and 11.
Solution: 2
If the number is divisible by 9 and 11 simultaneously, it will be divisible by the LCM of (9,11) or 99. Hence statement 1 alone is sufficient to answer the question. Now, take a number which is divisible by 99, e.g. 198. The number obtained after reversing the digits is 891, which is divisible by 9 and 11 as well. Hence statement 2 alone is sufficient to answer the question.
E.g. 4. What is the value of a^3 + b^3?
1. a^2 + b^2 = 22
2. ab = 3
Solution: 2.
From statements 1 and 2, we bet (a+b)=sqrt28 and –sqrt28. Though we can solve a^3+b^3, we will not have a unique solution for the same as there are two values of a+b. Hence we can not have a unique solution for a^3+b^3.
E.g. 5 Three friends P, Q and R are wearing hats, either black or white. Each
person can see the hats of the other two persons. What is the color of P’s?
hat?
1. P says that he can see one black hat and one white hat.
2. Q says that he can see one white hat and one black hat.
Solution: 4. Even if we combine the information given in statements 1 and 2, we can not find the color of P’s hat.
E.g. 6 What is the distance x between two cities A and B in integral number of Kms?
1. x satisfies the equation logx 2 = sqrt x
2. x < = 10 kms
Solution: 3.
Using statement 1 alone, we get values of x to be 4 and 16. satisfying the given equation. Using statement 2 simultaneously, we can rule out 16 to get the answer as 4. Hence both statements are required.
E. g 7 Mr. Mendel grew 100 flowering plants from black seeds and white seeds. Each seed giving rise to one plant. A plant gives flowers of only one color. From a black seed comes a plant giving red or blue flowers. From a white seed comes a plant giving red or white flowers. How many black seeds were used by Mr. Mendel?
1. The number of plants with white flowers was 10.
2. The number of plants with red flowers was 70.
Solution: 4.
It is given that white seed grows white or red flowers and black seed grows red or blue flowers. Now from statement 1 we know that out of 100 flowering plants, 10 are white flowering plants. Hence, there are at least 10 white seeds, but number of black seeds used can not be known. Using statement 2 together with it, we still can not find out number o black seeds as information about the color of remaining 20 flowers is still not known.
E. g 8 The average weight of students in a class is 50kgs. What is the number of students in the class?
1.The heaviest and the lightest members of the class weigh 60kg and 40kg respectively
2.Exclusion of the heaviest and the lightest members from the class does not change the average weight of the student.
Solution: 4.
Even with 1 and 2 taken together, the number of students can not be found out.
Tips and Tricks
1. After some practice, you will automatically memorize the order of the four options in a DS question. But questions with a different order of options have been asked in the past, so beware! Overlooking the order can be a big mistake if the test-makers jumble up the options order.
2. Once again, remember, you don’t have to find all conclusions from the given set of information; you have to find answer only to the question asked. E.g.1 illustrates the point well.
3. Be careful not to carry over any information from one numbered statement to another.
4. If a question asks for a numerical value (as opposed to a quantitative expression that includes variables), the question is answerable only if a numbered statement (1 and/ or 2) yields one and only one possible numerical answer–not a range of values.
5. Data Sufficiency questions are designed to test you primarily on quantitative concepts, not on your ability to manipulate numbers (that’s what Problem Solving questions are for). So if you find yourself doing a lot of calculations, you’re probably on the wrong track.
6. Just as in Problem Solving questions, in Data Sufficiency questions cast in a real-world setting you should make reasonable real-world assumptions. Don’t split hairs by looking for subtle meanings or ambiguous language. The test-makers are not out to trick you in this way.
7. Do not assume things which are not given in the question. Remember, there is an option which says ‘if the question cannot be answered even by using both statements together’.
Trained as we are to ‘crack questions and get answers’, it is relatively simpler to assume things so as to lead us to some answer. E.g. 2 illustrates this point well.
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Probability Basics
Probability is the likelihood or chance of an event occurring.
Probability = the number of ways of achieving success/ the total number of possible outcomes
For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P(heads) = ½ .
The probability of something which is certain to happen is 1.
The probability of something which is impossible to happen is 0.
The probability of something not happening is 1 minus the probability that it will happen.
Illustrations
Example
There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow?
The probability is the number of yellows in the bag divided by the total number of balls, i.e. 2/6 = 1/3.
Example
There is a bag full of coloured balls, red, blue, green and orange. Balls are picked out and replaced. John did this 1000 times and obtained the following results:
Number of blue balls picked out: 300
Number of red balls: 200
Number of green balls: 450
Number of orange balls: 50
a) What is the probability of picking a green ball?
b) If there are 100 balls in the bag, how many of them are likely to be green?
a) For every 1000 balls picked out, 450 are green. Therefore P(green) = 450/1000 = 0.45
b) The experiment suggests that 450 out of 1000 balls are green. Therefore, out of 100 balls, 45 are green (using ratios).
Example
Find a chance of throwing more than 15 in one throw with 3 dice.
1) 1/54 2)17/216 3)5/108 4) CANNOT BE DETERMINED
Probability = the number of ways of achieving success
the total number of possible outcomes
Total no of possible outcomes with 1 dice = 6
Total no of possible outcomes with 3 dice = 6x6x6 =216
Now the denominator is fixed. We have to find out the numerator.
