In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 stud

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Q. In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.

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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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This problem uses the Principle of Inclusion-Exclusion to find the number of students studying all three subjects. We first determine the total number of students studying at least one subject and then apply the formula, finding that 20 students study all three subjects.

When dealing with overlapping groups, like students studying different subjects, we often encounter the challenge of counting elements multiple times. For instance, a student studying both Mathematics and Physics would be counted in the 'Mathematics' group and again in the 'Physics' group. To get an accurate count of students studying at least one subject, or to find the size of specific overlaps, we use the Principle of Inclusion-Exclusion.

The core idea is to:

Include the sizes of all individual groups.
Exclude the sizes of all pairwise overlaps, because these elements were counted twice in step 1.
Include the size of the triple overlap, because elements in this region were initially counted thrice (in each individual group), then subtracted thrice (in each pairwise overlap), meaning they were effectively not counted at all. Adding them back once corrects this.

This principle ensures that every element belonging to at least one group is counted exactly once.

Let $M$ be the set of students studying Mathematics, $P$ be the set of students studying Physics, and $C$ be the set of students studying Chemistry. We are given the following information:

Total students in the survey: $N(\text{Total}) = 200$
Students studying Mathematics: $N(M) = 120$
Students studying Physics: $N(P) = 90$
Students studying Chemistry: $N(C) = 70$
Students studying Mathematics and Physics: $N(M \cap P) = 40$
Students studying Physics and Chemistry: $N(P \cap C) = 30$
Students studying Chemistry and Mathematics: $N(C \cap M) = 50$
Students studying none of these subjects: $N(\text{None}) = 20$

We need to find the number of students who study all three subjects, which is $N(M \cap P \cap C)$ .

Here is the step-by-step solution:

Determine the number of students studying at least one subject. The total number of students in the survey is 200. We know that 20 students study none of the subjects. This means the remaining students must be studying at least one subject.

N(M \cup P \cup C) = N(\text{Total}) - N(\text{None})

N(M \cup P \cup C) = 200 - 20

N(M \cup P \cup C) = 180

So, 180 students study at least one of Mathematics, Physics, or Chemistry.

2. Recall the Principle of Inclusion-Exclusion for three sets.

The formula for the union of three sets $A, B, C$ is:

> [!FORMULA]

> $N(A \cup B \cup C) = N(A) + N(B) + N(C) - N(A \cap B) - N(B \cap C) - N(C \cap A) + N(A \cap B \cap C)$ …

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

Showing the 12 most recent of 52 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The shaded region in the given Venn-diagram represents:$
(a) A ∪ B
(b) A ∩ B
(c) (A ∪ B)'
(d) (A ∩ B)'
›Reveal solutionSolution

The shaded region is everything in the universal set except A and B combined, which is exactly $(A \cup B)'$ .

The rectangle is the universal set $U$ , and the two overlapping circles are sets $A$ and $B$ . The description tells us the shading covers the rectangle except the two circles — i.e. every point that lies outside both $A$ and $B$ .

…

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CBSE 2026Set ANNUAL1 mark
$Q. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 6, 7}, then (A \cup B)' = ..............$
›Reveal solutionSolution

Find A∪B first, then take its complement in U.

Given $U=\{1,2,3,4,5,6,7,8,9\}$ , $A=\{2,4,6,8\}$ , $B=\{2,3,6,7\}$ .

First find $A\cup B$ (all elements in A or B or both):

$A\cup B = \{2,3,4,6,7,8\}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$X = \{1, 3, 5\}$
(a) $\{1, 2, 3, 4, 5\}$
(b) $\{1, 2, 3, 5\}$
(c) $\{1, 3\}$
(d) $\phi$
›Reveal solutionSolution

$X \cap Y$ consists of elements present in both $X$ and $Y$ , which gives $\{1,3\}$ .

Given $X = \{1,3,5\}$ and $Y=\{1,2,3\}$ . The intersection $X \cap Y$ contains only those elements that belong to BOTH sets.

…

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CBSE 2026Set ANNUAL1 mark
$\{2, 6, 10\}$
›Reveal solutionSolution

Sets are disjoint when their intersection is empty; comparing the elements of $\{2,6,10\}$ and $\{3,7,11\}$ shows no overlap.

Set $A=\{2,6,10\}$ and set $B=\{3,7,11\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{1, 2, 3\}, B = \{2, 3, 7\} \Rightarrow A \cup B =$
(a) $\{1, 2, 3\}$
(b) $\{1, 3, 7\}$
(c) $\{1, 2, 3, 7\}$
(d) $\{1, 2, 7\}$
›Reveal solutionSolution

$A \cup B = \{1, 2, 3, 7\}$ : the union lists every element that is in $A$ or in $B$ (or both), each written once.

