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GMAT Sets and Number PropertiesWhen you think about Set Questions, the traditional form of representing numbers in Sets comes to our mind. In GMAT set questions however, several other concepts will be tested simultaneously. Let us look into an example where Set representation is used to  solve Probability and Number properties GMAT Question.

Q) Rhonda’s Chocolate factory is creating packets of chocolates with 12, 13, 14, 25, 35, 44, 66, 77 and 88 chocolates in each packet. The manager at the Factory arranged the packets in such a way that all the bright colored packets were in one group, and dark colored packets in other. If the bright colored packet group had packets with 12, 25, 77, and 88 chocolates, and dark colored packet group had the remaining, what is the probability that picking a pair from dark and bright colored packet group gives even number of chocolates?

Answers

a)    1/3
b)    1/2
c)    1/5
d)    2/3
e)    1/8

Solution

The length of the question should not intimidate you. This simple set question tests the theory of Number...

Categories : Sets

Set Notation - Rooster and Builder MethodSet Operation is one of the important topics covered in GMAT exam. Before we go into operations on set elements, let us look at some basic representations.

Set is represented by Capital Letters and small letters denote the members. There are two methods of representing a set – Rooster and Builder Method.

Rooster Method

In rooster method, we represent the set starting with the Set Name in Capital letter followed by an equal sign, a curly opening brace, members of the set and finally closing the set members with curly closing braces. For a Set representation of first five alphabets in English, the rooster representation will be:

A = {a, b, c, d, e}


Set Builder Method

Instead of listing the set members, the set builder method defines the properties of the members. This method is also called the Property method.

For Example, if we want to represent the set of all odd integers, we represent the set starting with the Set Name in capital letter followed by an equal sign, curly opening braces, property of the set and finally closing the set representation with curly closing braces....

Categories : Venn Diagrams, Sets

On your GMAT, you will encounter 1-3 questions that contain overlapping groups with specific characteristics. You will almost never see more than two characteristics (since you can’t draw 3D on your scratch paper). For illustration, let’s take a look at the following Data Sufficiency example:

Q) Of the 70 children who visited a certain doctor last week, how many had neither a cold nor a cough?

(1) 40 of the 70 children had a cold but not a cough.
(2) 20 of the 70 children had both a cold and a cough.

There are two characteristics (cough and cold) and two categories for each (yes and no), so there are four total categories, as indicated by this matrix:
 
Four Total Categories
I’ve filled in the given information from both statements, and the parenthetical information is inferred. This clearly lays out the 4 combinations of options. If we sum vertically, we can infer that there are 60 total children with colds. Because there are 70 total children, this also means that 10 do NOT have colds. The bottom-right quadrant cannot be found because we do not know how those 10 children get divided between the two empty boxes. Choice E – together the statements are insufficient...

Categories : Sets

GMAT Set TheoryA sample of 100 people was surveyed for the use of computers, cell phones and credit cards. 31 people had credit cards, 61 had computers and 28 had cell phones. If 11 people had none of the three and 32 people had more than one, how many had all the three?

A) 1
B) 2
C) 3
D) 4
E) 5


Solution: This is a question based on Set Theory.

Total = 100
None = 11
Net Total = 100 - 11

P(A) = Credit Cards = 31
P(B) = Computers = 61
P(C) = Cell Phones = 28

Using the formula, P(S) = P(A) + P(B) + P(C) - P(A?B) – P(B?C) – P(C?A) + P(A?B?C)


P (A?B) + P(B?C) + P(C?A) = 32 (More than One)

Plugging in values, we get,

100 – 11 = 31 + 61 + 28 – (P (A?B)...

Categories : Venn Diagrams

During the past month, a disease control center tested X individuals for two viruses. If 1/3 of those tested had virus C and, of those with virus C, 1/5 also had virus D, how many individuals did not have both virus C and D?

A. X/4
B. 4X/15
C. 10X/5
D. 14X/15
E. 4 / 5

Answer

The number of individuals that had virus C is 1/3 of X, or X/3.

The number of individuals that also had virus D is 1/5 of the number that had virus C.

Thus (1/5)C= (1/5)(X/3) = X/15.

If X/15 of the X individuals have virus C and D, then X - (X/15) did not have both virus C and D.

Neither C or D: X - (X/15) = 14X/15.

The correct answer is D.




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