Possible favourable outcomes are 6,6,6 ; 6,6,5 ; 6,5,5 and 6,6,4.
For 6,6,6: no of ways =1
For 6,6,5 no of ways = 3
For 6,5,5 no of ways = 6,5,5 or 5,6,5 or 5,5,6 so 3.
For 6,6,4 no of ways = 3
Total no of favourable ways = 3+ +3+1 =10
Thus Ans =10/216 = 5/108 ie 3rd option.
Some more Basic Fundas
In mathematics a probability of an event, A is represented by a real number in the range from 0 to 1 and written as P(A), p(A) or Pr(A). An impossible event has a probability of 0, and a certain event has a probability of 1.
The opposite or complement of an event A is the event [not A] (that is, the event of A not occurring); its probability is given by P(not A) = 1 – P(A).
There are two very common terms in Probability : Independent and Mutually Exclusive.
For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is , because of the 52 cards of a deck 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the “3 that are both” are included in each of the “13 hearts” and the “12 face cards” but should only be counted once.
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read “the probability of A, given B”. It is defined by
If P(B) = 0 then is undefined.
For Independent events , P(B/A)=P(B)
And P(A/B)=P(A)
Bayer’s Theorem
=P(B/A) x P(A)/P(B)
Possibility Spaces
When working out what the probability of two things happening is, a probability/ possibility space can be drawn.
Example
if you throw two dice, what is the probability that you will get: a) 8, b) 9, c) either 8or9?
a) The black blobs indicate the ways of getting 8 (a 2 and a 6, a 3 and a 5, …). There are 5 different ways. The probability space shows us that when throwing 2 dice, there are 36 different possibilities (36 squares). With 5 of these possibilities, you will get 8. Therefore P(8) = 5/36 .
b) The red blobs indicate the ways of getting 9. There are four ways, therefore P(9) = 4/36 = 1/9.
c) You will get an 8 or 9 in any of the ‘blobbed’ squares. There are 9 altogether, so P(8 or 9) = 9/36 = 1/4.
Probability Trees
Another way of representing 2 or more events is on a probability tree.
Example
There are 3 balls in a bag: red, yellow and blue. One ball is picked out, and not replaced, and then another ball is picked out.
The first ball can be red, yellow or blue. The probability is 1/3 for each of these. If a red ball is picked out, there will be two balls left, a yellow and blue. The probability the second ball will be yellow is 1/2 and the probability the second ball will be blue is 1/2. The same logic can be applied to the cases of when a yellow or blue ball is picked out first.
In this example, the question states that the ball is not replaced. If it was, the probability of picking a red ball (etc.) the second time will be the same as the first (i.e. 1/3).
In the above example, the probability of picking a red first is 1/3 and a yellow second is 1/2. The probability that a red AND then a yellow will be picked is 1/3 × 1/2 = 1/6 (this is shown at the end of the branch). The rule is:
• If two events A and B are independent (this means that one event does not depend on the other), then the probability of both A and B occurring is found by multiplying the probability of A occurring by the probability of B occurring.
The probability of picking a red OR yellow first is 1/3 + 1/3 = 2/3. The rule is:
• If we have two events A and B and it isn’t possible for both events to occur, then the probability of A or B occuring is the probability of A occurring + the probability of B occurring.
On a probability tree, when moving from left to right we multiply and when moving down we add.
Example
What is the probability of getting a yellow and a red in any order?
This is the same as: what is the probability of getting a yellow AND a red OR a red AND a yellow.
P(yellow and red) = 1/3 × 1/2 = 1/6
P(red and yellow) = 1/3 × 1/2 = 1/6
P(yellow and red or red and yellow) = 1/6 + 1/6 = 1/3
Example
In a Shooting Competition, the probability of hitting the target by A is 2/5, by B is 2/3 and by C is 3/5. If all of them fire independently at the same target, then find the probability that only one of them will hit the target.
P(A)=2/5 P(A*) = 3/5
P(B)=2/3 P(B*) = 1/3
P(C)=3/5 P(C*) = 2/5
Probability that only one of them hits the target
= Probability that A hits the Target but not B and C
+ Probability that B hits the Target but not A and C
+ Probability that C hits the Target but not B and A
= P(A @ B* @ C*) + P(B @ C* @ A*) + P(C @ A* @ B*)
Where @ represents intersection symbol.
= 2/5 x 1/3 x 2/5 + 2/3 x 3/5 x 2/5 + 3/5 x 3/5 x 1/3
= 1/3 Ans
Example
In a fruit Basket 40% of the fruits are mangoes and rest are apples.
If 25% of the mangoes are ripe and 10% of the apples are ripe, find the probability that a Ripe fruit randomly selected is a Mango.
Suppose there are 100 fruits.
No of mangoes = 40 and apples = 60
No of ripe mangoes = 10 and ripe apples = 6
So here we have multiple events taking place. One is the selection of fruit : Mango or Apple. Other is the Selection of Type : Unripe or Ripe.
Let say Selection of Mango as event X and selection of ripe fruit as event Y .
P(Y)= Prob of selection of a ripe fruit = 16/100
P(X@Y) = Probabilty of selection of a Ripe Mango = 10/100
But what is asked is P(X/Y) = P(X@Y)/P(Y) = 10/100 divided by 16/100 = 10/16
= 5/8.
