In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10%

NCERT Exemplar · Q27

Q. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find

(a) The number of families which buy newspaper A only.

(b) The number of families which buy none of A, B and C.

Punjab PsebLong· 5mImportance★★★★★

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Concept understanding — Set Operations

The idea in plain words

Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".

The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.

Why this works

Sets are just labelled buckets that hold distinct items. The universal set $U$ is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Step by step

Let's take two concrete sets so you can see each operation in action:

A = \{1, 2, 3\}, \quad B = \{3, 4, 5\}

Step 1: Union — gather everything from both, but don't repeat anything.

A \cup B = \{1, 2, 3, 4, 5\}

Step 2: Intersection — only what's common to both.

A \cap B = \{3\}

Step 3: Difference ( $A$ minus $B$ ) — start with $A$ , remove anything that's also in $B$ .

A \setminus B = \{1, 2\}

Step 4: Complement — needs a universal set. Let $U = \{1, 2, 3, 4, 5\}$ . Then:

A^c = \{4, 5\}

Step 5: Symmetric Difference — combine the two differences:

A \triangle B = (A \setminus B) \cup (B \setminus A) = \{1, 2\} \cup \{4, 5\} = \{1, 2, 4, 5\}

A common slip …

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Use the principle of inclusion-exclusion to find the number of families buying only A and those buying none. The key is to subtract overlaps carefully: only A = 40% – (5% + 4% – 2%) = 33% → 3300 families; none = 100% – (40% + 20% + 10% – 5% – 3% – 4% + 2%) = 100% – 60% = 40% → 4000 families.

We are given percentages of families buying newspapers A, B, C in a town of 10,000 families. The question asks for the number of families that buy only A, and those that buy none of the three. This is a classic problem of set theory and probability — specifically, the principle of inclusion-exclusion for three sets.

The intuition: When we add the percentages for A, B, and C, we count families that buy two newspapers twice, and families that buy all three three times. To get the correct total for "at least one", we subtract the double-counted overlaps and then add back the triple-counted ones. For "only A", we start with all A families and subtract those that also buy B or C, being careful not to subtract the A∩B∩C families twice.

Let’s define:

$n(A) = 40\%$ of 10,000 = 4000 families
$n(B) = 20\%$ = 2000 families
$n(C) = 10\%$ = 1000 families
$n(A \cap B) = 5\%$ = 500 families
$n(B \cap C) = 3\%$ = 300 families
$n(A \cap C) = 4\%$ = 400 families
$n(A \cap B \cap C) = 2\%$ = 200 families

We work entirely in percentages first, then convert to numbers at the end.

Step-by-step solution

1. Find the number of families that buy only A.

Families buying A include those that may also buy B or C. To get only A, we subtract those that buy A and at least one other newspaper.

The families that buy A and B (including those that also buy C) are $n(A \cap B) = 5\%$ . Similarly, $n(A \cap C) = 4\%$ . But the intersection $A \cap B \cap C$ (2%) is counted in both of these. So the total percentage of families that buy A and at least one of B or C is:

n(A \cap B) + n(A \cap C) - n(A \cap B \cap C) = 5\% + 4\% - 2\% = 7\%

Therefore, only A = $n(A) - 7\% = 40\% - 7\% = 33\%$ .

Number of families = $33\%$ of 10,000 = 3300.

Tip

A common shortcut: only A = $n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$ . The plus sign for the triple intersection corrects the double subtraction. This works because the triple intersection was subtracted twice (once in each pairwise intersection), so we add it back once. …

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Q.In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

Step-by-step solution

Step 1 — Compute the union $B\cup C$

Step 2 — Compute the difference $A-(B\cup C)$

Q.In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find

Step-by-step solution

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