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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?

1. $X ⊂ Y$
2. $X ⊃ Y$
3. $X = Y$
4. $X - Y ≠ \phi$ and $Y - X ≠ \phi$
edited | 772 views
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People are explaining it via Venn Diagram. But multiple Venn Diagram are possible. And covering all of them in limited time is difficult . So use  algebra of sets.
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I feel Venn diag in some cases help to get a clearer view of what is actually happening. I don't find multiple cases in this question.
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I don't find multiple cases in this question.

Because you are looking at a particular Venn diagram. But based on different sets different kind of Venn diagrams are possible.

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@Chhotu if we take intersections to be not null, the Venn diagram will be unique, isn't it? can you please explain how multiple Venn diagrams are coming.?

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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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This might help ..

Answer is $C$ using Venn diagram.

edited

Option C is ans

hope it might help.....

Let E,F, and G belongs to same universe of discourse U, then we can write  E-F=E ∩ F' =EF' .

X = (E∩F) - (F∩G)  = (E∩F) ∩ (F∩G)' =EF (F' + G') =EFG' =(E ∩F ∩G' )

Y=(E−(E∩G))−(E−F) =E (EG)' - (EF') = E(E'+G') - (EF') = EG' - EF'= EG' (EF')' = EG'(E'+F) =EFG' = (E∩F∩G')

We can clearly see that ,X=Y.

## Option (C) X=Y   is the correct answer.

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