Brain Teaser
Bill and Ben takes turns tossing a coin. Whoever gets the head
first is the winner. If bill has the first toss, what is the
probability that he will win?
Ans : 2/3
Solution : Prob that A wins = prob of a head in 1st turn
+ prob of a head in 3rd turn
+ prob of a head in 5th turn
+ prob of a head in 7th turn
+ ….so on upto infinity
= 1/2 + (1/2)^3 + (1/2)^5 + …..
= ½ [ 1/ (1-{(1/2)^2})]
= 2/3.
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Narsee Monjee Institute of Management & Higher Studies (NMIMS)
Programs offered:
1) MBA
2) MBA – Pharm. Management
3) MBA – Capital Markets
4) MBA – Acturial Science
5) MBA – Global Business
6) MBA – Retail Managemnt
7) MBA – Services Management
8) MBA – Banking
9) PGDBM – Family Business
10) Joint Doctoral Programme
11) Part Time MBA
12) Diploma Programmes
13) Enterprise Training Programme for Women
14) Advanced Modes of Learning (DAML)
15) Management Programme for Executives
Full Time MBA
This is a two –year full time programme, spread over six trimesters and leading to the degree in Master of Business Administration. The most notable features of the programme are its flexibility and objectives of providing a genuinely broad based education. Students are given an opportunity to improve their communication skills in a well equipped, modern language lab with computers and cassettes. Special classes in French, German, and Spanish are also offered to prepare students for international careers. Annual Seminar, CEO series of guest lectures, business simulation games, personal growth lab, psychological profiling, personality development workshops (MBTI), create all round development for students. Final year students have a variety of specializations to choose from, depending on their aptitude and preferences and have the freedom to choose from a large number of electives every trimester. They can choose non – credit courses and cross – disciplinary free choice subjects, to suit their individual preferences.
NMIMS currently has overseas links with the University of EUROMED Marseille in France, ESC ROUEN in France and with ESSCA at both its campuses, i.e. Angers in France and Budapest in Hungary. The exchange covers faculty members as well as students from these places. Several scholarships schemes and research assistantships are also available.
Full Time MBA (Pharmaceuticals Management)
This course is a general management course concentrating on Pharmaceutical Marketing Strategy, distribution, promotion, pricing, product and promotion -mix. Courses on International Marketing, Managing Multinationals and Innovation Management are also offered. Mandarin (Chinese) or Spanish language is compulsory. All students have to do rigorous field work, analysis and reading in this course. This course is not meant for Bio-technology, clinical trials or Veternarians. This course is for Science and Pharmacy Graduates. Doctors may apply for positions in the Pharmaceutical industry. Biotechnologists, Boichemists and Veternarians may not find this course useful.
The distinct feature of the course is that the students undertake regular fieldwork and meet Doctors, Retailers, Stockist, etc. and work on special projects.
Intake Capacity
There are total 30 seats
Full Time MBA (Capital Markets)
Unique Features of the Programme
• Program is in collaboration with BSE
• First Program in India with Trading Room (proposed and supported by BSE) facility having Bloomberg and Reuters
• First Program in India with visit to International Financial Centres like Singapore and Hong Kong (proposed and supported by BSE)
• Program focuses on all major Financial Markets: Equity, Fixed Income, Forex, Commodity and Derivatives
• Skill set development for careers in Trading in Corporate and Bank Treasury
• Skill set development for careers with large Sell-Side Investment Banks (Research/Advisory on IPOs & M&A)
• Skill set development for Buy-Side players like Pension Funds, Hedge Funds and Mutual Funds
• Skill set development for large Private Equity player
MBA – Actuarial Science Program Introduction
Actuaries are experts in assessing the financial impact of tomorrow’s uncertain events. They enable financial decisions to be made with more confidence by:
• analyzing the past
• modeling the future
• assessing risks involved
• and communicating what the results mean in financial terms
The actuary’s skills in analysis and modeling of problems in finance, risk management and product design are used extensively in the areas of insurance, pensions, investment and more recently in wider fields such as project management, banking and health care. Within these sectors, actuaries perform a wide variety of roles such as design and pricing of product, financial management and corporate planning. Actuarial skills are valuable for any business managing long-term financial projects both in the public and private sectors.
The programme prepares the students for a role in various fields such as:
• Life Insurance
• General Insurance
• Health Insurance
• Re-insurance companies
• Pension funds
• Consultancy
• Investments
• Government
• Academics
• Risk Management
With the opening-up of economies all over the world, the role of the actuary is ever expanding.
Intake Capacity
30 seats
Full Time MBA (Global Business)
MBA (Global Business) is a six trimester General Management Programme with focus on International Business for developing a competent cadre of Global Managers for International Positions to meet the country’s growing requirements in view of an upsurge in Indian MNCs on world centrestage.The course tries to bridge the gap between developments in International Markets on the one hand and the consequential impact of the same on India’s domestic market on the other that calls for an increasingly professional approach and sensitivity to international business environment.
Intake Capacity
30 Students
Full Time MBA (Retail Management)
The retailing sector in India is highly fragmented (12 million retail outlets) and predominantly consists of small independent, owner-managed shops. Besides, the country is also dotted with low-cost kiosks and p ushcarts. There has been a boom in retail trade in India in last few years, owing to a gradual increase in the disposable incomes of the middle class households. The size of the retailing industry in India is expected to be around Rs 9,30,000 crore. Out of this, organized retailing industry is around Rs. 35,000 crore, which contributes to less than 2-3 per cent of the industry in India. According to a report by the ministry of commerce & industry, the retail sector in India is expected to grow 7% per annum.