For sets $A$ and $B$ , the union is $A \cup B = \{x : x \in A \text{ or } x \in B\}$ — combine both sets and remove duplicate entries.

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{3, 5, 7\}, Y = \{2, 3, 5\} \Rightarrow X \cap Y =$
(a) $\{3, 2\}$
(b) $\{3, 7\}$
(c) $\{5, 7\}$
(d) $\{3, 5\}$
›Reveal solutionSolution

$X \cap Y = \{3, 5\}$ : the intersection keeps only elements that belong to both sets.

For sets $X$ and $Y$ , $X \cap Y = \{x : x \in X \text{ and } x \in Y\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{1, 2\}, Y = \{2, 3, 5\}, Z = \{4, 6\} \Rightarrow X \cup Y \cup Z =$
(a) $\{1, 2, 3, 5, 6\}$
(b) $\{2, 3, 4, 5, 6\}$
(c) $\{1, 2, 3, 4, 5, 6\}$
(d) $\{1, 2\}$
›Reveal solutionSolution

$X \cup Y \cup Z = \{1, 2, 3, 4, 5, 6\}$ : list every element appearing in at least one of the three sets, once each.

Given $X = \{1, 2\}$ , $Y = \{2, 3, 5\}$ , $Z = \{4, 6\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{1, 2, 3, 6\}, Y = \{4, 5, 6\}, Z = \{4, 2, 3, 6\} \Rightarrow (X \cup Y) \cap Z =$
(a) $\{2, 3, 4, 6\}$
(b) $\{1, 5\}$
(c) $\{1, 2, 5\}$
(d) $\{1, 2, 3, 5\}$
›Reveal solutionSolution

$(X \cup Y) \cap Z = \{2, 3, 4, 6\}$ , found by first taking the union, then intersecting with $Z$ .

Given $X = \{1, 2, 3, 6\}$ , $Y = \{4, 5, 6\}$ , $Z = \{4, 2, 3, 6\}$ .

Step 1: $X \cup Y = \{1, 2, 3, 4, 5, 6\}$ (combine both, 6 counted once).

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{a, b, c, d\}, Y = \{c, a, r\}, Z = \{r, o, b\} \Rightarrow (X \cap Y) \cup Z =$
(a) $\{a, b, c, o, r\}$
(b) $\{c, a, r, b\}$
(c) $\{r, o, b, c\}$
(d) $\phi$
›Reveal solutionSolution

$(X \cap Y) \cup Z = \{a, b, c, o, r\}$ , found by first taking the intersection, then the union with $Z$ .

Given $X = \{a, b, c, d\}$ , $Y = \{c, a, r\}$ , $Z = \{r, o, b\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{x : x - 2 = 0\}, B = \{x : 2x = 6\} \Rightarrow A \cup B =$
(a) $\{2, 6\}$
(b) $\{-2, 6\}$
(c) $\{2, 3\}$
(d) $\{2, -3\}$
›Reveal solutionSolution

$A \cup B = \{2, 3\}$ .

$A = \{x : x - 2 = 0\} = \{2\}$ . $B = \{x : 2x = 6\} = \{3\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{x : x^2 + 5x + 6 = 0\}, B = \{x : x^2 + 8x + 15 = 0\} \Rightarrow$
(a) $A \subset B$
(b) $B \subset A$
(c) $A = B$
(d) $A \cap B = \{-3\}$
›Reveal solutionSolution

$A = \{-2, -3\}$ , $B = \{-3, -5\}$ , and their only common element is $-3$ , so $A \cap B = \{-3\}$ .

$A = \{x : x^2 + 5x + 6 = 0\}$ : factorising, $(x+2)(x+3) = 0 \Rightarrow x = -2, -3$ , so $A = \{-2, -3\}$ .

$B = \{x : x^2 + 8x + 15 = 0\}$ : factorising, $(x+3)(x+5) = 0 \Rightarrow x = -3, -5$ , so $B = \{-3, -5\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(a) $\{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$
(b) $\{4, 6, 8, 10, 12, 14\}$
(c) $\{8, 10, 12, 14\}$
(d) $\phi$
›Reveal solutionSolution

$A' = U - A = \{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$ .

Given $U = \{1, 2, \ldots, 15\}$ and $A = \{1, 2, 3, 5, 15\}$ . The complement $A' = U - A$ consists of every element of $U$ not in $A$ .

…

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