GATE CSE 2006 | Question: 22

Question

Let $E, F$ and $G$ be finite sets. Let

• $X = (E ∩ F) - (F ∩ G)$ and
• $Y = (E - (E ∩ G)) - (E - F)$.

Which one of the following is true?

• A. $X ⊂ Y$
• B. $X ⊃ Y$
• C. $X = Y$
• D. $X - Y ≠ \emptyset$ and $Y - X ≠ \emptyset$

Comments

Let $E\ =\ \{ \ 1,2,3,4,5,6,7,8,9,10\ \},\ F\ =\ \{\ 5,10\ \},\ G\ =\ \{\ 2,4,6,8,10\ \}\\E\ \cap\ F\ =\ \{\ 5,10\ \}\\F\ \cap\ G\ =\ \{\ 10\ \}\\X\ =\ \{\ 5\ \}\ \\E\ \cap\ G\ =\ \{\ 2,4,6,8,10\ \}\\E\ -\ E\ \cap\ G\ =\ \{ \ 1,3,5,7,9\ \}\\E\ -\ F\ =\ \{\ 1,2,3,4,6,7,8,9\}\\Y\ =\ (E\ -\ (E\ \cap\ G\ ))\ -\ (E\ -\ F) =\ \{\ 5 \ \}\\ \therefore\ X\ = \ Y. So\ answer\ is\ (C)$

Note that this example also eliminates all the other options.

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For these type of questions, this method seems simple and fast

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Answer 1

X=(E∩F)−(F∩G)
Y=(E−(E∩G))−(E−F)
Let E={1,2,3,4,5}   positive integers

F={2,3,5,7}  prime numbers

G={1,3,5} odd numbers

E∩ F ={2,3,5} ,F∩G={3,5}   so X={2}

E∩G={1,3,5} ,E-{E∩G}={2,4},E-F={1,4}  

Y={2} so X and Y are same so option C is right

Answer 2

As the question gives options into the hands of student:

 So, CASE 1 not equal to CASE 2.

CASE 1 => X=Y

CASE 2 => X, not equal Y and X is a subset of Y.

So what should be chosen?

And why we presumption for all set merger and there must be an intersection point.  

Comments

y is phi

Answer 3

BEST METHOD 

the solution can be obtained for boolean algebra as follows:
X=(E∩F)−(F∩G)
=EF−FG
=EF∩(FG)′
=EF.(F′+G′)
=EFF′+EFG′
=EFG′
 

 

Similarly, Y=(E−(E∩G))−(E−F)
=(E−EG)−(E.F′)
=E.(EG)′−EF′
=E.(E′+G′)−EF′
=EG′−EF′
=EG′.(EF′)′
=EG′.(E′+F)
=EE′G′+EFG′
=EFG′
 

Therefore, X=Y