Unique features of the programme
The course has a 4 months Summer Internships, offering the students an opportunity to gain hands on experience in the industry.
The Summer Internship also helps students to focus on the area where they would likely to be placed.
Intake Capacity
There are total 30 seats.
Full Time MBA (Services Management)
As we move toward greater economic development, like other economies across the world, India’s economy is becoming more service oriented. With the increasing importance of the service sector there would be a greater need for business leaders with thorough insights in managing service processes. Since the management of services poses specific challenges that require special attention, conventional insights and know-how emerging from traditional business environments may not apply, too well, to the services sector. Keeping this in mind, MBA (Services Management) has been designed to cater to the increasing need for people equipped with the necessary commercial and personal skills needed in service management functions.
Unique Features of the Programme
• Active industry interface
• External experts from industry for teaching specialized subjects
• Strong focus on practical learning
• Regular updation of syllabus based on industry inputs
• Rigorous Student Selection process
• Industry exposure through live projects
Intake Capacity
In each year, thirty students are admitted in the MBA (Services Management) Programme. The small batch size allows paying individual attention to each learner.
Full Time MBA (Banking)
Unique Features of the Programme
MBA (Banking) Degree would provide a broad – based articulated platform to the participants of the course in the matter of total dynamics of Commercial, developmental and investment banking as are being practiced today and also those likely to emerge in the medium term of next 5 – 7 years. In the process of being groomed both in theoretical and practical aspects of domestic and international banking, the degree holders are expected to perform with absolute dexterity wherever they are placed and, in particular, any finance – related activities.
The programme prepares the students for a role in various fields such as:
• Domestic Banking
• International Banking
• NBFC Function
• Tresaury / Fund Management Operations
• Consultancy
• Investments
• Academics
• Risk Management, in particular, Operational Risk Management
Intake Capacity
There are total 30 seats.
Post Graduate Diploma in Business Management (Family Business)
The Family Business Program at NMIMS University was initiated in 1999, to help meet the unique needs of family business owners and managers. The objective of the program is to assist family-owned businesses in understanding the changing dynamics of competition as well as organization, brought about by liberalization, privatization and globalization. The Program intends to build a partnership between family business owners, managers and NMIMS University. Also to develop family business successors as enterprising and knowledgeable owners of the businesses of their forefathers. Inculcate entrepreneurial qualities in them so that they contribute to the efficiency and growth of their enterprises.
http://www.nmims.edu/fulltime.mba/
Joint Doctoral Programme in Management Studies
SVKM’s NMIMS University, Mumbai, Xavier Institute of Management & Entrepreneurship, Bangalore, and T.A. Pai Institute of Management Education, Manipal have been offering post-graduate programmes in Management for the past several years. Developing a talent base for innovative thinking and research in management & related disciplines, both for academic and industry positions, has been engaging the attention of these institutions for sometime now.
Recognizing the acute shortage of Ph.Ds in management to serve the academia, research and industry, these three leading Business Schools of India –
• NMIMS, Mumbai
• TAPMI, Manipal and
• XIME, Bangalore
have come together to offer a high quality Ph.D programme. This is the first of its kind in the country where three Business Schools together offer a Ph.D programme.
http://www.nmims.edu/jointdoctoral.programme/
Part Time MBA (Masters in Business Administration)
The Part-Time MBA programme is spread over nine trimesters and leads to a degree in Masters of Business Administration. One of the major features that differentiate this programme from many other part – time programme is the fact that it follows the trimester system and is therefore able to offer greater academic input to the participants. This also helps the programme to be in step with ever changing industry requirement.
In-take Capacity
Marketing: 120
Finance: 120
HRM: 60
Systems: 60
http://www.nmims.edu/parttime.mba/
Part Time MBA in Social Entrepreneurship
NMIMS is committed to nurturing the social entrepreneurs of tomorrow and strengthen the social entrepreneurs of today. In this context, NMIMS has designed a uniquely architectured One Year Diploma Programme in Social Entrepreneurship and a Three Year Part Time MBA Programme in Social Entrepreneurship.
Programmes in Social Entrepreneurship aim at:
• Providing innovative business education to potential and current social entrepreneurs.
• Connect practitioners, academics and students working in this field across nations, sectors and disciplines.
• Help social sector practitioners and volunteers to incubate new ideas and spin out new social ventures.
• Support the dissemination of new ideas and the adoption of good business models and practice.
http://www.nmims.edu/parttime.mba.SE/
Part Time DBM (Diploma In Business Management)
The DIPLOMA IN BUSINESS MANAGEMENT (D.B.M.) is designed for those persons who recognise that in order to further their own careers they would benefit from an integrated understanding of how business management operates through organisation, management process and functions.
An ideal programme that covers all-important areas of business management in just three trimesters and provides excellent opportunities for personality development.
DBM | DFM | DHRM | P.G.D.Ed.M. | F-MBA | Diploma in Social Entrepreneurship
Enterprise Training Programme for Women (ETW)
An Indian Adaptation of The Scottish Enterprise Foundation of the University of Stirling, Scotland Subsidised by Maharashtra State Finance Corporation (MSFC) and Small Industries Development Bank of India (SIDBI).
