Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.

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Accepted Answer

We need non-negative integers x, y (W = whole numbers, starting at 0) with $x^2+y^2=100$ . Checking systematically: $x=0 \Rightarrow y^2=100 \Rightarrow y=10$ , giving $(0,10)$ . $x=6 \Rightarrow y^2=100-36=64 \Rightarrow y=8$ , giving $(6,8)$ . $x=8 \Rightarrow y^2=100-64=36 \Rightarrow y=6$ , giving $(8,6)$ . $x=10 \Rightarrow y^2=0 \Rightarrow y=0$ , giving $(10,0)$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set jan-20261 markMCQ
$A=\{2,\ 3\}$
›Reveal solutionSolution

$B\cap C=\{5\}$ , so $A\times\{5\}=\{(2,5),(3,5)\}$ .

Step 1 — Intersection $B\cap C$

Keep only the elements common to both sets:
$B\cap C=\{4,5\}\cap\{5,6\}=\{5\}.$
Step 2 — Cartesian product $A\times(B\cap C)$

Pair each element of $A=\{2,3\}$ with each element of $\{5\}$ , writing each ordered pair as (element from $A$ , element from $B\cap C$ ):

$A\times\{5\}=\{(2,5),\ (3,5)\}.$ …
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CA Foundation 2025Set jan-20251 markMCQ
$A = \{a, b, p\}$
›Reveal solutionSolution

$n(A \cup C) = 6$ , $n(B) = 2$ , so $n[(A \cup C) \times B] = 6 \times 2 = 12$ .

Step 1 — Form the union
$A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}$
The element $p$ is common and counted only once, so $n(A \cup C) = 6$ .

Step 2 — Note the size of $B$
$B = \{2, 3\} \;\Rightarrow\; n(B) = 2$
Step 3 — Cardinality of the Cartesian product
$n(X \times Y) = n(X) \times n(Y) = 6 \times 2 = 12$
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CA Foundation 2023Set jun-20231 markMCQ
$A=\{a, b, c\}, B=\{b, c, d\}$
›Reveal solutionSolution

A − B = {a}, B ∩ C = {c, d}, so {a} × {c, d} = {(a, c), (a, d)}.

Step 1 — Set difference

A = {a, b, c}, B = {b, c, d}: elements of A not in B ⇒ A − B = {a}.

Step 2 — Intersection

B = {b, c, d}, C = {a, d, c}: common elements ⇒ B ∩ C = {c, d}.

Step 3 — Cartesian product
$\{a\} \times \{c, d\} = \{(a, c), (a, d)\}$
Watch out

Ordered pairs: the first coordinate must come from A − B, so (c, a) and (d, a) are wrong. …

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Q.Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.

Concept understanding — Cartesian Product of Sets

Step 1 — Intersection $B\cap C$

Step 2 — Cartesian product $A\times(B\cap C)$

Step 1 — Form the union

Step 2 — Note the size of $B$

Step 3 — Cardinality of the Cartesian product

Step 1 — Set difference

Step 2 — Intersection

Step 3 — Cartesian product

Q.Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.

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