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CAT 2018 Slot 2 — QA Question 34

SetsEasy

For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is

Answer & solution

Correct answer: 7

  • A

    6

  • B

    8

  • C

    9

  • 7

Solution

Easy

The operation A Δ BA\,\Delta\,B is the symmetric difference: elements in exactly one of the two sets. Compute P Δ QP\,\Delta\,Q and R Δ SR\,\Delta\,S first, then take the symmetric difference of those two results and count.

P={1,2,3,4}P=\{1,2,3,4\}, Q={2,3,5,6}Q=\{2,3,5,6\}, R={1,3,7,8,9}R=\{1,3,7,8,9\}, S={2,4,9,10}S=\{2,4,9,10\}.

1

Compute P Δ QP\,\Delta\,Q. Keep elements in exactly one of P,QP,Q. Common: {2,3}\{2,3\}.

P Δ Q={1,4}∪{5,6}={1,4,5,6}\begin{aligned} &P\,\Delta\,Q = \{1,4\}\cup\{5,6\} = \{1,4,5,6\} \end{aligned}
2

Compute R Δ SR\,\Delta\,S. Common element: {9}\{9\}.

R Δ S={1,3,7,8}∪{2,4,10}={1,2,3,4,7,8,10}\begin{aligned} &R\,\Delta\,S = \{1,3,7,8\}\cup\{2,4,10\} = \{1,2,3,4,7,8,10\} \end{aligned}
3

Combine the two results. Take (P Δ Q) Δ (R Δ S)(P\,\Delta\,Q)\,\Delta\,(R\,\Delta\,S). Common to both: {1,4}\{1,4\}, which drop out.

{1,4,5,6} Δ {1,2,3,4,7,8,10}⇒ ={5,6}∪{2,3,7,8,10}⇒ ={2,3,5,6,7,8,10}⇒ number of elements=7\begin{aligned} &\{1,4,5,6\}\,\Delta\,\{1,2,3,4,7,8,10\}\\ &\Rightarrow\ = \{5,6\}\cup\{2,3,7,8,10\}\\ &\Rightarrow\ = \{2,3,5,6,7,8,10\}\\ &\Rightarrow\ \text{number of elements} = 7 \end{aligned}
7\textbf{7}

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CAT 2018 Slot 2 QA Q34 — Sets solution | TheCATExam