Women in India have traditionally been seen almost exclusively as wives, mothers and homemakers. The ascription of such stereotypes has had the effect of minimalizing their direct economic participation. This has led to a feeling of loss of autonomy in women and has also affected the overall growth of the economy. NMIMS Deemed University has been most active in planning strategies for the development of women in the fields of management and entrepreneurship. As a management institute, NMIMS is uniquely equipped to train women in planning, managing and running a business.
It is a proven fact that in undertaking any economic activity, women face problems that are quirt different from those that confront men. Especially in societies like India, where the home-maker/mother/wife role of the woman has traditionally been given priority, setting up a business or even taking up a job has to fit in to the framework of her traditionally assigned role. As such, acquiring capital, balancing family and business issues, networking with people outside the home, – these are all issues that a woman has to find answers to in a manner quite different from a man.
http://www.nmims.edu/etw.programme/
Advanced Modes of Learning
Management of businesses, organizations, countries and society is today a lot more complex than it was a few decades back. And it promises to become even more involved over the next few years. The skills and knowledge needed to handle these increasing complexities are themselves a lot more difficult to attain. This has lead to the exponential growth in need for good management education.
The Department of Advanced Modes of Learning (DAML) of NMIMS offers management programs that have been devised to take full advantage of these features of remote delivery of education. NMIMS faculty is now at your service, available to you without you having to leave your city and at timings convenient to you. We are sure that you, dear students, will benefit greatly from these programs. These programs will certainly put your career on the fast-track and secure your future.
Courses offered by DAML:
PGDBA PGDGM PGDM EPGM
Post Graduate Diploma In Human Resources (PGDHR)
Management Programme For Executives – MPE
A Unique 22 month Weekend HR Intensive Course for Working Executives
The Department of Human Resources and Behavioural Sciences is launching a Post Graduate Diploma in Human Resources (PGDHR) for working executives interested in the area of Human Resources. This forum will enable executives and managers from various industries to share their perspectives and enhance their knowledge on contemporary HR issues and challenges.
Unique Features of the Programme
• The pedagogy is interactive and largely based on case studies, supplemented by lectures, group activity, workshops, industry speakers and presentations.
• Unique learning experience offered by experienced Full-time, Visiting faculty and Guest speakers from academia and industry
• Emphasis on experiential learning and skills enhancement through workshops conducted by professional experts.
• Specialized Reading Material and handouts for all the subjects
• Access to well stocked library with 43,000 books, national, international magazines and journals, periodicals and research reports.
• Well planned, spacious class rooms, seminar and conference halls air conditioned to facilitate learning and interactive participation.
• Equipped with state-of-the art audio-visual aids.
http://www.nmims.edu/pgdhr/
Management Programme for Executives (MPE)
MANAGEMENT PROGRAMME FOR EXECUTIVES
POST GRADUATE DIPLOMA IN BUSINESS MANAGEMENT (PGDBM)
This is a Week-end Post Graduate Diploma in Business Management Programme meant for Working Executives with minimum three years of Work experience after Graduation in an executive capacity.
Completing this programme has a profound impact on the career of working professionals resulting in a changed outlook towards a wide range of areas. It also positively influences self perception and strengthens the executives’ confidence to take on new roles and objectives for their future career. Personal and professional growth is bound to happen as a consequence of the development of leadership and key decision making skills in this programme.
MPE is chosen by executives and managers for various reasons. Some of the executives wish to develop their skill sets and enhance their career within the scope of their current organization. Some wish to change industry & function and others wish to start up their own venture.
http://www.nmims.edu/executive.programme/
Campus:
SVKM’s NMIMS University is conveniently located at JVPD Scheme, Mumbai at a walking distance from Vile Parle railway station. The institute has 40,000 square feet of built up area. The classrooms, the seminar and the conference halls have been fully air – conditioned for long hours of teaching and interactive participation with a spacious well – planned campus. All classrooms and seminar halls are equipped with state-of-the-art audio-video aids and catering facilities.
Classroom
All classrooms in the institute are air-conditioned and have ceiling mounted LCDs. Most
Classrooms have a capacity of 60. In addition, the institute has a well equipped seminar hall and a conference hall.
Asian Paints Computer Centre
The Computer center is equipped with branded personal computers adequately supported by 3mbps shared leased line for Internet connectivity. It is also equipped with a wide range of licensed system software and applications software. The entire campus is connected with Wi-Fi network.
Library – The R. M. Desai Library Learning Resource Centre
The well furnished and air conditioned NMIMS library has a stock of more than 42,000 volumes and subscribes to about 308 Indian and Foreign periodicals spanning all aspects of management. The library also houses 521 video’s, 319 audio’s, 187 CD’s, 274 VCD’s, annual reports, management games, news clippings, newsletters and case studies. The open access system facilities free use of books on the shelf.
The Audio Interactive Language Lab
NMIMS has pioneered the initiative of scientifically assisting students in preparing for Group Discussions or Job Interviews by developing an Audio Interactive Language Lab on the campus. This move is mainly directed towards making the students display the right internal personality driven message to their recruiters during a corporate placement, group discussions or interviews that are lining -up in the campus.
