IPMAT Indore 2022 · Quantitative Ability · Question 36

Step 1 — pin down A and B. From \left( 1,4 \right)\in A\times B we get 1\in A and 4\in B. From \left( 4,1 \right)\in A\times B we get 4\in A and 1\in B. So

\{1,4\}\subseteq A \text{and}\{1,4\}\subseteq B

Step 2 — use the size constraint. We are told \left| A\times B \right|=4, and \left| A\times B \right|=\left| A \right|\,\left| B \right|. Since both A and B already contain at least two elements,

\left| A \right|\ge 2, \left| B \right|\ge 2, \left| A \right|\left| B \right|=4 \Rightarrow \left| A \right|=\left| B \right|=2

Therefore

A=B=\{1,4\}

Step 3 — evaluate the options. Since A and B are the same set,

A\times B=B\times A=\{\left( 1,1 \right),\left( 1,4 \right),\left( 4,1 \right),\left( 4,4 \right)\}

(d) A\times B=B\times A ✔ — true, because A=B.

(b) A\times B\ne B\times A ✘ — the opposite.

(c) ⌀\in A\times B ✘ — the elements of A\times B are ordered pairs; the empty set is a subset, not a member.

(a) None of these ✘ — (d) is true.

Answer: Option 4 — A\times B=B\times A

Shortcut. The size condition does all the work. Both sets must contain 1 and 4, so each has at least 2 elements; and \left| A \right|\left| B \right|=4 forces both to have exactly 2. Hence A=B=\{1,4\}, and once the sets are equal the products are trivially equal.

Note the general fact being tested: A\times B=B\times A holds only when A=B (or one of them is empty). The question engineers that equality through the counting constraint.

Option (c) is a type error worth flagging to students — confusing \in with \subseteq. The empty set is a subset of every set but an element of almost none.