Fees and Expenses:
FEES FOR ALL TWO YEAR’S FULL TIME – MBA PROGRAMMES
1st Year Rs.1,75,000/- (Open)
Rs.4,50,000/- (Management)
2nd Year Rs.1,67,100/- (Open)
Rs.4,42,100/- (Management)
Refund of fees in case of Withdrawal of Admission: ‘Withdrawal of Admission’ means voluntary withdrawal by the candidate for any reason.
The schedule of refund of fees will be as follows:-
Application for Withdrawal made, from the date of Payment of fees
First 15 days 16 to 30 days Above 30 days
Tuition Other Tuition Other Tuition Other
Fees Fees Fees Fees Fees Fees
85% 100% 65% 100% NIL 30%
HOSTEL FEES: –
First 30 days – Full refund
Above 30 days – Administrative & Processing fess of Rs. 3000/- will be deducted
Library Fees :
Students can avail the library facility on payment of refundable Library deposit of Rs. 1,000/- and library fees of Rs. 1000/- per year.
Faculty:
NMIMS faculty will go to a DiRECWAY Global Education studio in Mumbai and deliver lectures in front of cameras and other high tech audio-video equipment.
These lectures will get beamed via satellites instantaneously to all the classrooms. Students will be able to see the faculty, hear the faculty and interact with the faculty just as they do in class.
The key benefits of using this system are:
1. It offers the highest level of interaction.
2. Built in class-management tools allow the faculty to guide the class effectively.
3. Built in measurement and certification tools allow the faculty to monitor each student’s progress accurately and continuously.
http://www.nmims.edu/ourteam/lecturer.htm
Ranking:
MBA ranking
India Today 2007: #17
Outlook 2007: #10
Business Today 2005: #10
Business Today 2004: #17
Placements and Recruiters:
MBA Placement 2008
Highest Rs. 24 lacs p.a
Mean Rs. 11.40 lacs p.a
Median Rs. 10 lacs p.a
http://www.nmims.edu/placements/brochure2008.htm
Notable Alumni:
http://125.18.15.117/alumni/INDEX.asp
For more information and FREE online practice tests visit www.tenaday.in
For more information and FREE online practice tests, visit: www.tenaday.in
HCF And LCM
These are very simple and important topics. The trend has been to ask questions from this section quite frequently nowadays. Although the questions asked will not be direct, it will be a indirect question and thus the biggest challenge in an exam for this topic lies in identification of the question and that it is asked from this topic. Solving a problem after identification is easy.
Note: Calculation speed is very important for this topic esp. division with prime nos. like 2,3 ,7 etc for factorization.
Let’s start with very basic concepts and then go to HCF and LCM.
FACTOR
A factor of a given number is every number that divides exactly into that number.
Example
Write down all factors of 10.
10 = 2 x 5, so numbers 2 and 5 are factors of 10.
Also 10 = 10 x 1, so 10 and 1 are factors of 10.
The factors of 10 are 1, 2, 5, and 10.
NOTE: Number 1 and the number itself are always factors of any number.
PRIME AND COMPOSITE NUMBERS
A prime number has exactly 2 factors, the number itself and 1.
In other words, the prime number can be divided only by 1 and by itself.
NOTE: 0 and 1 are not prime numbers.
Example: 5 is a prime number, because the only factors it has are 1 and 5.
The prime numbers less than 20 are 2,3,5,7,11,13,17,19
Example:
Find all prime factors of 30.
Solution:
All the factors of 30 are 30, 15, 10, 6, 5, 3, 2, 1
But only 5, 3 and 2 are prime numbers.
Thefore all prime factors of 30 are 2, 3 and 5.
A composite number has at least one more factor that the number itself or 1.
In fact, all whole numbers that are not prime are composite except for 1 and 0, which are not prime and not composite.
The composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, and 18.
DIVISIBILITY RULES
The simple divisibility rules will help you to find factors of a number.
The number is divisible by:
• 2 if the last digit is 0, 2, 4, 6, or 8 (example: 12346);
• 3 if the sum of digits in the number are divisible by 3
(example: 1236, because 1+2+3+6 = 12 = 3 x 4);
• 4 if the last 2 digits are divisible by 4
(example: 897544, because 44 = 4 x 11);
• 5 if the last digit is 0 or 5
(example: 178965 or 40980);
• 6 if it is divisible by 2 and 3;
• 8 if the last 3 digits are divisible by 8
(example: 124987080, because 080 = 8 x 10;
• 9 if the sum of digits is divisible by 9
(example: 234612, because 2+3+4+6+1+2 = 18 = 9 x 2);
• 10 if the last digit is 0
(example: 99990 );
• 11 if the difference between sum of digits in odd and sum of digits in even places is 0 or a multiple of 11.
• 12 if the no is divisible by both 3 and 4.
• 19 if the no of tens added to twice the no of ones is divisible by 19.
(Example: 228; no of tens =22; no of ones = 8; the required sum = 22+ 2×8 =38; hence, 228 is divisible by 19)
• 25 if the no formed by the last two digits (ie digits at tens and ones place) is divisible by 25.
(Example: 31243482325; no formed by last two digits =25;
hence 31243482325 is divisible by 25)
• 100 if the last 2 digits are 0
(example 987600);
• 125 if the no formed by last 3 digits is divisible by 125.
(Note: there are some complex divisibility tests for 7, 13 and 17 too but I recommend not to follow them as dividing the no given would be easier and faster than applying those divisibility tests.)
NOTE: If a number is divisible by 2 factors, it is also divisible by the product of these factors.
Example 1: Number 18 is divisible by 2 and 3, so it must be divisible by 2 x 3 = 6.
Example 2: Number 945 is divisible by 9 (why?) and by 5 (why?), so it must be divisible by 9 x 5 = 45. (Can you check it?)
COMMON FACTORS
When two (or more) numbers have the same factor, that factor is called a common factor.
Example
Find all the common factors of 12 and 18.
Factors of 12 are 1, 2, 3, 4, 6, 12.
Factors of 18 are 1, 2, 3, 6, 18.
The common factors of 12 and 18 are 1, 2, 3 and 6.
HIGHEST COMMON FACTOR (H.C.F)
The Highest Common Factor (H.C.F) of two (or more) numbers is the largest number that divides evenly into both numbers. It is also known as G.C.D. (Greatest common divisor)
In other words the H.C.F is the largest of all the common factors.
The common factors or of 12 and 18 are 1, 2, 3 and 6.
The largest common factor is 6, so this is the H.C.F. of 12 and 18.
It is very easy to find a H.C.F. of small numbers, like 6 and 9 (it is 3) or 8 and 4 (it is 4).
The best way is to keep finding the factors of the smaller number, starting from the largest factor. The first factor of the smaller number that is also a factor of the larger number is a H.C.F.
LEAST COMMON MULTIPLE (L.C.M.)
A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24…
The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.
Concept – Prime Factorisation
A prime factorisation of a natural number can be expressed in the exponential form.
For example:
(i) 48 = 2 x 2 x 2x 2 x3 = 24 x 3
(ii) 420 = 2 x 2 x 3 x 5 x 7 = 22 x 3 x 5 x 7
Methods to Find HCF
Example. Find the H.C.F. of 72, 126 and 270.
Using Prime factorisation method
72 = 2 x 2 x 2 x 3 x 3 = 2 3 x 32
126 = 2 x 3 x 3 x 7 = 2 1 x 32 x 71
270 = 2 x 3 x 3 x 3 x 5 = 21 x 33 x 51
H.C.F. of the given numbers = the product of common factors with least index = 21 x 32
Using Division method
First find H.C.F. of 72 and 126
72|126|1
72
54| 72|1
54
18| 54| 3
54
0
H.C.F. of 72 and 126 = 18
Similarly calculate H.C.F. of 18 and 270 as 18
Hence H.C.F. of the given three numbers = 18
Co-prime numbers: Two natural numbers are called co-prime numbers if they have no common factor other than 1.
in other words, two natural numbers are co-prime if their H.C.F. is 1.
Some examples of co-prime numbers are: 4, 9; 8, 21; 27, 50.
Methods to Find LCM
Method 1 Simply list the multiples of each number (multiply by 2, 3, 4, etc.) then look for the smallest number that appears in each list.
Example: Find the least common multiple for 5, 6, and 15.
Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,…
Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,…
Multiples of 15 are 30, 45, 60, 75, 90,….
Now, when you look at the list of multiples, you can see that 30 is the smallest number that appears in each list.Therefore, the least common multiple of 5, 6 and 15 is 30.
Method 2 To use this method factor each of the numbers into primes. Then for each different prime number in all of the factorizations, do the following…
1. Count the number of times each prime number appears in each of the factorizations.
2. For each prime number, take the largest of these counts.
3. Write down that prime number as many times as you counted for it in step 2.
The least common multiple is the product of all the prime numbers written down.
Example: Find the least common multiple of 5, 6 and 15.
Factor into primes
Prime factorization of 5 is 5
Prime factorization of 6 is 2 x 3
Prime factorization of 15 is 3 x 5
Notice that the different primes are 2, 3 and 5.
Now, we do
Step #1 – Count the number of times each prime number appears in each of the factorizations…
The count of primes in 5 is one 5
The count of primes in 6 is one 2 and one 3
The count of primes in 15 is one 3 and one 5
Step #2 – For each prime number, take the largest of these counts. So we have…
The largest count of 2s is one
The largest count of 3s is one
The largest count of 5s is one
Step #3 – Since we now know the count of each prime number, you simply – write down that prime number as many times as you counted for it in step 2.
Here they are…2, 3, 5
Step #4 – The least common multiple is the product of all the prime numbers written down.
2 x 3 x 5 = 30
Therefore, the least common multiple of 5, 6 and 15 is 30.
Relation between L.C.M. and H.C.F. of two natural numbers
The product of L.C.M. and H.C.F. of two natural numbers = the product of the numbers.
Note. In particular, if Two natural numbers are co-prime then their L.C.M. = The product of the numbers.
HCF AND LCM of Fractions
1. HCF of Fractions = (HCF of Numerators/LCM of Denominators)
2. LCM of Fractions = (LCM of Numerators/HCF of Denominators)
Examples
1. Find the G.C.D of 12x2y3z2, 18x3y2z4, and 24xy4z3
(1) 6xy2z2
(2) 6x3y4z3
(3) 24xy2z2
(4) 18x2y2z3
Correct Choice is (1) and Correct Answer is 6xy2z2
________________________________________
Explanatory Answer
G.C.D of 12, 18 and 24 is 6.
The common factors are x, y, z and their highest powers common to all are 1, 2 and 2 respectively.
Therefore, G.C.D = 6xy2z2
2.Find the L.C.M. of 72, 240, 196.
Solution
Using Prime factorisation method
72 = 2×2×2×3×3 = 2³×3²
240 = 2×2×2×2×3×5 = 24×3×5
196 = 2×2×7×7 = 2²×7²
L.C.M. of the given numbers = product of all the prime factors of each of the given number with greatest index of common prime factors
= 24×3²×5×7² = 16×9×5×49 = 35280.
Using Division method
2 | 72, 240, 196
2 | 36, 120, 98
2 | 18, 60 , 49
3 | 9 , 30 , 49
| 3 , 10 , 49
L.C.M. of the given numbers
= product of divisors and the remaining numbers
= 2×2×2×3×3×10×49
= 72×10×49 = 35280.
3.Find the H.C.F. of 72, 126 and 270.
Solution
Using Prime factorisation method
72 = 2×2×2×3×3 = 2³×3²
126 = 2×3×3×7 = 21×3²×71
270 = 2×3×3×3×5 = 21×3³×51
H.C.F. of the given numbers = the product of common factors with least index
= 21×3² = 2×3×3 = 18
Using Division method
First find H.C.F. of 72 and 126
72|126|1
72
54| 72|1
54
18| 54| 3
54
0
H.C.F. of 72 and 126 = 18
Similarly calculate H.C.F. of 18 and 270 as 18
Hence H.C.F. of the given three numbers = 18
4. Arrange the fractions 2/15, 3/10 , 5/21 in ascending order of their respective magnitudes.
Solution
5 | 15, 10, 21
3 | 3, 2, 21
| 1, 2 , 7
LCM of 15,10,21 = 5 x 3 x 2 x 7 = 210
Thus,
2/15 = (2×14)/(15×14) = 28/210
3/10 = (3×21)/(10×21) = 63/210
5/21 = (5×10)/(21×10) = 50/210
Thus now comparing these fractions is simple. Greater the Numerator, greater is the Fraction.
Thus ,
28/ 210 ; 50/210 ; 63/210 in ascending order
Or
2/15 ; 5/21 ; 3/10 in ascending order.
5. What is the smallest no which when increased by 3 is divisible by 27,35,25 and 21?
1] 4722
2] 4725
3] 4728
4] 4731
Solution:
The smallest no that is divisible by 27, 35, 25 and 21 = LCM of these nos.
So our answer is LCM – 3.
The LCM can be obtained by any of the techniques described above.
LCM = 4725
Ans 4722 ie Option 1.
6. What is the smallest no which when decreased by 5 is divisible by 36,48,21 and 28?
1] 1008
2] 1003
3] 1013
4] 1018
Solution
Same as Q 5.
In this case ans = LCM + 5
Ans option 3 – 1013
7. If A381 is divisible by 11, find the value of A?
1] 5 2] 6 3] 7 4] 8
Solution
As per the divisibility test of 11, we have A+8 – 3 – 1 should be divisible by 11.
So A + 4 should be a multiple of 11.
Thus A+4=11k where k is an integer.
Put k=0, A= -4 Not Possible.
k=-1, A= -15 Not Possible.
k=1 , A=7 ans . Option[3]
No other value of k is feasible.
8. A no ‘A’ is not divisible by 3. which of the following is definitely divisible by 3.
1] A^2 +1 2] A^2 – A 3] A^2 – 1 4] A^2 + A
Solution:
Consider, Option [3], (A+1)(A-1). Now as A is not divisible by 3 one of these (A+1) or (A-1) will be divisible by 3.
Thus Ans Option[3].
9. What is the greatest no which when divided by 6,7,8,9,10 leaves remainders as 4,5,6,7,8 respectively.
1] 997920 2] 997918 3] 999999 4] 997922
Solution
Actually for CAT u can directly mark option 2 by seeing that on dividing the no by 10 the remainder is 8. By divisibility test we know only, option 2 will leave a remainder 8 with 10.
Proper solution for this is :
6-4 = 2 7-5 = 2 8-6 = 2 9-7 = 2 10-8 = 2
LCM of 6,7,8,9,10 =2520
Greatest no of six digits = 999999
Greatest six digit no that is a multiple of 2520 = 2520 x 396 = 997920
Subtract 2 from this no to get the required answer – 997918 and which will give remainders as 4,5,6,7,8 when divided by 6,7,8,9,10.
Hence [2].
10. The HCF and LCM of two nos is given. It is possible to find out the two nos uniquely if
I. Either the sum or the difference between the two nos is known.
II. HCF of the two nos = LCM of the two Nos.
III. (LCM/HCF) = prime no.
1] I and II only.
2] II only
3] II and III only
4] I, II and III
Solution:
I We know HCF X LCM = Product of the nos = AB(Lets say)
Given that A-B or A+B is known.
Thus we have 2 equations and two variables, thus the soln can be uniquely determined.
II When HCF = LCM the two nos are equal. So again can be solved.
III LCM/HCF = prime no is known.
Then one of the nos = LCM and the other = HCF.
Thus III is also true
So As I , II , III all are true, ANS is option [4].
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